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1lgxgzye9 | maths | 3d-geometry | lines-in-space | <p>The shortest distance between the lines $${{x + 2} \over 1} = {y \over { - 2}} = {{z - 5} \over 2}$$ and $${{x - 4} \over 1} = {{y - 1} \over 2} = {{z + 3} \over 0}$$ is :</p> | [{"identifier": "A", "content": "8"}, {"identifier": "B", "content": "7"}, {"identifier": "C", "content": "6"}, {"identifier": "D", "content": "9"}] | ["D"] | null | Given, the lines are
<br/><br/>$$
\begin{aligned}
\frac{x+2}{1} & =\frac{y}{-2}=\frac{z-5}{2} ~~~~..........(i)\\\\
\text { and } \frac{x-4}{1} & =\frac{y-1}{2}=\frac{z+3}{0} ~~~~..........(ii)
\end{aligned}
$$
<br/><br/>Formula for shortest distance between two skew-lines,
<br/><br/>$$
\begin{aligned}
S D & =\left|\frac{\left|\begin{array}{ccc}
x_2-x_1 & y_2-y_1 & z_2-z_1 \\
a_1 & b_1 & c_1 \\
a_2 & b_2 & c_2
\end{array}\right|}{\left|\begin{array}{ccc}
\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\
a_1 & b_1 & c_1 \\
a_2 & b_2 & c_2
\end{array}\right|}\right|=\left|\frac{\left|\begin{array}{ccc}
6 & 1 & -8 \\
1 & -2 & 2 \\
1 & 2 & 0
\end{array}\right|}{\left|\begin{array}{ccc}
\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\
1 & -2 & 2 \\
1 & 2 & 0
\end{array}\right|}\right| \\\\
& =\left|\frac{6(-4)-1(-2)-8(4)}{|-4 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}|}\right| \\\\
& =\left|\frac{-24+2-32}{\sqrt{36}}\right| \\\\
& =\left|\frac{-54}{6}\right|=|-9|=9
\end{aligned}
$$ | mcq | jee-main-2023-online-10th-april-morning-shift |
1lgzydtfy | maths | 3d-geometry | lines-in-space | <p>The shortest distance between the lines $$\frac{x-4}{4}=\frac{y+2}{5}=\frac{z+3}{3}$$ and $$\frac{x-1}{3}=\frac{y-3}{4}=\frac{z-4}{2}$$ is :</p> | [{"identifier": "A", "content": "$$3 \\sqrt{6}$$"}, {"identifier": "B", "content": "$$6 \\sqrt{2}$$"}, {"identifier": "C", "content": "$$6 \\sqrt{3}$$"}, {"identifier": "D", "content": "$$2 \\sqrt{6}$$"}] | ["A"] | null | The given lines are
<br/><br/>$$\frac{x-4}{4}=\frac{y+2}{5}=\frac{z+3}{3}$$ and $$\frac{x-1}{3}=\frac{y-3}{4}=\frac{z-4}{2}$$
<br/><br/>$$
\begin{aligned}
& \text { So, } \vec{b}_1=4 \hat{i}+5 \hat{j}+3 \hat{k} \\\\
& \vec{b}_2=3 \hat{i}+4 \hat{j}+2 \hat{k} \\\\
& \vec{a}_1=4 \hat{i}-2 \hat{j}-3 \hat{k} \\\\
&\vec{a}_2=\hat{i}+3 \hat{j}+4 \hat{k}
\end{aligned}
$$
<br/><br/>$$
\begin{aligned}
& \therefore \vec{b}_1 \times \vec{b}_2=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
4 & 5 & 3 \\
3 & 4 & 2
\end{array}\right| \\\\
& =(10-12) \hat{i}-(8-9) \hat{j}+(16-15) \hat{k} \\\\
& =-2 \hat{i}+\hat{j}+\hat{k}
\end{aligned}
$$
<br/><br/>$$
\begin{aligned}
& \text { Shortest distance, } d=\left|\frac{\left(\vec{a}_2-\vec{a}_1\right) \cdot\left(\vec{b}_1 \times \vec{b}_2\right)}{\left|\vec{b}_1 \times \vec{b}_2\right|}\right| \\\\
& =\left|\frac{(3 \hat{i}-5 \hat{j}-7 \hat{k}) \cdot(-2 \hat{i}+\hat{j}+\hat{k})}{\sqrt{4+1+1}}\right| \\\\
& =\left|\frac{-6-5-7}{\sqrt{6}}\right|=\frac{18}{\sqrt{6}}=3 \sqrt{6} \text { units }
\end{aligned}
$$ | mcq | jee-main-2023-online-8th-april-morning-shift |
1lh2197tv | maths | 3d-geometry | lines-in-space | <p>One vertex of a rectangular parallelopiped is at the origin $$\mathrm{O}$$ and the lengths of its edges along $$x, y$$ and $$z$$ axes are $$3,4$$ and $$5$$ units respectively. Let $$\mathrm{P}$$ be the vertex $$(3,4,5)$$. Then the shortest distance between the diagonal OP and an edge parallel to $$\mathrm{z}$$ axis, not passing through $$\mathrm{O}$$ or $$\mathrm{P}$$ is :</p> | [{"identifier": "A", "content": "$$\\frac{12}{\\sqrt{5}}$$"}, {"identifier": "B", "content": "$$12 \\sqrt{5}$$"}, {"identifier": "C", "content": "$$\\frac{12}{5}$$"}, {"identifier": "D", "content": "$$\\frac{12}{5 \\sqrt{5}}$$"}] | ["C"] | null | <img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lnq8po6g/d220c54e-806e-4b56-ac43-a5312272c160/0b7f4280-6aad-11ee-822f-a9d01571f2aa/file-6y3zli1lnq8po6h.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1lnq8po6g/d220c54e-806e-4b56-ac43-a5312272c160/0b7f4280-6aad-11ee-822f-a9d01571f2aa/file-6y3zli1lnq8po6h.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh; vertical-align: baseline;" alt="JEE Main 2023 (Online) 6th April Morning Shift Mathematics - 3D Geometry Question 47 English Explanation">
<br><br>Equation of $O P$ is
<br><br>$$
\begin{aligned}
&\frac{x-0}{3-0}=\frac{y-0}{4-0} =\frac{z-0}{5-0} \\\\
&\Rightarrow \frac{x}{3} =\frac{y}{4}=\frac{z}{5}
\end{aligned}
$$
<br><br>Equation of edge parallel to $Z$-axis is
<br><br>$$
\frac{x-3}{0}=\frac{y-0}{0}=\frac{z-5}{1}
$$
<br><br>$\therefore$ Shortest distance
<br><br>$$
=\frac{\left|\begin{array}{ccc}
x_2-x_1 & y_2-y_1 & z_2-z_1 \\
a_1 & b_1 & c_1 \\
a_2 & b_2 & c_2
\end{array}\right|}{\sqrt{\begin{array}{r}
\left(a_1 b_2-a_2 b_1\right)^2+\left(b_1 c_2-b_2 c_1\right)^2
+\left(c_1 a_2-c_2 a_1\right)^2
\end{array}}}
$$
<br><br>Here,
<br><br>$$
\begin{aligned}
& x_1=0, y_1=0, z_1=0 \\\\
& x_2=3, y_2=0, z_2=5 \\\\
& a_1=3, b_1=4, c_1=5 \\\\
& a_2=0, b_2=0, c_2=1
\end{aligned}
$$
<br><br>$$
\therefore\left|\begin{array}{ccc}
x_2-x_1 & y_2-y_1 & z_2-z_1 \\
a_1 & b_1 & c_1 \\
a_2 & b_2 & c_2
\end{array}\right|=\left|\begin{array}{ccc}
3 & 0 & 5 \\
3 & 4 & 5 \\
0 & 0 & 1
\end{array}\right|=4(3)=12
$$
<br><br>$\begin{aligned} & \text { and } \sqrt{\left(a_1 b_2-a_2 b_1\right)^2+\left(b_1 c_2-b_2 c_1\right)^2+\left(c_1 a_2-c_2 a_1\right)^2} \\\\ & =\sqrt{0+(4-0)^2+(0-3)^2} \\\\ & =\sqrt{16+9}=5 \\\\ & \therefore \text { Required shortest distance }=\frac{12}{5} \\\\ & \end{aligned}$ | mcq | jee-main-2023-online-6th-april-morning-shift |
1lh2z2249 | maths | 3d-geometry | lines-in-space | <p>If the lines $$\frac{x-1}{2}=\frac{2-y}{-3}=\frac{z-3}{\alpha}$$ and $$\frac{x-4}{5}=\frac{y-1}{2}=\frac{z}{\beta}$$ intersect, then the magnitude of the minimum value of $$8 \alpha \beta$$ is _____________.</p> | [] | null | 18 | Given, lines
<br/><br/>$$
\begin{aligned}
& \frac{x-1}{2} =\frac{2-y}{-3}=\frac{z-3}{\alpha} \\\\
&\Rightarrow \frac{x-1}{2} =\frac{y-2}{3}=\frac{z-3}{\alpha}=\lambda .........(i)
\end{aligned}
$$
<br/><br/>Any point on the line (i)
<br/><br/>$$
x=2 \lambda+1, y=3 \lambda+2, z=\alpha \lambda+3
$$
<br/><br/>and line $\frac{x-4}{5}=\frac{y-1}{2}=\frac{z}{\beta}=\mu$ ............(ii)
<br/><br/>Any point on line (ii)
<br/><br/>$$
\Rightarrow x=5 \mu+4, y=2 \mu+1, z=\beta \mu
$$
<br/><br/>Since, given lines intersects
<br/><br/>$$
\begin{aligned}
& \therefore 2 \lambda+1=5 \mu+4 ..........(iii)\\\\
& 3 \lambda+2=2 \mu+1 ............(iv)\\\\
& \text { and } \alpha \lambda+3=\beta \mu ..........(iv)
\end{aligned}
$$
<br/><br/>On solving (iii) and (iv), we get
<br/><br/>$$
\lambda=-1, \mu=-1
$$
<br/><br/>On putting value of $\lambda$ and $\mu$ in (v), we get
<br/><br/>$$
\begin{array}{cc}
& \alpha(-1)+3=-\beta \\\\
&\Rightarrow \alpha=\beta+3
\end{array}
$$
<br/><br/>Now,
<br/><br/>$$
\begin{aligned}
& 8 \alpha \beta=8 (\beta+3)(\beta) \\\\
&= 8\left(\beta^2+3 \beta\right) \\\\
& =8\left(\beta^2+3 \beta+\frac{9}{4}-\frac{9}{4}\right) \\\\
& =8\left\{\left(\beta+\frac{3}{2}\right)^2-\frac{9}{4}\right\}
\end{aligned}
$$
<br/><br/>$$=8\left(\beta+\frac{3}{2}\right)^2-18$$
<br/><br/>Here, minimum value $=-18$
<br/><br/>$\therefore$ Magnitude of the minimum value of $8 \alpha \beta$ is 18 . | integer | jee-main-2023-online-6th-april-evening-shift |
lsamoe4e | maths | 3d-geometry | lines-in-space | Let $\mathrm{P}$ and $\mathrm{Q}$ be the points on the line $\frac{x+3}{8}=\frac{y-4}{2}=\frac{z+1}{2}$ which are at a distance of 6 units from the point $\mathrm{R}(1,2,3)$. If the centroid of the triangle PQR is $(\alpha, \beta, \gamma)$, then $\alpha^2+\beta^2+\gamma^2$ is : | [{"identifier": "A", "content": "18"}, {"identifier": "B", "content": "24"}, {"identifier": "C", "content": "26"}, {"identifier": "D", "content": "36"}] | ["A"] | null | Any point on line $\frac{x+3}{8}=\frac{y-4}{2}=\frac{z+1}{2}$ <br/><br/>can be taken as $(8 \lambda-3,2 \lambda+4,2 \lambda-1)$
<br/><br/> If at a distance of 6 units from $R(1,2,3)$
<br/><br/>$$
\begin{aligned}
& \Rightarrow(8 \lambda-3-1)^2+(2 \lambda+4-2)^2+(2 \lambda-1-3)^2=36 \\\\
& \left.\Rightarrow \lambda^2-\lambda=0 \text { \{on simplification }\right\} \\\\
& \Rightarrow \lambda=0, \lambda=1
\end{aligned}
$$
<br/><br/>Here $P \& Q$ are $(-3,4,-1)$ and $(5,6,1)$ Centroid of $\triangle P Q R$
<br/><br/>$$
\begin{aligned}
& (\alpha, \beta, \gamma) \equiv\left(\frac{5-3+1}{3}, \frac{6+4+2}{3}, \frac{1-1+3}{3}\right) \\\\
& \Rightarrow \alpha=1, \beta=4, \gamma=1 \\\\
& \Rightarrow \alpha^2+\beta^2+\gamma^2=18
\end{aligned}
$$ | mcq | jee-main-2024-online-1st-february-evening-shift |
lsamrrrt | maths | 3d-geometry | lines-in-space | If the mirror image of the point $P(3,4,9)$ in the line
<br/><br/>$\frac{x-1}{3}=\frac{y+1}{2}=\frac{z-2}{1}$ is $(\alpha, \beta, \gamma)$, then 14 $(\alpha+\beta+\gamma)$ is : | [{"identifier": "A", "content": "102"}, {"identifier": "B", "content": "138"}, {"identifier": "C", "content": "132"}, {"identifier": "D", "content": "108"}] | ["D"] | null | <img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lsohd1pb/c1eafb9f-de3f-4c37-82e0-908506938702/622ee6f0-ccb2-11ee-aa98-13f456b8f7af/file-6y3zli1lsohd1pc.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1lsohd1pb/c1eafb9f-de3f-4c37-82e0-908506938702/622ee6f0-ccb2-11ee-aa98-13f456b8f7af/file-6y3zli1lsohd1pc.png" loading="lazy" style="max-width: 100%;height: auto;display: block;margin: 0 auto;max-height: 40vh;vertical-align: baseline" alt="JEE Main 2024 (Online) 1st February Evening Shift Mathematics - 3D Geometry Question 41 English Explanation">
<br><br>$\begin{aligned} & \overrightarrow{\mathrm{PN}}. \overrightarrow{\mathrm{b}}=0\\\\ & 3(3 \lambda-2)+2(2 \lambda-5)+(\lambda-7)=0 \\\\ & 14 \lambda=23 \Rightarrow \lambda=\frac{23}{14}\end{aligned}$
<br><br>$\begin{aligned} & \mathrm{N}\left(\frac{83}{14}, \frac{32}{14}, \frac{51}{14}\right) \\\\ & \therefore \frac{\alpha+3}{2}=\frac{83}{14} \Rightarrow \alpha=\frac{62}{7}\end{aligned}$
<br><br>$\begin{aligned} & \frac{\beta+4}{2}=\frac{32}{14} \Rightarrow \beta=\frac{4}{7} \\\\ & \frac{\gamma+9}{2}=\frac{51}{14} \Rightarrow \gamma=\frac{-12}{7}\end{aligned}$
<br><br>Now, $14(\alpha+\beta+\gamma)=14\left(\frac{62+4-12}{7}\right)=108$ | mcq | jee-main-2024-online-1st-february-evening-shift |
lsaptbc8 | maths | 3d-geometry | lines-in-space | If the shortest distance between the lines <br/><br/>$\frac{x-\lambda}{-2}=\frac{y-2}{1}=\frac{z-1}{1}$ and $\frac{x-\sqrt{3}}{1}=\frac{y-1}{-2}=\frac{z-2}{1}$ is 1 , then the sum of all possible values of $\lambda$ is : | [{"identifier": "A", "content": "0"}, {"identifier": "B", "content": "$2 \\sqrt{3}$"}, {"identifier": "C", "content": "$3 \\sqrt{3}$"}, {"identifier": "D", "content": "$-2 \\sqrt{3}$"}] | ["B"] | null | <p>Given the two lines:</p>
<p>$$ L_1: \frac{x-\lambda}{-2} = \frac{y-2}{1} = \frac{z-1}{1} $$</p>
<p>$$ L_2: \frac{x-\sqrt{3}}{1} = \frac{y-1}{-2} = \frac{z-2}{1} $$</p>
<p>We observe that these lines are not parallel as their directional vectors are not proportional. The directional vector for $L_1$ is $(-2,1,1)$ and for $L_2$ is $(1,-2,1)$. The shortest distance between two skew (non-intersecting and non-parallel) lines in the three-dimensional space is along the line that is perpendicular to both lines. This implies we can find a vector that is perpendicular to both directional vectors by taking their cross product.</p>
<p>The directional vector for $L_1$ is $d_1 = \langle -2, 1, 1 \rangle$, and for $L_2$ is $d_2 = \langle 1, -2, 1 \rangle$. The cross product of $d_1$ and $d_2$, which will be perpendicular to both lines, is given by:</p>
<p>$$ d = d_1 \times d_2 = \left| \begin{matrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
-2 & 1 & 1 \\
1 & -2 & 1 \\
\end{matrix} \right| $$
<p>Expanding the determinate gives:</p>
<p>$$ d = \mathbf{i}((1)(1) - (1)(-2)) - \mathbf{j}((-2)(1) - (1)(1)) + \mathbf{k}((-2)(-2) - (1)(1)) $$</p>
<p>$$ d = \mathbf{i}(1 + 2) - \mathbf{j}(-2 - 1) + \mathbf{k}(4 - 1) $$</p>
<p>$$ d = 3\mathbf{i} + 3\mathbf{j} + 3\mathbf{k} $$</p>
<p>$$ d = \langle 3, 3, 3 \rangle $$</p>
<p>The shortest distance $D$ between the two lines can then be given by the formula:</p>
<p>$$ D = \frac{\left| (\mathbf{a_2} - \mathbf{a_1}) \cdot d \right|}{\|d\|} $$</p>
<p>Where $\mathbf{a_1}$ and $\mathbf{a_2}$ are position vectors to any points on line $L_1$ and line $L_2$, respectively, and '$\cdot$' denotes the dot product.</p>
<p>From the lines' equations, we can choose a point on each line (when the parameter is zero). Thus, for $L_1$, let's choose the point $A(\lambda, 2, 1)$, and for $L_2$, let's choose the point $B(\sqrt{3}, 1, 2)$. These points correspond to the vectors $\mathbf{a_1} = \langle \lambda, 2, 1 \rangle$ and $\mathbf{a_2} = \langle \sqrt{3}, 1, 2 \rangle$, respectively.</p>
<p>The vector $\mathbf{a_2} - \mathbf{a_1}$ is:</p>
<p>$$ \mathbf{a_2} - \mathbf{a_1} = \langle \sqrt{3}, 1, 2 \rangle - \langle \lambda, 2, 1 \rangle $$</p>
<p>$$ \mathbf{a_2} - \mathbf{a_1} = \langle \sqrt{3} - \lambda, 1 - 2, 2 - 1 \rangle $$</p>
<p>$$ \mathbf{a_2} - \mathbf{a_1} = \langle \sqrt{3} - \lambda, -1, 1 \rangle $$</p>
<p>We can now substitute this, along with $d$, into the distance formula:</p>
<p>$$ D = \frac{\left| (\langle \sqrt{3} - \lambda, -1, 1 \rangle) \cdot \langle 3, 3, 3 \rangle \right|}{\|\langle 3, 3, 3 \rangle\|} $$</p>
<p>$$ D = \frac{\left| 3(\sqrt{3} - \lambda) + 3(-1) + 3(1) \right|}{\sqrt{3^2 + 3^2 + 3^2}} $$</p>
<p>$$ D = \frac{\left| 3\sqrt{3} - 3\lambda - 3 + 3 \right|}{\sqrt{27}} $$</p>
<p>$$ D = \frac{\left| 3\sqrt{3} - 3\lambda \right|}{3\sqrt{3}} $$</p>
<p>$$ D = \frac{\left| \sqrt{3} - \lambda \right|}{\sqrt{3}} $$</p>
<p>Given that the shortest distance $D$ between the lines is 1, we can equate the above result to 1, and solve for $\lambda$:</p>
<p>$$ \frac{\left| \sqrt{3} - \lambda \right|}{\sqrt{3}} = 1 $$</p>
<p>$$ \left| \sqrt{3} - \lambda \right| = \sqrt{3} $$</p>
<p>This absolute value equation gives us two possible cases:</p>
<p>Case 1: $\sqrt{3} - \lambda = \sqrt{3}$, which gives $\lambda = 0$.</p>
<p>Case 2: $\sqrt{3} - \lambda = -\sqrt{3}$, which gives $\lambda = 2\sqrt{3}$.</p>
<p>Therefore, the sum of all possible values of $\lambda$ is:</p>
<p>$$ \lambda_{sum} = \lambda_1 + \lambda_2 = 0 + 2\sqrt{3} = 2\sqrt{3} $$</p>
<p>Hence, option B ($2 \sqrt{3}$) is the correct answer.</p> | mcq | jee-main-2024-online-1st-february-morning-shift |
lsaqawqo | maths | 3d-geometry | lines-in-space | Let the line of the shortest distance between the lines
<br/><br/>$$
\begin{aligned}
& \mathrm{L}_1: \overrightarrow{\mathrm{r}}=(\hat{i}+2 \hat{j}+3 \hat{k})+\lambda(\hat{i}-\hat{j}+\hat{k}) \text { and } \\\\
& \mathrm{L}_2: \overrightarrow{\mathrm{r}}=(4 \hat{i}+5 \hat{j}+6 \hat{k})+\mu(\hat{i}+\hat{j}-\hat{k})
\end{aligned}
$$
<br/><br/>intersect $\mathrm{L}_1$ and $\mathrm{L}_2$ at $\mathrm{P}$ and $\mathrm{Q}$ respectively. If $(\alpha, \beta, \gamma)$ is the mid point of the line segment $\mathrm{PQ}$, then $2(\alpha+\beta+\gamma)$ is equal to ____________. | [] | null | 21 | $\begin{array}{ll}L_1 \equiv \vec{r}=(1,2,3)+\lambda(1,-1,1) & \left(\vec{r}=\vec{a}_1+\lambda \vec{b}_1\right) \\\\ L_2 \equiv \vec{r}=(4,5,6)+\mu(1,1,-1) & \left(\vec{r}=\vec{a}_2+\lambda \vec{b}_2\right)\end{array}$
<br/><br/>$P \equiv(\lambda+1,-\lambda+2, \lambda+3)$
<br/><br/>$Q \equiv(\mu+4, \mu+5,-\mu+6)$
<br/><br/>$\overrightarrow{P Q}=(\mu-\lambda+3, \mu+\lambda+3,-\mu-\lambda+3)$
<br/><br/>$\overrightarrow{P Q} \cdot \vec{b}_1=0 \Rightarrow 3 \lambda+\mu=3$ ..........(i)
<br/><br/>$\overrightarrow{P Q} \cdot \vec{b}_2=0 \Rightarrow 3 \mu+\lambda=3$ ..........(ii)
<br/><br/>From (i) and (ii),
<br/><br/>$$
\begin{aligned}
& P \equiv\left(\frac{5}{2}, \frac{1}{2}, \frac{9}{2}\right) \& Q \equiv\left(\frac{5}{2}, \frac{7}{2}, \frac{15}{2}\right) \\\\
& \alpha=\frac{5}{2}, \beta=\frac{4}{2}, \gamma=\frac{12}{2} \\\\
& 2(\alpha+\beta+\gamma)=21
\end{aligned}
$$ | integer | jee-main-2024-online-1st-february-morning-shift |
lsbklvv1 | maths | 3d-geometry | lines-in-space | The distance, of the point $(7,-2,11)$ from the line <br/><br/>$\frac{x-6}{1}=\frac{y-4}{0}=\frac{z-8}{3}$ along the line $\frac{x-5}{2}=\frac{y-1}{-3}=\frac{z-5}{6}$, is : | [{"identifier": "A", "content": "12"}, {"identifier": "B", "content": "18"}, {"identifier": "C", "content": "21"}, {"identifier": "D", "content": "14"}] | ["D"] | null | <p>$$\mathrm{B}=(2 \lambda+7,-3 \lambda-2,6 \lambda+11)$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lt1c0zyx/3ef0ec08-6ff8-4ca7-97e0-1c5c0fde9c3c/b5ee7090-d3c3-11ee-a50b-bb659a2e1d74/file-1lt1c0zyy.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lt1c0zyx/3ef0ec08-6ff8-4ca7-97e0-1c5c0fde9c3c/b5ee7090-d3c3-11ee-a50b-bb659a2e1d74/file-1lt1c0zyy.png" loading="lazy" style="max-width: 100%;height: auto;display: block;margin: 0 auto;max-height: 40vh;vertical-align: baseline" alt="JEE Main 2024 (Online) 27th January Morning Shift Mathematics - 3D Geometry Question 38 English Explanation"></p>
<p>Point B lies on $$\frac{x-6}{1}=\frac{y-4}{0}=\frac{z-8}{3}$$</p>
<p>$$\frac{2 \lambda+7-6}{1}=\frac{-3 \lambda-2-4}{0}=\frac{6 \lambda+11-8}{3}$$</p>
<p>$$-3 \lambda-6=0$$</p><p>
</p><p>$$\lambda=-2$$</p>
<p>$$\mathrm{B} \Rightarrow(3,4,-1)$$</p>
<p>$$\mathrm{AB}=\sqrt{(7-3)^2+(4+2)^2+(11+1)^2}$$</p>
