Three Dimensional Geometry
157 JEE Maths previous year questions on Three Dimensional Geometry — options free on every question; 16 include the answer & explanation free, the rest unlock with PYQ Pass.
Let the line L be \(\frac{x-1}{1}=\frac{y-4}{3}=\frac{z-7}{5}\) and foot of perpendicular from (1,-2,-1) to L is \((\alpha ,\beta ,\gamma )\), then \(\alpha +\beta +\gamma\) is :
\(\frac{-102}{35}\)
\(\frac{x-1}{1}=\frac{y-4}{3}-\frac{z-7}{5}=\lambda (l\mathrm{et})\\ B(\lambda +1,3\lambda +4,5\lambda +7)\\ DR's\text{of}AB=⟨\lambda ,3\lambda +6,5\lambda +8⟩\\ Now,\lambda +9\lambda +18+25\lambda +40=0\\ \lambda =\frac{-58}{35}\\ B=(\frac{-58}{35}+1,3\left(\frac{-58}{35}\right)+4,5\left(\frac{-58}{35}\right)+7)\\ =\left(\frac{-23}{35},\frac{-34}{35},\frac{-45}{35}\right)=\left(\alpha ,\beta ,\gamma \right)\\ \mathrm{Now},\alpha +\beta +\gamma =\frac{-23-34-45}{35}=\frac{-102}{35}\\ \\\)
The square of the distance of the point of intersection of the lines \(\vec{r}=(\hat{i}+\hat{j}-\hat{k})+\lambda(a \hat{i}-\hat{j}), a \neq 0\) and \(\vec{r}=(4 \hat{i}-\hat{k})+\mu(2 \hat{i}+a \hat{k})\) from the origin is:
[JEE Main 2026, 5 Apr (Shift 1)]
\(17\)
Let the point of intersection be \(A\)
\( \therefore A(1+\lambda a, 1-\lambda,-1) \equiv(4+2 \mu, 0,-1+\mu a) \)
\( \therefore 1-\lambda=0 \Rightarrow \lambda=1 \)
\( \mu a=0 \Rightarrow \mu=0 \ (\because a \neq 0) \)
\( 1+\lambda a=4+2 \mu \Rightarrow 2 \mu-a=-3\)
\( \Rightarrow a=3 \)
\( \therefore A(4,0,-1) \)
\( \therefore d^2=(\sqrt{16+1})^2=17\)
The perpendicular distance, of the line \(\frac{x-1}{2}=\frac{y+2}{-1}=\frac{z+3}{2}\) from the point \(P(2,-10,1)\), is:
[JEE Main 2025, 22 Jan (Shift 2)]
\(3\sqrt{5}\)
\(\frac{x-1}{2}=\frac{y+2}{-1}=\frac{z+3}{2}=\lambda \\ \text{Let}A\text{is the point on the line,}\\ A(2\lambda +1,-\lambda -2,2\lambda -3)\text{and}P(2,-10,1)\\ ∵\vec{\mathrm{PA}}\cdot \vec{\mathrm{n}}=0\\ \Rightarrow (2\lambda -1)2+(-\lambda +8)(-1)+(2\lambda -4)2=0\\ \Rightarrow 4\lambda -2+\lambda -8+4\lambda -8=0\\ \Rightarrow 9\lambda -18=0\Rightarrow \lambda =2\\ ∴\mathrm{A}(5,-4,1)\\ ∴\mathrm{AP}=\sqrt{{3}^{2}+{6}^{2}+{0}^{2}}=\sqrt{45}=3\sqrt{5}\)
The shortest distance between the lines\(\frac{x-4}{1}=\frac{y-3}{2}=\frac{z-2}{-3}\) and \(\frac{x+2}{2}=\frac{y-6}{4}=\frac{z-5}{-5}\) is:
[JEE Main 2026, 6 Apr (Shift 2)]
\(3\sqrt{5}\)
Shortest distance \(=\frac{\left|\left(\overrightarrow{a_2}-\overrightarrow{a_1}\right) \cdot\left(\overrightarrow{b_1} \times \overrightarrow{b_2}\right)\right|}{\left|\overrightarrow{b_1} \times \overrightarrow{b_2}\right|}\)
\(\overrightarrow{b_1} \times \overrightarrow{b_2}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & -3 \\ 2 & 4 & -5\end{array}\right|\)
\(=\hat{i}(-10+12)-\hat{j}(-5+6)+\hat{k}(4-4)\)
\(=2 \hat{i}-\hat{j}+0 \hat{k}\)
\(\left|\overrightarrow{b_1} \times \overrightarrow{b_2}\right|=\sqrt{2^2+(-1)^2+0^2}=\sqrt{5}\)
\(\overrightarrow{a_2}-\overrightarrow{a_1}=(-2-4) \hat{i}+(6-3) \hat{j}+(5-2) \hat{k}=-6 \hat{i}+3 \hat{j}+3 \hat{k}\)
\(\left(\overrightarrow{a_2}-\overrightarrow{a_1}\right) \cdot\left(\overrightarrow{b_1} \times \overrightarrow{b_2}\right)\)
\(=(-6 \hat{i}+3 \hat{j}+3 \hat{k})\cdot (2 \hat{i}-\hat{j}+0 \hat{k})=-15\)
Hence shortest distance
\(d=\frac{15}{\sqrt{5}}=3 \sqrt{5}\)
Let \(L\), be the line \(\frac{x+1}{2}=\frac{y+1}{3}=\frac{z+3}{6}\) and let \(S\) be the set of all points \((a, b, c)\) on \(L\), whose distance from the line \(\frac{x+1}{2}=\frac{y+1}{3}=\frac{z−9}{0}\) along the line \(L\) is \(7\). Then \(\sum _{(a,\text{ }b,\text{ }c)\in S}(a+b+c)\) is equal
[JEE Main 2026, 22 Jan (Shift 2)]
34
\(\frac{x+1}{2}=\frac{y+1}{3}=\frac{z+3}{6}=\lambda\)
\( x=2 \lambda-1, y=3 \lambda-1 \) & \(z=6 \lambda-3\)
\(\frac{x+1}{2}=\frac{y+1}{3}=\frac{z-9}{0}\)
\( x=2 \mu-1, y=3 \mu-1 \) & \(z=9\)
Let \(M\) is the point of intersection of both given lines
\(\Rightarrow 2\lambda −1=2\mu −1,3\lambda −1=3\mu −1,6\lambda −3=9\)
\(\Rightarrow \lambda =2=\mu\)
\(\Rightarrow \text{M }\left(3,5,9\right)\)
Now let point \(P\) be \(\left(2\text{ K}−1,3\text{ K}−1,6\text{ K}−3\right)\) on \(\text{L}\) such that \(\text{PM}=7\)
\(\Rightarrow \sqrt{{(2\text{K}−4)}^{2}+{(3\text{K}−6)}^{2}+{(6\text{K}−12)}^{2}}=7\)
\(\Rightarrow 49{\text{ K}}^{2}+196−196\text{K}=49\)
\(\Rightarrow {\text{K}}^{2}+4−4\text{K}=1\)
\(\Rightarrow {\text{K}}^{2}−4\text{K}+3=0\)
\(\Rightarrow \text{K}=1,3\)
So points are \(P\left(1,2,3\right)\) or \(P\left(5,8,15\right)\)
So sum of all co-ordinates of \(P = 34\)
Let \(\vec{\mathrm{a}}=\hat{\mathrm{i}}+2\hat{\mathrm{j}}+\hat{\mathrm{k}}\text{ and }\vec{\mathrm{b}}=2\hat{\mathrm{i}}+7\hat{\mathrm{j}}+3\hat{\mathrm{k}}.\) Let \({\mathrm{L}}_{1}:\vec{\mathrm{r}}=(-\hat{\mathrm{i}}+2\hat{\mathrm{j}}+\hat{\mathrm{k}})+\lambda \vec{\mathrm{a}},\lambda \in \mathrm{R}\text{ and }\)\({L}_{2}:\vec{\mathrm{r}}=(\hat{\mathrm{j}}+\hat{\mathrm{k}})+\mu \vec{\mathrm{b}},\mu \in \mathrm{R}\) be two lines. If the line \(L_3\) passes through the point of intersection of \(L_1\) and \(L_2\), and is parallel to \(\vec{a}+\vec{b}\), then \(L_3\) passes through the point:
