3D Geometry
453 Questions
Start JEE Mains Test
2020
Q301
JEE Mains
MCQ
14 Mar 2026
The mirror image of the point (1, 2, 3) in a plane
is
$\left( { - {7 \over 3}, - {4 \over 3}, - {1 \over 3}} \right)$. Which of the following points lies on this plane ?
$\left( { - {7 \over 3}, - {4 \over 3}, - {1 \over 3}} \right)$. Which of the following points lies on this plane ?
A.
(1, –1, 1)
B.
(–1, –1, –1)
C.
(–1, –1, 1)
D.
(1, 1, 1)
2020
Q302
JEE Mains
MCQ
14 Mar 2026
The shortest distance between the lines
${{x - 3} \over 3} = {{y - 8} \over { - 1}} = {{z - 3} \over 1}$ and
${{x + 3} \over { - 3}} = {{y + 7} \over 2} = {{z - 6} \over 4}$ is :
${{x - 3} \over 3} = {{y - 8} \over { - 1}} = {{z - 3} \over 1}$ and
${{x + 3} \over { - 3}} = {{y + 7} \over 2} = {{z - 6} \over 4}$ is :
A.
3
B.
${7 \over 2}\sqrt {30} $
C.
$3\sqrt {30} $
D.
$2\sqrt {30} $
2020
Q303
JEE Mains
MCQ
14 Mar 2026
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 :
A.
(4, 3, 2)
B.
(6, 5, - 2)
C.
(3, 4, -2)
D.
(6, 5, 2)
2020
Q304
JEE Mains
Numerical
14 Mar 2026
If the equation of a plane P, passing through the intersection of the planes,
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...........
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...........
Correct Answer: 3
Explanation:
Equation of plane P is
$(x + 4y - z + 7) + \lambda (3x + y + 5z - 8) = 0$
$ \Rightarrow x(1 + 3\lambda ) + y(4 + \lambda ) + z( - 1 + 5\lambda ) + (7 - 8\lambda ) = 0$
${{1 + 3\lambda } \over a} = {{4 + \lambda } \over b} = {{5\lambda - 1} \over 6} = {{7 - 8\lambda } \over { - 15}}$
$ \therefore $ 15 - 75$\lambda $ = 42 - 48$\lambda $
$ \Rightarrow $ -27 = 27$\lambda $
$ \Rightarrow $ $\lambda $ = -1
$ \therefore $ Plane is $(x + 4y - z + 7) - 1 (3x + y + 5z - 8) = 0$
$ \Rightarrow $ $2x - 3y + 6z - 15 = 0$
Distance of (3, 2, -1) from the plane P
= ${{\left| {6 - 6 - 6 - 15} \right|} \over 7} = {{21} \over 7} = 3$
$(x + 4y - z + 7) + \lambda (3x + y + 5z - 8) = 0$
$ \Rightarrow x(1 + 3\lambda ) + y(4 + \lambda ) + z( - 1 + 5\lambda ) + (7 - 8\lambda ) = 0$
${{1 + 3\lambda } \over a} = {{4 + \lambda } \over b} = {{5\lambda - 1} \over 6} = {{7 - 8\lambda } \over { - 15}}$
$ \therefore $ 15 - 75$\lambda $ = 42 - 48$\lambda $
$ \Rightarrow $ -27 = 27$\lambda $
$ \Rightarrow $ $\lambda $ = -1
$ \therefore $ Plane is $(x + 4y - z + 7) - 1 (3x + y + 5z - 8) = 0$
$ \Rightarrow $ $2x - 3y + 6z - 15 = 0$
Distance of (3, 2, -1) from the plane P
= ${{\left| {6 - 6 - 6 - 15} \right|} \over 7} = {{21} \over 7} = 3$
2020
Q305
JEE Mains
Numerical
14 Mar 2026
Let a plane P contain two lines
$\overrightarrow r = \widehat i + \lambda \left( {\widehat i + \widehat j} \right)$, $\lambda \in R$ and
$\overrightarrow r = - \widehat j + \mu \left( {\widehat j - \widehat k} \right)$, $\mu \in R$
If Q($\alpha $, $\beta $, $\gamma $) is the foot of the perpendicular drawn from the point M(1, 0, 1) to P, then 3($\alpha $ + $\beta $ + $\gamma $) equals _______.