<p>$$=\sqrt{16+36+144}$$</p>
<p>$$=\sqrt{196}=14$$</p> | mcq | jee-main-2024-online-27th-january-morning-shift |
lsbkxlh7 | maths | 3d-geometry | lines-in-space | If the shortest distance between the lines <br/><br/>$\frac{x-4}{1}=\frac{y+1}{2}=\frac{z}{-3}$ and $\frac{x-\lambda}{2}=\frac{y+1}{4}=\frac{z-2}{-5}$ is $\frac{6}{\sqrt{5}}$, then the sum of all possible values of $\lambda$ is : | [{"identifier": "A", "content": "10"}, {"identifier": "B", "content": "5"}, {"identifier": "C", "content": "7"}, {"identifier": "D", "content": "8"}] | ["D"] | null | <p>$$\begin{aligned}
& \frac{x-4}{1}=\frac{y+1}{2}=\frac{z}{-3} \\
& \frac{x-\lambda}{2}=\frac{y+1}{4}=\frac{z-2}{-5}
\end{aligned}$$</p>
<p>the shortest distance between the lines</p>
<p>$$=\left|\frac{(\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}) \cdot\left(\overrightarrow{\mathrm{d}_1} \times \overrightarrow{\mathrm{d}_2}\right)}{\left|\overrightarrow{\mathrm{d}_1} \times \overrightarrow{\mathrm{d}_2}\right|}\right|$$</p>
<p>$$=\left|\frac{\left|\begin{array}{ccc}\lambda-4 & 0 & 2 \\ 1 & 2 & -3 \\ 2 & 4 & -5\end{array}\right|}{\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & -3 \\ 2 & 4 & -5\end{array}\right|}\right|$$</p>
<p>$$=\left|\frac{(\lambda-4)(-10+12)-0+2(4-4)}{|2 \hat{i}-1 \hat{j}+0 \hat{k}|}\right|$$</p>
<p>$$\begin{aligned}
& \frac{6}{\sqrt{5}}=\left|\frac{2(\lambda-4)}{\sqrt{5}}\right| \\
& 3=|\lambda-4| \\
& \lambda-4= \pm 3 \\
& \lambda=7,1
\end{aligned}$$</p>
<p>Sum of all possible values of $$\lambda$$ is $$=8$$</p> | mcq | jee-main-2024-online-27th-january-morning-shift |
jaoe38c1lscn35vz | maths | 3d-geometry | lines-in-space | <p>Let the image of the point $$(1,0,7)$$ in the line $$\frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3}$$ be the point $$(\alpha, \beta, \gamma)$$. Then which one of the following points lies on the line passing through $$(\alpha, \beta, \gamma)$$ and making angles $$\frac{2 \pi}{3}$$ and $$\frac{3 \pi}{4}$$ with $$y$$-axis and $$z$$-axis respectively and an acute angle with $$x$$-axis ?</p> | [{"identifier": "A", "content": "$$(1,-2,1+\\sqrt{2})$$\n"}, {"identifier": "B", "content": "$$(3,-4,3+2 \\sqrt{2})$$\n"}, {"identifier": "C", "content": "$$(3,4,3-2 \\sqrt{2})$$\n"}, {"identifier": "D", "content": "$$(1,2,1-\\sqrt{2})$$"}] | ["C"] | null | <p>$$\mathrm{L}_1=\frac{\mathrm{x}}{1}=\frac{\mathrm{y}-1}{2}=\frac{\mathrm{z}-2}{3}=\lambda$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lt1toimr/45988a05-1bb2-47f1-9316-43cb63fa6ccd/bf271320-d408-11ee-b9d5-0585032231f0/file-1lt1toimt.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lt1toimr/45988a05-1bb2-47f1-9316-43cb63fa6ccd/bf271320-d408-11ee-b9d5-0585032231f0/file-1lt1toimt.png" loading="lazy" style="max-width: 100%;height: auto;display: block;margin: 0 auto;max-height: 40vh;vertical-align: baseline" alt="JEE Main 2024 (Online) 27th January Evening Shift Mathematics - 3D Geometry Question 35 English Explanation"></p>
<p>$$\begin{aligned}
& \mathrm{M}(\lambda, 1+2 \lambda, 2+3 \lambda) \\
& \overrightarrow{\mathrm{PM}}=(\lambda-1) \hat{\mathrm{i}}+(1+2 \lambda) \hat{\mathrm{j}}+(3 \lambda-5) \hat{\mathrm{k}}
\end{aligned}$$</p>
<p>$$\overrightarrow{\mathrm{PM}}$$ is perpendicular to line $$\mathrm{L}_1$$</p>
<p>$$\begin{aligned}
& \overrightarrow{\mathrm{PM}} \overrightarrow{\mathrm{b}}=0 \quad(\overrightarrow{\mathrm{b}}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}) \\
& \Rightarrow \lambda-1+4 \lambda+2+9 \lambda-15=0 \\
& 14 \lambda=14 \Rightarrow \lambda=1 \\
& \therefore \mathrm{M}=(1,3,5) \\
& \overrightarrow{\mathrm{Q}}=2 \overrightarrow{\mathrm{M}}-\overrightarrow{\mathrm{P}}[\mathrm{M} \text { is midpoint of } \overrightarrow{\mathrm{P}} \& \overrightarrow{\mathrm{Q}}]
\end{aligned}$$</p>
<p>$$\begin{aligned}
& \overrightarrow{\mathrm{Q}}=2 \hat{\mathrm{i}}+6 \hat{\mathrm{j}}+10 \hat{\mathrm{k}}-\hat{\mathrm{i}}-7 \hat{\mathrm{k}} \\
& \overrightarrow{\mathrm{Q}}=\hat{\mathrm{i}}+6 \hat{\mathrm{j}}+3 \hat{\mathrm{k}} \\
& \therefore(\alpha, \beta, \gamma)=(1,6,3)
\end{aligned}$$</p>
<p>Required line having direction cosine $$(l, \mathrm{~m}, \mathrm{n})$$</p>
<p>$$\begin{aligned}
& l^2+m^2+n^2=1 \\
& \Rightarrow l^2+\left(-\frac{1}{2}\right)^2+\left(-\frac{1}{\sqrt{2}}\right)^2=1
\end{aligned}$$</p>
<p>$$l^2=\frac{1}{4}$$</p>
<p>$$\therefore l=\frac{1}{2}$$ [Line make acute angle with $$\mathrm{x}$$-axis]</p>
<p>Equation of line passing through $$(1,6,3)$$ will be</p>
<p>$$\overrightarrow{\mathrm{r}}=(\hat{\mathrm{i}}+6 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})+\mu\left(\frac{1}{2} \hat{\mathrm{i}}-\frac{1}{2} \hat{\mathrm{j}}-\frac{1}{\sqrt{2}} \hat{\mathrm{k}}\right)$$</p>
<p>Option (3) satisfying for $$\mu=4$$</p> | mcq | jee-main-2024-online-27th-january-evening-shift |
jaoe38c1lscogowm | maths | 3d-geometry | lines-in-space | <p>The lines $$\frac{x-2}{2}=\frac{y}{-2}=\frac{z-7}{16}$$ and $$\frac{x+3}{4}=\frac{y+2}{3}=\frac{z+2}{1}$$ intersect at the point $$P$$. If the distance of $$\mathrm{P}$$ from the line $$\frac{x+1}{2}=\frac{y-1}{3}=\frac{z-1}{1}$$ is $$l$$, then $$14 l^2$$ is equal to __________.</p> | [] | null | 108 | <p>$$\begin{aligned}
& \frac{\mathrm{x}-2}{1}=\frac{\mathrm{y}}{-1}=\frac{\mathrm{z}-7}{8}=\lambda \\
& \frac{\mathrm{x}+3}{4}=\frac{\mathrm{y}+2}{3}=\frac{\mathrm{z}+2}{1}=\mathrm{k} \\
& \Rightarrow \lambda+2=4 \mathrm{k}-3 \\
& -\lambda=3 \mathrm{k}-2 \\
& \Rightarrow \mathrm{k}=1, \lambda=-1 \\
& 8 \lambda+7=\mathrm{k}-2 \\
& \therefore \mathrm{P}=(1,1,-1)
\end{aligned}$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lt1xgjzw/7313cfc0-b5bf-4cc6-9a03-217f83a625f3/86406ed0-d417-11ee-b9d5-0585032231f0/file-1lt1xgjzx.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lt1xgjzw/7313cfc0-b5bf-4cc6-9a03-217f83a625f3/86406ed0-d417-11ee-b9d5-0585032231f0/file-1lt1xgjzx.png" loading="lazy" style="max-width: 100%;height: auto;display: block;margin: 0 auto;max-height: 40vh;vertical-align: baseline" alt="JEE Main 2024 (Online) 27th January Evening Shift Mathematics - 3D Geometry Question 36 English Explanation"></p>
<p>Projection of $$2 \hat{i}-2 \hat{k}$$ on $$2 \hat{i}+3 \hat{j}+\hat{k}$$ is</p>
<p>$$\begin{aligned}
& =\frac{4-2}{\sqrt{4+9+1}}=\frac{2}{\sqrt{14}} \\
& \therefore l^2=8-\frac{4}{14}=\frac{108}{14} \\
& \Rightarrow 14 l^2=108
\end{aligned}$$</p> | integer | jee-main-2024-online-27th-january-evening-shift |
jaoe38c1lsd3h061 | maths | 3d-geometry | lines-in-space | <p>Let $$(\alpha, \beta, \gamma)$$ be the mirror image of the point $$(2,3,5)$$ in the line $$\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$$. Then, $$2 \alpha+3 \beta+4 \gamma$$ is equal to</p> | [{"identifier": "A", "content": "32"}, {"identifier": "B", "content": "31"}, {"identifier": "C", "content": "33"}, {"identifier": "D", "content": "34"}] | ["C"] | null | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lsjwr9eg/7ae9597b-3dbe-4745-92d2-f1eee5f94c62/ab25ed80-ca2e-11ee-8854-3b5a6c9e9092/file-6y3zli1lsjwr9eh.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1lsjwr9eg/7ae9597b-3dbe-4745-92d2-f1eee5f94c62/ab25ed80-ca2e-11ee-8854-3b5a6c9e9092/file-6y3zli1lsjwr9eh.png" loading="lazy" style="max-width: 100%;height: auto;display: block;margin: 0 auto;max-height: 40vh;vertical-align: baseline" alt="JEE Main 2024 (Online) 31st January Evening Shift Mathematics - 3D Geometry Question 34 English Explanation"></p>
<p>$$\begin{aligned}
& \because \overrightarrow{\mathrm{PR}} \perp(2,3,4) \\
& \therefore \overrightarrow{\mathrm{PR}} \cdot(2,3,4)=0 \\
& (\alpha-2, \beta-3, \gamma-5) \cdot(2,3,4)=0 \\
& \Rightarrow 2 \alpha+3 \beta+4 \gamma=4+9+20=33
\end{aligned}$$</p> | mcq | jee-main-2024-online-31st-january-evening-shift |
jaoe38c1lsd4flem | maths | 3d-geometry | lines-in-space | <p>The shortest distance, between lines $$L_1$$ and $$L_2$$, where $$L_1: \frac{x-1}{2}=\frac{y+1}{-3}=\frac{z+4}{2}$$ and $$L_2$$ is the line, passing through the points $$\mathrm{A}(-4,4,3), \mathrm{B}(-1,6,3)$$ and perpendicular to the line $$\frac{x-3}{-2}=\frac{y}{3}=\frac{z-1}{1}$$, is</p> | [{"identifier": "A", "content": "$$\\frac{141}{\\sqrt{221}}$$\n"}, {"identifier": "B", "content": "$$\\frac{24}{\\sqrt{117}}$$\n"}, {"identifier": "C", "content": "$$\\frac{42}{\\sqrt{117}}$$\n"}, {"identifier": "D", "content": "$$\\frac{121}{\\sqrt{221}}$$"}] | ["A"] | null | <p>$$\begin{aligned}
& \mathrm{L}_2=\frac{\mathrm{x}+4}{3}=\frac{\mathrm{y}-4}{2}=\frac{\mathrm{z}-3}{0} \\
& \therefore \mathrm{S} . \mathrm{D}=\frac{\left|\begin{array}{ccc}
\mathrm{x}_2-\mathrm{x}_1 & \mathrm{y}_2-\mathrm{y}_1 & \mathrm{z}_2-\mathrm{z}_1 \\
2 & -3 & 2 \\
3 & 2 & 0
\end{array}\right|}{\left|\overrightarrow{\mathrm{n}_1} \times \overrightarrow{\mathrm{n}_2}\right|} \\
& =\frac{\left|\begin{array}{ccc}
5 & -5 & -7 \\
2 & -3 & 2 \\
3 & 2 & 0
\end{array}\right|}{\left|\overrightarrow{\mathrm{n}_1} \times \overrightarrow{\mathrm{n}_2}\right|} \\
& =\frac{141}{|-4 \hat{\mathrm{i}}+6 \hat{\mathrm{j}}+13 \hat{\mathrm{k}}|} \\
& =\frac{141}{\sqrt{16+36+169}} \\
& =\frac{141}{\sqrt{221}}
\end{aligned}$$</p> | mcq | jee-main-2024-online-31st-january-evening-shift |
jaoe38c1lsd5a4ux | maths | 3d-geometry | lines-in-space | <p>A line passes through $$A(4,-6,-2)$$ and $$B(16,-2,4)$$. The point $$P(a, b, c)$$, where $$a, b, c$$ are non-negative integers, on the line $$A B$$ lies at a distance of 21 units, from the point $$A$$. The distance between the points $$P(a, b, c)$$ and $$Q(4,-12,3)$$ is equal to __________.</p> | [] | null | 22 | <p>$$\begin{aligned}
& \frac{x-4}{12}=\frac{x+6}{4}=\frac{z+2}{6} \\
& \frac{x-4}{\frac{6}{7}}=\frac{y+6}{\frac{2}{7}}=\frac{z+2}{\frac{3}{7}}=21 \\
& \left(21 \times \frac{6}{7}+4, \frac{2}{7} \times 21-6, \frac{3}{7} \times 21-2\right) \\
& =(22,0,7)=(a, b, c) \\
& \therefore \sqrt{324+144+16}=22
\end{aligned}$$</p> | integer | jee-main-2024-online-31st-january-evening-shift |
jaoe38c1lse5wwlu | maths | 3d-geometry | lines-in-space | <p>Let $$\mathrm{Q}$$ and $$\mathrm{R}$$ be the feet of perpendiculars from the point $$\mathrm{P}(a, a, a)$$ on the lines $$x=y, z=1$$ and $$x=-y, z=-1$$ respectively. If $$\angle \mathrm{QPR}$$ is a right angle, then $$12 a^2$$ is equal to _________.</p> | [] | null | 12 | <p>$$\begin{aligned}
& \frac{x}{1}=\frac{y}{1}=\frac{z-1}{0}=r \rightarrow Q(r, r, 1) \\
& \frac{x}{1}=\frac{y}{-1}=\frac{z+1}{0}=k \rightarrow R(k,-k,-1) \\
& \overline{P Q}=(a-r) \hat{i}+(a-r) \hat{j}+(a-1) \hat{k} \\
& a=r+a-r=0 \\
& 2 a=2 r \rightarrow a=r \\
& \overline{P R}=(a-k) i+(a+k) \hat{j}+(a+1) \hat{k} \\
& a-k-a-k=0 \Rightarrow k=0 \\
& A s, P Q \perp P R \\
& (a-r)(a-k)+(a-r)(a+k)+(a-1)(a+1)=0 \\
& a=1 \text { or }-1 \\
& 12 a^2=12
\end{aligned}$$</p> | integer | jee-main-2024-online-31st-january-morning-shift |
jaoe38c1lsf08uic | maths | 3d-geometry | lines-in-space | <p>Let $$P Q R$$ be a triangle with $$R(-1,4,2)$$. Suppose $$M(2,1,2)$$ is the mid point of $$\mathrm{PQ}$$. The distance of the centroid of $$\triangle \mathrm{PQR}$$ from the point of intersection of the lines $$\frac{x-2}{0}=\frac{y}{2}=\frac{z+3}{-1}$$ and $$\frac{x-1}{1}=\frac{y+3}{-3}=\frac{z+1}{1}$$ is</p> | [{"identifier": "A", "content": "69"}, {"identifier": "B", "content": "$$\\sqrt{99}$$"}, {"identifier": "C", "content": "$$\\sqrt{69}$$"}, {"identifier": "D", "content": "9"}] | ["C"] | null | <p>Centroid $$G$$ divides MR in $$1: 2$$</p>
<p>$$\mathrm{G}(1,2,2)$$</p>
<p>Point of intersection $$A$$ of given lines is $$(2,-6,0)$$</p>
<p>$$\mathrm{AG}=\sqrt{69}$$</p> | mcq | jee-main-2024-online-29th-january-morning-shift |
jaoe38c1lsf0lnlm | maths | 3d-geometry | lines-in-space | <p>A line with direction ratios $$2,1,2$$ meets the lines $$x=y+2=z$$ and $$x+2=2 y=2 z$$ respectively at the points $$\mathrm{P}$$ and $$\mathrm{Q}$$. If the length of the perpendicular from the point $$(1,2,12)$$ to the line $$\mathrm{PQ}$$ is $$l$$, then $$l^2$$ is __________.</p> | [] | null | 65 | <p>Let $$\mathrm{P}(\mathrm{t}, \mathrm{t}-2, \mathrm{t})$$ and $$\mathrm{Q}(2 \mathrm{~s}-2, \mathrm{~s}, \mathrm{~s})$$</p>
<p>D.R's of PQ are 2, 1, 2</p>
<p>$$\begin{aligned}
& \frac{2 \mathrm{~s}-2-\mathrm{t}}{2}=\frac{\mathrm{s}-\mathrm{t}+2}{1}=\frac{\mathrm{s}-\mathrm{t}}{2} \\
& \Rightarrow \mathrm{t}=6 \text { and } \mathrm{s}=2 \\
& \Rightarrow \mathrm{P}(6,4,6) \text { and } \mathrm{Q}(2,2,2) \\
& \mathrm{PQ}: \frac{\mathrm{x}-2}{2}=\frac{\mathrm{y}-2}{1}=\frac{\mathrm{z}-2}{2}=\lambda
\end{aligned}$$</p>
<p>Let $$\mathrm{F}(2 \lambda+2, \lambda+2,2 \lambda+2)$$</p>
<p>$$\mathrm{A}(1,2,12)$$</p>
<p>$$\overrightarrow{\mathrm{AF}} \cdot \overrightarrow{\mathrm{PQ}}=0$$</p>
<p>$$\therefore \lambda=2$$</p>
<p>So $$F(6,4,6)$$ and $$A F=\sqrt{65}$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lt30pn5h/00c97644-b936-4207-92b6-5fa8af4c0084/0701ea50-d4b1-11ee-8384-811001421c41/file-1lt30pn5i.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lt30pn5h/00c97644-b936-4207-92b6-5fa8af4c0084/0701ea50-d4b1-11ee-8384-811001421c41/file-1lt30pn5i.png" loading="lazy" style="max-width: 100%;height: auto;display: block;margin: 0 auto;max-height: 40vh;vertical-align: baseline" alt="JEE Main 2024 (Online) 29th January Morning Shift Mathematics - 3D Geometry Question 29 English Explanation"></p> | integer | jee-main-2024-online-29th-january-morning-shift |
jaoe38c1lsfkztkt | maths | 3d-geometry | lines-in-space | <p>Let O be the origin, and M and $$\mathrm{N}$$ be the points on the lines $$\frac{x-5}{4}=\frac{y-4}{1}=\frac{z-5}{3}$$ and $$\frac{x+8}{12}=\frac{y+2}{5}=\frac{z+11}{9}$$ respectively such that $$\mathrm{MN}$$ is the shortest distance between the given lines. Then $$\overrightarrow{O M} \cdot \overrightarrow{O N}$$ is equal to _________.</p> | [] | null | 9 | <p>$$\mathrm{L}_1: \frac{\mathrm{x}-5}{4}=\frac{\mathrm{y}-4}{1}=\frac{\mathrm{z}-5}{3}=\lambda \quad \operatorname{drs}(4,1,3)=\mathrm{b}_1$$</p>
<p>$$\begin{aligned}
& \mathrm{M}(4 \lambda+5, \lambda+4,3 \lambda+5) \\
& \mathrm{L}_2: \frac{\mathrm{x}+8}{12}=\frac{\mathrm{y}+2}{5}=\frac{\mathrm{z}+11}{9}=\mu \\
& \mathrm{N}(12 \mu-8,5 \mu-2,9 \mu-11) \\
& \overrightarrow{\mathrm{MN}}=(4 \lambda-12 \mu+13, \lambda-5 \mu+6,3 \lambda-9 \mu+16) \quad \text{..... (1)}
\end{aligned}$$</p>
<p>Now</p>
<p>$$\overrightarrow{\mathrm{b}}_1 \times \overrightarrow{\mathrm{b}}_2=\left|\begin{array}{ccc}
\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\
4 & 1 & 3 \\
12 & 5 & 9
\end{array}\right|=-6 \hat{\mathrm{i}}+8 \hat{\mathrm{k}} \quad \text{.... (2)}$$</p>
<p>Equation (1) and (2)</p>
<p>$$\therefore \frac{4 \lambda-12 \mu+13}{-6}=\frac{\lambda-5 \mu+6}{0}=\frac{3 \lambda-9 \mu+16}{8}$$</p>
<p>I and II</p>
<p>$$\lambda-5 \mu+6=0 \quad \text{.... (3)}$$</p>
<p>I and III</p>
<p>$$\lambda-3 \mu+4=0 \quad \text{.... (4)}$$</p>
<p>Solve (3) and (4) we get</p>
<p>$$\begin{aligned}
\lambda= & -1, \mu=1 \\
\therefore \quad & \mathrm{M}(1,3,2) \\
& \mathrm{N}(4,3,-2) \\
\therefore \quad & \overrightarrow{\mathrm{OM}} \cdot \overrightarrow{\mathrm{ON}}=4+9-4=9
\end{aligned}$$</p> | integer | jee-main-2024-online-29th-january-evening-shift |
1lsg4acyi | maths | 3d-geometry | lines-in-space | <p>Let $$L_1: \vec{r}=(\hat{i}-\hat{j}+2 \hat{k})+\lambda(\hat{i}-\hat{j}+2 \hat{k}), \lambda \in \mathbb{R}$$,</p>
<p>$$L_2: \vec{r}=(\hat{j}-\hat{k})+\mu(3 \hat{i}+\hat{j}+p \hat{k}), \mu \in \mathbb{R} \text {, and } L_3: \vec{r}=\delta(\ell \hat{i}+m \hat{j}+n \hat{k}), \delta \in \mathbb{R}$$</p>
<p>be three lines such that $$L_1$$ is perpendicular to $$L_2$$ and $$L_3$$ is perpendicular to both $$L_1$$ and $$L_2$$. Then, the point which lies on $$L_3$$ is</p> | [{"identifier": "A", "content": "$$(1,7,-4)$$\n"}, {"identifier": "B", "content": "$$(1,-7,4)$$\n"}, {"identifier": "C", "content": "$$(-1,7,4)$$\n"}, {"identifier": "D", "content": "$$(-, 1-7,4)$$"}] | ["C"] | null | <p>$$\mathrm{L}_1 \perp \mathrm{L}_2 \quad \mathrm{~L}_3 \perp \mathrm{L}_1, \mathrm{~L}_2$$</p>
<p>$$\begin{aligned}
& 3-1+2 \mathrm{P}=0 \\
& \mathrm{P}=-1 \\
& \left|\begin{array}{ccc}
\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\
1 & -1 & 2 \\
3 & 1 & -1
\end{array}\right|=-\hat{\mathrm{i}}+7 \hat{\mathrm{j}}+4 \hat{\mathrm{k}} \\
& \therefore(-\delta, 7 \delta, 4 \delta) \text { will lie on } \mathrm{L}_3
\end{aligned}$$</p>
<p>For $$\delta=1$$ the point will be $$(-1,7,4)$$</p> | mcq | jee-main-2024-online-30th-january-evening-shift |
1lsg550xe | maths | 3d-geometry | lines-in-space | <p>Let a line passing through the point $$(-1,2,3)$$ intersect the lines $$L_1: \frac{x-1}{3}=\frac{y-2}{2}=\frac{z+1}{-2}$$ at $$M(\alpha, \beta, \gamma)$$ and $$L_2: \frac{x+2}{-3}=\frac{y-2}{-2}=\frac{z-1}{4}$$ at $$N(a, b, c)$$. Then, the value of $$\frac{(\alpha+\beta+\gamma)^2}{(a+b+c)^2}$$ equals __________.</p> | [] | null | 196 | <p>$$\begin{aligned}
& \mathrm{M}(3 \lambda+1,2 \lambda+2,-2 \lambda-1) \quad \therefore \alpha+\beta+\gamma=3 \lambda+2 \\
& \mathrm{~N}(-3 \mu-2,-2 \mu+2,4 \mu+1) \quad \therefore \mathrm{a}+\mathrm{b}+\mathrm{c}=-\mu+1
\end{aligned}$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lsoxte3w/356f1335-5223-47e5-b5cc-ef41c9b2c27b/bac74ec0-ccf2-11ee-a330-494dca5e9a63/file-6y3zli1lsoxte3x.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1lsoxte3w/356f1335-5223-47e5-b5cc-ef41c9b2c27b/bac74ec0-ccf2-11ee-a330-494dca5e9a63/file-6y3zli1lsoxte3x.png" loading="lazy" style="max-width: 100%;height: auto;display: block;margin: 0 auto;max-height: 40vh;vertical-align: baseline" alt="JEE Main 2024 (Online) 30th January Evening Shift Mathematics - 3D Geometry Question 25 English Explanation"></p>