[JEE Main 2025, 29 Jan (Shift 1)]
\((8, 26, 12) \)
Given
\({L}_{1}:\vec{\mathrm{r}}=(-\hat{\mathrm{i}}+2\hat{\mathrm{j}}+\hat{\mathrm{k}})+\lambda (\hat{\mathrm{i}}+2\hat{\mathrm{j}}+\hat{\mathrm{k}})\\ \Rightarrow \vec{\mathrm{r}}=(\lambda -1)\hat{\mathrm{i}}+2(\lambda +1)\hat{\mathrm{j}}+(\lambda +1)\hat{\mathrm{k}}\)
Next,
\({L}_{2}:\vec{\mathrm{r}}=(\hat{\mathrm{j}}+\hat{\mathrm{k}})+\mu (2\hat{\mathrm{i}}+7\hat{\mathrm{j}}+3\hat{\mathrm{k}})\\ \Rightarrow \vec{\mathrm{r}}=2\mu \hat{\mathrm{i}}+(1+7\mu )\hat{\mathrm{j}}+(1+3\mu )\hat{\mathrm{k}}\)
for point of intersection of \({L}_{1}\text{and}{L}_{2}\)
\(\lambda -1=2\mu \text{and}2(\lambda +1)=1+7\mu \\ \text{On solving, }\lambda =3\text{ and }\mu =1\\ \Rightarrow \vec{\mathrm{a}}+\vec{\mathrm{b}}=3\hat{\mathrm{i}}+9\hat{\mathrm{j}}+4\hat{\mathrm{k}}\\ \text{hence, }\\ {\mathrm{L}}_{3}:\vec{\mathrm{r}}=2\hat{\mathrm{i}}+8\hat{\mathrm{j}}+4\hat{\mathrm{k}}+\alpha (3\hat{\mathrm{i}}+9\hat{\mathrm{j}}+4\hat{\mathrm{k}})\\ \text{from options, we can see}{\mathrm{L}}_{3}\mathrm{passes}\\ \mathrm{through}(8,26,12)\)
Let the direction cosines of two lines satisfy the equations: \(4l+m-n=0\) and \(2mn+10nl+3lm=0\). Then the cosine of the acute angle between these lines is:
[JEE Main 2026, 23 Jan (Shift 1)]
\(\frac{10}{3\sqrt{38}}\)
Direction cosines of two lines satisfy the equation
\(4ℓ+m-n=0\\ \Rightarrow n=4ℓ+m...\left(1\right)\\ 2mn+10nℓ+3ℓm=0\\ \Rightarrow n\left(2m+10ℓ\right)+3ℓm=0...\left(2\right)\)
from eq. (1) and (2):
\(\Rightarrow (4ℓ+m)(2m+10ℓ)+3ℓm=0\\ \Rightarrow 8ℓm+40{ℓ}^{2}+2{m}^{2}+10ℓm+3ℓm=0\)
\(\Rightarrow 40{ℓ}^{2}+21ℓm+2{m}^{2}=0\\ \Rightarrow (8ℓ+m)(5ℓ+2m)=0\)
Case-1: \(8ℓ+m=0\Rightarrow m=-8ℓ\)
So direction ratio of \({L}_{1}\) is \(ℓ,-8ℓ,-4ℓ\)
Case-2: \(5ℓ+2m=0\Rightarrow m=\frac{-5}{2}ℓ\)
direction ratio of \({L}_{2}\) is \(ℓ,\frac{-5ℓ}{2},\frac{3ℓ}{2}\)
\(\cos \theta =\left|\frac{{ℓ}^{2}+20{ℓ}^{2}-6{ℓ}^{2}}{\sqrt{{ℓ}^{2}+64{ℓ}^{2}+16{ℓ}^{2}}\sqrt{{ℓ}^{2}+\frac{25{ℓ}^{2}}{4}+\frac{9{ℓ}^{2}}{4}}}\right|\\ =\frac{15{ℓ}^{2}}{\left(9ℓ\right)\frac{\sqrt{38}ℓ}{2}}=\frac{10}{3\sqrt{38}}\\ =\frac{10}{3\sqrt{38}}\)
Let the values of \(\lambda\) for which the shortest distance between the lines \(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}\) and \(\frac{x-\lambda }{3}=\frac{y-4}{4}=\frac{z-5}{5}\) is \(\frac{1}{\sqrt{6}}\) be \({\lambda }_{1}\) and \({\lambda }_{2}\). Then the radius of the circle passing through the points \((0,0),\left({\lambda }_{1},{\lambda }_{2}\right)\) and \(\left({\lambda }_{2},{\lambda }_{1}\right)\) is
[JEE Main 2025, 8 Apr (Shift 1)]
\(\frac{5\sqrt{2}}{3}\)
\(\Rightarrow \text{ }\vec{p}\text{ }\times \text{ }\vec{q}=\text{ }\left|\begin{matrix}\hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 4 \\ 3 & 4 & 5\end{matrix}\right|\text{ }=−\hat{i}+2\hat{j}-k\)
\(A\text{ }\equiv \text{ }\left(1,\text{ }2,\text{ }3\right)\text{ }B\text{ }\equiv \text{ }\left(\lambda ,\text{ }4,\text{ }5\right)\)
Shortest Distance
\(=\text{ }\left|\frac{\vec{AB}\text{ }⋅\text{ }\left(\vec{p}\text{ }\times \vec{q}\right)}{\left|\vec{p}\text{ }\times \text{ }\vec{q}\right|}\right|\text{ }.\frac{1}{\sqrt{6}}\)
\(=\text{ }\left|\frac{\left(\left(\lambda −1\right)\text{ }\hat{i}\text{ }−\text{ }2\hat{j}\text{ }+\text{ }2\hat{k}\right)\text{ }⋅\text{ }\left(−\hat{i}\text{ }+\text{ }2\hat{j}\text{ }−\text{ }\hat{k}\right)}{\sqrt{6}}\right|\)
\(\Rightarrow \text{ }\left|−\text{ }\lambda +1+4−2\right|=1\)
\(\Rightarrow \text{ }\left|\text{ }\lambda −3\right|=1\)
\(\Rightarrow \text{ }\lambda =3\text{ }\pm \text{ }1=\text{ }4\text{ },\text{ }2\text{ }\)
Radius of circle passing through points (0, 0), (4, 2) & (2, 4)
\(=\frac{abc}{4\Delta }\)
\(=\frac{\sqrt{20}\times \sqrt{20}\times \sqrt{8}}{4\times \frac{1}{2}\text{ }\left|\begin{matrix}1 & 1 & 1 \\ 0 & 4 & 2 \\ 0 & 2 & 4\end{matrix}\right|}\)
\(=\frac{20\times 2\sqrt{2}}{2\times 12}=\frac{5\sqrt{2}}{3}\)
Let the acute angle bisector of the two planes \(x-2 y-2 z+1=0\) and \(2 x-3 y-6 z+1=0\) be the plane \(P\). Then which of the following points lies on \(P\) ?
\(\left(-2,0,-\frac{1}{2}\right)\)
\({P}_{1}:x-2y-2z+1=0\\ {P}_{2}:2x-3y-6z+1=0\\ \left|\frac{x-2y-2z+1}{\sqrt{1+4+4}}\right|=\left|\frac{2x-3y-6z+1}{\sqrt{{2}^{2}+{3}^{2}+{6}^{2}}}\right|\\ \frac{x-2y-2z+1}{3}=\pm \frac{2x-3y-6z+1}{7}\\ \text{ Since }{a}_{1}{a}_{2}+{b}_{1}{b}_{2}+{c}_{1}{c}_{2}=20>0\\ \text{∴Negative sign will give acute bisector}\\ 7x-14y-14z+7=-[6x-9y-18z+3]\\ \Rightarrow 13x-23y-32z+10=0\\ \left(-2,0,-\frac{1}{2}\right)\text{ satisfy it. }\)
\({\mathrm{L}}_{1}=\frac{x-1}{1}=\frac{y-2}{-1}=\frac{z-1}{2},{L}_{2}:\frac{x+1}{-1}=\frac{y-2}{2}=\frac{z}{1}\)
Let the line \({\mathrm{L}}_{3}\) passes through the point \((\alpha ,\beta ,\gamma )\) perpendicular to \({L}_{1}\&{L}_{2}\) and \({L}_{3}\) intersect line \({L}_{1}\) then \(|5\alpha -11\beta -8\gamma |\).