$\overrightarrow r = \widehat i + \lambda \left( {\widehat i + \widehat j} \right)$, $\lambda \in R$ and
$\overrightarrow r = - \widehat j + \mu \left( {\widehat j - \widehat k} \right)$, $\mu \in R$
If Q($\alpha $, $\beta $, $\gamma $) is the foot of the perpendicular drawn from the point M(1, 0, 1) to P, then 3($\alpha $ + $\beta $ + $\gamma $) equals _______.
Correct Answer: 5
Explanation:
Given lines,
$\overrightarrow r = \widehat i + \lambda (\widehat i + \widehat j)$ parallel to $(\widehat i + \widehat j)$
Let, $\overrightarrow {{n_1}} = (\widehat i + \widehat j)$
and $\overrightarrow r = - \widehat j + \mu (\widehat j - \widehat k)$ parallel to $(\widehat j - \widehat k)$
Let, $\overrightarrow {{n_2}} = (\widehat j - \widehat k)$
$ \therefore $ Normal of plane, $\overrightarrow n = \overrightarrow {{n_1}} \times \overrightarrow {{n_2}} $
$\overrightarrow n = \left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr 2 & 1 & 0 \cr 0 & 1 & { - 1} \cr } } \right|$
$ = - \widehat i + \widehat j + \widehat k$
Line $\overrightarrow r = \widehat i + \lambda (\widehat i + \widehat j)$ is on the plane so, point on the line (1, 0, 0) will be also on the plane.
$ \therefore $ Equation of the plane,
$ - 1(x - 1) + 1(y - 0) + 1(z - 0) = 0$
$ \Rightarrow x - y - z - 1 = 0$
Foot of perpendicular from (x1, y1, z1) on the plane,
${{x - {x_1}} \over a} = {{y - {y_1}} \over b} = {{z - {z_1}} \over c} = - {{(a{x_1} + b{y_1} + c{z_1} + d)} \over {{a^2} + {b^2} + {c^2}}}$
Here foot of perpendicular is drawn from M(1, 0, 1),
$ \therefore $ ${{x - 1} \over 1} = {{y - 0} \over { - 1}} = {{z - 1} \over { - 1}} = - {{(1 - 0 - 1 - 1)} \over 3}$
$ \therefore $ $x - 1 = {1 \over 3} \Rightarrow x = {4 \over 3}$
${y \over { - 1}} = {1 \over 3} \Rightarrow y = - {1 \over 3}$
${{z - 1} \over { - 1}} = {1 \over 3} \Rightarrow z = {2 \over 3}$
According to the question,
$x = \alpha $, $y = \beta $, $z = \gamma $
$ \therefore $ $\alpha = {4 \over 3}$, $\beta = - {1 \over 3}$, $\gamma = {2 \over 3}$
$ \therefore $ $3(\alpha + \beta + \gamma ) = 3\left( {{4 \over 3} - {1 \over 3} + {2 \over 3}} \right) = 5$
$\overrightarrow r = \widehat i + \lambda (\widehat i + \widehat j)$ parallel to $(\widehat i + \widehat j)$
Let, $\overrightarrow {{n_1}} = (\widehat i + \widehat j)$
and $\overrightarrow r = - \widehat j + \mu (\widehat j - \widehat k)$ parallel to $(\widehat j - \widehat k)$
Let, $\overrightarrow {{n_2}} = (\widehat j - \widehat k)$
$ \therefore $ Normal of plane, $\overrightarrow n = \overrightarrow {{n_1}} \times \overrightarrow {{n_2}} $
$\overrightarrow n = \left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr 2 & 1 & 0 \cr 0 & 1 & { - 1} \cr } } \right|$
$ = - \widehat i + \widehat j + \widehat k$
Line $\overrightarrow r = \widehat i + \lambda (\widehat i + \widehat j)$ is on the plane so, point on the line (1, 0, 0) will be also on the plane.