<p>$$\begin{aligned}
& \frac{3 \lambda+2}{-3 \mu-1}=\frac{2 \lambda}{-2 \mu}=\frac{-2 \lambda-4}{4 \mu-2} \\
& 3 \lambda \mu+2 \mu=3 \lambda \mu+\lambda \\
& 2 \mu=\lambda \\
& 2 \lambda \mu-\lambda=\lambda \mu+2 \mu \\
& \lambda \mu=\lambda+2 \mu \\
&\Rightarrow \lambda \mu=2 \lambda
\end{aligned}$$</p>
<p>$$\begin{aligned}
\Rightarrow \quad & \mu=2 \quad(\lambda \neq 0) \\
\therefore \quad & \lambda=4 \\
& \alpha+\beta+\gamma=14 \\
& a+b+c=-1 \\
& \frac{(\alpha+\beta+\gamma)^2}{(a+b+c)^2}=196
\end{aligned}$$</p> | integer | jee-main-2024-online-30th-january-evening-shift |
1lsg95okx | maths | 3d-geometry | lines-in-space | <p>Let $$(\alpha, \beta, \gamma)$$ be the foot of perpendicular from the point $$(1,2,3)$$ on the line $$\frac{x+3}{5}=\frac{y-1}{2}=\frac{z+4}{3}$$. Then $$19(\alpha+\beta+\gamma)$$ is equal to :</p> | [{"identifier": "A", "content": "99"}, {"identifier": "B", "content": "102"}, {"identifier": "C", "content": "101"}, {"identifier": "D", "content": "100"}] | ["C"] | null | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lsqnidnf/3617c118-3469-40cf-b021-2c95fa43dbe8/fe8826b0-cde3-11ee-a0d3-7b75c4537559/file-6y3zli1lsqnidng.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1lsqnidnf/3617c118-3469-40cf-b021-2c95fa43dbe8/fe8826b0-cde3-11ee-a0d3-7b75c4537559/file-6y3zli1lsqnidng.png" loading="lazy" style="max-width: 100%;height: auto;display: block;margin: 0 auto;max-height: 40vh;vertical-align: baseline" alt="JEE Main 2024 (Online) 30th January Morning Shift Mathematics - 3D Geometry Question 22 English Explanation"></p>
<p>Let foot $$P(5 k-3,2 k+1,3 k-4)$$</p>
<p>DR's $$\rightarrow$$ AP : $$5 \mathrm{k}-4,2 \mathrm{k}-1,3 \mathrm{k}-7$$</p>
<p>DR's $$\rightarrow$$ Line: $$5,2,3$$</p>
<p>Condition of perpendicular lines $$(25 k-20)+(4 k-2)+(9 k-21)=0$$</p>
<p>Then $$\mathrm{k}=\frac{43}{38}$$</p>
<p>Then $$19(\alpha+\beta+\gamma)=\mathbf{1 0 1}$$</p> | mcq | jee-main-2024-online-30th-january-morning-shift |
1lsgclxib | maths | 3d-geometry | lines-in-space | <p>If $$\mathrm{d}_1$$ is the shortest distance between the lines $$x+1=2 y=-12 z, x=y+2=6 z-6$$ and $$\mathrm{d}_2$$ is the shortest distance between the lines $$\frac{x-1}{2}=\frac{y+8}{-7}=\frac{z-4}{5}, \frac{x-1}{2}=\frac{y-2}{1}=\frac{z-6}{-3}$$, then the value of $$\frac{32 \sqrt{3} \mathrm{~d}_1}{\mathrm{~d}_2}$$ is :</p> | [] | null | 16 | <p>$$\mathrm{L}_1: \frac{\mathrm{x}+1}{1}=\frac{\mathrm{y}}{1 / 2}=\frac{\mathrm{z}}{-1 / 12}, \mathrm{~L}_2: \frac{\mathrm{x}}{1}=\frac{\mathrm{y}+2}{1}=\frac{\mathrm{z}-1}{\frac{1}{6}}$$</p>
<p>$$\mathrm{d}_1=$$ shortest distance between $$\mathrm{L}_1 ~\& \mathrm{~L}_2$$</p>
<p>$$\begin{aligned}
& =\left|\frac{\left(\vec{a}_2-\vec{a}_1\right) \cdot\left(\vec{b}_1 \times \vec{b}_2\right)}{\left|\left(\vec{b}_1 \times \vec{b}_2\right)\right|}\right| \\
& d_1=2 \\
& \mathrm{~L}_3: \frac{\mathrm{x}-1}{2}=\frac{\mathrm{y}+8}{-7}=\frac{\mathrm{z}-4}{5}, \mathrm{~L}_4: \frac{\mathrm{x}-1}{2}=\frac{\mathrm{y}-2}{1}=\frac{\mathrm{z}-6}{-3} \\
& \mathrm{~d}_2=\text { shortest distance between } \mathrm{L}_3 ~\& \mathrm{~L}_4 \\
& \mathrm{~d}_2=\frac{12}{\sqrt{3}} \text { Hence } \\
& =\frac{32 \sqrt{3} \mathrm{~d}_1}{\mathrm{~d}_2}=\frac{32 \sqrt{3} \times 2}{\frac{12}{\sqrt{3}}}=16
\end{aligned}$$</p> | integer | jee-main-2024-online-30th-january-morning-shift |
luxwcs1z | maths | 3d-geometry | lines-in-space | <p>Consider the line $$\mathrm{L}$$ passing through the points $$(1,2,3)$$ and $$(2,3,5)$$. The distance of the point $$\left(\frac{11}{3}, \frac{11}{3}, \frac{19}{3}\right)$$ from the line $$\mathrm{L}$$ along the line $$\frac{3 x-11}{2}=\frac{3 y-11}{1}=\frac{3 z-19}{2}$$ is equal to</p> | [{"identifier": "A", "content": "6"}, {"identifier": "B", "content": "3"}, {"identifier": "C", "content": "5"}, {"identifier": "D", "content": "4"}] | ["B"] | null | <p>$$L: \frac{x-1}{1}=\frac{y-2}{1}=\frac{z-3}{2}=\mu$$</p>
<p>Measured along $$L_2: \frac{x-\frac{11}{3}}{\frac{2}{3}}=\frac{y-\frac{11}{3}}{\frac{1}{3}}=\frac{z-\frac{19}{3}}{\frac{2}{3}}=\lambda$$</p>
<p>Any point on $$L_1:(\mu+1, \mu+2,2 \mu+3)$$</p>
<p>Any point on $$L_2\left(\frac{2}{3} \lambda+\frac{11}{3}, \frac{\lambda}{3}+\frac{11}{3}, \frac{2}{3} \lambda+\frac{19}{3}\right)$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lw1mpu79/db1f9115-bd83-4afa-bbc1-7bdbc6f490b3/2a21d550-0f52-11ef-91cd-f19f7dc20f18/file-1lw1mpu7a.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lw1mpu79/db1f9115-bd83-4afa-bbc1-7bdbc6f490b3/2a21d550-0f52-11ef-91cd-f19f7dc20f18/file-1lw1mpu7a.png" loading="lazy" style="max-width: 100%;height: auto;display: block;margin: 0 auto;max-height: 40vh;vertical-align: baseline" alt="JEE Main 2024 (Online) 9th April Evening Shift Mathematics - 3D Geometry Question 20 English Explanation"></p>
<p>Now</p>
<p>$$\begin{aligned}
& \mu+1=\frac{2}{3} \lambda+\frac{11}{3} \\
& \frac{\mu+2=\frac{\lambda}{3}+\frac{11}{3}}{\lambda=-3} \\
& \mu=\frac{2}{3} \\
\end{aligned}$$</p>
<p>$$\begin{aligned}
& \text { Point on } L=\left(\frac{5}{3}, \frac{8}{3}, \frac{13}{3}\right) \\
& d=\sqrt{\left(\frac{11}{3}-\frac{5}{3}\right)^2+\left(\frac{8}{3}-\frac{11}{3}\right)^2+\left(\frac{19}{3}-\frac{13}{3}\right)^2} \\
& d=\sqrt{4+1+4} \\
& d=3
\end{aligned}$$</p> | mcq | jee-main-2024-online-9th-april-evening-shift |
luxwden8 | maths | 3d-geometry | lines-in-space | <p>The square of the distance of the image of the point $$(6,1,5)$$ in the line $$\frac{x-1}{3}=\frac{y}{2}=\frac{z-2}{4}$$, from the origin is __________.</p> | [] | null | 62 | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lw1my772/a7f8567c-57ab-46bd-a498-b733b7e930a6/12a4b0e0-0f53-11ef-91cd-f19f7dc20f18/file-1lw1my773.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lw1my772/a7f8567c-57ab-46bd-a498-b733b7e930a6/12a4b0e0-0f53-11ef-91cd-f19f7dc20f18/file-1lw1my773.png" loading="lazy" style="max-width: 100%;height: auto;display: block;margin: 0 auto;max-height: 40vh;vertical-align: baseline" alt="JEE Main 2024 (Online) 9th April Evening Shift Mathematics - 3D Geometry Question 19 English Explanation"></p>
<p>$$\begin{aligned}
& \overrightarrow{P A}=(3 \lambda-5) \hat{i}+(2 \lambda-1) \hat{j}+(4 \lambda-3) \hat{k} \\
& (3 \lambda-5) 3+(2 \lambda-1) 2+(4 \lambda-3) 4=0 \\
& \Rightarrow 9 \lambda-15+4 \lambda-2+16 \lambda-12=0 \\
& \Rightarrow 29 \lambda=29 \\
& \therefore \lambda=1 \\
& \therefore \quad A(4,2,6) \\
& \therefore \quad P^{\prime}: \text { mirror image of } P \\
& \Rightarrow P^{\prime}(2,3,7) \\
& \left(O P^{\prime}\right)^2=4+9+49 \\
& =62 \\
&
\end{aligned}$$</p> | integer | jee-main-2024-online-9th-april-evening-shift |
luy6z5hh | maths | 3d-geometry | lines-in-space | <p>The shortest distance between the lines $$\frac{x-3}{4}=\frac{y+7}{-11}=\frac{z-1}{5}$$ and $$\frac{x-5}{3}=\frac{y-9}{-6}=\frac{z+2}{1}$$ is:</p> | [{"identifier": "A", "content": "$$\\frac{185}{\\sqrt{563}}$$\n"}, {"identifier": "B", "content": "$$\\frac{187}{\\sqrt{563}}$$\n"}, {"identifier": "C", "content": "$$\\frac{178}{\\sqrt{563}}$$\n"}, {"identifier": "D", "content": "$$\\frac{179}{\\sqrt{563}}$$"}] | ["B"] | null | <p>Given lines are</p>
<p>$$\frac{x-3}{4}=\frac{y-(-7)}{-11}=\frac{z-1}{5}$$ and</p>
<p>$$\frac{x-5}{3}=\frac{y-9}{-6}=\frac{z-(-2)}{1}$$</p>
<p>Shortest distance between two lines,</p>
<p>$$\begin{aligned}
& d=\frac{\left|\left(\vec{a}_2-\vec{a}_1\right) \cdot\left(\vec{b}_1 \times \vec{b}_2\right)\right|}{\left|\left(\vec{b}_1 \times \vec{b}_2\right)\right|} \\
& \vec{a}_2-\vec{a}_1=2 \hat{i}+16 \hat{j}-3 \hat{k} \text { and } \\
& \vec{b}_1 \times \vec{b}_2=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
4 & -11 & 5 \\
3 & -6 & 1
\end{array}\right|=19 \hat{i}+11 \hat{j}+9 \hat{k} \\
& \therefore \quad d=\frac{187}{\sqrt{563}}
\end{aligned}$$</p> | mcq | jee-main-2024-online-9th-april-morning-shift |
luy6z4qf | maths | 3d-geometry | lines-in-space | <p>Let the line $$\mathrm{L}$$ intersect the lines $$x-2=-y=z-1,2(x+1)=2(y-1)=z+1$$ and be parallel to the line $$\frac{x-2}{3}=\frac{y-1}{1}=\frac{z-2}{2}$$. Then which of the following points lies on $$\mathrm{L}$$ ?</p> | [{"identifier": "A", "content": "$$\\left(-\\frac{1}{3}, 1,-1\\right)$$\n"}, {"identifier": "B", "content": "$$\\left(-\\frac{1}{3},-1,1\\right)$$\n"}, {"identifier": "C", "content": "$$\\left(-\\frac{1}{3},-1,-1\\right)$$\n"}, {"identifier": "D", "content": "$$\\left(-\\frac{1}{3}, 1,1\\right)$$"}] | ["A"] | null | <p>$$\begin{aligned}
& L_1: \frac{x-2}{1}=\frac{y}{-1}=\frac{z-1}{1}=\lambda \\
& L_2: \frac{x+1}{(1 / 2)}=\frac{y-1}{(1 / 2)}=\frac{z+1}{1}=\mu
\end{aligned}$$</p>
<p>Any point of $$L_1$$ and $$L_2$$ will be $$(\lambda+2,-\lambda, \lambda+1)$$ and
$$\left(\frac{\mu}{2}-1, \frac{\mu}{2}+1, \mu-1\right)$$</p>
<p>Now Dr of line $$<\lambda-\frac{\mu}{2}+3,-\lambda-\frac{\mu}{2}-1, \lambda-\mu+2>$$</p>
<p>Now $$\frac{\lambda-\frac{\mu}{3}+3}{3}=\frac{-\lambda-\frac{\mu}{2}-1}{1}=\frac{\lambda-\mu+2}{2}$$</p>
<p>$$\left. {\matrix{
{\lambda - {\mu \over 3} + 3 = - 3\lambda - {{3\mu } \over 2} - 3\,\,\,...(1)} \cr
{2\left( {\lambda - {\mu \over 3} + 3} \right) = 3(\lambda - \mu + 2)\,\,...(2)} \cr
} } \right\}\lambda = {{ - 4} \over 3},\mu = {{ - 2} \over 3}$$</p>
<p>$$\therefore \quad$$ Points will be $$\left(\frac{2}{3}, \frac{4}{3}, \frac{-1}{3}\right)$$ and $$\left(\frac{-4}{3}, \frac{2}{3}, \frac{-5}{3}\right)$$</p>
<p>$$\therefore \quad L$$ will be $$\frac{x-\frac{2}{3}}{3}=\frac{y-\frac{4}{3}}{1}=\frac{z+\frac{1}{3}}{2}$$</p>
<p>$$\therefore \quad\left(\frac{-1}{3}, 1,-1\right)$$ will satisfy $$L$$</p> | mcq | jee-main-2024-online-9th-april-morning-shift |
lv0vxcns | maths | 3d-geometry | lines-in-space | <p>Let the point, on the line passing through the points $$P(1,-2,3)$$ and $$Q(5,-4,7)$$, farther from the origin and at a distance of 9 units from the point $$P$$, be $$(\alpha, \beta, \gamma)$$. Then $$\alpha^2+\beta^2+\gamma^2$$ is equal to :</p> | [{"identifier": "A", "content": "150"}, {"identifier": "B", "content": "155"}, {"identifier": "C", "content": "160"}, {"identifier": "D", "content": "165"}] | ["B"] | null | <p>Line through $$P Q$$</p>
<p>$$\frac{x-1}{4}=\frac{y+2}{-2}=\frac{z-3}{4}$$</p>
<p>Any point on $$P Q$$. be $$R(4 \lambda+1,-2 \lambda-2,4 \lambda+3)$$</p>
</p>$$P R=9$$ unit</p>
<p>$$(P R)^2=81$$</p>
<p>$$(4 \lambda+1-1)^2+(-2 \lambda-2+2)^2+(4 \lambda+3-3)^2=81$$</p>
<p>$$16 \lambda^2+4 \lambda^2+16 \lambda^2=81$$</p>
<p>$$36 \lambda^2=81$$</p>
<p>$$\lambda= \pm \frac{9}{6}= \pm \frac{3}{2}$$</p>
<p>$$\therefore R$$ can be $$(7,-5,9)$$ or $$(-5,1,-3)$$</p>
<p>Distance from origin for both points be $$\sqrt{49+25+81}$$ and $$\sqrt{25+1+9}=\sqrt{35}$$</p>
<p>$$\therefore$$ Distance of $$(7,-5,9)$$ is farthest from origin</p>
<p>$$\therefore(\alpha, \beta, \gamma)=(7,-5,9)$$</p>
<p>Now $$7^2+(-5)^2+9^2=155$$</p> | mcq | jee-main-2024-online-4th-april-morning-shift |
lv2er3nv | maths | 3d-geometry | lines-in-space | <p>Let $$\mathrm{P}$$ be the point of intersection of the lines $$\frac{x-2}{1}=\frac{y-4}{5}=\frac{z-2}{1}$$ and $$\frac{x-3}{2}=\frac{y-2}{3}=\frac{z-3}{2}$$. Then, the shortest distance of $$\mathrm{P}$$ from the line $$4 x=2 y=z$$ is </p> | [{"identifier": "A", "content": "$$\\frac{3 \\sqrt{14}}{7}$$\n"}, {"identifier": "B", "content": "$$\\frac{5 \\sqrt{14}}{7}$$\n"}, {"identifier": "C", "content": "$$\\frac{\\sqrt{14}}{7}$$\n"}, {"identifier": "D", "content": "$$\\frac{6 \\sqrt{14}}{7}$$"}] | ["A"] | null | <p>$$\begin{aligned}
& L_1: \frac{x-2}{1}=\frac{y-4}{5}=\frac{z-2}{1} \\
& L_2: \frac{x-3}{2}=\frac{y-2}{3}=\frac{z-3}{2}
\end{aligned}$$</p>
<p>Point of intersection of $$L_1$$ and $$L_2$$ is $$(-1,1,-1)$$</p>
<p>Distance of point $$P$$ from $$L_3: 4 x=2 y=z$$</p>
<p>$$L_3: \frac{x}{\frac{1}{4}}=\frac{y}{\frac{1}{2}}=\frac{z}{1}$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lwhgb1g0/feeaa3f8-43fb-402d-a95e-980c0569abe3/a234b100-1805-11ef-b156-f754785ad3ce/file-1lwhgb1g1.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lwhgb1g0/feeaa3f8-43fb-402d-a95e-980c0569abe3/a234b100-1805-11ef-b156-f754785ad3ce/file-1lwhgb1g1.png" loading="lazy" style="max-width: 100%;height: auto;display: block;margin: 0 auto;max-height: 40vh;vertical-align: baseline" alt="JEE Main 2024 (Online) 4th April Evening Shift Mathematics - 3D Geometry Question 15 English Explanation"></p>
<p>Any point on $$L_3$$ be</p>
<p>$$\begin{aligned}
& \left(\frac{\lambda}{4}, \frac{\lambda}{2}, \lambda\right) \\
& P R:\left\langle\frac{\lambda}{4}+1, \frac{\lambda}{2}-1, \lambda+1\right\rangle \\
& \because P R \perp\left\langle\frac{1}{4}, \frac{1}{2}, 1\right\rangle
\end{aligned}$$</p>
<p>$$\begin{aligned}
& \Rightarrow\left(\frac{\lambda}{4}+1\right) \frac{1}{4}+\frac{1}{2}\left(\frac{\lambda}{2}-1\right)+\lambda+1=0 \\
& \quad \frac{\lambda}{16}+\frac{1}{4}+\frac{\lambda}{4}-\frac{1}{2}+\lambda+1=0 \\
& \Rightarrow \quad \lambda=\frac{-4}{7} \\
& \therefore \quad R\left(\frac{-1}{7}, \frac{-2}{7}, \frac{-4}{7}\right) \\
& \text { Now } R P: \sqrt{\left(\frac{-1}{7}+1\right)^2+\left(\frac{-2}{7}-1\right)^2+\left(\frac{-4}{7}+1\right)^2} \\
& \quad=\sqrt{\frac{36}{49}+\frac{81}{49}+\frac{9}{49}}=\frac{\sqrt{126}}{7}=\frac{3 \sqrt{14}}{7}
\end{aligned}$$</p> | mcq | jee-main-2024-online-4th-april-evening-shift |
lv2er9tw | maths | 3d-geometry | lines-in-space | <p>Consider a line $$\mathrm{L}$$ passing through the points $$\mathrm{P}(1,2,1)$$ and $$\mathrm{Q}(2,1,-1)$$. If the mirror image of the point $$\mathrm{A}(2,2,2)$$ in the line $$\mathrm{L}$$ is $$(\alpha, \beta, \gamma)$$, then $$\alpha+\beta+6 \gamma$$ is equal to __________.</p> | [] | null | 6 | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lwhk80oi/a0341b0b-ecfa-449a-9d8f-e828635e1020/f2b47520-1814-11ef-a08c-4f00e8f9a4da/file-1lwhk80oj.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lwhk80oi/a0341b0b-ecfa-449a-9d8f-e828635e1020/f2b47520-1814-11ef-a08c-4f00e8f9a4da/file-1lwhk80oj.png" loading="lazy" style="max-width: 100%;height: auto;display: block;margin: 0 auto;max-height: 40vh;vertical-align: baseline" alt="JEE Main 2024 (Online) 4th April Evening Shift Mathematics - 3D Geometry Question 14 English Explanation"></p>
<p>$$\begin{aligned}
& A(2,2,2) \\
& P Q: \frac{x-1}{1}=\frac{y-2}{-1}=\frac{z-1}{-2}
\end{aligned}$$</p>
<p>General point,</p>
<p>$$(k+1,-k+2,-2 k+1)$$</p>
<p>$$\begin{aligned}
& \overrightarrow{O A}=(k+1-2) \hat{i}+(-k+2-2) \hat{j}+(-2 k+1-2) \hat{k} \\
& \overrightarrow{O A}=(k-1) \hat{i}-k \hat{j}+(-2 k-1) \hat{k} \\
& \overrightarrow{P Q}=1 \hat{i}-\hat{j}-2 \hat{k} \\
& \overrightarrow{O A} \cdot \overrightarrow{P Q}=0 \\
& (k-1)+k+2(2 k+1)=0 \\
& k-1+k+4 k+2=0 \\
& 6 k+1=0 \\
& k=\frac{-1}{6} \\
& 0\left(\frac{-1}{6}+1, \frac{+1}{6}+2,-2\left(\frac{-1}{6}\right)+1\right) \\
& 0\left(\frac{5}{6}, \frac{13}{6}, \frac{-8}{6}\right) \\
& 0\left(\frac{5}{6}, \frac{13}{6}, \frac{8}{6}\right)=\left(\frac{\alpha+2}{2}, \frac{\beta+2}{2}, \frac{\gamma+2}{2}\right)
\end{aligned}$$</p>
<p>$$\begin{array}{l|l}
\alpha+2=\frac{10}{6} & \beta+2=\frac{26}{6} \\
\alpha=\frac{10}{6}-2 & \beta=\frac{26-12}{6} \\
\alpha=\frac{-2}{6} & \beta=\frac{14}{6} \\
\alpha =-\frac{1}{3} & \beta=\frac{7}{3}
\end{array}$$</p>
<p>$$\begin{aligned}
& \gamma+2=\frac{16}{6} \\
& \gamma=\frac{16-12}{6} \\
& \gamma=\frac{4}{6} \\
& \Rightarrow \alpha+\beta+6 \gamma \\
& \Rightarrow \frac{-1}{3}+\frac{7}{3}+6 \times \frac{4}{6} \\
& \Rightarrow 2+4=6
\end{aligned}$$</p> | integer | jee-main-2024-online-4th-april-evening-shift |
lv3vef5l | maths | 3d-geometry | lines-in-space | <p>If the shortest distance between the lines $$\frac{x-\lambda}{2}=\frac{y-4}{3}=\frac{z-3}{4}$$ and $$\frac{x-2}{4}=\frac{y-4}{6}=\frac{z-7}{8}$$ is $$\frac{13}{\sqrt{29}}$$, then a value of $$\lambda$$ is :</p> | [{"identifier": "A", "content": "$$\\frac{13}{25}$$\n"}, {"identifier": "B", "content": "1"}, {"identifier": "C", "content": "$$-$$1\n"}, {"identifier": "D", "content": "$$-\\frac{13}{25}$$"}] | ["B"] | null | <p>$$\begin{aligned}
& \vec{l}_1=2 \hat{i}+3 \hat{j}+4 \hat{k} \\
& \vec{l}_2=4 \hat{i}+6 \hat{j}+8 \hat{k}
\end{aligned}$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lw4lkbz1/158cd7a8-1619-430b-bf76-8b2ddd399871/01d314d0-10f4-11ef-8553-fdfc6347789d/file-1lw4lkbz2.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lw4lkbz1/158cd7a8-1619-430b-bf76-8b2ddd399871/01d314d0-10f4-11ef-8553-fdfc6347789d/file-1lw4lkbz2.png" loading="lazy" style="max-width: 100%;height: auto;display: block;margin: 0 auto;max-height: 40vh;vertical-align: baseline" alt="JEE Main 2024 (Online) 8th April Evening Shift Mathematics - 3D Geometry Question 13 English Explanation"></p>
<p>$$S . D .=\frac{|(2 \hat{i}+3 \hat{j}+4 \hat{k}) \times((\lambda-2) \hat{i}-4 \hat{k})|}{|2 \hat{i}+3 \hat{j}+4 \hat{k}|}$$</p>
<p>$$\begin{aligned}
& \left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
2 & 3 & 4 \\
\lambda-2 & 0 & -4
\end{array}\right|=-12 \hat{i}-\hat{j}(-8-4 \lambda+8)+\hat{k}(6-3 \lambda) \\
& =-12 \hat{i}+4 \lambda \hat{j}+(6-3 \lambda) \hat{k} \\
& \frac{\sqrt{144+16 \lambda^2+(6-3 \lambda)^2}}{\sqrt{29}}=\frac{13}{\sqrt{29}} \\
& 144+16 \lambda^2+(6-3 \lambda)^2=169 \\