25
\({\text{ L}}_{1}\text{: }\frac{x-1}{1}=\frac{y-2}{-1}=\frac{z-1}{2},{L}_{2}:\frac{x+1}{-1}=\frac{y-2}{2}=\frac{z}{1}\\ D'ratioof{L}_{1}\times {L}_{2}=\left|\begin{matrix}\hat{ı} & \hat{ȷ} & \hat{k} \\ 1 & -1 & 2 \\ -1 & 2 & 1\end{matrix}\right|=-5\hat{ı}-3\hat{ȷ}+\hat{k}\\ {L}_{3}:\frac{x-\alpha }{-5}=\frac{y-\beta }{-3}=\frac{z-\gamma }{1}=\mu \\ \left(5\mu +\alpha ,3\mu +\beta ,-\mu +\gamma \right)\\ Pat{L}_{1},\left(\lambda +1,-\lambda +2,2\lambda +1\right)\\ 5\mu +\alpha =\lambda +13\mu +\beta =-\lambda +2-\mu +\gamma =2\lambda +1\\ 5\mu -\lambda =1-\alpha 3\mu +\lambda =2-\beta \\ \mu =\frac{3-\alpha -\beta }{8}then\lambda =\frac{7+3\alpha -5\beta }{8}\\ -\left(\frac{3-\alpha -\beta }{8}\right)+\gamma =2\left(\frac{7+3\alpha -5\beta }{8}\right)+1\\ -3+\alpha +\beta +8\gamma =14+6\alpha -10\beta +8\\ \left|5\alpha -11\beta -8\gamma \right|=\left|-25\right|=25\)
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
[JEE Main 2024, 29 Jan (Shift 1)]
\(\sqrt{69}\)
G is centroid of \(\triangle PQR\)
\(∴G\equiv \left(\frac{2\times 2−1}{3},\frac{2\times 1+4}{3},\frac{2\times 2+2}{3}\right)\)
\(G\equiv (1,2,2)\)
Let \(x\) is the point of intersection of lines
\({l}_{1}:\frac{x−2}{0}=\frac{y}{2}=\frac{z+3}{−1}\)
\({k}_{2}:\frac{x−1}{1}=\frac{y+3}{−3}=\frac{z+1}{1},\text{ then }x\equiv (2,−6,0)\)
\(∴\)distance between \(G\) and \(x=\sqrt{69}\)
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
[JEE Main 2024, 29 Jan (Shift 1)]
\(\sqrt{69}\)
G is centroid of \(\triangle PQR\)
\(∴G\equiv \left(\frac{2\times 2−1}{3},\frac{2\times 1+4}{3},\frac{2\times 2+2}{3}\right)\)
\(G\equiv (1,2,2)\)
Let \(x\) is the point of intersection of lines
\({l}_{1}:\frac{x−2}{0}=\frac{y}{2}=\frac{z+3}{−1}\)
\({k}_{2}:\frac{x−1}{1}=\frac{y+3}{−3}=\frac{z+1}{1},\text{ then }x\equiv (2,−6,0)\)
\(∴\)distance between \(G\) and \(x=\sqrt{69}\)
The perpendicular distance, of the line \(\frac{x-1}{2}=\frac{y+2}{-1}=\frac{z+3}{2}\) from the point \(P(2,-10,1)\), is:
[JEE Main 2025, 22 Jan (Shift 2)]
\(3\sqrt{5}\)
Given \(P(2,-10,1)\)
\(\frac{x-1}{2}=\frac{y+2}{-1}=\frac{z+3}{2}=\lambda\)
Let \(A\) is the point on the line,
\(A\left(2\lambda +1,-\lambda -2,2\lambda -3\right)\)
\(∵\vec{\mathrm{PA}}\cdot \vec{\mathrm{n}}=0\\ \Rightarrow \left(2\lambda -1\right)2+\left(-\lambda +8\right)\left(-1\right)+\left(2\lambda -4\right)2=0\\ \Rightarrow 4\lambda -2+\lambda -8+4\lambda -8=0\\ \Rightarrow 9\lambda -18=0\Rightarrow \lambda =2\\ ∴\mathrm{A}\left(5,-4,1\right)\\ ∴\mathrm{AP}=\sqrt{{3}^{2}+{6}^{2}+{0}^{2}}=\sqrt{45}=3\sqrt{5}\)
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?
\((3,4,3-2 \sqrt{2})\)
Given point P(1,0,7).
Given line:
\(\frac{\left(x−1\right)}{1}=\frac{y}{\left(−3\right)}=\frac{\left(z−8\right)}{1}\)
A point on line is A(1,0,8).
Direction ratios of line = (1,−3,1).
Find foot of perpendicular H from P to line.
Using reflection formula:
\({P}^{'}=2H−P\)
After applying angle conditions given in question, required point obtained is the option corresponding to (B).
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?
\((3,4,3-2 \sqrt{2})\)
Given point P(1,0,7).
Given line:
\(\frac{\left(x−1\right)}{1}=\frac{y}{\left(−3\right)}=\frac{\left(z−8\right)}{1}\)
A point on line is A(1,0,8).
Direction ratios of line = (1,−3,1).
Find foot of perpendicular H from P to line.
Using reflection formula:
\({P}^{'}=2H−P\)
After applying angle conditions given in question, required point obtained is the option corresponding to (B).
The shortest distance between the lines \(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-1}{4}\) and \(\frac{x+2}{7}=\frac{y-2}{8}=\frac{z+1}{2}\) is (22 Jan, Shift I, Memory Based)
\(\frac{88}{\sqrt{1277}}\)
\(\begin{aligned}& A \equiv(1,2,1), C(-2,2,-1) \\& \text { where } \hat{\mathrm{n}}=\left|\begin{array}{lll}\hat{\mathrm{i}} & \hat{j} & \hat{k} \\2 & 3 & 4 \\7 & 8 & 2\end{array}\right|=-26 \hat{\mathrm{i}}+24 \hat{\mathrm{j}}-5 \hat{\mathrm{k}} \\& \\& =\left|\frac{(3 \hat{\mathrm{i}}+2 \hat{k}) \cdot(26 \hat{\mathrm{i}}-24 \hat{\mathrm{j}}+5 \hat{k})}{\sqrt{1277}}\right| \\& =\frac{78+10}{\sqrt{1277}}=\frac{88}{\sqrt{1277}}\end{aligned}\)
Let a straight line L pass through the point \(\mathrm{P}(2,-1,3)\) and be perpendicular to the lines \(\frac{x-1}{2}=\frac{y+1}{1}=\frac{z-3}{-2}\)and \(\frac{x-3}{1}=\frac{y-2}{3}=\frac{z+2}{4}\). If the line \(L\) intersects the \(y z\)-plane at the point \(Q\), then the distance between the points \(P\)and \(Q\) is :
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The shortest distance between the lines \(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-1}{4}\) and \(\frac{x+2}{7}=\frac{y-2}{8}=\frac{z+1}{2}\) is (22 Jan, Shift I, Memory Based)
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\(\text { Perpendicular distance from the point } P(-2,0,2) \text { to the line } \frac{x+1}{2}=\frac{y-1}{-1}=\frac{z+3}{2}\) (22 Jan, Shift II, Memory Based)
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If the square of the shortest distance between the lines \(\frac{x-2}{1}=\frac{y-1}{2}=\frac{z+3}{-3}\) and \(\frac{x+1}{2}=\frac{y+3}{4}=\frac{z+5}{-5}\) is \(\frac{m}{n}\), where \(m, n\) are coprime numbers, then \(m+n\) is equal to:
[JEE Main 2025, 23 Jan (Shift 2)]
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Let the vertices Q and R of the triangle PQR lie on the line \(\frac{\mathrm{x}+3}{5}=\frac{\mathrm{y}-1}{2}=\frac{\mathrm{z}+4}{3},\mathrm{QR}=5\) and the coordinates of the point P be \((0,2,3)\). If the area of the triangle PQR is\(\frac{m}{n}\) then:
[JEE Main 2025, 2 Apr (Shift 1)]
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Let \(\text{P}\left(\alpha ,\beta ,\gamma \right)\) be the point on the line \(\frac{x−1}{2}=\frac{y+1}{−3}=z\) at a distance \(4\sqrt{14}\) from the point \((1, –1, 0)\) and nearer to the origin. Then the shortest distance, between the lines \(\frac{x−\alpha }{1}=\frac{y−\beta }{2}=\frac{z−\gamma }{3}\) and \(\frac{x+5}{2}=\frac{y−10}{1}=\frac{z−3}{1}\), is equal to
[JEE Main 2026, 22 Jan (Shift 1)]
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Let the values of \(\lambda\) for which the shortest distance between the lines \(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}\) and \(\frac{x-\lambda }{3}=\frac{y-4}{4}=\frac{z-5}{5}\) is \(\frac{1}{\sqrt{6}}\) be \({\lambda }_{1}\) and \({\lambda }_{2}\). Then the radius of the circle passing through the points \((0,0),\left({\lambda }_{1},{\lambda }_{2}\right)\) and \(\left({\lambda }_{2},{\lambda }_{1}\right)\) is
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Let \({\mathrm{L}}_{1}:\frac{\mathrm{x}-1}{1}=\frac{\mathrm{y}-2}{-1}=\frac{\mathrm{z}-1}{2}\) and \({L}_{2}:\frac{\mathrm{x}+1}{-1}=\frac{\mathrm{y}-2}{2}=\frac{\mathrm{z}}{1}\)be two lines. Let \(L_3\) be a line passing through the point \((\alpha ,\beta ,\gamma )\) and be perpendicular to both \(L_1\) and \(L_2\). If \(L_3\) intersects \({L}_1\), then \(|5\alpha -11\beta -8\gamma |\) equals :
[JEE Main 2025, 29 Jan (Shift 1)]
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Let the direction cosines of two lines satisfy the equations: \(4l+m-n=0\) and \(2mn+10nl+3lm=0\). Then the cosine of the acute angle between these lines is:
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Let the foot of perpendicular from the point \((\lambda ,2,3)\) on the line \(\frac{x-4}{1}=\frac{y-9}{2}=\frac{z-5}{1}\) be the point \((1,\mu ,2)\). Then the distance between the lines \(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z+4}{6}\) and \(\frac{x-\lambda }{2}=\frac{y-\mu }{3}=\frac{z+5}{6}\) is equal to:
[JEE Main 2026, 8 Apr (Shift 2)]
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Let a line \(L\) be perpendicular to both the line \({L}_{1}:\frac{x+1}{3}=\frac{y+3}{5}=\frac{z+5}{7}\) and \({L}_{2}:\frac{x-2}{1}=\frac{y-4}{4}=\frac{z-6}{7}\). If \(\theta\) is the acute angle between the lines \(L\) and \({L}_{3}:\frac{x-\frac{8}{7}}{2}=\frac{y-\frac{4}{7}}{1}=\frac{z}{2}\), then \(\tan \theta\) is equal to:
[JEE Main 2026, 6 Apr (Shift 1)]
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The sum of all values of \(\alpha\), for which the shortest distance between the lines \(\frac{x+1}{\alpha }=\frac{y−2}{−1}=\frac{z−4}{−\alpha }\) and \(\frac{x}{\alpha }=\frac{y−1}{2}=\frac{z−1}{2\alpha }\) is \(\sqrt{2}\), is
[JEE Main 2026, 24 Jan (Shift 2)]
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The distance of the line \(\frac{x-2}{2}=\frac{y-6}{3}=\frac{z-3}{4}\) from the point \((1,4,0)\) along the line \(\frac{x}{1}=\frac{y-2}{2}=\frac{z+3}{3}\) is :
[JEE Main 2025, 23 Jan (Shift 2)]
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If the square of the shortest distance between the lines \(\frac{x-2}{1}=\frac{y-1}{2}=\frac{z+3}{-3}\) and \(\frac{x+1}{2}=\frac{y+3}{4}=\frac{z+5}{-5}\) is \(\frac{m}{n}\), where \(m, n\) are coprime numbers then \(\mathrm{m}+\mathrm{n}\) is equal to?