$ \therefore $ Equation of the plane,
$ - 1(x - 1) + 1(y - 0) + 1(z - 0) = 0$
$ \Rightarrow x - y - z - 1 = 0$
Foot of perpendicular from (x1, y1, z1) on the plane,
${{x - {x_1}} \over a} = {{y - {y_1}} \over b} = {{z - {z_1}} \over c} = - {{(a{x_1} + b{y_1} + c{z_1} + d)} \over {{a^2} + {b^2} + {c^2}}}$
Here foot of perpendicular is drawn from M(1, 0, 1),
$ \therefore $ ${{x - 1} \over 1} = {{y - 0} \over { - 1}} = {{z - 1} \over { - 1}} = - {{(1 - 0 - 1 - 1)} \over 3}$
$ \therefore $ $x - 1 = {1 \over 3} \Rightarrow x = {4 \over 3}$
${y \over { - 1}} = {1 \over 3} \Rightarrow y = - {1 \over 3}$
${{z - 1} \over { - 1}} = {1 \over 3} \Rightarrow z = {2 \over 3}$
According to the question,
$x = \alpha $, $y = \beta $, $z = \gamma $
$ \therefore $ $\alpha = {4 \over 3}$, $\beta = - {1 \over 3}$, $\gamma = {2 \over 3}$
$ \therefore $ $3(\alpha + \beta + \gamma ) = 3\left( {{4 \over 3} - {1 \over 3} + {2 \over 3}} \right) = 5$
2020
Q306
JEE Mains
Numerical
14 Mar 2026
If the distance between the plane,
23x – 10y – 2z + 48 = 0 and the plane
containing the lines ${{x + 1} \over 2} = {{y - 3} \over 4} = {{z + 1} \over 3}$
and ${{x + 3} \over 2} = {{y + 2} \over 6} = {{z - 1} \over \lambda }\left( {\lambda \in R} \right)$
is equal to ${k \over {\sqrt {633} }}$, then k is equal to ______.
containing the lines ${{x + 1} \over 2} = {{y - 3} \over 4} = {{z + 1} \over 3}$
and ${{x + 3} \over 2} = {{y + 2} \over 6} = {{z - 1} \over \lambda }\left( {\lambda \in R} \right)$
is equal to ${k \over {\sqrt {633} }}$, then k is equal to ______.
Correct Answer: 3
Explanation:
Required distance = perpendicular distance of plane 23x – 10y – 2z + 48 = 0 either from (–1, 3, –1) or (–3, –2, 1)
$ \Rightarrow $ $\left| {{{ - 23 - 30 + 2 + 48} \over {\sqrt {{{\left( {23} \right)}^2} + {{\left( {10} \right)}^2} + {{\left( 2 \right)}^2}} }}} \right|$ = ${k \over {\sqrt {633} }}$
$ \Rightarrow $ $\left| {{3 \over {\sqrt {633} }}} \right|$ = ${k \over {\sqrt {633} }}$
$ \Rightarrow $ k = 3
$ \Rightarrow $ $\left| {{{ - 23 - 30 + 2 + 48} \over {\sqrt {{{\left( {23} \right)}^2} + {{\left( {10} \right)}^2} + {{\left( 2 \right)}^2}} }}} \right|$ = ${k \over {\sqrt {633} }}$
$ \Rightarrow $ $\left| {{3 \over {\sqrt {633} }}} \right|$ = ${k \over {\sqrt {633} }}$
$ \Rightarrow $ k = 3
2020
Q307
JEE Mains
Numerical
14 Mar 2026
The projection of the line segment joining the
points (1, –1, 3) and (2, –4, 11) on the line
joining the points (–1, 2, 3) and (3, –2, 10)
is ____________.