& \Rightarrow 16 \lambda^2+9 \lambda^2+36-36 \lambda+144-169=0 \\
\end{aligned}$$</p>
<p>$$\begin{aligned}
& \Rightarrow 25 \lambda^2-36 \lambda+11=0 \\
& \Rightarrow 25 \lambda^2-25 \lambda-11 \lambda+11=0 \\
& \Rightarrow(25 \lambda-11)(\lambda-1)=0 \\
& \Rightarrow \lambda=1, \frac{11}{25}
\end{aligned}$$</p> | mcq | jee-main-2024-online-8th-april-evening-shift |
lv3vefgs | maths | 3d-geometry | lines-in-space | <p>Let $$\mathrm{P}(\alpha, \beta, \gamma)$$ be the image of the point $$\mathrm{Q}(1,6,4)$$ in the line $$\frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3}$$. Then $$2 \alpha+\beta+\gamma$$ is equal to ________</p> | [] | null | 11 | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lw4mrnop/6ea8698b-9387-47c8-8aca-29425e7a83aa/b6aa6990-10f8-11ef-b50f-395513bc7af2/file-1lw4mrnoq.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lw4mrnop/6ea8698b-9387-47c8-8aca-29425e7a83aa/b6aa6990-10f8-11ef-b50f-395513bc7af2/file-1lw4mrnoq.png" loading="lazy" style="max-width: 100%;height: auto;display: block;margin: 0 auto;max-height: 40vh;vertical-align: baseline" alt="JEE Main 2024 (Online) 8th April Evening Shift Mathematics - 3D Geometry Question 12 English Explanation"></p>
<p>$$\begin{aligned}
& \overrightarrow{Q R} \cdot(\hat{i}+2 \hat{j}+3 \hat{k})=0 \\
& (t-1)+(2 t-5) \times 2+(3 t-2) \times 3=0 \Rightarrow t=\frac{17}{14} \\
& \Rightarrow R \equiv\left(\frac{17}{14}, \frac{48}{14}, \frac{79}{14}\right) \\
& \Rightarrow \frac{\alpha+1}{2}=\frac{17}{14}, \frac{\beta+6}{2}=\frac{48}{14}, \frac{\gamma+4}{2}=\frac{79}{14} \\
& 2 \alpha+\beta+\gamma=\frac{68}{14}-2+\frac{96}{14}-6+\frac{158}{14}-4=11
\end{aligned}$$</p> | integer | jee-main-2024-online-8th-april-evening-shift |
lv5gsxy6 | maths | 3d-geometry | lines-in-space | <p>If the shortest distance between the lines</p>
<p>$$\begin{array}{ll}
L_1: \vec{r}=(2+\lambda) \hat{i}+(1-3 \lambda) \hat{j}+(3+4 \lambda) \hat{k}, & \lambda \in \mathbb{R} \\
L_2: \vec{r}=2(1+\mu) \hat{i}+3(1+\mu) \hat{j}+(5+\mu) \hat{k}, & \mu \in \mathbb{R}
\end{array}$$</p>
<p>is $$\frac{m}{\sqrt{n}}$$, where $$\operatorname{gcd}(m, n)=1$$, then the value of $$m+n$$ equals</p> | [{"identifier": "A", "content": "384"}, {"identifier": "B", "content": "387"}, {"identifier": "C", "content": "390"}, {"identifier": "D", "content": "377"}] | ["B"] | null | <p>$$\begin{aligned}
& L_1: \vec{r}=(2+\lambda) \hat{i}+(1-3 \lambda) \hat{j}+(3+4 \lambda) \hat{k} \\
& L_1=2 \hat{i}+\hat{j}+3 \hat{k}+\lambda(\hat{i}-3 \hat{j}+4 \hat{k}) \\
& L_2: \vec{r}=2 \hat{i}+3 \hat{j}+5 \hat{k}+\mu(2 \hat{i}+3 \hat{j}+\hat{k}) \\
& \vec{a}_1=2 \hat{i}+\hat{j}+3 \hat{k} \\
& \vec{a}_2=2 \hat{i}+3 \hat{j}+5 \hat{k} \\
& \vec{a}_2-\vec{a}_1=2 \hat{j}+2 \hat{k} \\
& \vec{b}_1=\hat{i}-3 \hat{j}+4 \hat{k}, \vec{b}_2=2 \hat{i}+3 \hat{j}+\hat{k} \\
& \vec{b}_1 \times \vec{b}_2=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
1 & -3 & 4 \\
2 & 3 & 1
\end{array}\right|
\end{aligned}$$</p>
<p>$$\begin{aligned}
& \hat{i}(-3-12)-\hat{j}(1-8)+\hat{k}(3+6) \\
& =-15 \hat{i}+7 \hat{j}+9 \hat{k} \\
& \left|\vec{b}_1 \times \vec{b}_2\right|=\sqrt{225+49+81} \\
& \left|\frac{\left(\vec{a}_2-\vec{a}_1\right) \cdot\left(\vec{b}_1 \times \vec{b}_2\right)}{\left|\vec{b}_1 \times \vec{b}_2\right|}\right|=\frac{14+18}{\sqrt{355}}=\frac{32}{\sqrt{355}} \\
& m+n=387
\end{aligned}$$</p> | mcq | jee-main-2024-online-8th-april-morning-shift |
lv7v3jvb | maths | 3d-geometry | lines-in-space | <p>Let $$\mathrm{d}$$ be the distance of the point of intersection of the lines $$\frac{x+6}{3}=\frac{y}{2}=\frac{z+1}{1}$$ and $$\frac{x-7}{4}=\frac{y-9}{3}=\frac{z-4}{2}$$ from the point $$(7,8,9)$$. Then $$\mathrm{d}^2+6$$ is equal to :</p> | [{"identifier": "A", "content": "75"}, {"identifier": "B", "content": "78"}, {"identifier": "C", "content": "72"}, {"identifier": "D", "content": "69"}] | ["A"] | null | <p>$$\begin{aligned}
& P_1:(3 k-6,2 k, k-1) \\
& P_2(4 \alpha+7,3 \alpha+9,2 \alpha+4) \\
& P_1 \equiv P_2 \\
& 3 k-6=4 \alpha+7 \Rightarrow 3 k-4 \alpha=13 \\
& 2 k=3 \alpha+9 \Rightarrow 2 k-3 \alpha=9 \\
& \therefore k=3, \alpha=-1 \\
& \therefore P_1:(3,6,2)
\end{aligned}$$</p>
<p>Distance of $$(3,6,2)$$ and $$(7,8,9)$$</p>
<p>$$\begin{aligned}
& =\sqrt{16+4+49}=\sqrt{69}=d \\
& d^2+6=69+6=75
\end{aligned}$$</p> | mcq | jee-main-2024-online-5th-april-morning-shift |
lv9s1zwj | maths | 3d-geometry | lines-in-space | <p>Let $$(\alpha, \beta, \gamma)$$ be the image of the point $$(8,5,7)$$ in the line $$\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-2}{5}$$. Then $$\alpha+\beta+\gamma$$ is equal to :</p> | [{"identifier": "A", "content": "16"}, {"identifier": "B", "content": "20"}, {"identifier": "C", "content": "18"}, {"identifier": "D", "content": "14"}] | ["D"] | null | <p>$$(x, y, z) \equiv(2 \lambda+1,3 \lambda+1,5 \lambda+2)$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lwenv6x5/39fb95d9-cd80-4c68-91b8-1f5c0166f573/d8931b90-167c-11ef-9070-f523f4c6bd4b/file-1lwenv6x6.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lwenv6x5/39fb95d9-cd80-4c68-91b8-1f5c0166f573/d8931b90-167c-11ef-9070-f523f4c6bd4b/file-1lwenv6x6.png" loading="lazy" style="max-width: 100%;height: auto;display: block;margin: 0 auto;max-height: 40vh;vertical-align: baseline" alt="JEE Main 2024 (Online) 5th April Evening Shift Mathematics - 3D Geometry Question 7 English Explanation"></p>
<p>DR of $$P Q$$ :</p>
<p>$$(2 \lambda-7,3 \lambda-6,5 \lambda-5)$$</p>
<p>DR of line : $$(2,3,5)$$</p>
<p>$$\begin{array}{ll}
\therefore & 2(2 \lambda-7)+3(3 \lambda-6)+5(5 \lambda-5)=0 \\
\Rightarrow & \lambda=\frac{3}{2} \\
\therefore \quad & Q\left(4, \frac{7}{2}, \frac{19}{2}\right) \\
\therefore \quad & (\alpha, \beta, y) \equiv(0,2,12) \quad\left(Q \text { is mid point of } P \& P^{\prime}\right) \\
& \alpha+\beta+y \equiv 14
\end{array}$$</p> | mcq | jee-main-2024-online-5th-april-evening-shift |
lv9s20kw | maths | 3d-geometry | lines-in-space | <p>Let the point $$(-1, \alpha, \beta)$$ lie on the line of the shortest distance between the lines $$\frac{x+2}{-3}=\frac{y-2}{4}=\frac{z-5}{2}$$ and $$\frac{x+2}{-1}=\frac{y+6}{2}=\frac{z-1}{0}$$. Then $$(\alpha-\beta)^2$$ is equal to _________.</p> | [] | null | 25 | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lweuubn4/2752e8cb-54f7-4c91-8212-691518b0463e/20511700-1698-11ef-ad74-71e0a1829759/file-1lweuubn5.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lweuubn4/2752e8cb-54f7-4c91-8212-691518b0463e/20511700-1698-11ef-ad74-71e0a1829759/file-1lweuubn5.png" loading="lazy" style="max-width: 100%;height: auto;display: block;margin: 0 auto;max-height: 40vh;vertical-align: baseline" alt="JEE Main 2024 (Online) 5th April Evening Shift Mathematics - 3D Geometry Question 6 English Explanation"></p>
<p>$$\overrightarrow{A B} \perp \vec{L}_1 \text { and } \overrightarrow{A B} \perp \vec{L}_2$$</p>
<p>$$\begin{aligned}
& \overrightarrow{A B}=(-3 m-2+n+2,4 m+2-2 n+6,2 m+5-1) \\
& =(-3 m+n, 4 m-2 n+8,2 m+4) \\
& \overrightarrow{A B} \perp \vec{L}_1 \\
& \Rightarrow-3(-3 m+n)+4(4 m-2 n+8)+2(2 m+4)=0 \\
& (9 m+16 m+4 m)+(-3 n-8 n)+32+8=0 \\
& \Rightarrow 29 m-11 n+40=0 \quad \ldots(1) \\
& \overrightarrow{A B} \perp \vec{L}_2 \\
& \Rightarrow-1(-3 m+n)+2(4 m-2 n+8)+0(2 m+4)=0 \\
& \Rightarrow 3 m-n+8 m-4 n+16=0 \\
& \Rightarrow 11 m-5 n+16=0 \Rightarrow m=-1, n=1 \\
& \Rightarrow A \equiv(1,-2,3), \quad B \equiv(-3,-4,1) \\
& A B \text { line } \Rightarrow \frac{x-1}{2}=\frac{y+2}{1}=\frac{z-3}{1} \\
& \Rightarrow \alpha=-2, \quad \beta=3 \\
& \Rightarrow(\alpha-\beta)^2=25
\end{aligned}$$</p> | integer | jee-main-2024-online-5th-april-evening-shift |
lvb294m1 | maths | 3d-geometry | lines-in-space | <p>Let $$\mathrm{P}(\alpha, \beta, \gamma)$$ be the image of the point $$\mathrm{Q}(3,-3,1)$$ in the line $$\frac{x-0}{1}=\frac{y-3}{1}=\frac{z-1}{-1}$$ and $$\mathrm{R}$$ be the point $$(2,5,-1)$$. If the area of the triangle $$\mathrm{PQR}$$ is $$\lambda$$ and $$\lambda^2=14 \mathrm{~K}$$, then $$\mathrm{K}$$ is equal to :</p> | [{"identifier": "A", "content": "18"}, {"identifier": "B", "content": "81"}, {"identifier": "C", "content": "72"}, {"identifier": "D", "content": "36"}] | ["B"] | null | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lwad7urv/960c0402-e139-4dd9-89a4-53df9b260717/11b135b0-1420-11ef-90c1-6758c45d5635/file-1lwad7urw.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lwad7urv/960c0402-e139-4dd9-89a4-53df9b260717/11b135b0-1420-11ef-90c1-6758c45d5635/file-1lwad7urw.png" loading="lazy" style="max-width: 100%;height: auto;display: block;margin: 0 auto;max-height: 40vh;vertical-align: baseline" alt="JEE Main 2024 (Online) 6th April Evening Shift Mathematics - 3D Geometry Question 5 English Explanation"></p>
<p>$$\text { Line } L \Rightarrow \frac{x-0}{1}=\frac{y-3}{1}=\frac{z-1}{-1}=\mu$$</p>
<p>Point $$M(\mu, \mu+3,-\mu+1)$$</p>
<p>Direction vector of $$Q M=(\mu-3, \mu+6,-\mu)$$</p>
<p>Direction vector of line $$L=(1,1,-1)$$</p>
<p>Both direction vector are perpendicular so</p>
<p>$$\begin{aligned}
& 1 \times(\mu-3)+1 \times(\mu+6)-1 \times(-\mu)=0 \\
& \mu=-1
\end{aligned}$$</p>
<p>Point $$M(-1,2,2)$$ midpoint of $$P Q$$</p>
<p>So point $$P(\alpha, \beta, \gamma)=(-5,7,3)$$</p>
<p>Area of $$\triangle P Q R=\frac{1}{2}|\overrightarrow{P Q} \times \overrightarrow{Q R}|$$</p>
<p>$$\begin{aligned}
& \overrightarrow{P Q}=(8,-10,-2) \\
& \overrightarrow{Q R}=(-1,8,-2) \\
& \overrightarrow{P Q} \times \overrightarrow{Q R}=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
8 & -10 & -2 \\
-1 & 8 & -2
\end{array}\right|=36 \hat{i}+18 \hat{j}+54 \hat{k}
\end{aligned}$$</p>
<p>Area of $$\triangle P Q R=\lambda$$</p>
<p>$$\lambda=\frac{1}{2} \sqrt{(36)^2+(18)^2+(54)^2}$$</p>
<p>$$\begin{aligned}
& \lambda^2=\frac{1}{4}(4536) \\
& \lambda^2=1134 \\
& \lambda^2=14 K \text { (given) } \\
& 14 K=1134 \\
& K=\frac{1134}{14}=81 \\
& K=81
\end{aligned}$$</p> | mcq | jee-main-2024-online-6th-april-evening-shift |
lvb29516 | maths | 3d-geometry | lines-in-space | <p>If the shortest distance between the lines $$\frac{x-\lambda}{3}=\frac{y-2}{-1}=\frac{z-1}{1}$$ and $$\frac{x+2}{-3}=\frac{y+5}{2}=\frac{z-4}{4}$$ is $$\frac{44}{\sqrt{30}}$$, then the largest possible value of $$|\lambda|$$ is equal to _________.</p> | [] | null | 43 | <p>$$\begin{aligned}
& L_1: \frac{x-\lambda}{3}=\frac{y-2}{-1}=\frac{z-1}{1} \\
& L_2: \frac{x+2}{-3}=\frac{y+5}{2}=\frac{z-4}{4}
\end{aligned}$$</p>
<p>$$\begin{aligned}
& n_1 \times n_2=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
3 & -1 & 1 \\
-3 & 2 & 4
\end{array}\right| \\
& =-6 \hat{i}-15 \hat{j}+3 \hat{k} \\
& d=\left|\frac{[(\lambda+2) \hat{i}+7 \hat{j}-3 \hat{k}][-6 \hat{i}-15 \hat{j}+3 \hat{k}]}{|-6 \hat{i}-15 \hat{j}+3 \hat{k}|}\right|=\frac{44}{\sqrt{30}} \\
& \left|\frac{-6 \lambda-12-105-9}{\sqrt{270}}\right|=\frac{44}{\sqrt{30}} \\
& |6 \lambda+126|=132 \\
& |\lambda+21|=22 \\
& \lambda+21= \pm 22 \\
& |\lambda|_{\max }=43
\end{aligned}$$</p> | integer | jee-main-2024-online-6th-april-evening-shift |
lvc58e18 | maths | 3d-geometry | lines-in-space | <p>The shortest distance between the lines $$\frac{x-3}{2}=\frac{y+15}{-7}=\frac{z-9}{5}$$ and $$\frac{x+1}{2}=\frac{y-1}{1}=\frac{z-9}{-3}$$ is</p> | [{"identifier": "A", "content": "$$8 \\sqrt{3}$$\n"}, {"identifier": "B", "content": "$$6 \\sqrt{3}$$\n"}, {"identifier": "C", "content": "$$5 \\sqrt{3}$$\n"}, {"identifier": "D", "content": "$$4 \\sqrt{3}$$"}] | ["D"] | null | <p>Given two lines are represented as:</p>
<p>$ \frac{x-3}{2} = \frac{y+15}{-7} = \frac{z-9}{5} $</p>
<p>and</p>
<p>$ \frac{x+1}{2} = \frac{y-1}{1} = \frac{z-9}{-3} $</p>
<p>The formula for the shortest distance between two lines is:</p>
<p>$ d = \frac{|\left(\vec{a}_2 - \vec{a}_1\right) \cdot \left(\vec{b}_1 \times \vec{b}_2\right)|}{|\vec{b}_1 \times \vec{b}_2|} $</p>
<p>From the given lines:</p>
<p>$ \vec{a}_1 = (3, -15, 9) $</p>
<p>$ \vec{a}_2 = (-1, 1, 9) $</p>
<p>$ \vec{b}_1 = (2, -7, 5) $</p>
<p>$ \vec{b}_2 = (2, 1, -3) $</p>
<p>Calculate the difference:</p>
<p>$ \vec{a}_2 - \vec{a}_1 = (-1-3, 1+15, 9-9) = (-4, 16, 0) $</p>
<p>Next, compute the cross product:</p>
<p>$ \vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -7 & 5 \\ 2 & 1 & -3 \end{vmatrix} = (16\hat{i} + 16\hat{j} + 16\hat{k}) $</p>
<p>The magnitude of the cross product is:</p>
<p>$ |\vec{b}_1 \times \vec{b}_2| = |(16\hat{i} + 16\hat{j} + 16\hat{k})| = 16\sqrt{3} $</p>
<p>Calculate the dot product:</p>
<p>$ (\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2) = (-4\hat{i} + 16\hat{j} + 0\hat{k}) \cdot (16\hat{i} + 16\hat{j} + 16\hat{k}) = -64 + 256 + 0 = 192 $</p>
<p>Finally, find the shortest distance:</p>
<p>$ d = \frac{|192|}{16\sqrt{3}} = \frac{192}{16\sqrt{3}} = 4\sqrt{3} $</p>
<p>Thus, the shortest distance is:</p>
<p>$ \boxed{4 \sqrt{3}} $</p> | mcq | jee-main-2024-online-6th-april-morning-shift |
lvc58e9z | maths | 3d-geometry | lines-in-space | <p>Let $$P$$ be the point $$(10,-2,-1)$$ and $$Q$$ be the foot of the perpendicular drawn from the point $$R(1,7,6)$$ on the line passing through the points $$(2,-5,11)$$ and $$(-6,7,-5)$$. Then the length of the line segment $$P Q$$ is equal to _________.</p> | [] | null | 13 | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lwd5pdpk/204a0778-5148-47b3-a659-091fdb696421/09fd8280-15a9-11ef-99cc-63deb36dcc4a/file-1lwd5pdpl.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lwd5pdpk/204a0778-5148-47b3-a659-091fdb696421/09fd8280-15a9-11ef-99cc-63deb36dcc4a/file-1lwd5pdpl.png" loading="lazy" style="max-width: 100%;height: auto;display: block;margin: 0 auto;max-height: 40vh;vertical-align: baseline" alt="JEE Main 2024 (Online) 6th April Morning Shift Mathematics - 3D Geometry Question 1 English Explanation"></p>
<p>$$\begin{aligned}
& P(10,-2,-1) \\
& M N: \frac{x-2}{8}=\frac{y+5}{-12}=\frac{z-11}{16}
\end{aligned}$$</p>
<p>General point</p>
<p>$$(8k + 2, - 12k - 5,16k + 11)$$</p>
<p>$$\overrightarrow {RQ} = (8k + 2 - 1)\widehat i + ( - 12k - 5 - 7)\widehat j + (16k + 11 - 6)\widehat k$$</p>
<p>$$\overrightarrow {RQ} = (8k + 1)\widehat i - (12k + 12)\widehat j + (16k + 5)\widehat k$$</p>
<p>$$\overrightarrow {RQ} \,.\,\overrightarrow {MN} = 0$$ (as both are perpendicular)</p>
<p>$$8(8 k+1)+12(12 k+12)+16(16 k+5)=0$$</p>
<p>$$\begin{aligned}
& 64 k+8+144 k+144+256 k+80=0 \\
& 464 k=-232 \\
& k=\frac{-232}{464}=\frac{-1}{2} \\
& Q(-4+2,6-5,-8+11) \\
& Q(-2,1,3) \\
& P Q=\sqrt{(10+2)^2+(-3)^2+(4)^2} \\
& P Q=\sqrt{12^2+3^2+4^2} \\
& P Q=\sqrt{169}=13
\end{aligned}$$</p> | integer | jee-main-2024-online-6th-april-morning-shift |
ZZ2mbSw3xnje5rne | maths | 3d-geometry | plane-in-space | The $$d.r.$$ of normal to the plane through $$(1, 0, 0), (0, 1, 0)$$ which makes an angle $$\pi /4$$ with plane $$x+y=3$$ are : | [{"identifier": "A", "content": "$$1,\\sqrt 2 ,1$$ "}, {"identifier": "B", "content": "$$1,1,\\sqrt 2 $$ "}, {"identifier": "C", "content": "$$1, 1, 2$$"}, {"identifier": "D", "content": "$$\\sqrt 2 ,1,1$$ "}] | ["B"] | null | Equation of plane through $$\left( {1,0,0} \right)$$ is
<br><br>$$a\left( {x - 1} \right) + by + cz = 0\,\,\,\,\,\,\,\,\,\,...\left( i \right)$$
<br><br>$$(i)$$ passes through $$\left( {0,1,0} \right).$$
<br><br>$$ - a + b = 0 \Rightarrow b = a;$$
<br><br>Also, $$\cos {45^ \circ }$$
<br><br>$$ = {{a + a} \over {\sqrt {2\left( {2{a^2} + {c^2}} \right)} }} \Rightarrow 2a = \sqrt {2{a^2} + {c^2}} $$
<br><br>$$ \Rightarrow 2{a^2} = {c^2}$$
<br><br>$$ \Rightarrow c = \sqrt 2 a.$$
<br><br>So $$d.r$$ of normal area $$a,$$ $$a\sqrt {2a} $$ i.e. $$1,$$ $$1,\sqrt 2 .$$ | mcq | aieee-2002 |
tklE0iEL4MC3JSAh | maths | 3d-geometry | plane-in-space | The shortest distance from the plane $$12x+4y+3z=327$$ to the sphere <br/><br/>$${x^2} + {y^2} + {z^2} + 4x - 2y - 6z = 155$$ is | [{"identifier": "A", "content": "$$39$$ "}, {"identifier": "B", "content": "$$26$$ "}, {"identifier": "C", "content": "$$11{4 \\over {13}}$$ "}, {"identifier": "D", "content": "$$13$$"}] | ["D"] | null | Shortest distance $$=$$ perpendicular distance between the plane and sphere $$=$$ distance of plane from center of sphere $$-$$ radius
<br><br>$$ = \left| {{{ - 2 \times 12 + 4 \times 1 + 3 \times 3 - 327} \over {\sqrt {144 + 9 + 16} }}} \right| - \sqrt {4 + 1 + 9 + 155} $$
<br><br>$$ = 26 - 13 = 13$$ | mcq | aieee-2003 |
CFmX8122sUhvE9Dl | maths | 3d-geometry | plane-in-space | Two systems of rectangular axes have the same origin. If a plane cuts then at distances $$a,b,c$$ and $$a', b', c'$$ from the origin then | [{"identifier": "A", "content": "$${1 \\over {{a^2}}} + {1 \\over {{b^2}}} + {1 \\over {{c^2}}} - {1 \\over {a{'^2}}} - {1 \\over {b{'^2}}} - {1 \\over {c{'^2}}} = 0$$ "}, {"identifier": "B", "content": "$$\\,{1 \\over {{a^2}}} + {1 \\over {{b^2}}} + {1 \\over {{c^2}}} + {1 \\over {a{'^2}}} + {1 \\over {b{'^2}}} + {1 \\over {c{'^2}}} = 0$$ "}, {"identifier": "C", "content": "$${1 \\over {{a^2}}} + {1 \\over {{b^2}}} - {1 \\over {{c^2}}} + {1 \\over {a{'^2}}} - {1 \\over {b{'^2}}} - {1 \\over {c{'^2}}} = 0$$ "}, {"identifier": "D", "content": "$${1 \\over {{a^2}}} - {1 \\over {{b^2}}} - {1 \\over {{c^2}}} + {1 \\over {a{'^2}}} - {1 \\over {b{'^2}}} - {1 \\over {c{'^2}}} = 0$$ "}] | ["A"] | null | Equation of planes be $${x \over a} + {y \over b} + {z \over c} = 1\,\,\& \,\,{x \over {a'}} + {y \over {b'}} + {z \over {c'}} = 1$$
<br><br>So the distance from (0, 0, 0) to both the plane is same.