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Let a triangle \(PQR\) be such that \(P\) and \(Q\) lie on the line \(\frac{x+3}{8}=\frac{y-4}{2}=\frac{z+1}{2}\) and are at a distance of \(6\) units from \(\mathrm{R}(1,2,3)\). If \((\alpha, \beta, \gamma)\) is the centroid of \(\triangle \mathrm{PQR}\), then \(\alpha+\beta+\gamma\) is equal to:
[JEE Main 2026, 5 Apr (Shift 2)]
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The line \({\mathrm{L}}_{1}\) is parallel to the vector \(\vec{\mathrm{a}}=-3\hat{\mathrm{i}}+2\hat{\mathrm{j}}+4\hat{\mathrm{k}}\) and passes through the point \((7,6,2)\) and the line\({L}_{2}\) is parallel to the vector \(\vec{\mathrm{b}}=2\hat{\mathrm{i}}+\hat{\mathrm{j}}+3\hat{\mathrm{k}}\) and passes through the point \((5,3,4)\). The shortest distance between the lines \({L}_{1}\) and \({\mathrm{L}}_{2}\) is :
[JEE Main 2025, 2 Apr (Shift 2)]
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Let \(L\), be the line \(\frac{x+1}{2}=\frac{y+1}{3}=\frac{z+3}{6}\) and let \(S\) be the set of all points \((a, b, c)\) on \(L\), whose distance from the line \(\frac{x+1}{2}=\frac{y+1}{3}=\frac{z−9}{0}\) along the line \(L\) is \(7\). Then \(\sum _{(a,\text{ }b,\text{ }c)\in S}(a+b+c)\) is equal
[JEE Main 2026, 22 Jan (Shift 2)]
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The square of the distance of the point \((-2,-8,6)\) from the line \(\frac{x-1}{1}=\frac{y-1}{2}=\frac{z}{-1}\) along the line \(\frac{x+5}{1}=\frac{y+5}{-1}=\frac{z}{2}\)is equal to:
[JEE Main 2026, 4 Apr (Shift 1)]
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Let the lines \({L}_{1}:\vec{r}=\hat{i}+2\hat{j}+3\hat{k}\)
\(+\text{ }\lambda \left(2\hat{i}+3\hat{j}+4\hat{k}\right),\lambda \in R\) and
\({L}_{2}:\vec{r}\text{ }=\text{ }\left(4\hat{i}\text{ }+\text{ }\hat{j}\right)\text{ }+\text{ }\mu \left(5\hat{i}\text{ }+\text{ }2\hat{j}+\hat{k}\right),\)
\(\mu \text{ }\in \text{ }R,\) intersect at the point R.
Let P and Q be the points lying on lines
\({L}_{1}\) and \({L}_{2}\), respectively, such that
\(\left|\vec{PR}\right|=\sqrt{29}\) and \(\left|\vec{PQ}\right|=\sqrt{\frac{47}{3}}.\)
If the point P lies in the first octant, then \(27{\left(QR\right)}^{2}\) is equal to
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Let a line pass through two distinct points \(P(-2,-1,3)\) and Q, and be parallel to the vector \(3\hat{i}+2\hat{j}+2\hat{k}\). If the distance of the point Q from the point \(R(1,3,3)\) is 5 , then the square of the area of \(\triangle \mathrm{PQR}\) is equal to:
[JEE Main 2025, 22 Jan (Shift 2)]
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Let the line passing through the point \((–1, 2, 1)\) and parallel to the line \(\frac{x-1}{2}=\frac{y+1}{3}=\frac{z}{4}\)intersect the line \(\frac{x+2}{3}=\frac{y-3}{2}=\frac{z-4}{1}\)at the point P. Then the distance of P from the point \(Q(4, – 5, 1)\) is :
[JEE Main 2025, 24 Jan (Shift 1)]
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Let the values of \(p\), for which the shortest distance between the lines \(\frac{x+1}{3}=\frac{y}{4}=\frac{z}{5}\) and \(\vec{\mathrm{r}}=(\mathrm{p}\hat{\mathrm{i}}+2\hat{\mathrm{j}}+\hat{\mathrm{k}})+\lambda (2\hat{\mathrm{i}}+3\hat{\mathrm{j}}+4\hat{\mathrm{k}})\) is \(\frac{1}{\sqrt{6}}\), be \(\mathrm{a},\mathrm{b}\), \((\mathrm{a}<\mathrm{b})\). Then the length of the latus rectum of the ellipse \(\frac{{x}^{2}}{{a}^{2}}+\frac{{y}^{2}}{{b}^{2}}=1\) is :-
[JEE Main 2025, 4 Apr (Shift 1)]
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Let the acute angle bisector of the two planes \(x-2 y-2 z+1=0\) and \(2 x-3 y-6 z+1=0\) be the plane \(P\). Then which of the following points lies on \(P\) ?
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\({\mathrm{L}}_{1}=\frac{x-1}{1}=\frac{y-2}{-1}=\frac{z-1}{2},{L}_{2}:\frac{x+1}{-1}=\frac{y-2}{2}=\frac{z}{1}\)
Let the line \({\mathrm{L}}_{3}\) passes through the point \((\alpha ,\beta ,\gamma )\) perpendicular to \({L}_{1}\&{L}_{2}\) and \({L}_{3}\) intersect line \({L}_{1}\) then \(|5\alpha -11\beta -8\gamma |\).
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\({\mathrm{L}}_{1}=\frac{x-1}{1}=\frac{y-2}{-1}=\frac{z-1}{2},{L}_{2}:\frac{x+1}{-1}=\frac{y-2}{2}=\frac{z}{1}\)
Let the line \({\mathrm{L}}_{3}\) passes through the point \((\alpha ,\beta ,\gamma )\) perpendicular to \({L}_{1}\&{L}_{2}\) and \({L}_{3}\) intersect line \({L}_{1}\) then \(|5\alpha -11\beta -8\gamma |\).
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Let \(\text{P}\left(\alpha ,\beta ,\gamma \right)\) be the point on the line \(\frac{x−1}{2}=\frac{y+1}{−3}=z\) at a distance \(4\sqrt{14}\) from the point \((1, –1, 0)\) and nearer to the origin. Then the shortest distance, between the lines \(\frac{x−\alpha }{1}=\frac{y−\beta }{2}=\frac{z−\gamma }{3}\) and \(\frac{x+5}{2}=\frac{y−10}{1}=\frac{z−3}{1}\), is equal to
[JEE Main 2026, 22 Jan (Shift 1)]
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If the square of the shortest distance between the lines \(\frac{x-2}{1}=\frac{y-1}{2}=\frac{z+3}{-3}\) and \(\frac{x+1}{2}=\frac{y+3}{4}=\frac{z+5}{-5}\) is \(\frac{m}{n}\), where \(m, n\) are coprime numbers, then \(m+n\) is equal to:
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If the point of intersection of the lines \(\frac{x+1}{3}=\frac{y+a}{5}=\frac{z+b+1}{7}\) and \(\frac{x-2}{1}=\frac{y-b}{4}=\frac{z-2a}{7}\) lies on \(xy-\)plane, then the value of \(a+b\) is:
[JEE Main 2026, 2 Apr (Shift 1)]
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If the image of the point \(\mathrm{P}(1,0,3)\) in the line joining the points \(\mathrm{A}(4,7,1)\) and \(\mathrm{B}(3,5,3)\) is \(Q(\alpha ,\beta ,\gamma )\), then \(\alpha +\beta +\gamma\) is equal to
[JEE Main 2025, 2 Apr (Shift 2)]
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\(A\) and \(C\) are two points on the line \(\frac{x-6}{3}=\frac{y-7}{2}=\frac{z-7}{-2}\) such that \(A C=6 . \quad B\) is \((1,1,-2)\). Find area of \(\triangle A B C\). (24 Jan, Shift I, Memory Based)
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The perpendicular distance of point \(P(3,2,5)\) from the line \(\vec{r}=2 \hat{\imath}-\hat{\jmath}+\hat{k}+\lambda(4 \hat{\imath}-\hat{\jmath}+5 \hat{k})\) is (22 Jan, Shift II, Memory Based)
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\(\text { Perpendicular distance from the point } P(-2,0,2) \text { to the line } \frac{x+1}{2}=\frac{y-1}{-1}=\frac{z+3}{2}\) (22 Jan, Shift II, Memory Based)
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The sum of all values of \(\alpha\), for which the shortest distance between the lines \(\frac{x+1}{\alpha }=\frac{y−2}{−1}=\frac{z−4}{−\alpha }\) and \(\frac{x}{\alpha }=\frac{y−1}{2}=\frac{z−1}{2\alpha }\) is \(\sqrt{2}\), is
[JEE Main 2026, 24 Jan (Shift 2)]
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The perpendicular distance of point \(P(3,2,5)\) from the line \(\vec{r}=2 \hat{\imath}-\hat{\jmath}+\hat{k}+\lambda(4 \hat{\imath}-\hat{\jmath}+5 \hat{k})\) is (22 Jan, Shift II, Memory Based)
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Let the line L be \(\frac{x-1}{1}=\frac{y-4}{3}=\frac{z-7}{5}\) and foot of perpendicular from (1,-2,-1) to L is \((\alpha ,\beta ,\gamma )\), then \(\alpha +\beta +\gamma\) is :
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\({\mathrm{L}}_{1}=\frac{x-1}{1}=\frac{y-2}{-1}=\frac{z-1}{2},{L}_{2}:\frac{x+1}{-1}=\frac{y-2}{2}=\frac{z}{1}\)
Let the line \({\mathrm{L}}_{3}\) passes through the point \((\alpha ,\beta ,\gamma )\) perpendicular to \({L}_{1}\&{L}_{2}\) and \({L}_{3}\) intersect line \({L}_{1}\) then \(|5\alpha -11\beta -8\gamma |\).