Correct Answer: 8
Explanation:
Let A (1, – 1, 3), B(2, – 4, 11), C (–1, 2, 3) & D (3, –2, 10)
$ \therefore $ $\overrightarrow {AB} = \widehat i - 3\widehat j + 8\widehat k$
$ \Rightarrow $ $\overrightarrow {CD} = 4\widehat i - 4\widehat j + 7\widehat k$
Projection of $\overrightarrow {AB} $ on $\overrightarrow {CD} $ = ${{\overrightarrow {AB} .\overrightarrow {CD} } \over {\left| {\overrightarrow {CD} } \right|}}$
= ${{4 + 12 + 56} \over {\sqrt {16 + 16 + 49} }}$
= ${{72} \over 9}$
= 8
$ \therefore $ $\overrightarrow {AB} = \widehat i - 3\widehat j + 8\widehat k$
$ \Rightarrow $ $\overrightarrow {CD} = 4\widehat i - 4\widehat j + 7\widehat k$
Projection of $\overrightarrow {AB} $ on $\overrightarrow {CD} $ = ${{\overrightarrow {AB} .\overrightarrow {CD} } \over {\left| {\overrightarrow {CD} } \right|}}$
= ${{4 + 12 + 56} \over {\sqrt {16 + 16 + 49} }}$
= ${{72} \over 9}$
= 8
2020
Q308
JEE Mains
Numerical
14 Mar 2026
If the foot of the perpendicular drawn from the point (1, 0, 3) on a line passing through ($\alpha $, 7, 1)
is
$\left( {{5 \over 3},{7 \over 3},{{17} \over 3}} \right)$, then $\alpha $ is equal to______.
Correct Answer: 4
Explanation:
Direction Ratio of PQ are
= (${5 \over 3} - 1$, ${7 \over 3} - 0$, ${{17} \over 3} - 3$)
= (2, 7, 8)
Direction ratio of line QA are
= ($\alpha - {5 \over 3}$, $7 - {7 \over 3}$, 1 - ${{17} \over 3}$)
= (3$\alpha $ – 5, 14, –14)
PQ is perpendicular to line QA
$ \therefore $ $\overrightarrow {PQ} .\overrightarrow {QA} $ = 0
$ \Rightarrow $ 2(3$\alpha $ – 5) + 7.14 + (–14).8 = 0
$ \Rightarrow $ $\alpha $ = 4
2020
Q309
JEE Advanced
MSQ
14 Mar 2026
Let $\alpha $2 + $\beta $2 + $\gamma $2 $ \ne $ 0 and $\alpha $ + $\gamma $ = 1. Suppose the point (3, 2, $-$1) is the mirror image of the point (1, 0, $-$1) with respect to the plane $\alpha $x + $\beta $y + $\gamma $z = $\delta $. Then which of the following statements is/are TRUE?
A.
$\alpha $ + $\beta $ = 2
B.
$\delta $ $-$ $\gamma $ = 3
C.
$\delta $ + $\beta $ = 4
D.
$\alpha $ + $\beta $ + $\gamma $ = $\delta $
2020
Q310
JEE Advanced
MSQ
14 Mar 2026
Let L1 and L2 be the following straight lines.
${L_1}:{{x - 1} \over 1} = {y \over { - 1}} = {{z - 1} \over 3}$ and ${L_2}:{{x - 1} \over { - 3}} = {y \over { - 1}} = {{z - 1} \over 1}$.
Suppose the straight line
$L:{{x - \alpha } \over l} = {{y - 1} \over m} = {{z - \gamma } \over { - 2}}$
lies in the plane containing L1 and L2 and passes through the point of intersection of L1 and L2. If the line L bisects the acute angle between the lines L1 and L2, then which of the following statements is/are TRUE?
${L_1}:{{x - 1} \over 1} = {y \over { - 1}} = {{z - 1} \over 3}$ and ${L_2}:{{x - 1} \over { - 3}} = {y \over { - 1}} = {{z - 1} \over 1}$.