<br><br>$$ \therefore $$ $$\left| {{{ - 1} \over {\sqrt {{1 \over {{a^2}}} + {1 \over {{b^2}}} + {1 \over {{c^2}}}} }}} \right| = \left| {{{ - 1} \over {\sqrt {{1 \over {a{'^2}}} + {1 \over {b{'^2}}} + {1 \over {c{'^2}}}} }}} \right|$$
<br><br>$$ \Rightarrow $$ $${1 \over {{a^2}}} + {1 \over {{b^2}}} + {1 \over {{c^2}}} - {1 \over {{{a'}^{2}}}} - {1 \over {{{b'}^{2}}}} - {1 \over {{{c'}^{2}}}} = 0$$ | mcq | aieee-2003 |
HYn9nj1UWdOirDBu | maths | 3d-geometry | plane-in-space | The radius of the circle in which the sphere
<br/><br>$${x^2} + {y^2} + {z^2} + 2x - 2y - 4z - 19 = 0$$ is cut by the plane
<br/><br>$$x+2y+2z+7=0$$ is </br></br> | [{"identifier": "A", "content": "$$4$$"}, {"identifier": "B", "content": "$$1$$"}, {"identifier": "C", "content": "$$2$$ "}, {"identifier": "D", "content": "$$3$$"}] | ["D"] | null | <img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265346/exam_images/c0dqhctziwbekfugrolb.webp" loading="lazy" alt="AIEEE 2003 Mathematics - 3D Geometry Question 321 English Explanation">
<br><br>Center of sphere $$ = \left( { - 1,1,2} \right)$$
<br><br>Radius of sphere $$\sqrt {1 + 1 + 4 + 19} = 5$$
<br><br>Perpendicular distance from center to the plane
<br><br>$$OC = d = \left| {{{ - 1 + 2 + 4 + 7} \over {\sqrt {1 + 4 + 4} }}} \right| = {{12} \over 3} = 4.$$
<br><br>$$A{C^2} = A{O^2} - O{C^2} = {5^2} - {4^2} = 9$$
<br><br>$$ \Rightarrow AC = 3$$ | mcq | aieee-2003 |
wXNJAE9OixV0o1Ul | maths | 3d-geometry | plane-in-space | The intersection of the spheres
<br/>$${x^2} + {y^2} + {z^2} + 7x - 2y - z = 13$$ and
<br/>$${x^2} + {y^2} + {z^2} - 3x + 3y + 4z = 8$$
<br/>is the same as the intersection of one of the sphere and the plane | [{"identifier": "A", "content": "$$2x-y-z=1$$ "}, {"identifier": "B", "content": "$$x-2y-z=1$$"}, {"identifier": "C", "content": "$$x-y-2z=1$$ "}, {"identifier": "D", "content": "$$x-y-z=1$$ "}] | ["A"] | null | The equation of spheres are
<br><br>$${S_1}:{x^2} + {y^2} + {z^2} + 7x - 2y - z - 13 = 0$$ and
<br><br>$${S_2}:{x^2} + {y^2} + {z^2} - 3x + 3y + 4z - 8 = 0$$
<br><br>Their plane of intersection is $${S_1} - {S_2} = 0$$
<br><br>$$ \Rightarrow 10x - 5y - 5z - 5 = 0$$
<br><br>$$ \Rightarrow 2x - y - z = 1$$ | mcq | aieee-2004 |
wUkvqWxVsZ5qPXxv | maths | 3d-geometry | plane-in-space | Distance between two parallel planes
<br/><br>$$\,2x + y + 2z = 8$$ and $$4x + 2y + 4z + 5 = 0$$ is :</br> | [{"identifier": "A", "content": "$${9 \\over 2}$$ "}, {"identifier": "B", "content": "$${5 \\over 2}$$"}, {"identifier": "C", "content": "$${7 \\over 2}$$"}, {"identifier": "D", "content": "$${3 \\over 2}$$"}] | ["C"] | null | The planes are $$2x + y + 2x - 8 = 0\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)$$
<br><br>and $$4x + 2y + 4z + 5 = 0$$
<br><br>or $$2x + y + 2z + {5 \over 2} = 0\,\,\,\,\,\,\,\,\,\,...\left( 2 \right)$$
<br><br>$$\therefore$$ Distance between $$\left( 1 \right)$$ and $$\,\left( 2 \right)$$
<br><br>$$ = \left| {{{{5 \over 2} + 8} \over {\sqrt {{2^2} + {1^2} + {2^2}} }}} \right|$$
<br><br>$$ = \left| {{{21} \over {2\sqrt 9 }}} \right|$$
<br><br>$$ = {7 \over 2}$$ | mcq | aieee-2004 |
gsaP2wxbhrwI4MNS | maths | 3d-geometry | plane-in-space | The plane $$x+2y-z=4$$ cuts the sphere $${x^2} + {y^2} + {z^2} - x + z - 2 = 0$$ in a circle of radius | [{"identifier": "A", "content": "$$3$$ "}, {"identifier": "B", "content": "$$1$$ "}, {"identifier": "C", "content": "$$2$$ "}, {"identifier": "D", "content": "$${\\sqrt 2 }$$"}] | ["B"] | null | <img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265266/exam_images/iljyyh25rrxltbqfibh1.webp" loading="lazy" alt="AIEEE 2005 Mathematics - 3D Geometry Question 311 English Explanation">
<br><br>Perpendicular distance of center $$\left( {{1 \over 2},0, - {1 \over 2}} \right)$$
<br><br>from $$x+2y-2=4$$ is given by
<br><br>$${{\left| {{1 \over 2} + {1 \over 2} - 4} \right| = \sqrt {{3 \over 2}} } \over {\sqrt 6 }}$$
<br><br>radius of sphere
<br><br>$$ = \sqrt {{1 \over 4} + {1 \over 4} + 2} = \sqrt {{5 \over 2}} $$
<br><br>$$\therefore$$ radius of circle
<br><br>$$ = \sqrt {{5 \over 2} - {3 \over 2}} = 1.$$ | mcq | aieee-2005 |
jSTkPlfBJ5I6fd65 | maths | 3d-geometry | plane-in-space | <b>Statement-1 :</b> The point $$A(3, 1, 6)$$ is the mirror image of the point $$B(1, 3, 4)$$ in the plane $$x-y+z=5.$$
<br/><br><b>Statement-2 :</b> The plane $$x-y+z=5$$ bisects the line segment joining $$A(3, 1, 6)$$ and $$B(1, 3, 4).$$ </br> | [{"identifier": "A", "content": "Statement - 1 is true, Statement - 2 is true; Statement - 2 is <b>not</b> a correct explanation for Statement - 1."}, {"identifier": "B", "content": "Statement - 1 is true, Statement - 2 is false."}, {"identifier": "C", "content": "Statement - 1 is false , Statement - 2 is true."}, {"identifier": "D", "content": "Statement - 1 is true, Statement - 2 is true; Statement - 2 is a correct explanation for Statement - 1."}] | ["A"] | null | $$A\left( {3,1,6} \right);\,\,B = \left( {1,3,4} \right)$$
<br><br>Mid-point of $$AB = \left( {2,2,5} \right)$$ lies on the plane.
<br><br>and d.r's of $$AB=(2,-2,2)$$
<br><br>d.r's of normal to plane $$ = \left( {1, - 1,1} \right).$$
<br><br>Direction ratio of $$AB$$ and normal to the plane are proportional therefore,
<br><br>$$AB$$ is perpendicular to the plane
<br><br>$$\therefore$$ $$A$$ is image of $$B$$
<br><br>Statement-$$2$$ is correct but it is not correct explanation. | mcq | aieee-2010 |
DuqJJduwK99t03Hr | maths | 3d-geometry | plane-in-space | A equation of a plane parallel to the plane $$x-2y+2z-5=0$$ and at a unit distance from the origin is : | [{"identifier": "A", "content": "$$x-2y+2z-3=0$$"}, {"identifier": "B", "content": "$$x-2y+2z+1=0$$"}, {"identifier": "C", "content": "$$x-2y+2z-1=0$$ "}, {"identifier": "D", "content": "$$x-2y+2z+5=0$$"}] | ["A"] | null | Given equation of a plane is
<br><br>$$x - 2y + 2z - 5 = 0$$
<br><br>So, Equation of parallel plane is given by
<br><br>$$x - 2y + 2z + d = 0$$
<br><br>Now, it is given that distance from origin to the parallel planes is $$1.$$
<br><br>$$\therefore$$ $$\left| {{d \over {\sqrt {{1^2} + {2^2} + {2^3}} }}} \right| = 1 \Rightarrow d = \pm 3$$
<br><br>So equation of required plane
<br><br>$$x - 2y + 2z \pm 3 = 0$$ | mcq | aieee-2012 |
1rZ38hZcB3hCW5WP | maths | 3d-geometry | plane-in-space | Distance between two parallel planes $$2x+y+2z=8$$ and $$4x+2y+4z+5=0$$ is : | [{"identifier": "A", "content": "$${3 \\over 2}$$ "}, {"identifier": "B", "content": "$${5 \\over 2}$$"}, {"identifier": "C", "content": "$${7 \\over 2}$$"}, {"identifier": "D", "content": "$${9 \\over 2}$$"}] | ["C"] | null | $$2x + y + 2z - 8 = 0\,\,\,\,\,\,\,\,\,\,\,\,...\left( {Plane\,\,1} \right)$$
<br><br>$$2x + y + 2z + {5 \over 2} = 0\,\,\,\,\,\,\,\,\,\,\,...\left( {Plane\,\,2} \right)$$
<br><br>Distance between Plane $$1$$ and $$2$$
<br><br>$$ = \left| {{{ - 8 - {5 \over 2}} \over {\sqrt {{2^2} + {1^2} + {2^2}} }}} \right|$$
<br><br>$$ = \left| {{{ - 21} \over 6}} \right| = {7 \over 2}$$ | mcq | jee-main-2013-offline |
Mb2JUKkOpNuTwuvd2BgWO | maths | 3d-geometry | plane-in-space | The distance of the point (1, − 2, 4) from the plane passing through the point
(1, 2, 2) and perpendicular to the planes x − y + 2z = 3 and 2x − 2y + z + 12 = 0, is : | [{"identifier": "A", "content": "$$2\\sqrt 2 $$ "}, {"identifier": "B", "content": "2"}, {"identifier": "C", "content": "$$\\sqrt 2 $$"}, {"identifier": "D", "content": "$${1 \\over {\\sqrt 2 }}$$ "}] | ["A"] | null | Equation of plane passing through point (1, 2, 2) is,
<br><br>a(x $$-$$ 1) + b(y $$-$$ 2) + c(z $$-$$ 2) = 0 . . .(1)
<br><br>This plane is perpendicular to the plane
<br><br>x $$-$$ y + 2z = 3 and 2x $$-$$ 2y + z + 12 = 0
<br><br>When two planes,
<br><br>a<sub>1</sub>x + b<sub>1</sub>y + c<sub>1</sub>z + d<sub>1</sub> = 0
<br><br>and a<sub>2</sub>x + b<sub>2</sub>y + c<sub>2</sub>z + d<sub>2</sub> = 0
<br><br>are perpendicular to each other then
<br><br>a<sub>1</sub>a<sub>2</sub> + b<sub>1</sub>b<sub>2</sub> + c<sub>1</sub>c<sub>2</sub> = 0
<br><br>So, we can say,
<br><br>a $$ \times $$ 1 + b $$ \times $$ ($$-$$ 1) + c $$ \times $$ 2 = 0
<br><br>$$ \Rightarrow $$ a $$-$$ b + 2c = 0 . . . (2)
<br><br>and, a $$ \times $$ 2 + b($$-$$2) + c(1) = 0
<br><br>$$ \Rightarrow $$ 2a $$-$$ 2b + c = 0 . . .(3)
<br><br>Solving (2) and (3)
<br><br>$${a \over { - 1 + 4}}$$ = $${b \over {4 - 1}}$$ = $${c \over { - 2 + 2}}$$ = $$\lambda $$
<br><br>$$ \Rightarrow $$ a = 3$$\lambda $$, b = 3$$\lambda $$, c = 0
<br><br>Putting the values of a, b, c in equation (1) we get,
<br><br>3$$\lambda $$ (x $$-$$ 1) + 3$$\lambda $$ (y $$-$$ 2) = 0
<br><br>$$ \Rightarrow $$ 3(x $$-$$ 1) + 3(y $$-$$ 2) = 0
<br><br>$$ \Rightarrow $$ 3x + 3y $$-$$ 9 = 0
<br><br>$$ \Rightarrow $$ x + y $$-$$ 3 = 0
<br><br>$$ \therefore $$ Distance of point (1, $$-$$2, 4) from plane x + y $$-$$ 3 = 0 is,
<br><br>= $$\left| {{{1 - 2 - 3} \over {\sqrt {1 + 1} }}} \right|$$ = $${4 \over {\sqrt 2 }}$$ = 2$$\sqrt 2 $$ | mcq | jee-main-2016-online-9th-april-morning-slot |
HsclIbPpXGLLYwNuOmjqV | maths | 3d-geometry | plane-in-space | If x = a, y = b, z = c is a solution of the system of linear equations
<br/><br/>x + 8y + 7z = 0
<br/><br/>9x + 2y + 3z = 0
<br/><br/>x + y + z = 0
<br/><br/>such that the point (a, b, c) lies on the plane x + 2y + z = 6, then 2a + b + c equals : | [{"identifier": "A", "content": "$$-$$ 1"}, {"identifier": "B", "content": "0"}, {"identifier": "C", "content": "1"}, {"identifier": "D", "content": "2"}] | ["C"] | null | Given,
<br><br>x + 8y + 7z = 0
<br><br>9x + 2y + 3z = 0
<br><br>x + y + z = 0
<br><br>Solving those equations, we get
<br><br>x = $$\lambda $$, y = 6$$\lambda $$, z = -7$$\lambda $$
<br><br>This point lies on the plane x + 2y + z = 6
<br><br>$$ \therefore $$ $$\lambda $$ + 2(6$$\lambda $$) + (-7$$\lambda $$) = 0
<br><br>$$ \Rightarrow $$ $$\lambda $$ = 1
<br><br>$$ \therefore $$ x = 1, y = 6, z = -7
<br><br>$$ \therefore $$ a = 1, b = 6, c = -7
<br><br>So, 2a + b + c = 2(1) + 6 + (-7) = 1 | mcq | jee-main-2017-online-9th-april-morning-slot |
QOo7bAqiG06GOKThuvm05 | maths | 3d-geometry | plane-in-space | If a variable plane, at a distance of 3 units from the origin, intersects the coordinate
axes at A, B and C, then the locus of the centroid of $$\Delta $$ABC is :
| [{"identifier": "A", "content": "$${1 \\over {{x^2}}} + {1 \\over {{y^2}}} + {1 \\over {{z^2}}} = 1$$ "}, {"identifier": "B", "content": "$${1 \\over {{x^2}}} + {1 \\over {{y^2}}} + {1 \\over {{z^2}}} = 3$$ "}, {"identifier": "C", "content": "$${1 \\over {{x^2}}} + {1 \\over {{y^2}}} + {1 \\over {{z^2}}} = {1 \\over 9}$$ "}, {"identifier": "D", "content": "$${1 \\over {{x^2}}} + {1 \\over {{y^2}}} + {1 \\over {{z^2}}} = 9$$ "}] | ["A"] | null | Suppose centroid be (h, k, $$l$$)
<br><br>$$ \therefore $$ x $$-$$ intp $$=$$ 3h, y $$-$$ intp $$=$$ 3k, z $$-$$ intp $$=$$ 3$$l$$
<br><br>Equation $${x \over {3h}} + {y \over {3k}} + {z \over {3l}} = 1$$
<br><br>$$ \therefore $$ Distance from (0, 0, 0)
<br><br>$$\left| {{{ - 1} \over {\sqrt {{1 \over {9{h^2}}} + {1 \over {9{k^2}}} + {1 \over {9{l^2}}}} }}} \right| = 3$$
<br><br>$$ \Rightarrow $$ $${1 \over {{x^2}}} + {1 \over {{y^2}}} + {1 \over {{z^2}}} = 1$$ | mcq | jee-main-2017-online-9th-april-morning-slot |
By2W9q22rMEe3evacfhKc | maths | 3d-geometry | plane-in-space | The sum of the intercepts on the coordinate axes of the plane passing through the point ($$-$$2, $$-2,$$ 2) and containing the line joining the points (1, $$-$$1, 2) and (1, 1, 1) is : | [{"identifier": "A", "content": "4"}, {"identifier": "B", "content": "$$-$$ 4"}, {"identifier": "C", "content": "$$-$$ 8"}, {"identifier": "D", "content": "12"}] | ["B"] | null | Equation of plane passing through three given points is :
<br><br>$$\left| {\matrix{
{x - {x_1}} & {y - {y_1}} & {z - {z_1}} \cr
{{x_2} - {x_1}} & {{y_2} - {y_1}} & {{z_2} - {z_1}} \cr
{{x_3} - {x_1}} & {{y_3} - {y_1}} & {{z_3} - {z_1}} \cr
} } \right| = 0$$
<br><br>$$ \Rightarrow $$ $$\left| {\matrix{
{x + 2} & {y + 2} & {z - 2} \cr
{1 + 2} & { - 1 + 2} & {2 - 2} \cr
{1 + 2} & {1 + 2} & {1 - 2} \cr
} } \right| = 0$$
<br><br>$$ \Rightarrow $$ $$\left| {\matrix{
{x + 2} & {y + 2} & {z - 2} \cr
3 & 1 & 0 \cr
3 & {30} & { - 1} \cr
} } \right| = 0$$
<br><br>$$ \Rightarrow $$ $$ - x + 3y + 6z - 8 = 0$$
<br><br>$$ \Rightarrow $$ $${x \over 8} - {{3y} \over 8} - {{6z} \over 8} + {8 \over 8} = 0$$
<br><br>$$ \Rightarrow $$ $${x \over 8} - {y \over {{8 \over 3}}} - {z \over {{8 \over 6}}} = - 1$$
<br><br>$$ \Rightarrow $$ $${x \over { - 8}} + {y \over {{8 \over 3}}} + {z \over {{8 \over 6}}} = 1$$
<br><br>$$ \therefore $$ Sum of intercepts $$ = - 8 + {8 \over 3} + {8 \over 6} = - 4$$ | mcq | jee-main-2018-online-16th-april-morning-slot |
EJOo70wLbGOQ1WD7CS7IU | maths | 3d-geometry | plane-in-space | A plane bisects the line segment joining the points (1, 2, 3) and ($$-$$ 3, 4, 5) at rigt angles. Then this plane also passes through the point : | [{"identifier": "A", "content": "($$-$$ 3, 2, 1)"}, {"identifier": "B", "content": "(3, 2, 1)"}, {"identifier": "C", "content": "($$-$$ 1, 2, 3)"}, {"identifier": "D", "content": "(1, 2, $$-$$ 3)"}] | ["A"] | null | Since the plane bisects the line joining the points (1, 2, 3) and ($$-$$3, 4, 5) then the plane passes through the midpoint of the line which is :
<br><br>$$\left( {{{1 - 3} \over 2},{{2 + 4} \over 2},{{5 + 3} \over 2}} \right)$$ $$ \equiv $$ $$\left( {{{ - 2} \over 2},{6 \over 2},{8 \over 2}} \right)$$ $$ \equiv $$ ($$-$$1, 3, 4).
<br><br>As plane cuts the line segment at right angle, so the direction cosines of the normal of the plane are ($$-$$ 3 $$-$$ 1, 4 $$-$$ 2, 5 $$-$$ 3) = ($$-$$ 4, 2, 2)
<br><br>So the equation of the plane is : $$-$$ 4x + 2y + 2z = $$\lambda $$
<br><br>As plane passes through ($$-$$ 1, 3, 4)
<br><br>So, $$-$$ 4($$-$$ 1) + 2(3) + 2(4) = $$\lambda $$ $$ \Rightarrow $$ $$\lambda $$ = 18
<br><br>Therefore, equation of plane is : $$-$$ 4x + 2y + 2z = 18
<br><br>Now, only ($$-$$ 3, 2, 1) satiesfies the given plane as
<br><br>$$-$$ 4($$-$$ 3) + 2(2) + 2(1) = 18 | mcq | jee-main-2018-online-15th-april-evening-slot |
pFJKQ1n9aFLfryRPPa1Lz | maths | 3d-geometry | plane-in-space | A variable plane passes through a fixed point (3,2,1) and meets x, y and z axes at A, B and C respectively. A plane is drawn parallel to yz -plane through A, a second plane is drawn parallel zx-plane through B and a third plane is drawn parallel to xy-plane through C. Then the locus of the point of intersection of these three planes, is : | [{"identifier": "A", "content": "$${x \\over 3} + {y \\over 2} + {z \\over 1} = 1$$ "}, {"identifier": "B", "content": "x + y + z = 6"}, {"identifier": "C", "content": "$${1 \\over x} + {1 \\over y} + {1 \\over z} = {{11} \\over 6}$$"}, {"identifier": "D", "content": "$${3 \\over x} + {2 \\over y} + {1 \\over z} = 1$$"}] | ["D"] | null | If a, b, c are the intercepts of the variable plane on the x,y,z axes respectively, then the equation of the plane is <br/><br/>$${x \over a} + {y \over b} + {z \over c} = 1$$
<br><br>And the point of intersection of the planes parallel to the xy, yz and zx planes is $$\left( {a,b,c} \right)$$.
<br><br>As the point (3, 2, 1) lies on the variables plane,
<br><br>so $${3 \over a} + {2 \over b} + {1 \over c} = 1$$
<br><br>Therefore, the required locus is
<br><br>$${3 \over x} + {2 \over y} + {1 \over z} = 1$$ | mcq | jee-main-2018-online-15th-april-morning-slot |
LfSBYkVJfNOMRk2S | maths | 3d-geometry | plane-in-space | If L<sub>1</sub> is the line of intersection of the planes 2x - 2y + 3z - 2 = 0, x - y + z + 1 = 0 and L<sub>2</sub> is the line of
intersection of the planes x + 2y - z - 3 = 0, 3x - y + 2z - 1 = 0, then the distance of the origin from the
plane, containing the lines L<sub>1</sub> and L<sub>2</sub>, is : | [{"identifier": "A", "content": "$${1 \\over {\\sqrt 2 }}$$"}, {"identifier": "B", "content": "$${1 \\over {4\\sqrt 2 }}$$"}, {"identifier": "C", "content": "$${1 \\over {3\\sqrt 2 }}$$"}, {"identifier": "D", "content": "$${1 \\over {2\\sqrt 2 }}$$"}] | ["C"] | null | <img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265244/exam_images/mtsizoslixc1qq9ocqig.webp" loading="lazy" alt="JEE Main 2018 (Offline) Mathematics - 3D Geometry Question 287 English Explanation">
<br><br>L<sub>1</sub> is the line of intersection of plane 1 and plane 2.
<br><br>L<sub>2</sub> is the line of intersection of plane $$3$$ and plane 4.
<br><br>Line L<sub>1</sub> and L<sub>2</sub> are present on plane 5.
<br><br>Now we have to find the distance of origin from the plane 5.
<br><br>The plane 5 is passing through the Line L<sub>1</sub>,
<br><br>and L<sub>1</sub> is the line of intersection of first two planes.