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Let a line \(L\) passing through the point \((1,1,1)\) be perpendicular to both the vectors \(2\hat{i}+2\hat{j}+\hat{k}\) and \(\hat{i}+2\hat{j}+2\hat{k}\). If \(P(a,b,c)\) is the foot of perpendicular from the origin on the line \(L\), then the value of \(34(a+b+c)\) is:
[JEE Main 2026, 2 Apr (Shift 1)]
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The shortest distance between the lines \(\vec{r}=\left(\frac{1}{3}\hat{i}+2\hat{j}+\frac{8}{3}\hat{k}\right)+\lambda (2\hat{i}-5\hat{j}+6\hat{k})\) and \(\vec{r}=\left(-\frac{2}{3}\hat{i}-\frac{1}{3}\hat{k}\right)+\mu (\hat{j}-\hat{k}),\lambda ,\mu \in R\), is:
[JEE Main 2026, 4 Apr (Shift 2)]
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\(A\) and \(C\) are two points on the line \(\frac{x-6}{3}=\frac{y-7}{2}=\frac{z-7}{-2}\) such that \(A C=6 . \quad B\) is \((1,1,-2)\). Find area of \(\triangle A B C\). (24 Jan, Shift I, Memory Based)
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The length of the perpendicular drawn from the point \(\ (3,-1,11) \) to the line \(\ \frac{x}{2}=\frac{y-2}{3}=\frac{z-3}{4} \) is :
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The distance of the point \(P(4,6,-2)\) from the line passing through the point \((-3,2,3)\) and parallel to a line with direction ratios \(3,3,-1\) is equal to:
[JEE Main 2023, 25 Jan (Shift 1)]
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The distance of the point \((-1,9,-16)\) from the plane \(2 x+3 y\) \(-z=5\) measured parallel to the line \(\frac{x+4}{3}=\frac{2-y}{4}=\frac{z-3}{12}\)
[JEE Main 2023, 24 Jan (Shift 1)]
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Let the plane \(P\) pass through the intersection of the planes \(2 x+3 y-z=2\) and \(x+2 y+3 z=6\), and be perpendicular to the plane \(2 x+y-z+1=0\). If \(d\) is the distance of \(P\) from the point \((-7,1,1)\), then \(d^2\) is equal to
[JEE Main 2023, 1 Feb (Shift 2)]
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Shortest distance between the lines \(\frac{x-1}{2}=\frac{y+8}{-7}=\frac{z-4}{5}\) and \(\frac{x-1}{2}=\frac{y-2}{1}=\frac{z-6}{-3}\) is
[JEE Main 2023, 29 Jan (Shift 2)]
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If the equation of the plane containing the line \(x+2 y+\) \(3 z-4=0=2 x+y-z+5\) and perpendicular to the plane \(\vec{r}=(\hat{i}-\hat{j})+\lambda(\hat{i}+\hat{j}+\hat{k})+\mu(\hat{i}-2 \hat{j}+3 \hat{k})\) is \(a x+b y+c z\) \(=4\), then \((a-b+c)\) is equal to
[JEE Main 2023, 8 Apr (Shift 1)]
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The equation of the plane passing through the line of intersection of the planes \(\vec{r} \cdot(\hat{i}+\hat{j}+\hat{k})=1\) and \(\vec{r} \cdot(2 \hat{i}+3 \hat{j}-\hat{k})+4=0\) and parallel to the \(x\)-axis is
[JEE Main 2021, 27 Aug (Shift 2)]
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Let the plane containing the line of intersection of the planes \(P_1: x+(\lambda+4) y+z=1\) and
\(P_2: 2 x+y+z=2\) pass through the points \((0,1,0)\) and \((1,0,1)\). Then the distance of the point \((2 \lambda, \lambda,-\lambda)\) from the plane \(P_2\) is
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The equation of the line through the point \((0,1,2)\) and perpendicular to the line \(\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{-2}\) is:
[JEE Main 2021, 25 Feb (Shift 1)]
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If equation of the plane that contains the point \((-2,3,5)\) and is perpendicular to each of the planes \(2 x+4 y+5 z\) \(=8\) and \(3 x-2 y+3 z=5\) is \(\alpha x+\beta y+\gamma z+97=0\) then \(\alpha+\beta+\gamma=\)
[JEE Main 2023, 11 Apr (Shift 1)]
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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
\[\begin{aligned}& 3+\frac{x-11}{(y-19)^2(z-12)^2}+\frac{y-9}{(x-11)^2(z-12)^2} \\&+\frac{z-12}{(x-11)^2(y-19)^2}-\frac{x+y+z}{14(x-11)(y-19)(z-12)}\end{aligned}\] is equal to
[JEE Main 2021, 16 Mar (Shift 2)]
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The shortest distance between the lines x + 1 = 2y = –12z and x = y + 2 = 6z – 6 is
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The equation of the plane passing through the point (1, \(2,-3)\) and perpendicular to the planes \(3 x+y-2 z=5\) and \(2 x-5 y-z=7\), is:
[JEE Main 2021, 24 Feb (Shift 1)]
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The line, that is coplanar to the line \(\frac{x+3}{-3}=\frac{y-1}{1}=\frac{z-5}{5}\), is
[JEE Main 2023, 13 Apr (Shift 2)]
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Let \(P\) be the plane, passing through the point \((1,-1,-5)\) and perpendicular to the line joining the points \((4,1,-3)\) and \((2,4,3)\). Then the distance of \(P\) from the point \((3,-2,2)\) is
[JEE Main 2023, 31 Jan (Shift 2)]
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Let \(P\) be the plane passing through the points \((5,3,0)\), \((13,3,-2)\) and \((1,6,2)\). For \(\alpha \in N\), if the distances of the points \(A(3,4, \alpha)\) and \(B(2, \alpha, a)\) from the plane \(P\) are 2 and 3 respectively, then the positive value of \(a\) is
[JEE Main 2023, 11 Apr (Shift 2)]
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Let two vertices of triangle \(A B C\) be \((2,4,6)\) and \((0,-2\), \(-5)\), and its centroid be \((2,1,-1)\). If the image of third vertex in the plane \(x+2 y+4 z=11\) is \((\alpha, \beta, \gamma)\), then \(\alpha \beta+\) \(\beta \gamma+\gamma \alpha\) is equal to
[JEE Main 2023, 10 Apr (Shift 1)]
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If the lines \(\frac{x-1}{1}=\frac{y-2}{2}=\frac{z+3}{1}\) and \(\frac{x-a}{2}=\frac{y+2}{3}=\frac{z-3}{1}\) intersects at the point \(P\), then the distance of the point \(P\) from the plane \(z=a\) is:
[JEE Main 2023, 29 Jan (Shift 2)]
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Consider the line \(L\) given by the equation
\(\frac{x-3}{2}=\frac{y-1}{1}=\frac{z-2}{1}\). Let \(Q\) be the mirror image of the point \((2,3,-1)\) with respect to \(L\). Let a plane \(P\) be such that it passes through \(Q\), and the line \(L\) is perpendicular to \(P\). Then which of the following points is on the plane \(P\) ?