Suppose the straight line
$L:{{x - \alpha } \over l} = {{y - 1} \over m} = {{z - \gamma } \over { - 2}}$
lies in the plane containing L1 and L2 and passes through the point of intersection of L1 and L2. If the line L bisects the acute angle between the lines L1 and L2, then which of the following statements is/are TRUE?
A.
$\alpha $ $-$ $\gamma $ = 3
B.
l + m = 2
C.
$\alpha $ $-$ $\gamma $ = 1
D.
l + m = 0
2019
Q311
JEE Mains
MCQ
14 Mar 2026
The length of the perpendicular drawn from the point (2, 1, 4) to the plane containing the lines
$\overrightarrow r = \left( {\widehat i + \widehat j} \right) + \lambda \left( {\widehat i + 2\widehat j - \widehat k} \right)$ and $\overrightarrow r = \left( {\widehat i + \widehat j} \right) + \mu \left( { - \widehat i + \widehat j - 2\widehat k} \right)$ is :
$\overrightarrow r = \left( {\widehat i + \widehat j} \right) + \lambda \left( {\widehat i + 2\widehat j - \widehat k} \right)$ and $\overrightarrow r = \left( {\widehat i + \widehat j} \right) + \mu \left( { - \widehat i + \widehat j - 2\widehat k} \right)$ is :
A.
${1 \over 3}$
B.
${1 \over {\sqrt 3 }}$
C.
3
D.
${\sqrt 3 }$
2019
Q312
JEE Mains
MCQ
14 Mar 2026
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 :
A.
(1, –4, 1)
B.
(1, 4, –1)
C.
(2, 4, 1)
D.
(2, –4, 1)
2019
Q313
JEE Mains
MCQ
14 Mar 2026
If the line ${{x - 2} \over 3} = {{y + 1} \over 2} = {{z - 1} \over { - 1}}$
intersects the plane 2x + 3y – z + 13 = 0 at a point P and the plane
3x + y + 4z = 16 at a point Q, then PQ is equal to :
A.
$2\sqrt 7 $
B.
14
C.
$2\sqrt {14} $
D.
$\sqrt {14} $
2019
Q314
JEE Mains
MCQ
14 Mar 2026
A perpendicular is drawn from a point on the line ${{x - 1} \over 2} = {{y + 1} \over { - 1}} = {z \over 1}$ to the plane x + y + z = 3 such that the
foot of the perpendicular Q also lies on the plane x – y + z = 3. Then the co-ordinates of Q are :
A.
(4, 0, – 1)
B.
(2, 0, 1)
C.
(1, 0, 2)
D.
(– 1, 0, 4)
2019
Q315
JEE Mains
MCQ
14 Mar 2026
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 :
A.
13
B.
9
C.
5
D.
15
2019
Q316
JEE Mains
MCQ
14 Mar 2026
If the length of the perpendicular from the point ($\beta $, 0, $\beta $) ($\beta $ $ \ne $ 0) to the line,
${x \over 1} = {{y - 1} \over 0} = {{z + 1} \over { - 1}}$ is $\sqrt {{3 \over 2}} $, then $\beta $ is equal to :
${x \over 1} = {{y - 1} \over 0} = {{z + 1} \over { - 1}}$ is $\sqrt {{3 \over 2}} $, then $\beta $ is equal to :
A.
2
B.
1
C.
-2
D.
-1
2019
Q317
JEE Mains
MCQ
14 Mar 2026
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 :
A.
${{\sqrt {65} } \over 2}$
B.
$2\sqrt {13} $
C.
${{\sqrt {91} } \over 2}$
D.
${{\sqrt {91} } \over 4}$
2019
Q318
JEE Mains
MCQ
14 Mar 2026
The vertices B and C of a $\Delta $ABC lie on the line,
${{x + 2} \over 3} = {{y - 1} \over 0} = {z \over 4}$ such that BC = 5 units.
Then the area (in sq. units) of this triangle, given that the point A(1, –1, 2), is :
${{x + 2} \over 3} = {{y - 1} \over 0} = {z \over 4}$ such that BC = 5 units.