<br><br>$$\therefore$$ Equation of plane 5,
<br><br>$$(2x-2y+3z-2)$$ + $$\lambda \left( {x - y + z + 1} \right)0$$
<br><br>Now find a point on L<sub>2</sub> which is line of intersection of
<br><br>x + 2y $$-$$ z $$-$$ 3 = 0 $$\,\,\,\,$$ .....(1)
<br><br>and 3x $$-$$ y + 2z $$-$$ 1 = 0 $$\,\,\,\,$$ .....(2)
<br><br>Put x = 0 at both (1) and (2),
<br><br>2y $$-$$ z = 3 $$\,\,\,\,.....$$(3)
<br><br>$$-$$y + 2z = 1 $$\,\,\,\,....$$ (4)
<br><br>by solving (3) and (4) we get
<br><br>$$y = {7 \over 3},\,z = {5 \over 3}$$
<br><br>So, the point on line L<sub>2</sub> is $$ = \left( {0,{7 \over 3},{5 \over 3}} \right)$$
<br><br>This point is also present on the plane $$5.$$
<br><br>So, putting this point on equation of plane $$5,$$
<br><br>$$ - {{14} \over 5} + 5 - 2 + \lambda \left( { - {7 \over 3} + {5 \over 3}} \right) = 0$$
<br><br>$$ \Rightarrow - {5 \over 3} + {\lambda \over 3} = 0$$
<br><br>$$ \Rightarrow \lambda = 5$$
<br><br>$$\therefore$$ equation of plane 5 is
<br><br>2x $$-$$ 2y + 3z $$-$$ 2 + $$\lambda $$(x $$-$$ y + z + 1) = 0
<br><br>$$ \Rightarrow $$ 2x $$-$$ 2y + 3z $$-$$ 2 + 5 (x $$-$$ y + z + 1) = 0
<br><br>$$ \Rightarrow $$ 7x $$-$$ 7y + 8z + 3 = 0
<br><br>Distance of this plane 5 from (0, 0, 0)
<br><br>$$ = {3 \over {\sqrt {49 + 49 + 64} }}$$
<br><br>$$ = {3 \over {\sqrt {162} }}$$
<br><br>$$ = {3 \over {9\sqrt 2 }}$$
<br><br>$$ = {1 \over {3\sqrt 2 }}$$ | mcq | jee-main-2018-offline |
h2kYrw078v0a3uPgCCURP | maths | 3d-geometry | plane-in-space | The perpendicular distance from the origin to the plane containing the two lines, <br/><br/>$${{x + 2} \over 3} = {{y - 2} \over 5} = {{z + 5} \over 7}$$ and <br/><br/>$${{x - 1} \over 1} = {{y - 4} \over 4} = {{z + 4} \over 7},$$ is : | [{"identifier": "A", "content": "$$6\\sqrt {11} $$"}, {"identifier": "B", "content": "$${{11} \\over {\\sqrt 6 }}$$"}, {"identifier": "C", "content": "11"}, {"identifier": "D", "content": "11$$\\sqrt 6 $$"}] | ["B"] | null | $$\left| {\matrix{
i & j & k \cr
3 & 5 & 7 \cr
1 & 4 & 7 \cr
} } \right|$$
<br><br>= $$\widehat i$$(35 $$-$$ 28) $$-$$ $$\widehat j$$(21.7) + $$\widehat k$$(12 $$-$$ 5)
<br><br>= 7$$\widehat i$$ $$-$$ 14$$\widehat j$$ + 7$$\widehat k$$
<br><br>= $$\widehat i$$ $$-$$ 2$$\widehat j$$ + $$\widehat k$$
<br><br>1(x + 2) $$-$$ 2(y $$-$$ 2) + 1 (z + 15) = 0
<br><br>x $$-$$ 2y + z + 11 = 0
<br><br>$${{11} \over {\sqrt {4 + 1 + 1} }} = {{11} \over {\sqrt 6 }}$$ | mcq | jee-main-2019-online-12th-january-morning-slot |
ZNjWs09KjUBWLXz3C53rsa0w2w9jxadan6d | maths | 3d-geometry | plane-in-space | A plane which bisects the angle between the two given planes 2x – y + 2z – 4 = 0 and x + 2y + 2z – 2 = 0,
passes through the point : | [{"identifier": "A", "content": "(1, \u20134, 1)"}, {"identifier": "B", "content": "(1, 4, \u20131)"}, {"identifier": "C", "content": "(2, 4, 1)"}, {"identifier": "D", "content": "(2, \u20134, 1)"}] | ["D"] | null | Planes bisecting the given planes are<br><br>
$${{2x - y + 2z - 4} \over 3} = \pm {{x + 2y + 2z - 2} \over 3}$$<br><br>
$$ \Rightarrow $$ x - 3y = 2 or 3x + y + 4z = 6<br><br>
Out of the four given points in question's options only (2, -4, 1) lies on the plane 3x + y + 4z = 6 | mcq | jee-main-2019-online-12th-april-evening-slot |
pOnUJRdMYw4ui1JSKs3rsa0w2w9jx1zs3oi | maths | 3d-geometry | plane-in-space | If the plane 2x – y + 2z + 3 = 0 has the distances
$${1 \over 3}$$
and
$${2 \over 3}$$
units from the planes 4x – 2y + 4z + $$\lambda $$ = 0 and
2x – y + 2z + $$\mu $$ = 0, respectively, then the maximum value of $$\lambda $$ + $$\mu $$ is equal to : | [{"identifier": "A", "content": "13"}, {"identifier": "B", "content": "9"}, {"identifier": "C", "content": "5"}, {"identifier": "D", "content": "15"}] | ["A"] | null | Distance formula<br><br>
(i) $${{\left| {\lambda - 6} \right|} \over {\sqrt {16 + 4 + 16} }} = \left| {{{\lambda - 6} \over 6}} \right| = {1 \over 3}$$<br><br>
$$ \Rightarrow $$ $$\left| {\lambda - 6} \right| = 2$$<br><br>
$$ \Rightarrow $$ $$\lambda = 8,4$$<br><br>
(ii) $${{\left| {\mu - 3} \right|} \over {\sqrt {4 + 4 + 1} }} = {2 \over 3}$$<br><br>
$$ \Rightarrow $$ $$\left| {\mu - 3} \right| = 2$$<br><br>
$$ \Rightarrow $$ $$\mu = 5,1$$<br><br>
$$ \therefore $$ $${\left( {\mu + \lambda } \right)_{\max }} = 13$$
| mcq | jee-main-2019-online-10th-april-evening-slot |
ACCrs5Q2N2AyKcnuNg3rsa0w2w9jwxv0uci | maths | 3d-geometry | plane-in-space | If Q(0, –1, –3) is the image of the point P in the plane 3x – y + 4z = 2 and R is the point (3, –1, –2), then the
area (in sq. units) of $$\Delta $$PQR is : | [{"identifier": "A", "content": "$${{\\sqrt {65} } \\over 2}$$"}, {"identifier": "B", "content": "$$2\\sqrt {13} $$"}, {"identifier": "C", "content": "$${{\\sqrt {91} } \\over 2}$$"}, {"identifier": "D", "content": "$${{\\sqrt {91} } \\over 4}$$"}] | ["C"] | null | Image of Q in plane<br><br>
$${{\left( {x - 0} \right)} \over 3} = {{\left( {y + 1} \right)} \over { - 1}} = {{z + 3} \over { + 4}} = {{ - 2(1 - 12 - 2)} \over {9 + 1 + 16}} = 1$$<br><br>
x = 3, y = –2, z = 1<br><br>
P(3, –2, 1), Q(0, –1, –3), R(3, –1, –2)<br><br>
Now area of $$\Delta $$PQR is<br><br>
$${1 \over 2}\left| {\overline {PQ} \times \overline {QR} } \right| = {1 \over 2}\left| {\left| {\matrix{
{\widehat i} & {\widehat j} & {\widehat k} \cr
3 & { - 1} & 4 \cr
3 & 0 & 1 \cr
} } \right|} \right|$$<br><br>
$$ \Rightarrow {1 \over 2}\left| {\left\{ {\widehat i( - 1) - \widehat j(3 - 12) + \widehat k(3)} \right\}} \right|$$<br><br>
$$ \Rightarrow {1 \over 2}\sqrt {(1 + 81 + 9)} $$<br><br>
$$ \Rightarrow {{\sqrt {91} } \over 2}$$ | mcq | jee-main-2019-online-10th-april-morning-slot |
yodXktCxgD0IJ4xePL18hoxe66ijvwvwdsy | maths | 3d-geometry | plane-in-space | Let P be the plane, which contains the line of
intersection of the planes, x + y + z – 6 = 0 and
2x + 3y + z + 5 = 0 and it is perpendicular to the
xy-plane. Then the distance of the point (0, 0, 256)
from P is equal to : | [{"identifier": "A", "content": "205$$\\sqrt5$$"}, {"identifier": "B", "content": "63$$\\sqrt5$$"}, {"identifier": "C", "content": "11/$$\\sqrt5$$"}, {"identifier": "D", "content": "17/$$\\sqrt5$$"}] | ["C"] | null | P<sub>1</sub> : x + y + z – 6 = 0
<br><br>P<sub>2</sub> : 2x + 3y + z + 5 = 0
<br><br>Equation of plane which passes through the line of intersection of P<sub>1</sub> and P<sub>2</sub> is
<br><br>P<sub>1</sub> + $$\lambda $$P<sub>2</sub> = 0
<br><br>$$ \Rightarrow $$ (x + y + z – 6) + $$\lambda $$(2x + 3y + z + 5) = 0
<br><br>$$ \Rightarrow $$ (1 + 2$$\lambda $$)x + (1 + 3$$\lambda $$)y + (1 + $$\lambda $$) + (5$$\lambda $$ - 6) = 0
<br><br>As the above plane is perpendicular to xy plane
<br><br>$$ \therefore $$ $$\overrightarrow n .\widehat k = 0$$
<br><br>$$ \Rightarrow $$ $$\left[ {\left( {2 + \lambda } \right)\widehat i + \left( {3 + \lambda } \right)\widehat j + \left( {1 + \lambda } \right)\widehat k} \right].\widehat k$$ = 0
<br><br>$$ \Rightarrow $$ 1 + $$\lambda $$ = 0
<br><br>$$ \Rightarrow $$ $$\lambda $$ = -1
<br><br>So, equation of plane
<br><br>- x - 2y - 11 = 0
<br><br>$$ \Rightarrow $$ x + 2y + 11 = 0
<br><br>Distance of the point (0, 0, 256) from this plane
<br><br>= $$\left| {{{0 + 0 + 11} \over {\sqrt 5 }}} \right|$$ = $${{11} \over {\sqrt 5 }}$$ | mcq | jee-main-2019-online-9th-april-evening-slot |
oW3MTlGq7uoYSy1HduycQ | maths | 3d-geometry | plane-in-space | A plane passing through the points (0, –1, 0)
and (0, 0, 1) and making an angle $${\pi \over 4}$$ with the
plane y – z + 5 = 0, also passes through the
point | [{"identifier": "A", "content": "$$\\left( {\\sqrt 2 ,1,4} \\right)$$"}, {"identifier": "B", "content": "$$\\left(- {\\sqrt 2 ,1,4} \\right)$$"}, {"identifier": "C", "content": "$$\\left( -{\\sqrt 2 ,-1,-4} \\right)$$"}, {"identifier": "D", "content": "$$\\left( {\\sqrt 2 ,-1,4} \\right)$$"}] | ["A"] | null | Let ax + by + cz = 1 be the equation of the plane
<br><br>it passed through point (0, –1, 0).
<br><br>$$ \therefore $$ -b = 1
<br><br>$$ \Rightarrow $$ b = -1
<br><br>Also it passes through point (0, 0, 1)
<br><br>$$ \therefore $$ c = 1
<br><br>So the plane is ax - y + z = 1.
<br><br>This plane an angle $${\pi \over 4}$$ with the
plane y – z + 5 = 0.
<br><br>Normal to the plane ax - y + z = 1 is
<br><br>$${\overrightarrow a }$$ = $$a\widehat i - \widehat j + \widehat k$$
<br><br>Normal to the plane y – z + 5 = 0 is
<br><br>$${\overrightarrow b }$$ = $$\widehat j - \widehat k$$
<br><br>cos $$\theta $$ = $$\left| {{{\overrightarrow a .\overrightarrow b } \over {\left| {\overrightarrow a } \right|\left| {\overrightarrow b } \right|}}} \right|$$
<br><br>$$ \Rightarrow $$ $${1 \over {\sqrt 2 }}$$ = $$\left| {{{\overrightarrow a .\overrightarrow b } \over {\left| {\overrightarrow a } \right|\left| {\overrightarrow b } \right|}}} \right|$$
<br><br>$$ \Rightarrow $$ $${{\left| {0 - 1 - 1} \right|} \over {\sqrt {{a^2} + 1 + 1} \sqrt {{1^2} + {1^2}} }}$$ = $${1 \over {\sqrt 2 }}$$
<br><br>$$ \Rightarrow $$ $${a^2} + 2 = 4$$
<br><br>$$ \Rightarrow $$ a = $$ \pm $$ $$\sqrt 2 $$
<br><br>$$ \therefore $$ Equation of plane
<br><br>$$ \pm $$ $$\sqrt 2 $$x - y + z = 1
<br><br>Now by checking each options you can see
<br><br>equation - $$\sqrt 2 $$x - y + z = 1 satisfy by the point $$\left( {\sqrt 2 ,1,4} \right)$$. | mcq | jee-main-2019-online-9th-april-morning-slot |
1h4HQ5Q8R2RpdCkkTSLdT | maths | 3d-geometry | plane-in-space | The magnitude of the projection of the vector
$$\mathop {2i}\limits^ \wedge + \mathop {3j}\limits^ \wedge + \mathop k\limits^ \wedge $$ on the vector perpendicular to the plane
containing the vectors $$\mathop {i}\limits^ \wedge + \mathop {j}\limits^ \wedge + \mathop k\limits^ \wedge $$ and $$\mathop {i}\limits^ \wedge + \mathop {2j}\limits^ \wedge + \mathop {3k}\limits^ \wedge $$ , is : | [{"identifier": "A", "content": "$${{\\sqrt 3 } \\over 2}$$"}, {"identifier": "B", "content": "$$\\sqrt 6 $$"}, {"identifier": "C", "content": "$$\\sqrt {3 \\over 2} $$"}, {"identifier": "D", "content": "3$$\\sqrt 6 $$"}] | ["C"] | null | Let vector $$\overrightarrow p $$ is perpendicular to the both vectors $$\mathop {i}\limits^ \wedge + \mathop {j}\limits^ \wedge + \mathop k\limits^ \wedge $$ and $$\mathop {i}\limits^ \wedge + \mathop {2j}\limits^ \wedge + \mathop {3k}\limits^ \wedge $$.
<br><br>$$ \therefore $$ $$\overrightarrow p $$ = ($$\mathop {i}\limits^ \wedge + \mathop {j}\limits^ \wedge + \mathop k\limits^ \wedge $$) $$ \times $$ ($$\mathop {i}\limits^ \wedge + \mathop {2j}\limits^ \wedge + \mathop {3k}\limits^ \wedge $$)
<br><br>= $$\left| {\matrix{
{\widehat i} & {\widehat j} & {\widehat k} \cr
1 & 1 & 1 \cr
1 & 2 & 3 \cr
} } \right|$$
<br><br>= $$\widehat i - 2\widehat j + \widehat k$$
<br><br>Now a vector $$\overrightarrow a $$ = $$\mathop {2i}\limits^ \wedge + \mathop {3j}\limits^ \wedge + \mathop k\limits^ \wedge $$ is given and we have to findout projection of vector $$\overrightarrow a $$ on $$\overrightarrow p $$.
<br><br>$$ \therefore $$ Projection of vector $$\overrightarrow a $$ on $$\overrightarrow p $$
<br><br>= $$\left| {\overrightarrow a } \right|\cos \theta $$
<br><br>= $$\left| {\overrightarrow a } \right| \times {{\overrightarrow a .\overrightarrow p } \over {\left| {\overrightarrow a } \right|\left| {\overrightarrow p } \right|}}$$
<br><br>= $${{\overrightarrow a .\overrightarrow p } \over {\left| {\overrightarrow p } \right|}}$$
<br><br>= $${{\left( {2\widehat i + 3\widehat j + \widehat k} \right).\left( {\widehat i - 2\widehat j + \widehat k} \right)} \over {\sqrt {1 + 4 + 1} }}$$
<br><br>= $${{2 - 6 + 1} \over {\sqrt 6 }}$$
<br><br>= $${{ - 3} \over {\sqrt 6 }}$$
<br><br>Magnitude of projection of vector $$\overrightarrow a $$ on $$\overrightarrow p $$
<br><br>= $$\left| {{{ - 3} \over {\sqrt 6 }}} \right|$$ = $${{3 \over {\sqrt 6 }}}$$ = $${{\sqrt 3 } \over {\sqrt 2 }}$$
| mcq | jee-main-2019-online-8th-april-morning-slot |
A3dJQnt8WpOkVMARcRG7E | maths | 3d-geometry | plane-in-space | Let S be the set of all real values of $$\lambda $$ such that a plane passing through the points (–$$\lambda $$<sup>2</sup>, 1, 1), (1, –$$\lambda $$<sup>2</sup>, 1) and (1, 1, – $$\lambda $$<sup>2</sup>) also passes through the point (–1, –1, 1). Then S is equal to : | [{"identifier": "A", "content": "{1, $$-$$1}"}, {"identifier": "B", "content": "{3, $$-$$ 3}"}, {"identifier": "C", "content": "$$\\left\\{ {\\sqrt 3 } \\right\\}$$"}, {"identifier": "D", "content": "$$\\left\\{ {\\sqrt 3 , - \\sqrt 3 } \\right\\}$$"}] | ["D"] | null | All four points are coplanar so
<br><br>$$\left| {\matrix{
{1 - {\lambda ^2}} & 2 & 0 \cr
2 & { - {\lambda ^2} + 1} & 0 \cr
2 & 2 & { - {\lambda ^2} - 1} \cr
} } \right| = 0$$
<br><br>($$\lambda $$<sup>2</sup> + 1)<sup>2</sup> (3 $$-$$ $$\lambda $$<sup>2</sup>) = 0
<br><br>$$\lambda $$ = $$ \pm $$$$\sqrt 3 $$ | mcq | jee-main-2019-online-12th-january-evening-slot |
vCpFMyMOEky4bJBVl2v9s | maths | 3d-geometry | plane-in-space | If the point (2, $$\alpha $$, $$\beta $$) lies on the plane which passes through the points (3, 4, 2) and (7, 0, 6) and is perpendicular to the plane 2x – 5y = 15, then 2$$\alpha $$ – 3$$\beta $$ is equal to | [{"identifier": "A", "content": "12"}, {"identifier": "B", "content": "7"}, {"identifier": "C", "content": "17"}, {"identifier": "D", "content": "5"}] | ["B"] | null | Normal vector of plane
<br><br>$$ = \left| {\matrix{
{\widehat i} & {\widehat j} & {\widehat k} \cr
2 & { - 5} & 0 \cr
4 & { - 4} & 5 \cr
} } \right|$$
<br><br>$$ = - 4\left( {5\widehat i + 2\widehat j - 3\widehat k} \right)$$
<br><br>equation of plane is
<br><br>5(x $$-$$ 7) + 2y $$-$$ 3(z $$-$$ 6) = 0
<br><br>5x + 2y $$-$$ 3z = 17
<br><br>5 $$ \times $$ 2 + 2$$\alpha $$ $$-$$ 3$$\beta $$ = 17
<br><br>$$ \therefore $$ 2$$\alpha $$ $$-$$ 3$$\beta $$ = 17 $$-$$ 10 = 7 | mcq | jee-main-2019-online-11th-january-evening-slot |
y1O9ASDQ4I3iqhyAwRaOV | maths | 3d-geometry | plane-in-space | The direction ratios of normal to the plane through the points (0, –1, 0) and (0, 0, 1) and making an angle $${\pi \over 4}$$ with the plane y $$-$$ z + 5 = 0 are : | [{"identifier": "A", "content": "2, $$-$$1, 1"}, {"identifier": "B", "content": "$$2\\sqrt 3 ,1, - 1$$"}, {"identifier": "C", "content": "$$\\sqrt 2 ,1, - 1$$"}, {"identifier": "D", "content": "$$\\sqrt 2 , - \\sqrt 2 $$"}] | ["C"] | null | Let the equation of plane be
<br><br>a(x $$-$$ 0) + b(y + 1) + c(z $$-$$ 0) = 0
<br><br>It passes through (0, 0, 1) then
<br><br>b + c = 0 . . . . (1)
<br><br>Now cos $${\pi \over 4}$$ = $${{a\left( 0 \right) + b\left( 1 \right) + c\left( { - 1} \right)} \over {\sqrt 2 \sqrt {{a^2} + {b^2} + {c^2}} }}$$
<br><br>$$ \Rightarrow $$ a<sup>2</sup> $$=$$ $$-$$ 2bc and b $$=$$ $$-$$ c
<br><br>we get a<sup>2</sup> $$=$$ 2c<sup>2</sup>
<br><br>$$ \Rightarrow $$ a $$=$$ $$ \pm $$ $$\sqrt 2 $$ c
<br><br>$$ \Rightarrow $$ direction ratio (a, b, c) = ($$\sqrt 2 $$, $$-$$1, 1) or ($$\sqrt 2 $$, 1, $$-$$ 1) | mcq | jee-main-2019-online-11th-january-morning-slot |
FZp6IAeQMHnGO3LU7gCrN | maths | 3d-geometry | plane-in-space | The plane which bisects the line segment joining the points (–3, –3, 4) and (3, 7, 6) at right angles, passes through which one of the following points ? | [{"identifier": "A", "content": "(2, 1, 3)"}, {"identifier": "B", "content": "(4, $$-$$ 1, 2)"}, {"identifier": "C", "content": "(4, 1, $$-$$ 2)"}, {"identifier": "D", "content": "($$-$$ 2, 3, 5)"}] | ["C"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264937/exam_images/uqar8tycbulapjs4smas.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 10th January Evening Slot Mathematics - 3D Geometry Question 261 English Explanation">
<br>p : 3(x $$-$$ 0) + 5(y $$-$$ 2) + 1 (z $$-$$ 5) = 0
<br><br>3x + 5y + z = 15
<br><br>On checking all the options, the option (4, 1, − 2)
satisfy the equation of plane. | mcq | jee-main-2019-online-10th-january-evening-slot |
X8MEKBrNXBZseySo39pJw | maths | 3d-geometry | plane-in-space | Let A be a point on the line $$\overrightarrow r = \left( {1 - 3\mu } \right)\widehat i + \left( {\mu - 1} \right)\widehat j + \left( {2 + 5\mu } \right)\widehat k$$ and B(3, 2, 6) be a point in the space. Then the value of $$\mu $$ for which the vector $$\overrightarrow {AB} $$ is parallel to the plane x $$-$$ 4y + 3z = 1 is - | [{"identifier": "A", "content": "$${1 \\over 8}$$"}, {"identifier": "B", "content": "$${1 \\over 2}$$"}, {"identifier": "C", "content": "$${1 \\over 4}$$"}, {"identifier": "D", "content": "$$-$$ $${1 \\over 4}$$"}] | ["C"] | null | Let point A is
<br><br>$$\left( {1 - 3\mu } \right)\widehat i + \left( {\mu - 1} \right)\widehat j + \left( {2 + 5\mu } \right)\widehat k$$
<br><br>and point B is (3, 2, 6)
<br><br>then $$\overrightarrow {AB} = \left( {2 + 3\mu } \right)\widehat i + \left( {3 - \mu } \right)\widehat j + \left( {4 - 5\mu } \right)\widehat k$$
<br><br>which is parallel to the the plane x $$-$$ 4y + 3z = 1
<br><br>$$ \therefore $$ 2 + 3$$\mu $$ $$-$$ 12 + 4$$\mu $$ + 12 $$-$$ 15$$\mu $$ = 0
<br><br>8$$\mu $$ = 2
<br><br>$$\mu $$ = $${1 \over 4}$$ | mcq | jee-main-2019-online-10th-january-morning-slot |
xUxj1ys4ZXlUo5ebGRM5g | maths | 3d-geometry | plane-in-space | The plane passing through the point (4, –1, 2) and parallel to the lines $${{x + 2} \over 3} = {{y - 2} \over { - 1}} = {{z + 1} \over 2}$$ and $${{x - 2} \over 1} = {{y - 3} \over 2} = {{z - 4} \over 3}$$ also passes through the point - | [{"identifier": "A", "content": "(1, 1, $$-$$ 1)"}, {"identifier": "B", "content": "(1, 1, 1)"}, {"identifier": "C", "content": "($$-$$ 1, $$-$$ 1, $$-$$1)"}, {"identifier": "D", "content": "($$-$$ 1, $$-$$ 1, 1)"}] | ["B"] | null | Let $$\overrightarrow n $$ be the normal vector to the plane passing through (4, $$-$$1, 2) and parallel to the lines L<sub>1</sub> & L<sub>2</sub>
<br><br>then $$\overrightarrow n $$ = $$\left| {\matrix{
{\widehat i} & {\widehat j} & {\widehat k} \cr
3 & { - 1} & 2 \cr
1 & 2 & 3 \cr
} } \right|$$
<br><br>$$ \therefore $$ $$\overrightarrow n $$ = $$ - 7\widehat i - 7\widehat j + 7\widehat k$$
<br><br>$$ \therefore $$ Equation of plane is
<br><br>$$-$$ 1(x $$-$$ 4) $$-$$ 1(y + 1) + 1(z $$-$$ 2) = 0
<br><br>$$ \therefore $$ x + y $$-$$ z $$-$$ 1 = 0
<br><br>Now check options | mcq | jee-main-2019-online-10th-january-morning-slot |
8wJsTP566s6V7Wx2nqpZD | maths | 3d-geometry | plane-in-space | The equation of the plane containing the straight line $${x \over 2} = {y \over 3} = {z \over 4}$$ and perpendicular to the plane containing the straight lines $${x \over 3} = {y \over 4} = {z \over 2}$$ and $${x \over 4} = {y \over 2} = {z \over 3}$$ is : | [{"identifier": "A", "content": "x $$-$$ 2y + z = 0"}, {"identifier": "B", "content": "3x + 2y $$-$$ 3z = 0"}, {"identifier": "C", "content": "x + 2y $$-$$ 2z = 0"}, {"identifier": "D", "content": "5x + 2y $$-$$ 4z = 0"}] | ["A"] | null | Vector $$ \bot $$ to given plane = $$\left| {\matrix{
{\widehat i} & {\widehat j} & {\widehat k} \cr
3 & 4 & 2 \cr
4 & 2 & 3 \cr
} } \right|$$
<br><br>= $$\widehat i\left( {12 - 4} \right) - \widehat j\left( {9 - 8} \right) + \widehat k\left( {6 - 16} \right)$$
<br><br>= $$8\widehat i - \widehat j - 10\widehat k\,$$ . . . . (1)
<br><br>Vector parallel to given line
<br><br>= $$2\widehat i + 3\widehat j + 4\widehat k\,\,\,\,$$ . . . (2)
<br><br>Vector $$ \bot \,\,\,$$ to both (1) & (2) vector
<br><br>= $$\left| {\matrix{
{\widehat i} & {\widehat j} & {\widehat k} \cr
8 & { - 1} & { - 10} \cr
2 & 3 & 4 \cr
} } \right|$$
<br><br>= $$\widehat i\left( { - 4 + 30} \right) - \widehat j\left( {32 + 20} \right) + \widehat k\left( {24 + 2} \right)$$
<br><br>= $$26\widehat i - 52\widehat j + 26\widehat k$$
<br><br>Dr's of normal of required plane is
<br><br>(26, $$-$$52, 26) $$ \Rightarrow $$ (1, $$-$$2, 1)
<br><br>Equation of plane whose Dr's of Normal is (1, $$-$$2, 1) and passes through origin
<br><br>1.(x $$-$$ 0) $$-$$ 2(y $$-$$ 0) + 1.(z $$-$$ 0) = 0
<br><br>x $$-$$ 2y + z = 0 | mcq | jee-main-2019-online-9th-january-evening-slot |
OOzulklPdGI4Zf272LTtO | maths | 3d-geometry | plane-in-space | A tetrahedron has vertices P(1, 2, 1), Q(2, 1, 3), R(–1, 1, 2) and O(0, 0, 0). The angle between the faces OPQ and PQR is : | [{"identifier": "A", "content": "cos<sup>$$-$$1</sup>$$\\left( {{{17} \\over {31}}} \\right)$$"}, {"identifier": "B", "content": "cos<sup>$$-$$1</sup>$$\\left( {{{9} \\over {35}}} \\right)$$"}, {"identifier": "C", "content": "cos<sup>$$-$$1</sup>$$\\left( {{{19} \\over {35}}} \\right)$$"}, {"identifier": "D", "content": "cos<sup>$$-$$1</sup>$$\\left( {{7 \\over {31}}} \\right)$$"}] | ["C"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265437/exam_images/kiidub3irc2kt9mrj7ft.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 12th January Morning Slot Mathematics - 3D Geometry Question 255 English Explanation">
<br>$$\overrightarrow {OP} \times \overrightarrow {OQ} = \left( {\widehat i + 2\widehat j + \widehat k} \right) \times \left( {2\widehat i + \widehat j + 3\widehat k} \right)$$
<br><br>= 5$$\widehat i$$ $$-$$ $$\widehat j$$ $$-$$ 3$$\widehat k$$
<br><br>$$\overrightarrow {PQ} \times \overrightarrow {PR} = \left( {\widehat i - \widehat j + 2\widehat k} \right) \times \left( { - 2\widehat i - \widehat j + \widehat k} \right)$$
<br><br>= $$\widehat i$$ $$-$$ 5$$\widehat j$$ $$-$$ 3$$\widehat k$$
<br><br>cos$$\theta $$ = $${{5 + 5 + 9} \over {{{\left( {\sqrt {25 + 9 + 1} } \right)}^2}}} = {{19} \over {35}}$$ | mcq | jee-main-2019-online-12th-january-morning-slot |
iWc51afvvP3KvJRMFwjgy2xukf467nsf | maths | 3d-geometry | plane-in-space | The plane which bisects the line joining, the
points (4, –2, 3) and (2, 4, –1) at right angles
also passes through the point : | [{"identifier": "A", "content": "(4, 0, 1)"}, {"identifier": "B", "content": "(0, \u20131, 1)"}, {"identifier": "C", "content": "(0, 1, \u20131)"}, {"identifier": "D", "content": "(4, 0, \u20131)"}] | ["D"] | null | Direction ratios of normal to plane are < 2, –6, 4 >
<br><br> Also plane passes through (3, 1, 1)
<br><br>$$ \therefore $$ Equation of plane
<br><br/>2(x–3)–6(y–1)+4(z–1) = 0
<br><br>$$ \Rightarrow $$ x – 3y + 2z = 2
<br><br>By checking all options we can see this equation passes through (4, 0, –1) | mcq | jee-main-2020-online-3rd-september-evening-slot |
48HxDRJEsiZs9PgCGajgy2xukf8zwdb2 | maths | 3d-geometry | plane-in-space | If the equation of a plane P, passing through the intersection of the planes, <br/>x + 4y - z + 7 = 0
and 3x + y + 5z = 8 is ax + by + 6z = 15 for some a, b $$ \in $$ R, then the distance of the point
(3, 2, -1) from the plane P is........... | [] | null | 3 | Equation of plane P is<br><br>$$(x + 4y - z + 7) + \lambda (3x + y + 5z - 8) = 0$$<br><br>$$ \Rightarrow x(1 + 3\lambda ) + y(4 + \lambda ) + z( - 1 + 5\lambda ) + (7 - 8\lambda ) = 0$$<br><br>$${{1 + 3\lambda } \over a} = {{4 + \lambda } \over b} = {{5\lambda - 1} \over 6} = {{7 - 8\lambda } \over { - 15}}$$
<br><br>$$ \therefore $$ 15 - 75$$\lambda $$ = 42 - 48$$\lambda $$
<br><br>$$ \Rightarrow $$ -27 = 27$$\lambda $$
<br><br>$$ \Rightarrow $$ $$\lambda $$ = -1
<br><br>$$ \therefore $$ Plane is $$(x + 4y - z + 7) - 1 (3x + y + 5z - 8) = 0$$
<br><br>$$ \Rightarrow $$ $$2x - 3y + 6z - 15 = 0$$
<br><br>Distance of (3, 2, -1) from the plane P
<br><br>= $${{\left| {6 - 6 - 6 - 15} \right|} \over 7} = {{21} \over 7} = 3$$ | integer | jee-main-2020-online-4th-september-morning-slot |
4fJqBCT7tmvq1cvtwz7k9k2k5hk0mnw | maths | 3d-geometry | plane-in-space | The mirror image of the point (1, 2, 3) in a plane
is<br/><br> $$\left( { - {7 \over 3}, - {4 \over 3}, - {1 \over 3}} \right)$$. Which of the following
points lies on this plane ?</br> | [{"identifier": "A", "content": "(1, \u20131, 1)"}, {"identifier": "B", "content": "(\u20131, \u20131, \u20131)"}, {"identifier": "C", "content": "(\u20131, \u20131, 1)"}, {"identifier": "D", "content": "(1, 1, 1)"}] | ["A"] | null | Let A(1, 2, 3), B$$\left( { - {7 \over 3}, - {4 \over 3}, - {1 \over 3}} \right)$$
<br><br>$$ \therefore $$ Midpoint of AB = M = $$\left( {{{{{ - 7} \over 3} + 1} \over 2},{{{{ - 4} \over 3} + 2} \over 2},{{{{ - 1} \over 3} + 3} \over 2}} \right)$$
<br><br>= $$\left( {{{ - 2} \over 3},{1 \over 3},{4 \over 3}} \right)$$
<br><br>DR of AM = $$\left( {1 + {2 \over 3},2 - {1 \over 3},3 - {4 \over 3}} \right)$$
<br><br>= $$\left( {{5 \over 3},{5 \over 3},{5 \over 3}} \right)$$
<br><br>= (1, 1, 1)
<br><br>Equation of plane
<br><br>$$a\left( {x + {2 \over 3}} \right) + b\left( {y - {1 \over 3}} \right) + c\left( {z - {4 \over 3}} \right)$$ = 0
<br><br>$$ \Rightarrow $$ $$1\left( {x + {2 \over 3}} \right) + 1\left( {y - {1 \over 3}} \right) + 1\left( {z - {4 \over 3}} \right)$$ = 0
<br><br>$$ \Rightarrow $$ x + y + z = 1
<br><br>$$ \therefore $$ (1, –1, 1) lies on the plane. | mcq | jee-main-2020-online-8th-january-evening-slot |
Ou0sxBmmwtNWG6HOVU7k9k2k5e374ev | maths | 3d-geometry | plane-in-space | Let P be a plane passing through the points (2, 1, 0), (4, 1, 1) and (5, 0, 1) and R be any point
(2, 1, 6). Then the image of R in the plane P is :
| [{"identifier": "A", "content": "(4, 3, 2) "}, {"identifier": "B", "content": "(6, 5, - 2) "}, {"identifier": "C", "content": "(3, 4, -2)"}, {"identifier": "D", "content": "(6, 5, 2)"}] | ["B"] | null | Plane passing through (2, 1, 0), (4, 1, 1) and
(5, 0, 1) is
<br><br>$$\left| {\matrix{
{x - 2} & {y - 1} & {z - 0} \cr
{4 - 2} & {1 - 1} & {1 - 0} \cr
{5 - 2} & {0 - 1} & {1 - 0} \cr
} } \right|$$ = 0
<br><br>$$ \Rightarrow $$ x + y – 2z = 3
<br><br>$$ \therefore $$ Image of R(2, 1, 6) in this plane is
<br><br>$${{x - 2} \over 1} = {{y - 1} \over 1} = {{z - 6} \over { - 2}} = - 2{{\left( {2 + 1 - 12 - 3} \right)} \over {1 + 1 + 4}}$$
<br><br>$$ \therefore $$ (x, y, z) = (6, 5, –2) | mcq | jee-main-2020-online-7th-january-morning-slot |
T0VNJtB7BfAmHEDeS51klregikd | maths | 3d-geometry | plane-in-space | The equation of the plane passing through the point (1, 2, -3) and perpendicular to the
planes <br/><br>3x + y - 2z = 5 and 2x - 5y - z = 7, is :</br> | [{"identifier": "A", "content": "6x - 5y + 2z + 10 =0"}, {"identifier": "B", "content": "3x - 10y - 2z + 11 = 0"}, {"identifier": "C", "content": "6x - 5y - 2z - 2 = 0"}, {"identifier": "D", "content": "11x + y + 17z + 38 = 0"}] | ["D"] | null | Given, equation of planes are
<br/><br/>3x + y - 2z = 5
<br/><br/>2x - 5y - z = 7
<br/><br/>and point ( 1, 2, 3).