[JEE Main 2021, 20 Jul (Shift 2)]
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The shortest distance between the lines \(\frac{x-5}{1}=\frac{y-2}{2}=\frac{z-4}{-3}\) and \(\frac{x+3}{1}=\frac{y+5}{4}=\frac{z-1}{-5}\) is
[JEE Main 2023, 1 Feb (Shift 1)]
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If the equation of the plane passing through the line of intersection of the planes \(2 x-y+z=3,\) \(4 x-3 y+5 z+9=0\) and parallel to the line \(\frac{x+1}{-2}=\frac{y+3}{4}=\frac{z-2}{5}\) is \(a x+b y+\) \(c z+6=0\). then \(a+b+c\) is equal to
[JEE Main 2023, 6 Apr (Shift 1)]
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Let \(P\) be the plane, passing through the point \((1,-1,-5)\) and perpendicular to the line joining the points \((4,1,-3)\) and \((2,4,3)\). Then the distance of \(P\) from the point \((3,-2,2)\) is
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Let the plane \(P: 8 x+\alpha_1 y+\alpha_2 z+12=0\) be parallel to the line \(L: \frac{x+2}{2}=\frac{y-3}{3}=\frac{z+4}{5}\). If the intercept of \(P\) on the \(y\)-axis is 1 , then the distance between \(P\) and \(L\) is:
[JEE Main 2023, 31 Jan (Shift 2)]
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A line makes angles of \(\ 45^{\circ} \) and \(\ 60^{\circ} \) with the positive axes of \(\ X \) and \(\ Y \) respectively. The angle made by the same line with the positive axis of \(\ Z \) , is.
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Let the lines \(l_1: \frac{x+5}{3}=\frac{y+4}{1}=\frac{z-\alpha}{-2}\) and \(l_2: 3 x+2 y+z-2=0=x-3 y+2 z-13\) be coplanar. If the point \(P(a, b, c)\) on \(l_1\) is nearest to the point \(Q(-4,-3,2)\), then \(|a|+|b|+|c|\) is equal to
[JEE Main 2023, 12 Apr (Shift 1)]
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Let the foot of perpendicular from a point \(P(1,2\), \(-1)\) to the straight line \(L: \frac{x}{1}=\frac{y}{0}=\frac{z}{-1}\) be \(N\). Let a line be drawn from \(P\) parallel to the plane \(x+y+\) \(2 z=0\) which meets \(L\) at point \(Q\). If \(\alpha\) is the acuteangle between the lines \(P N\) and \(P Q\), then \(\cos \alpha\) is equal to:
[JEE Main 2021, 25 Jul (Shift 1)]
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A plane \(P\) contains the line of intersection of the plane \(\vec{r} \cdot(\hat{i}+\hat{j}+\hat{k})=6\) and \(\vec{r} \cdot(2 \hat{i}+3 \hat{j}+4 \hat{k})=-5\). If \(P\) passes through the point \((0,2,-2)\), then the square of distance of the point \((12,12,18)\) from the plane \(P\) is
[JEE Main 2023, 6 Apr (Shift 2)]
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The foot of perpendicular from the origin \(O\) to a plane \(P\) which meets the co-ordinate axes at the points \(A, B, C\) is \((2, a, 4), a \in N\). If the volume of the tetrahedron \(O A B C\) is \(144\) unit\(^3\), then which of the following points is NOT on \(P\) ?
[JEE Main 2023, 31 Jan (Shift 2)]
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The foot of perpendicular of the point \((2,0,5)\) on the line \(\frac{x+1}{2}=\frac{y-1}{5}=\frac{z+1}{-1}\) is \((\alpha, \beta, \gamma)\). Then which of the following is NOT correct?
[JEE Main 2023, 25 Jan (Shift 2)]
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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
[JEE Main 2023, 08 Apr (Shift 1)]
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One vertex of a rectangular parallelopiped is at the origin \(O\) and the lengths of its edges along \(x, y\) and \(z\) axes are \(3 , 4\) and \(5\) units respectively. Let \(P\) be the vertex \((3,4,5)\).Then the shortest distance between the diagonal \(O P\) and an edge parallel to \(z\) axis, not passing through \(O\) or \(P\) is:
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The shortest distance between the lines \(\frac{x+2}{1}=\frac{y}{-2}=\frac{z-5}{2}\) and \(\frac{x-4}{1}=\frac{y-1}{2}=\frac{z+3}{0}\) is
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For \(a, b \in Z\) and \(|a-b| \leq 10\), let the angle between the plane \(P: a x+y-z=b\) and the line \(l: x-1=a-y=z+1\) be \(\cos ^{-1}\left(\frac{1}{3}\right)\) If the distance of the point \((6,-6,4)\) from the plane \(P\) is \(3 \sqrt{6}\), then \(a^4+b^2\) is equal to
[JEE Main 2023, 08 Apr (Shift 2)]
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The plane which bisects the line joining, the points \( (4,-2,3) \) and \( (2,4,-1) \) at right angles also passes through the point:
[JEE Main 2020, 3 Sep (Shift 2)]
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If the shortest distance between the straight lines \(3(x-1)=6(y-2)=2(z-1)\) and \(4(x-2)=2(y-\lambda)=(z-3), \lambda \in R\) is \(\frac{1}{\sqrt{38}}\), then the integral value of \(\lambda\) is equal to:
[JEE Main 2021, 22 Jul (Shift 2)]
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The foot of the perpendicular drawn from the point \((4,2,3)\) to the line joining the points \((1,-2,3)\) and \((1,1,0)\) lies on the plane:
[JEE Main 2020, 3 Sep (Shift 1)]
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Consider the line \(L\) given by the equation
\(\frac{x-3}{2}=\frac{y-1}{1}=\frac{z-2}{1}\). Let \(Q\) be the mirror image of the point \((2,3,-1)\) with respect to \(L\). Let a plane \(P\) be such that it passes through \(Q\), and the line \(L\) is perpendicular to \(P\). Then which of the following points is one the plane \(P\) ?
[JEE Main 2021, 20 Jul (Shift 2)]
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Let the image of the point \(P(1,2,6)\) in the plane passing through the points \(A(1,2,0), B(1,4,1)\) and \(C(0,5,1)\) be \(Q(\alpha, \beta, \gamma)\). Then \(\left(\alpha^2+\beta^2+\gamma^2\right)\) is equal to:
[JEE Main 2023, 10 Apr (Shift 2)]
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The distance of the point \((-1,2,-2)\) from the line of intersection of the planes \(2 x+3 y+2 z=0\) and \(x-2 y+z=0\)
[JEE Main 2021, 31 Aug (Shift 2)]
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For \(a, b \in Z\) and \(|a-b| \leq 10\), let the angle between the plane \(P: a x+y-z=b\) and the line \(l: x-1=a-y=z+1\) be \(\cos ^{-1}\left(\frac{1}{3}\right)\) If the distance of the point \((6,-6,4)\) from the plane \(P\) is \(3 \sqrt{6}\), then \(a^4+b^2\) is equal to
[JEE Main 2023, 8 Apr (Shift 2)]
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If the mirror image of the point \((1,3,5)\) with respect to the plane \(4 x-5 y+2 z=8\) is \((\alpha, \beta, \gamma)\) then \(5(\alpha+\beta+\gamma)\) equals:
[JEE Main 2021, 26 Feb (Shift 2)]
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The plane \(2 x-y+z=4\) intersects the line segment joining the points \(A(a,-2,4)\) and \(B(2, b,-3)\) at the point \(C\) in the ratio 2: 1 and the distance of the point \(C\) from the origin is \(\sqrt{5}\). If \(a b<0\) and \(P\) is the point \((a-b, b, 2 b-a)\) then \(C P^2\) is equal to:
[JEE Main 2023, 29 Jan (Shift 2)]
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Let the plane containing the line of intersection of the planes \(P_1: x+(\lambda+4) y+z=1\) and \(P_2: 2 x+y+z=2\) pass through the points \((0,1,0)\) and \((1,0,1)\). Then the distance of the point \((2 \lambda, \lambda,-\lambda)\) from the plane \(P_2\) is
[JEE Main 2023, 24 Jan (Shift 2)]
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Let the equation of the plane, that passes through the point \(\left( 1,4, - 3 \right)\)and and contains the line of intersection of the planes \(3x - 2y + 4z - 7 = 0\) and \(x + 5y - 2z + 9 = 0\), be \(\alpha x\ + \beta y + \gamma z + 3 = 0\) then \(\alpha + \beta + \gamma\) is equal to?
[JEE Main 2021, 31 Aug (Shift 1)]
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The lines \(\vec{r}=(\hat{i}-\hat{j})+l(2 \hat{i}+\hat{k})\) and \(\vec{r}=(2 \hat{i}-\hat{j})+m(\hat{i}+\hat{j}-\hat{k})\)
[JEE Main 2020, 3 Sep (Shift 1)]
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Let \(L\) be the line of intersection of planes \(\vec{r} \cdot(\hat{i}-\hat{j}+2 \hat{k})=2\) and \(\vec{r} \cdot(2 \hat{i}+\hat{j}-\hat{k})=2\). If \(P(\alpha, \beta, \lambda)\) is the foot of perpendicular on \(L\) from the point \((1,2,0)\), then the value of \(35(\alpha+\beta+\lambda)\) is equal to:
[JEE Main 2021, 22 Jul (Shift 2)]
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A line makes angles of 45° and 60° with the positive axes of \(X\) and \(Y\) respectively. The angle made by the same line with the positive axis of \(Z,\) is
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Let \(\alpha\) be the angle between the lines whose direction cosines satisfy the equations \(l+m-n=0\) and \(l^2+m^2-n^2\) \(=0\). Then the value of \(\sin ^4 \alpha+\cos ^4 \alpha\) is:
[JEE Main 2021, 25 Feb (Shift 1)]
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Let the plane \(P: 4 x-y+z=10\) be rotated by an angle \(\pi / 2\) about its line of intersection with the plane \(x+y-z=4\). If \(\alpha\) is the distance of the point \((2,3,-4)\) from the new position of the plane \(P\), then \(35 \alpha\) is
[JEE Main 2023, 12 Apr (Shift 1)]
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Equation of a plane at a distance \(\sqrt{\frac{2}{21}}\) from the origin, which contains the line of intersection of the planes \(x-y-z-1\) \(=0\) and \(2 x+y-3 z+4=0\), is:
[JEE Main 2021, 27 Aug (Shift 1)]
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If for \(a>0\), the feet of perpendiculars from the points \(A\) \((a,-2 a, 3)\) and \(B(0,4,5)\) on the plane \(lx+m y+n z=0\) are points \(C(0,-a,-1)\) and \(D\) respectively, then the length of line segment \(C D\) is equal to:
[JEE Main 2021, 16 Mar (Shift 1)]
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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
\[\begin{aligned}& 3+\frac{x-11}{(y-19)^2(z-12)^2}+\frac{y-19}{(x-11)^2(z-12)^2} \\&+\frac{z-12}{(x-11)^2(y-19)^2}-\frac{x+y+z}{14(x-11)(y-19)(z-12)}\end{aligned}\]
is equal to:
[JEE Main 2021, 16 Mar (Shift 2)]
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Let a unit vector \(\overrightarrow{O P}\) make an angle \(\alpha, \beta, \gamma\) with the positive directions of the co-ordinate axes \(O X, O Y, O Z\) respectively, where \(\beta \in\left(0, \frac{\pi}{2}\right)\) if \(\overrightarrow{O P}\) is perpendicular to the plane through points \((1,2,3),(2,3,4)\) and \((1,5,7)\), then which one of the following is true?
[JEE Main 2023, 30 Jan (Shift 1)]
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Let \(P\) be the plane passing through the line \(\frac{x-1}{1}=\frac{y-2}{-3}=\frac{z+5}{7}\) and the point \((2,4,-3)\). If the image of the point \((-1,3,4)\) in the plane \(P\) is \((\alpha, \beta, \gamma)\), then \(\alpha+\beta+\gamma\) is equal to
[JEE Main 2023, 8 Apr (Shift 2)]
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Let \(P\) be the point of intersection of the line \(\frac{x+3}{3}=\frac{y+2}{1}=\frac{1-z}{2}\) and the plane \(x+y+z=2\). If the distance of the point \(P\) from the plane \(3 x-4 y+12 z=32\) is \(q\), then \(q\) and \(2 q\) are the roots of the equation
[JEE Main 2023, 10 Apr (Shift 1)]
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The Distance of the point \((7,-3,-4)\) from the plane passing through the points \((2,-3,1),(-1,1,-2)\) and \((3,-4,2)\) is:
[JEE Main 2023, 24 Jan (Shift 1)]
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If a plane passes through the points \((-1, k, 0),(2, k,-1)\), \((1,1,2)\) and is parallel to the line \(\frac{x-1}{1}=\frac{2 y+1}{2}=\frac{z+1}{-1}\), then the value of \(\frac{k^2+1}{(k-1)(k-2)}\) is
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If for some \(\alpha\) and \(\beta\) in \(R\), the intersection of the following three planes
\[\begin{aligned}& x+4 y-2 z=1 \\& x+7 y-5 z=\beta \\& x+5 y+\alpha z=5\end{aligned}\] is a line in \(R ^3\), then \(\alpha+\beta\) is equal to :
[JEE Main 2020, 9 Jan (Shift 1)]
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Consider the lines \(L_1\) and \(L_2\) given by
\(\begin{aligned}& L_1: \frac{x-1}{2}=\frac{y-3}{1}=\frac{z-2}{2} \\& L_2: \frac{x-2}{1}=\frac{y-2}{2}=\frac{z-3}{3}\end{aligned}\)
A line \(L_3\) having direction ratios \(1,-1,-2\), intersects \(L_1\) and \(L_2\) at the points \(P\) and \(Q\) respectively. Then the length of line segment \(P Q\) is
[JEE Main 2023, 25 Jan (Shift 1)]
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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 \(\quad\)
[JEE Main 2023, 8 Apr (Shift 1)]
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The lines \(\frac{x-2}{1}=\frac{y-3}{1}=\frac{z-4}{-k}\) and \(\frac{x-1}{k}=\frac{y-4}{2}=\frac{z-5}{1}\) are coplanar if
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Let the equation of plane passing through the line of intersection of the planes \(x+2 y+a z=2\) and \(x-y+z=3\) be \(5 x-11 y+b z=6 a-1\). For \(c \in Z\), if the distance of this plane from the point \((a,-c, c)\) is \(\frac{2}{\sqrt{a}}\), then \(\frac{a+b}{c}\) is equal
[JEE Main 2023, 13 Apr (Shift 1)]
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A vector \(\vec{v}\) in the first octant is inclined to the \(x\)-axis at \(60^{\circ}\), to the \(y\)-axis at \(45^{\circ}\) and to the \(z\)-axis at an acute angle. If a plane passing through the points \((\sqrt{2},-1,1)\) and \((a, b, c)\), is normal to \(\vec{v}\), then
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Let the position vectors of two points \(P\) and \(Q\) be \(3 \hat{i}-\hat{j}+2 \hat{k}\) and \(\hat{i}+2 \hat{j}-4 \hat{k}\) respectively. Let \(R\) and \(S\) be two points such that the direction ratios of lines \(P R\) and \(Q S\) are \((4,-1,2)\) and \((-2,1,-2)\), respectively. Let lines \(P R\) and \(Q S\) intersect at \(T\). If the vector \(\overrightarrow{T A}\) is perpendicular to both \(\overrightarrow{P R}\) and \(\overrightarrow{Q S}\) and the length of vector \(\overrightarrow{T A}\) is \(\sqrt{5}\) units, then the modulus of a position vector of \(A\) is:
[JEE Main 2021, 16 Mar (Shift 1)]
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Let the foot of perpendicular of the point \(P(3,-2,-9)\) on the plane passing through the points \((-1,-2,-3),(9,3,4)\), \((9,-2,1)\) be \(Q(\alpha, \beta, \gamma)\). Then the distance of \(Q\) from the origin is:
[JEE Main 2023, 15 Apr (Shift 1)]
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Let the equation of the plane, that passes through the point \((1,4,-3)\) and contains the line of intersection of the planes \(3 x-2 y+4 z-7=0\) and \(x+5 y-2 z+9=0\), be \(\alpha x+\beta y+\) \(\gamma z+3=0\), then \(\alpha+\beta+\gamma\) is equal to:
[JEE Main 2021, 31 Aug (Shift 1)]
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The plane, passing through the points \((0,-1,2)\) and \((-1,2,1)\) and parallel to the line passing through \((5,1,-7)\) and \((1,-1,-1)\) also passes through the point
[JEE Main 2023, 13 Apr (Shift 2)]
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Let a, b and c be distinct positive numbers. If the vectors \(a\hat{i}+a\hat{j}+c\hat{k},\hat{i}+\hat{k}\) and \(c\hat{i}+c\hat{j}+b\hat{k}\) are coplanar, then c is equal to:
[JEE Main 2021, 25 Jul (Shift 2)]
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The distance of line \(3 y-2 z-1=0=3 x-z+4\) from the point \((2,-1,6)\) is:
[JEE Main 2021, 1 Sep (Shift 2)]
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Let the shortest distance between the lines \(L: \frac{x-5}{-2}=\frac{y-\lambda}{0}=\frac{z+\lambda}{1}, \lambda \geq 0\) and \(L _1: x+1=y-1=4\) \(-z\) be \(2 \sqrt{6}\). If \((\alpha, \beta, \gamma)\) lies on \(L\), then which of the following is NOT possible?
[JEE Main 2023, 31 Jan (Shift 1)]
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Let the line passing through the points, \(P(2,-1,2)\) and \(Q(5,3,4)\) meet the plane \(x-y+z=4\) at the point \(R\). Then the distance of the point \(R\) from the plane \(x+2 y+3 z+2=0\) measured parallel to the line \(\frac{x-7}{2}=\frac{y+3}{2}=\frac{z-2}{1}\) is equal to
[JEE Main 2023, 11 Apr (Shift 2)]
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Let the foot of perpendicular from a point \(P(1,2\), \(-1)\) to the straight line \(L: \frac{x}{1}=\frac{y}{0}=\frac{z}{-1}\) be \(N\). Let a line be drawn from \(P\) parallel to the plane \(x+y+\) \(2 z=0\) which meets \(L\) at point \(Q\). If \(\alpha\) is the acute
angle between the lines \(P N\) and \(P Q\), then \(\cos \alpha\) is equal to:
[JEE Main 2021, 25 Jul (Shift 1)]
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\(A\) 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 \(\overline{ OP }\) on this plane is of length:
[JEE Main 2021, 25 Feb (Shift 2)]
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Consider the lines \(L_1\) and \(L_2\) given by
\[\begin{aligned}& L_1: \frac{x-1}{2}=\frac{y-3}{1}=\frac{z-2}{2} \\& L_2: \frac{x-2}{1}=\frac{y-2}{2}=\frac{z-3}{3}\end{aligned}\]
A line \(L_3\) having direction ratios \(1,-1,-2\), intersects \(L_1\) and \(L_2\) at the points \(P\) and \(Q\) respectively. Then the length of line segment \(P Q\) is
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The shortest distance between the lines \(x+1=2 y=-12 z\) and \(x=y+2=6 z-6\) is
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Let \(P\) be a plane \(l x+m y+n z=0\) containing the line, \(\frac{1-x}{1}=\frac{y+4}{2}=\frac{z+2}{3}\). If plane \(P\) divides the line segment \(A B\) joining points \(A(-3,-6,1)\) and \(B(2,4,-3)\) in ratio \(k: 1\), then the value of \(k\) is equal to:
[JEE Main 2021, 16 Mar (Shift 1)]
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The shortest distance between the lines \(x+1=2 y=-12 z\) and \(x=y+2=6 z-6\) is
[JEE Main 2023, 25 Jan (Shift 2)]
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For real numbers \(\alpha\) and \(\beta \neq 0\), if the point of intersection of the straight lines \(\frac{x-\alpha}{1}=\frac{y-1}{2}=\frac{z-1}{3}\) and \(\frac{x-4}{\beta}=\frac{y-6}{3}=\frac{z-7}{3}\) lies on the plane \(x+2 y-z=8\), then \(\alpha-\beta\) is equal to:
[JEE Main 2021, 27 Jul (Shift 2)]
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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 \(\quad\)
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The foot of perpendicular of the point \((2,0,5)\) on the line \(\frac{x+1}{2}=\frac{y-1}{5}=\frac{z+1}{-1}\) is \((\alpha, \beta, \gamma)\). Then.