<br/><br/>Normal vector of required plane = n = $$\left| {\matrix{
{\widehat i} & {\widehat j} & {\widehat k} \cr
3 & 1 & { - 2} \cr
2 & { - 5} & { - 1} \cr
} } \right|$$
<br/><br/>= $${\widehat i}$$(-1 - 10) - $${\widehat j}$$( -3 + 4) + $${\widehat k}$$( -15 - 2)
<br/><br/>= -11$${\widehat i}$$ - $${\widehat j}$$ - 17$${\widehat k}$$
<br/><br/>Now, the equation of plane passing through (1, 2, -3) having normal
vector -11$${\widehat i}$$ - $${\widehat j}$$ - 17$${\widehat k}$$ is
<br/><br/>-[11(x - 1) + (y - 2) + 17(z + 3)] = 0
<br/><br/>$$ \Rightarrow $$ 11x + y + 17z + 38 = 0 | mcq | jee-main-2021-online-24th-february-morning-slot |
t1IiOet4hCn5Cjas1c1klt7cbt1 | maths | 3d-geometry | plane-in-space | A plane passes through the points A(1, 2, 3), B(2, 3, 1) and C(2, 4, 2). If O is the origin and P is (2, $$-$$1, 1), then the projection of $$\overrightarrow {OP} $$ on this plane is of length : | [{"identifier": "A", "content": "$$\\sqrt {{2 \\over 7}} $$"}, {"identifier": "B", "content": "$$\\sqrt {{2 \\over 5}} $$"}, {"identifier": "C", "content": "$$\\sqrt {{2 \\over 3}} $$"}, {"identifier": "D", "content": "$$\\sqrt {{2 \\over 11}} $$"}] | ["D"] | null | A(1, 2, 3), B(2, 3, 1), C(2, 4, 2), O(0, 0, 0)<br><br>Equation of plane passing through A, B, C will be<br><br>$$\left| {\matrix{
{x - 1} & {y - 2} & {z - 3} \cr
{2 - 1} & {3 - 2} & {1 - 3} \cr
{2 - 1} & {4 - 2} & {2 - 3} \cr
} } \right| = 0$$<br><br>$$ \Rightarrow \left| {\matrix{
{x - 1} & {y - 2} & {z - 3} \cr
1 & 1 & { - 2} \cr
1 & 2 & { - 1} \cr
} } \right| = 0$$<br><br>$$ \Rightarrow (x - 1)( - 1 + 4) - (y - 2)( - 1 + 2) + (z - 3)(2 - 1) = 0$$<br><br><br>$$ \Rightarrow (x - 1)(3) - (y - 2)(1) + (z - 3)(1) = 0$$<br><br>$$ \Rightarrow 3x - 3 - y + 2 + z - 3 = 0$$<br><br>$$ \Rightarrow 3x - y + z - 4 = 0$$, is the required plane.<br><br>Now, O(0, 0, 0) & P(2, $$-$$1, 1)<br><br><picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266653/exam_images/bdmf1ojhb2oj7cfg67mp.webp"><source media="(max-width: 500px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265020/exam_images/klrgfdsmvp0eukgdixrm.webp"><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267761/exam_images/fvcqglwivzefc79a48mr.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 25th February Evening Shift Mathematics - 3D Geometry Question 212 English Explanation"></picture><br><br>Plane is $$3x - y + z - 4 = 0$$<br><br>O' & P' are foot of perpendiculars.<br><br>For O'<br><br>$${{x - 0} \over 3} = {{y - 0} \over { - 1}} = {{z - 0} \over 1} = {{ - (0 - 0 + 0 - 4)} \over {9 + 1 + 1}}$$<br><br>$${x \over 3} = {y \over { - 1}} = {z \over 1} = {4 \over {11}}$$<br><br>$$ \Rightarrow O'\left( {{{12} \over {11}},{{ - 4} \over {11}},{4 \over {11}}} \right)$$<br><br>for P'<br><br>$${{x - 2} \over 3} = {{y + 1} \over { - 1}} = {{z - 1} \over 1} = {{ - (3(2) - ( - 1) + 1 - 4)} \over {9 + 1 + 1}}$$<br><br>$${{x - 2} \over 3} = {{y + 1} \over { - 1}} = {{z - 1} \over 1} = \left( {{{ - 4} \over {11}}} \right)$$<br><br>$$P'\left( {{{ - 12} \over {11}} + 2,{4 \over {11}} - 1,{{ - 4} \over {11}} + 1} \right)$$<br><br>$$ \Rightarrow P'\left( {{{10} \over {11}},{{ - 7} \over {11}},{7 \over {11}}} \right)$$<br><br>$$O'P' = \sqrt {{{\left( {{{10} \over {11}} - {{12} \over {11}}} \right)}^2} + {{\left( {{{ - 7} \over {11}} + {4 \over {11}}} \right)}^2} + {{\left( {{7 \over {11}} - {4 \over {11}}} \right)}^2}} $$<br><br>$$ \Rightarrow O'P' = {1 \over {11}}\sqrt {4 + 9 + 9} $$<br><br>$$ \Rightarrow O'P' = {{\sqrt {22} } \over {11}}$$<br><br>$$ \Rightarrow O'P' = {{\sqrt 2 \times \sqrt {11} } \over {11}}$$<br><br>$$ \Rightarrow O'P' = \sqrt {{2 \over {11}}} $$ | mcq | jee-main-2021-online-25th-february-evening-slot |
MgtSscZbWDByJ9eKdx1kluh6hjj | maths | 3d-geometry | plane-in-space | Consider the three planes<br/><br/>P<sub>1</sub> : 3x + 15y + 21z = 9,<br/><br/>P<sub>2</sub> : x $$-$$ 3y $$-$$ z = 5, and <br/><br/>P<sub>3</sub> : 2x + 10y + 14z = 5<br/><br/>Then, which one of the following is true? | [{"identifier": "A", "content": "P<sub>1</sub> and P<sub>2</sub> are parallel."}, {"identifier": "B", "content": "P<sub>1</sub>, P<sub>2</sub> and P<sub>3</sub> all are parallel."}, {"identifier": "C", "content": "P<sub>1</sub> and P<sub>3</sub> are parallel."}, {"identifier": "D", "content": "P<sub>2</sub> and P<sub>3</sub> are parallel."}] | ["C"] | null | P<sub>1</sub> : 3x + 15y + 21z = 9,<br><br>P<sub>2</sub> : x $$-$$ 3y $$-$$ z = 5<br><br>P<sub>3</sub> : x + 5y + 7z = 5/2
<br><br>$$ \therefore $$ P<sub>1</sub> and P<sub>3</sub> are parallel. | mcq | jee-main-2021-online-26th-february-morning-slot |
XBPH37cUetzOYSGHad1kluh8adb | maths | 3d-geometry | plane-in-space | If (1, 5, 35), (7, 5, 5), (1, $$\lambda$$, 7) and (2$$\lambda$$, 1, 2) are coplanar, then the sum of all possible values of $$\lambda$$ is : | [{"identifier": "A", "content": "$$ - {{44} \\over 5}$$"}, {"identifier": "B", "content": "$$ - {{39} \\over 5}$$"}, {"identifier": "C", "content": "$${{44} \\over 5}$$"}, {"identifier": "D", "content": "$${{39} \\over 5}$$"}] | ["C"] | null | A(1, 5, 35), B(7, 5, 5), C(1, $$\lambda$$, 7), D(2$$\lambda$$, 1, 2)<br><br>$$\overrightarrow {AB} $$ = 6$$\widehat i$$ $$-$$ 30$$\widehat k$$,
<br><br>$$\overrightarrow {BC} $$ = $$-$$6$$\widehat i$$ ($$\lambda$$ $$-$$ 5)$$\widehat j$$ + 2$$\widehat k$$,
<br><br>$$\overrightarrow {CD} $$ = (2$$\lambda$$ $$-$$ 1)$$\widehat i$$ + (1 $$-$$ $$\lambda$$)$$\widehat j$$ $$-$$ 5$$\widehat k$$<br><br>Points are coplanar<br><br>$$ \Rightarrow 0 = \left| {\matrix{
6 & 0 & { - 30} \cr
{ - 6} & {\lambda - 5} & 2 \cr
{2\lambda - 1} & {1 - \lambda } & { - 5} \cr
} } \right|$$<br><br>= 6($$-$$5$$\lambda$$ + 25 $$-$$ 2 + 2$$\lambda$$) $$-$$ 30($$-$$6 + 6$$\lambda$$ $$-$$ (2$$\lambda$$<sup>2</sup> $$-$$ $$\lambda$$ $$-$$ 10$$\lambda$$ + 5))<br><br>= 6($$-$$3$$\lambda$$ + 23) $$-$$ 30($$-$$2$$\lambda$$<sup>2</sup> + 11$$\lambda$$ $$-$$ 5 $$-$$ 6 + 6$$\lambda$$)<br><br>= 6($$-$$3$$\lambda$$ + 23) $$-$$ 30($$-$$2$$\lambda$$<sup>2</sup> + 17$$\lambda$$ $$-$$11)<br><br>= 6($$-$$3$$\lambda$$ + 23 + 10$$\lambda$$<sup>2</sup> $$-$$ 85$$\lambda$$ + 55)<br><br>= 6(10$$\lambda$$<sup>2</sup> $$-$$ 88$$\lambda$$ + 78) = 12(5$$\lambda$$<sup>2</sup> $$-$$ 44$$\lambda$$ + 39)<br><br>$$ \Rightarrow $$ 0 = 12(5$$\lambda$$<sup>2</sup> $$-$$ 44$$\lambda$$ + 39)
<br><br>$$ \Rightarrow $$ 5$$\lambda$$<sup>2</sup> $$-$$ 44$$\lambda$$ + 39 = 0
<br><br>this quadratic equation has two values $$\lambda$$<sub>1</sub> and $$\lambda$$<sub>2</sub>
<br><br>$$ \therefore $$ $$\lambda$$<sub>1</sub> + $$\lambda$$<sub>2</sub> = $${{44} \over 5}$$ | mcq | jee-main-2021-online-26th-february-morning-slot |
c11i6Yyeu5j7HAV9um1kluwweez | maths | 3d-geometry | plane-in-space | If the mirror image of the point (1, 3, 5) with respect to the plane <br/><br/>4x $$-$$ 5y + 2z = 8 is ($$\alpha$$, $$\beta$$, $$\gamma$$), then 5($$\alpha$$ + $$\beta$$ + $$\gamma$$) equals : | [{"identifier": "A", "content": "39"}, {"identifier": "B", "content": "41"}, {"identifier": "C", "content": "47"}, {"identifier": "D", "content": "43"}] | ["C"] | null | Image of (1, 3, 5) in the plane 4x $$-$$ 5y + 2z = 8 is ($$\alpha$$, $$\beta$$, $$\gamma$$)<br><br>$$ \Rightarrow {{\alpha - 1} \over 4} = {{\beta - 3} \over { - 5}} = {{\gamma - 5} \over 2} = - {{(4(1) - 5(3) + 2(5) - 8)} \over {{4^2} + {5^2} + {2^2}}} = {2 \over 5}$$<br><br>$$ \therefore $$ $$\alpha = 1 + 4\left( {{2 \over 5}} \right) = {{13} \over 5}$$<br><br>$$\beta = 3 - 5\left( {{2 \over 5}} \right) = 1 = {5 \over 5}$$<br><br>$$\gamma = 5 + 2\left( {{2 \over 5}} \right) = {{29} \over 5}$$<br><br>Thus, $$5(\alpha + \beta + \gamma ) = 5\left( {{{13} \over 5} + {5 \over 5} + {{29} \over 5}} \right) = 47$$ | mcq | jee-main-2021-online-26th-february-evening-slot |
KTmD8ld7JILBD51VmY1kmhxegkb | maths | 3d-geometry | plane-in-space | If for a > 0, the feet of perpendiculars from the points A(a, $$-$$2a, 3) and B(0, 4, 5) on the plane lx + my + nz = 0 are points C(0, $$-$$a, $$-$$1) and D respectively, then the length of line segment CD is equal to : | [{"identifier": "A", "content": "$$\\sqrt {41} $$"}, {"identifier": "B", "content": "$$\\sqrt {55} $$"}, {"identifier": "C", "content": "$$\\sqrt {31} $$"}, {"identifier": "D", "content": "$$\\sqrt {66} $$"}] | ["D"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265417/exam_images/r4nxivfvhrrbvtvvir7b.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 16th March Morning Shift Mathematics - 3D Geometry Question 204 English Explanation">
<br><br>Let $$\phi $$ is the angle between $$\overrightarrow {AB} $$ and $$\overrightarrow n $$.