Which of the following is NOT correct?
[JEE Main 2023, 25 Jan (Shift 2)]
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The distance of the point \((1,1,9)\) from the point of intersection of the line \(\frac{x - 3}{1} = \frac{y - 4}{2} = \frac{z - 5}{2}\) and the plane \(x + y + z = 17\) is:
[JEE Main 2021, 24 Feb (Shift 1)]
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Let \((\alpha, \beta, \gamma)\) be the image of the point \(P(2,3,5)\) in the plane \(2 x+y-3 z=6\). Then \(\alpha+\beta+\gamma\) is equal to
[JEE Main 2023, 11 Apr (Shift 1)]
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The equation of the plane which contains the \(y\)-axis and passes through the point \((1,2,3)\) is :
[JEE Main 2021, 17 Mar (Shift 1)]
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Let P be the plane passing through the point (1,2,3) and the line of intersection of the planes \(\vec{r}\cdot (\hat{i}+\hat{j}+4\hat{k})=16\) and \(\vec{r}\cdot (-\hat{i}+\hat{j}+\hat{k})=6\).
Then which of the following points does NOT lie on P ?
[JEE Main 2021, 26 Aug (Shift 2)]
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Let \(P\) be the plane passing through the point \((1,2,3)\) and the line of intersection of the planes \(\vec{r} \cdot(\hat{i}+\hat{j}+4 \hat{k})=16\) and \(\vec{r} \cdot(-\hat{i}+\hat{j}+\hat{k})=6\). Then which of the following points does NOT lie on \(P\) ?
[JEE Main 2021, 26 Aug (Shift 2)]
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The vector equation of the plane passing through the intersection of the planes \(\vec{r} \cdot(\hat{i}+\hat{j}+\hat{k})=1\) and \(\vec{r} \cdot(\hat{i}-2 \hat{j})=-2\), and the point \((1,0,2)\) is:
[JEE Main 2021, 24 Feb (Shift 2)]
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Let the line \(\frac{x}{1}=\frac{6-y}{2}=\frac{z+8}{5}\) intersect the lines \(\frac{x-5}{4}=\frac{y-7}{3}=\frac{z+2}{1}\) and \(\frac{x+3}{6}=\frac{3-y}{3}=\frac{z-6}{1}\) at the points \(A\) and \(B\) respectively. Then the distance of the midpoint of the line segment \(A B\) from the plane \(2 x-2 y+z\) \(=14\) is
[JEE Main 2023, 10 Apr (Shift 2)]
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Consider the three planes \(P_{1}: 3 x+15 y+21 z=9\), \(P_{2}: x-3 y-z=5\) and \(P_{3}: 2 x+10 y+14 z=5\). Then, which one of the following is true?
[JEE Main 2021, 26 Feb (Shift 1)]
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Let the system of linear equations
\(\begin{aligned}& -x+2 y-9 z=7 \\& -x+3 y+7 z=9 \\& -2 x+y+5 z=8 \\& -3 x+y+13 z=\lambda\end{aligned}\)
has a unique solution \(x=\alpha, y=\beta, z=\gamma\). Then the distance of the point \((\alpha, \beta, \gamma)\) from the plane \(2 x-2 y+z=\lambda\) is
[JEE Main 2023, 15 Apr (Shift 1)]
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Let the equation of plane passing through the line of intersection of the planes \(x+2 y+a z=2\) and \(x-y+z=3\) be \(5 x-11 y+b z=6 a-1\). For \(c \in Z\), if the distance of this plane from the point \((a,-c, c)\) is \(\frac{2}{\sqrt{a}}\), then \(\frac{a+b}{c}\) is equal to:
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Let the equation of the plane, that passes through the point \(\left( 1,4, - 3 \right)\) and contains the line of intersection of the planes \(3x - 2y + 4z - 7 = 0\) and \(x + 5y - 2z + 9 = 0\), be \(\alpha x\ + \beta y + \gamma z + 3 = 0\) then \(\alpha + \beta + \gamma\) is equal to?
[JEE Main 2021, 31 Aug (Shift 1)]
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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:
[JEE Main 2021, 27 Jul (Shift 1)]
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The distance from the point \((3,4,5)\) to the point where the line \(\frac{x-3}{1}=\frac{y-4}{2}=\frac{z-5}{2}\) meets the plane \(\mathrm{x}+\mathrm{y}+\mathrm{z}=17\) is
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Let the system of linear equations \(-x+2 y-9 z=7\), \(-x+3 y+7 z=9\), \(-2 x+y+5 z=8\) and \(-3 x+y+13 z=\lambda\) has a unique solution \(x=\alpha, y=\beta, z=\gamma\). Then the distance of the point \((\alpha, \beta, \gamma)\) from the plane \(2 x-2 y+z=\lambda\) is
[JEE Main 2023, 15 Apr (Shift 1)]
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The lines \(x=ay–1=z–2\) and \(x=3y–2=bz–2,(ab\neq 0)\) are coplanar, if:
[JEE Main 2021, 20 Jul (Shift 2)]
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Let \(a, b \in R\). If the mirror image of the point \(P(a, 6,9)\) with respect to the line \(\frac{x-3}{7}=\frac{y-2}{5}=\frac{z-1}{-9}\) is (20, \(b,-a-9)\), then \(|a+b|\) is equal to:
[JEE Main 2021, 24 Feb (Shift 2)]
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Let \(S\) be the set of all values of \(\lambda\), for which the shortest distance between the lines \(\frac{x-\lambda}{0}=\frac{y-3}{4}=\frac{z+6}{1}\) and \(\frac{x+\lambda}{3}=\frac{y}{-4}=\frac{z-6}{0}\) is \(13\) . Then \(8\left|\sum_{\lambda \in S} \lambda\right|\) is equal to
[JEE Main 2023, 15 Apr (Shift 1)]
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Let \(P\) be the plane passing through the line \(\frac{x-1}{1}=\frac{y-2}{-3}=\frac{z+5}{7}\) and the point \((2,4,-3)\). If the image of the point \((-1,3,4)\) in the plane \(P\) is \((\alpha, \beta, \gamma)\), then \(\alpha+\beta+\gamma\) is equal to
[JEE Main 2023, 08 Apr (Shift 2)]
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If the equation of the plane containing the line x + 2y + 3z – 4 = 0 = 2x + y – z + 5 and perpendicular to the plane \(\vec{r}=\left(\hat{i}−\hat{j}\right)+\lambda \left(\hat{i}+\hat{j}+\hat{k}\right)+\mu \left(\hat{i}−2\hat{j}+3\hat{k}\right)\) is ax + by + cz = 4, then (a – b + c) is equal to
[JEE Main 2023, 08 Apr (Shift 1)]
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Let \(Q\) be the cube with the set of vertices \(\left\{\left(x_1, x_2, x_3\right)\right\}\) \(\in R ^3: x_1, x_2, x_3\{0,1\}\). Let \(F\) be the set of all \(F\) twelve lines containing the diagonals of the six faces of the cube \(Q\). Let \(S\) be the set of all four lines containing the main diagonals of the cube \(Q\); for instance, the line passing through the vertices \((0,0,0)\) and \((1,1,1)\) is in \(S\). For lines \(\ell_1\) and \(\ell_2\), let \(d\left(\ell_1, \ell_2\right)\) denote the shortest distance between them. Then the maximum value of \(d \left(\ell_1, \ell_2\right)\), as \(\ell_1\) varies over \(F\) and \(\ell_2\) varies over \(S\), is
[JEE Advanced 2023]
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The equation of the plane which contains the \(y\)-axis and passes through the point \((1,2,3)\) is:
[JEE Main 2021, 17 Mar (Shift 1)]
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A plane \(P\) contains the line \(x+2 y+3 z+1=0=x-y-z-6\), and is perpendicular to the plane \(-2 x+y+z+8=0\). Then which of the following points lies on \(P\) ?
[JEE Main 2021, 26 Aug (Shift 1)]
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