<br><br>CD = AR = | AB |sin$$\phi$$<br><br>CD = | AB | $$\sqrt {1 - {{\cos }^2}\phi } $$<br><br>CD = | AB | $$\sqrt {1 - {{\left( {{{\overrightarrow {AB} .\,\overrightarrow n } \over {|AB|}}} \right)}^2}} $$<br><br>$$ = \sqrt {{{(AB)}^2} - {{(\overrightarrow {AB} \,.\,\overrightarrow n )}^2}} $$<br><br>[ $$\cos \phi = {{\overrightarrow {AB} \,.\,\overrightarrow n } \over {|\overrightarrow n ||\overrightarrow {AB} |}}$$]<br><br>$$|\overrightarrow {AB} |\, = a\widehat i - (2a + 4)\widehat j - 2\widehat k$$<br><br>$$\overrightarrow {AB} \,.\,\overrightarrow n = la - (2a + 4) - 2n$$<br><br>C on plane<br><br>(0)l $$-$$ am $$-$$ n = 0 ..... (1)<br><br>Also, $$\overrightarrow {AC} $$ || $$\overrightarrow n $$<br><br>$${a \over l} = {{ - a} \over m} = {4 \over n}$$<br><br>m = $$-$$l & an + 4m = 0 ..... (2)<br><br>From (1) and (2)<br><br>a<sup>2</sup>m + an = 0<br><br>$$\underline {4m + an = 0} $$<br><br>(a<sup>2</sup> $$-$$ 4)m = 0 $$ \Rightarrow $$ a = 2<br><br>2m + n = 0 .... (1)<br><br>m + l = 0<br><br>l<sup>2</sup> + m<sup>2</sup> + n<sup>2</sup> = 1<br><br>m<sup>2</sup> + m<sup>2</sup> + 4m<sup>2</sup> = 1<br><br>m<sup>2</sup> = $${1 \over 6}$$<br><br>m = $${1 \over {\sqrt 6 }}$$<br><br>n = $${{ - 2} \over {\sqrt 6 }}$$<br><br>l = $${{ - 1} \over {\sqrt 6 }}$$<br><br>Now, $$\overrightarrow {AB} \,.\,\overrightarrow n = 2\left( {{{ - 1} \over {\sqrt 6 }}} \right) - 8\left( {{{ - 1} \over {\sqrt 6 }}} \right) - 2\left( {{{ - 2} \over {\sqrt 6 }}} \right)$$<br><br>$$ = {{ - 2 - 8 + 4} \over {\sqrt 6 }} = - \sqrt 6 $$<br><br>$$|\overrightarrow {AB} |\, = \sqrt {4 + 64 + 4} = \sqrt {72} $$<br><br>$$CD = \sqrt {72 - 6} $$<br><br>$$CD = \sqrt {66} $$ | mcq | jee-main-2021-online-16th-march-morning-shift |
LOKOg3XFtIRmJyfvtU1kmiwwdp4 | maths | 3d-geometry | plane-in-space | If (x, y, z) be an arbitrary point lying on a plane P which passes through the points (42, 0, 0), (0, 42, 0) and (0, 0, 42), then the value of the expression <br/>$$3 + {{x - 11} \over {{{(y - 19)}^2}{{(z - 12)}^2}}} + {{y - 19} \over {{{(x - 11)}^2}{{(z - 12)}^2}}} + {{z - 12} \over {{{(x - 11)}^2}{{(y - 19)}^2}}} - {{x + y + z} \over {14(x - 11)(y - 19)(z - 12)}}$$ is equal to : | [{"identifier": "A", "content": "3"}, {"identifier": "B", "content": "39"}, {"identifier": "C", "content": "$$-$$45"}, {"identifier": "D", "content": "0"}] | ["A"] | null | From intercept from, equation of plane is x + y + z = 42<br><br>$$ \Rightarrow $$ (x $$-$$ 11) + (y $$-$$ 19) + (z $$-$$ 12) = 0<br><br>let a = x $$-$$ 11, b = y $$-$$ 19, c = z $$-$$ 12<br><br>a + b + c = 0<br><br>Now, given expression is<br><br>$$3 + {a \over {{b^2}{c^2}}} + {b \over {{a^2}{c^2}}} + {c \over {{a^2}{b^2}}} - {{42} \over {14abc}}$$<br><br>$$3 + {{{a^3} + {b^3} + {c^3} - 3abc} \over {{a^2}{b^2}{c^2}}}$$<br><br>If a + b + c = 0<br><br>$$ \Rightarrow $$ a<sup>3</sup> + b<sup>3</sup> + c<sup>3</sup> = 3 abc<br><br>$$ \therefore $$ $$3 + {0 \over {{a^2}{b^2}{c^2}}}$$<br><br>$$ = 3$$ | mcq | jee-main-2021-online-16th-march-evening-shift |
eUwQOrxgcwYlcgZH1a1kmjaqpo0 | maths | 3d-geometry | plane-in-space | The equation of the plane which contains the y-axis and passes through the point (1, 2, 3) is : | [{"identifier": "A", "content": "x + 3z = 0"}, {"identifier": "B", "content": "3x $$-$$ z = 0"}, {"identifier": "C", "content": "x + 3z = 10"}, {"identifier": "D", "content": "3x + z = 6"}] | ["B"] | null | Let the equation of the plane is a (x $$-$$ 1) + b(y $$-$$ 2) + c(z $$-$$ 3) = 0<br><br>Y-axis lies on it.<br><br>D.R.'s of y-axis are 0, 1, 0<br><br>$$ \therefore $$ 0.a + 1.b + 0.c = 0 $$ \Rightarrow $$ b = 0<br><br>$$ \therefore $$ Equation of plane is a(x $$-$$ 1) + c(z $$-$$ 3) = 0<br><br>x = 0, z = 0 also satisfy it $$-$$a $$-$$3c = 0 $$ \Rightarrow $$ a = $$-$$3c<br><br>$$-$$3c (x $$-$$ 1) + c (z $$-$$ 3) = 0<br><br>$$-$$3 + 3 + z $$-$$ 3 = 0<br><br>3x $$-$$ z = 0 | mcq | jee-main-2021-online-17th-march-morning-shift |
lRRAykFzsnTCagmm7r1kmko3iaz | maths | 3d-geometry | plane-in-space | Let P be an arbitrary point having sum of the squares of the distances from the planes x + y + z = 0, lx $$-$$ nz = 0 and x $$-$$ 2y + z = 0, equal to 9. If the locus of the point P is x<sup>2</sup> + y<sup>2</sup> + z<sup>2</sup> = 9, then the value of l $$-$$ n is equal to _________. | [] | null | 0 | Let point P is ($$\alpha$$, $$\beta$$, $$\gamma$$)<br><br>$${\left( {{{\alpha + \beta + \gamma } \over {\sqrt 3 }}} \right)^2} + {\left( {{{l\alpha - n\gamma } \over {\sqrt {{l^2} + {n^2}} }}} \right)^2} + {\left( {{{\alpha - 2\beta + \gamma } \over {\sqrt 6 }}} \right)^2} = 9$$<br><br>Locus is $${{{{(x + y + z)}^2}} \over 3} + {{{{(\ln - nz)}^2}} \over {{l^2} + {n^2}}} + {{{{(x - 2y + z)}^2}} \over 6} = 9$$<br><br>$${x^2}\left( {{1 \over 2} + {{{l^2}} \over {{l^2} + {n^2}}}} \right) + {y^2} + {z^2}\left( {{1 \over 2} + {{{n^2}} \over {{l^2} + {n^2}}}} \right) + 2zx\left( {{1 \over 2} - {{\ln } \over {{l^2} + {n^2}}}} \right) - 9 = 0$$<br><br>Since its given that $${x^2} + {y^2} + {z^2} = 9$$<br><br>After solving l = n,<br><br>then, l $$-$$ n = 0 | integer | jee-main-2021-online-17th-march-evening-shift |
yDMFPVyWe7YnyrRy3A1kmllwl3e | maths | 3d-geometry | plane-in-space | Let the plane ax + by + cz + d = 0 bisect the line joining the points (4, $$-$$3, 1) and (2, 3, $$-$$5) at the right angles. If a, b, c, d are integers, then the <br/>minimum value of (a<sup>2</sup> + b<sup>2</sup> + c<sup>2</sup> + d<sup>2</sup>) is _________. | [] | null | 28 | Normal of plane = $$\overrightarrow {PQ} = - 2\widehat i + 6\widehat j - 6\widehat k$$<br><br>a = $$-$$2, b = 6, c = $$-$$6<br><br>& equation of plane is <br><br>$$-$$2x + 6y $$-$$ 6z + d = 0<br><br>$$ M(3,0, - 2)$$ is the midpoint of the line which present on the plane
<br>which satisfy the plane<br><br>$$ \therefore $$ d = $$-$$6<br><br><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265089/exam_images/sfsfdbvdnonrmbtx93fv.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 18th March Morning Shift Mathematics - 3D Geometry Question 195 English Explanation"><br><br>Now equation of plane is<br><br>$$-$$2x + 6y $$-$$ 6z $$-$$ 6 = 0<br><br>x $$-$$ 3y + 3z + 3 = 0<br><br>$$ \Rightarrow $$ (a<sup>2</sup> + b<sup>2</sup> + c<sup>2</sup> + d<sup>2</sup>)<sub>min</sub> = 1<sup>2</sup> + 9 + 9 + 9 = 28 | integer | jee-main-2021-online-18th-march-morning-shift |
rnZCDGXIib2fJxI3oU1kmlm0roj | maths | 3d-geometry | plane-in-space | The equation of the planes parallel to the plane x $$-$$ 2y + 2z $$-$$ 3 = 0 which are at unit distance from the point (1, 2, 3) is ax + by + cz + d = 0. If (b $$-$$ d) = k(c $$-$$ a), then the positive value of k is : | [] | null | 4 | The equation of the planes parallel to the plane x $$-$$ 2y + 2z $$-$$ 3 = 0
<br><br>$$x - 2y + 2z + \lambda = 0$$<br><br>Now given<br><br>$$d = {{\left| {1 - 4 + 6 + \lambda } \right|} \over {\sqrt 9 }} = 1$$<br><br>$$\left| {\lambda + 3} \right| = 3$$<br><br>$$\lambda + 3 = \pm 3 \Rightarrow \lambda = 0, - 6$$<br><br>So planes are : $$x - 2y + 2z - 6 = 0$$<br><br>and $$x - 2y + 2z = 0$$<br><br>$$b - d = - 2 + 6 = 4$$<br><br>$$c - a = 2 - 1 = 1$$<br><br>$$ \therefore $$ $$ {{b - d} \over {c - a}} = k$$<br><br>$$ \Rightarrow k = 4$$ | integer | jee-main-2021-online-18th-march-morning-shift |
wimTMaIOTLKy7lhiU31kmm3y91x | maths | 3d-geometry | plane-in-space | Let the mirror image of the point (1, 3, a) with respect to the plane $$\overrightarrow r .\left( {2\widehat i - \widehat j + \widehat k} \right) - b = 0$$ be ($$-$$3, 5, 2). Then, the value of | a + b | is equal to ____________. | [] | null | 1 | <p>Given equation of plane in vector form is $$\overrightarrow r \,.\,(2\widehat i - \widehat j + \widehat k) - b = 0$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l3b0s96m/cc9e3c18-ff43-4aab-b277-327a534afd5b/f4fd09e0-d659-11ec-9a06-bd4ec5b93eb4/file-1l3b0s96n.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l3b0s96m/cc9e3c18-ff43-4aab-b277-327a534afd5b/f4fd09e0-d659-11ec-9a06-bd4ec5b93eb4/file-1l3b0s96n.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2021 (Online) 18th March Evening Shift Mathematics - 3D Geometry Question 192 English Explanation"></p>
<p>Its Cartesian form will be</p>
<p>$$2x - y + z = b$$ ...... (i)</p>
<p>$$\because$$ R is the mid-point of PQ.</p>
<p>$$\therefore$$ $$R \equiv {{P + Q} \over 2} \Rightarrow R \equiv \left( { - 1,4,{{a + 2} \over 2}} \right)$$</p>
<p>$$\because$$ R lies on the plane (i).</p>
<p>$$\therefore$$ $$ - 2 - 4 + {{a + 2} \over 2} = b \Rightarrow a + 2 = 2b + 12$$</p>
<p>$$ \Rightarrow a = 2b + 10$$ ....... (ii)</p>
<p>$$\because$$ Direction ratio's of QP is $$(1 - ( - 3),3 - 5,a - 2)$$</p>
<p>i.e. $$(4, - 2,a - 2)$$</p>
<p>and direction ratios of normal to the given plane are (2, $$-$$1, 1)</p>
<p>$$\because$$ n and QP are parallel.</p>
<p>$$\therefore$$ $${2 \over 4} = {{ - 1} \over { - 2}} = {1 \over {a - 2}}$$</p>
<p>$$\therefore$$ $$a - 2 = 2 \Rightarrow a = 4$$</p>
<p>From Eq. (ii), b = $$-$$3</p>
<p>$$\therefore$$ $$|a + b| = |4 - 3| = |1| = 1$$</p> | integer | jee-main-2021-online-18th-march-evening-shift |
1ks090yaw | maths | 3d-geometry | plane-in-space | Let the plane passing through the point ($$-$$1, 0, $$-$$2) and perpendicular to each of the planes 2x + y $$-$$ z = 2 and x $$-$$ y $$-$$ z = 3 be ax + by + cz + 8 = 0. Then the value of a + b + c is equal to : | [{"identifier": "A", "content": "3"}, {"identifier": "B", "content": "8"}, {"identifier": "C", "content": "5"}, {"identifier": "D", "content": "4"}] | ["D"] | null | Normal of required plane $$\left( {2\widehat i + \widehat j - \widehat k} \right) \times \left( {\widehat i - \widehat j - \widehat k} \right)$$<br><br>$$ = - 2\widehat i + \widehat j - 3\widehat k$$<br><br>Equation of plane <br><br>$$ - 2(x + 1) + 1(y - 0) - 3(z + 2) = 0$$<br><br>$$ - 2x + y - 3z - 8 = 0$$<br><br>$$2x - y + 3z + 8 = 0$$<br><br>$$a + b + c = 4$$ | mcq | jee-main-2021-online-27th-july-morning-shift |
1ktd1xchr | maths | 3d-geometry | plane-in-space | Let P be the plane passing through the point (1, 2, 3) and the line of intersection of the planes $$\overrightarrow r \,.\,\left( {\widehat i + \widehat j + 4\widehat k} \right) = 16$$ and $$\overrightarrow r \,.\,\left( { - \widehat i + \widehat j + \widehat k} \right) = 6$$. Then which of the following points does NOT lie on P? | [{"identifier": "A", "content": "(3, 3, 2)"}, {"identifier": "B", "content": "(6, $$-$$6, 2)"}, {"identifier": "C", "content": "(4, 2, 2)"}, {"identifier": "D", "content": "($$-$$8, 8, 6)"}] | ["C"] | null | $$(x + y + 4z - 16) + \lambda ( - x + y + z - 6) = 0$$<br><br>Passes through (1, 2, 3)<br><br>$$ - 1 + \lambda ( - 2) \Rightarrow \lambda = - {1 \over 2}$$<br><br>$$2(x + y + 4z - 16) - ( - x + y + z - 6) = 0$$<br><br>$$3x + y + 7z - 26 = 0$$ | mcq | jee-main-2021-online-26th-august-evening-shift |
1kto1zhhj | maths | 3d-geometry | plane-in-space | Let the acute angle bisector of the two planes x $$-$$ 2y $$-$$ 2z + 1 = 0 and 2x $$-$$ 3y $$-$$ 6z + 1 = 0 be the plane P. Then which of the following points lies on P? | [{"identifier": "A", "content": "$$\\left( {3,1, - {1 \\over 2}} \\right)$$"}, {"identifier": "B", "content": "$$\\left( { - 2,0, - {1 \\over 2}} \\right)$$"}, {"identifier": "C", "content": "(0, 2, $$-$$4)"}, {"identifier": "D", "content": "(4, 0, $$-$$2)"}] | ["B"] | null | $${P_1}:x - 2y - 2z + 1 = 0$$<br><br>$${P_2}:2x - 3y - 6z + 1 = 0$$<br><br>$$\left| {{{x - 2y - 2z + 1} \over {\sqrt {1 + 4 + 4} }}} \right| = \left| {{{2x - 3y - 6z + 1} \over {\sqrt {{2^2} + {3^2} + {6^2}} }}} \right|$$<br><br>$${{x - 2y - 2z + 1} \over 3} = \pm {{2x - 3y - 6z + 1} \over 7}$$<br><br>Since $${a_1}{a_2} + {b_1}{b_2} + {c_1}{c_2} = 20 > 0$$<br><br>$$\therefore$$ Negative sign will give acute bisector<br><br>$$7x - 14y - 14z + 7 = - [6x - 9y - 18z + 3]$$<br><br>$$ \Rightarrow 13x - 23y - 32z + 10 = 0$$<br><br>$$\left( { - 2,0, - {1 \over 2}} \right)$$ satisfy it $$\therefore$$ Ans. (b) | mcq | jee-main-2021-online-1st-september-evening-shift |
1l544kl7n | maths | 3d-geometry | plane-in-space | <p>If the mirror image of the point (2, 4, 7) in the plane 3x $$-$$ y + 4z = 2 is (a, b, c), then 2a + b + 2c is equal to :</p> | [{"identifier": "A", "content": "54"}, {"identifier": "B", "content": "50"}, {"identifier": "C", "content": "$$-$$6"}, {"identifier": "D", "content": "$$-$$42"}] | ["C"] | null | <p>We know mirror image of point (x<sub>1</sub>, y<sub>1</sub>, z<sub>1</sub>) in the plane ax + by + cz = d</p>
<p>$${{x - {x_1}} \over a} = {{y - {y_1}} \over b} = {{z - {z_1}} \over c} = {{ - 2(a{x_1} + b{y_1} + c{z_1} - d)} \over {{a^2} + {b^2} + {c^2}}}$$</p>
<p>Here given point (2, 4, 7) and plane $$3x - y + 4z = 2$$ then mirror image is</p>
<p>$${{x - 2} \over 3} = {{y - 4} \over { - 1}} = {{z - 7} \over 4} = {{ - 2(6 - 4 + 28 - 2)} \over {9 + 1 + 16}}$$</p>
<p>$$ \Rightarrow {{x - 2} \over 3} = {{y - 4} \over { - 1}} = {{z - 7} \over 4} = - {{28} \over {13}}$$</p>
<p>$$\therefore$$ $$x = - {{58} \over {13}} = a$$</p>
<p>$$y = {{80} \over {13}} = b$$</p>
<p>$$z = - {{21} \over {13}} = c$$</p>
<p>$$\therefore$$ $$2a + b + 2c$$</p>
<p>$$ = 2\left( { - {{58} \over {13}}} \right) + {{80} \over {13}} + 2\left( { - {{21} \over {13}}} \right)$$</p>
<p>$$ = {{ - 116 + 80 - 42} \over {13}} = {{ - 78} \over {13}} = - 6$$</p> | mcq | jee-main-2022-online-29th-june-morning-shift |
1l5461cy3 | maths | 3d-geometry | plane-in-space | <p>Let d be the distance between the foot of perpendiculars of the points P(1, 2, $$-$$1) and Q(2, $$-$$1, 3) on the plane $$-$$x + y + z = 1. Then d<sup>2</sup> is equal to ___________.</p> | [] | null | 26 | <p>Foot of perpendicular from P</p>
<p>$${{x - 1} \over { - 1}} = {{y - 2} \over 1} = {{z + 1} \over 1} = {{ - ( - 1 + 2 - 1 - 1)} \over 3}$$</p>
<p>$$ \Rightarrow p' \equiv \left( {{2 \over 3},{7 \over 3},{{ - 2} \over 3}} \right)$$</p>
<p>and foot of perpendicular from Q</p>
<p>$${{x - 2} \over { - 1}} = {{y + 1} \over 1} = {{z - 3} \over 1} = {{ - ( - 2 - 1 + 3 - 1)} \over 3}$$</p>
<p>$$ \Rightarrow Q' \equiv \left( {{5 \over 3},{{ - 2} \over 3},{{10} \over 3}} \right)$$</p>
<p>$$P'Q' = \sqrt {{{(1)}^2} + {{(3)}^2} + {{(4)}^2}} = d = \sqrt {26} $$</p>
<p>$$ \Rightarrow {d^2} = 26$$</p> | integer | jee-main-2022-online-29th-june-morning-shift |
1l55ini45 | maths | 3d-geometry | plane-in-space | <p>Let the plane ax + by + cz = d pass through (2, 3, $$-$$5) and is perpendicular to the planes <br/>2x + y $$-$$ 5z = 10 and 3x + 5y $$-$$ 7z = 12. If a, b, c, d are integers d > 0 and gcd (|a|, |b|, |c|, d) = 1, then the value of a + 7b + c + 20d is equal to :</p> | [{"identifier": "A", "content": "18"}, {"identifier": "B", "content": "20"}, {"identifier": "C", "content": "24"}, {"identifier": "D", "content": "22"}] | ["D"] | null | <p>Equation of pane through point (2, 3, $$-$$5) and perpendicular to planes 2x + y $$-$$ 5z = 10 and 3x + 5y $$-$$ 7z = 12 is</p>
<p>$$\left| {\matrix{
{x - 2} & {y - 3} & {z + 5} \cr
2 & 1 & { - 5} \cr
3 & 5 & { - 7} \cr
} } \right| = 0$$</p>
<p>$$\therefore$$ Equation of plane is $$(x - 2)( - 7 + 25) - (y - 3)$$</p>
<p>$$( - 14 + 15) + (z + 5)\,.\,7 = 0$$</p>
<p>$$\therefore$$ $$18x - y + 7z + 2 = 0$$</p>
<p>$$ \Rightarrow 18x - y + 7z = - 2$$</p>
<p>$$\therefore$$ $$ - 18x + y - 7z = 2$$</p>
<p>On comparing with $$ax + by + cz = d$$ where d > 0 is a = $$-$$ 18, b = 1, c = $$-$$ 7, d = 2</p>
<p>$$\therefore$$ $$a + 7b + c + 20d = 22$$</p> | mcq | jee-main-2022-online-28th-june-evening-shift |
1l566zqfs | maths | 3d-geometry | plane-in-space | <p>The acute angle between the planes P<sub>1</sub> and P<sub>2</sub>, when P<sub>1</sub> and P<sub>2</sub> are the planes passing through the intersection of the planes $$5x + 8y + 13z - 29 = 0$$ and $$8x - 7y + z - 20 = 0$$ and the points (2, 1, 3) and (0, 1, 2), respectively, is :</p> | [{"identifier": "A", "content": "$${\\pi \\over 3}$$"}, {"identifier": "B", "content": "$${\\pi \\over 4}$$"}, {"identifier": "C", "content": "$${\\pi \\over 6}$$"}, {"identifier": "D", "content": "$${\\pi \\over 12}$$"}] | ["A"] | null | <p>Family of Plane's equation can be given by</p>
<p>$$(5 + 8\lambda )x + (8 - 7\lambda )y + (13 + \lambda )z - (29 + 20\lambda ) = 0$$</p>
<p>P<sub>1</sub> passes through (2, 1, 3)</p>
<p>$$ \Rightarrow (10 + 16\lambda ) + (8 - 7\lambda ) + (39 + 3\lambda ) - (29 + 20\lambda ) = 0$$</p>
<p>$$ \Rightarrow - 8\lambda + 28 = 0 \Rightarrow \lambda = {7 \over 2}$$</p>
<p>d.r, s of normal to P<sub>1</sub></p>
<p>$$\left\langle {33,{{ - 33} \over 2},{{33} \over 2}} \right\rangle $$ or $$\left\langle {1, - {1 \over 2},{1 \over 2}} \right\rangle $$</p>
<p>P<sub>2</sub> passes through (0, 1, 2)</p>
<p>$$ \Rightarrow 8 - 7\lambda + 26 + 2\lambda - (29 + 20\lambda ) = 0$$</p>
<p>$$ \Rightarrow 5 - 25\lambda = 0$$</p>
<p>$$ \Rightarrow \lambda = {1 \over 5}$$</p>
<p>d.r, s of normal to P<sub>2</sub></p>
<p>$$\left\langle {{{33} \over 5},{{33} \over 5},{{66} \over 5}} \right\rangle $$ or $$\left\langle {1,1,2} \right\rangle $$</p>
<p>Angle between normals</p>
<p>$$ = {{\left( {\widehat i - {1 \over 2}\widehat j + {1 \over 2}\widehat k} \right).\,\left( {\widehat i + \widehat j + 2\widehat k} \right)} \over {{{\sqrt 3 } \over 2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\sqrt 6 }}$$</p>
<p>$$\cos \theta = {{1 - {1 \over 2} + 1} \over 3} = {1 \over 2}$$</p>
<p>$$\theta = {\pi \over 3}$$</p> | mcq | jee-main-2022-online-28th-june-morning-shift |
1l57p9hh3 | maths | 3d-geometry | plane-in-space | <p>Let the mirror image of the point (a, b, c) with respect to the plane 3x $$-$$ 4y + 12z + 19 = 0 be (a $$-$$ 6, $$\beta$$, $$\gamma$$). If a + b + c = 5, then 7$$\beta$$ $$-$$ 9$$\gamma$$ is equal to ______________.</p> | [] | null | 137 | <p>$${{x - a} \over 3} = {{y - b} \over { - 4}} = {{z - c} \over {12}} = {{ - 2(3a - 4b + 12c + 19)} \over {{3^2} + {{( - 4)}^2} + {{12}^2}}}$$</p>
<p>$${{x - a} \over 3} = {{y - b} \over { - 4}} = {{z - c} \over {12}} = {{ - 6a + 8b - 24c - 38} \over {169}}$$</p>
<p>$$(x,y,z) \equiv (a - 6,\,\beta ,\gamma )$$</p>
<p>$${{(a - 6) - a} \over 3} = {{\beta - b} \over { - 4}} = {{\gamma - c} \over {12}} = {{ - 6a + 8b - 24c - 38} \over {169}}$$</p>
<p>$${{\beta - b} \over { - 4}} = - 2 \Rightarrow \beta = 8 + b$$</p>
<p>$${{\gamma - c} \over {12}} = - 2 \Rightarrow \gamma = - 24 + c$$</p>
<p>$${{ - 6a + 8b - 24c - 38} \over {169}} = - 2$$</p>
<p>$$ \Rightarrow 3a - 4b + 12c = 150$$ ..... (1)</p>
<p>$$a + b + c = 5$$</p>
<p>$$3a + 3b + 3c = 15$$ ...... (2)</p>
<p>Applying (1) - (2)</p>
<p>$$ - 7b + 9c = 135$$</p>
<p>$$7b - 9c = - 135$$</p>
<p>$$7\beta - 9\gamma = 7(8 + b) - 9( - 24 + c)$$</p>
<p>$$ = 56 + 216 + 7b - 9c$$</p>
<p>$$ = 56 + 216 - 135 = 137$$</p> | integer | jee-main-2022-online-27th-june-morning-shift |
1l58g6a8s | maths | 3d-geometry | plane-in-space | <p>If the plane $$2x + y - 5z = 0$$ is rotated about its line of intersection with the plane $$3x - y + 4z - 7 = 0$$ by an angle of $${\pi \over 2}$$, then the plane after the rotation passes through the point :</p> | [{"identifier": "A", "content": "(2, $$-$$2, 0)"}, {"identifier": "B", "content": "($$-$$2, 2, 0)"}, {"identifier": "C", "content": "(1, 0, 2)"}, {"identifier": "D", "content": "($$-$$1, 0, $$-$$2)"}] | ["C"] | null | <p>$${P_1}:2x + y - 52 = 0$$, $${P_2}:3x - y + 4z - 7 = 0$$</p>
<p>Family of planes P<sub>1</sub> and P<sub>2</sub></p>
<p>$$P:{P_1} + \lambda {P_2}$$</p>
<p>$$\therefore$$ $$P:(2 + 3\lambda )x + (1 - \lambda )y + ( - 5 + 4\lambda )z - 7\lambda = 0$$</p>
<p>$$\because$$ $$P \bot {P_1}$$</p>
<p>$$\therefore$$ $$4 + 6\lambda + 1 - \lambda + 25 - 20\lambda = 0$$</p>
<p>$$\lambda = 2$$</p>
<p>$$\therefore$$ $$P:8x - y + 32 - 14 = 0$$</p>
<p>It passes through the point (1, 0, 2)</p> | mcq | jee-main-2022-online-26th-june-evening-shift |
1l59k5k64 | maths | 3d-geometry | plane-in-space | <p>Let p be the plane passing through the intersection of the planes $$\overrightarrow r \,.\,\left( {\widehat i + 3\widehat j - \widehat k} \right) = 5$$ and $$\overrightarrow r \,.\,\left( {2\widehat i - \widehat j + \widehat k} \right) = 3$$, and the point (2, 1, $$-$$2). Let the position vectors of the points X and Y be $$\widehat i - 2\widehat j + 4\widehat k$$ and $$5\widehat i - \widehat j + 2\widehat k$$ respectively. Then the points :</p> | [{"identifier": "A", "content": "X and X + Y are on the same side of P"}, {"identifier": "B", "content": "Y and Y $$-$$ X are on the opposite sides of P"}, {"identifier": "C", "content": "X and Y are on the opposite sides of P"}, {"identifier": "D", "content": "X + Y and X $$-$$ Y are on the same side of P"}] | ["C"] | null | <p>Let the equation of required plane</p>
<p>$$\pi :(x + 3y - z - 5) + \lambda (2x - y + z - 3) = 0$$</p>
<p>$$\because$$ (2, 1, $$-$$2) lies on it so, $$2 + \lambda ( - 2) = 0$$</p>
<p>$$ \Rightarrow \lambda = 1$$</p>
<p>Hence, $$\pi :3x + 2y - 8 = 0$$</p>
<p>$$\because$$ $${\pi _x} = - 9$$, $${\pi _y} = 5$$, $${\pi _{x + y}} = 4$$</p>
<p>$${\pi _{x - y}} = - 22$$ and $${\pi _{y - x}} = 6$$</p>
<p>Clearly X and Y are on opposite sides of plane $$\pi$$</p> | mcq | jee-main-2022-online-25th-june-evening-shift |
1l5bagla7 | maths | 3d-geometry | plane-in-space | <p>Let the points on the plane P be equidistant from the points ($$-$$4, 2, 1) and (2, $$-$$2, 3). Then the acute angle between the plane P and the plane 2x + y + 3z = 1 is :</p> | [{"identifier": "A", "content": "$${\\pi \\over 6}$$"}, {"identifier": "B", "content": "$${\\pi \\over 4}$$"}, {"identifier": "C", "content": "$${\\pi \\over 3}$$"}, {"identifier": "D", "content": "$${5\\pi \\over 12}$$"}] | ["C"] | null | <p>Let P(x, y, z) be any point on plane P<sub>1</sub></p>
<p>Then $${(x + 4)^2} + {(y - 2)^2} + {(z - 1)^2} = {(x - 2)^2} + {(y + 2)^2} + {(z - 3)^2}$$</p>
<p>$$ \Rightarrow 12x - 8y + 4z + 4 = 0$$</p>
<p>$$ \Rightarrow 3x - 2y + z + 1 = 0$$</p>
<p>And $${P_2}:2x + y + 3z = 0$$</p>
<p>$$\therefore$$ angle between P<sub>1</sub> and P<sub>2</sub></p>
<p>$$\cos \theta =\left| {{{6 - 2 + 3} \over {14}}} \right| \Rightarrow \theta = {\pi \over 3}$$</p> | mcq | jee-main-2022-online-24th-june-evening-shift |
1l6f2mwg1 | maths | 3d-geometry | plane-in-space | <p>A plane $$E$$ is perpendicular to the two planes $$2 x-2 y+z=0$$ and $$x-y+2 z=4$$, and passes through the point $$P(1,-1,1)$$. If the distance of the plane $$E$$ from the point $$Q(a, a, 2)$$ is $$3 \sqrt{2}$$, then $$(P Q)^{2}$$ is equal to :</p> | [{"identifier": "A", "content": "9"}, {"identifier": "B", "content": "12"}, {"identifier": "C", "content": "21"}, {"identifier": "D", "content": "33"}] | ["C"] | null | <p>First plane, $${P_1} = 2x - 2y + z = 0$$, normal vector $$ \equiv {\overline n _1} = (2, - 2,1)$$</p>
<p>Second plane, $${P_2} \equiv x - y + 2z = 4$$, normal vector $$ \equiv {\overline n _2} = (1, - 1,2)$$</p>
<p>Plane perpendicular to P<sub>1</sub> and P<sub>2</sub> will have normal vector $${\overline n _3}$$</p>
<p>Where $${\overline n _3} = \left( {{{\overline n }_1} \times {{\overline n }_2}} \right)$$</p>
<p>Hence, $${\overline n _3} = ( - 3, - 3,0)$$</p>
<p>Equation of plane E through $$P(1, - 1,1)$$ and $${\overline n _3}$$ as normal vector</p>
<p>$$(x - 1,y + 1,z - 1)\,.\,( - 3, - 3,0) = 0$$</p>
<p>$$ \Rightarrow x + y = 0 \equiv E$$</p>
<p>Distance of $$PQ(a,a,2)$$ from $$E = \left| {{{2a} \over {\sqrt 2 }}} \right|$$</p>
<p>as given, $$\left| {{{2a} \over {\sqrt 2 }}} \right| = 3\sqrt 2 \Rightarrow a = \, \pm \,3$$</p>
<p>Hence, $$Q \equiv ( \pm \,3, \pm \,3,2)$$</p>
<p>Distance $$7Q = \sqrt {21} \Rightarrow {(PQ)^2} = 21$$</p> | mcq | jee-main-2022-online-25th-july-evening-shift |
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