Then the area (in sq. units) of this triangle, given that the point A(1, –1, 2), is :
A.
6
B.
$5\sqrt {17} $
C.
$\sqrt {34} $
D.
$2\sqrt {34} $
2019
Q319
JEE Mains
MCQ
14 Mar 2026
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 :
A.
205$\sqrt5$
B.
63$\sqrt5$
C.
11/$\sqrt5$
D.
17/$\sqrt5$
2019
Q320
JEE Mains
MCQ
14 Mar 2026
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
A.
$\left( {\sqrt 2 ,1,4} \right)$
B.
$\left(- {\sqrt 2 ,1,4} \right)$
C.
$\left( -{\sqrt 2 ,-1,-4} \right)$
D.
$\left( {\sqrt 2 ,-1,4} \right)$
2019
Q321
JEE Mains
MCQ
14 Mar 2026
If the line, ${{x - 1} \over 2} = {{y + 1} \over 3} = {{z - 2} \over 4}$ meets the plane,
x + 2y + 3z = 15 at a point P, then the distance of P from the origin is :
A.
${{\sqrt 5 } \over 2}$
B.
2$\sqrt 5$
C.
9/2
D.
7/2
2019
Q322
JEE Mains
MCQ
14 Mar 2026
If a point R(4, y, z) lies on the line segment joining
the points P(2, –3, 4) and Q(8, 0, 10), then the
distance of R from the origin is :
A.
$2 \sqrt {14}$
B.
$ \sqrt {53}$
C.
$2 \sqrt {21}$
D.
6
2019
Q323
JEE Mains
MCQ
14 Mar 2026
The vector equation of the plane through the line
of intersection of the planes x + y + z = 1 and 2x
+ 3y+ 4z = 5 which is perpendicular to the plane
x – y + z = 0 is :
A.
$\mathop r\limits^ \to \times \left( {\mathop i\limits^ \wedge - \mathop k\limits^ \wedge } \right) - 2 = 0$
B.
$\mathop r\limits^ \to . \left( {\mathop i\limits^ \wedge + \mathop k\limits^ \wedge } \right) + 2 = 0$
C.
$\mathop r\limits^ \to . \left( {\mathop i\limits^ \wedge - \mathop k\limits^ \wedge } \right) + 2 = 0$
D.
$\mathop r\limits^ \to \times \left( {\mathop i\limits^ \wedge - \mathop k\limits^ \wedge } \right) + 2 = 0$
2019
Q324
JEE Mains
MCQ
14 Mar 2026
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 :
A.
${{\sqrt 3 } \over 2}$
B.
$\sqrt 6 $
C.
$\sqrt {3 \over 2} $
D.
3$\sqrt 6 $
2019
Q325
JEE Mains
MCQ
14 Mar 2026
The equation of a plane containing the line of
intersection of the planes 2x – y – 4 = 0 and
y + 2z – 4 = 0 and passing through the point
(1, 1, 0) is :
A.
x – 3y – 2z = –2
B.
2x – z = 2
C.
x – y – z = 0
D.
x + 3y + z = 4
2019
Q326
JEE Mains
MCQ
14 Mar 2026
The length of the perpendicular from the point
(2, –1, 4) on the straight line,
${{x + 3} \over {10}}$= ${{y - 2} \over {-7}}$ = ${{z} \over {1}}$ is :
${{x + 3} \over {10}}$= ${{y - 2} \over {-7}}$ = ${{z} \over {1}}$ is :
A.
less than 2
B.
greater than 4
C.
greater than 2 but less than 3
D.
greater than 3 but less than 4
2019
Q327
JEE Mains
MCQ
14 Mar 2026
Let S be the set of all real values of $\lambda $ such that a plane passing through the points (–$\lambda $2, 1, 1), (1, –$\lambda $2, 1) and (1, 1, – $\lambda $2) also passes through the point (–1, –1, 1). Then S is equal to :
A.
{1, $-$1}
B.
{3, $-$ 3}
C.
$\left\{ {\sqrt 3 } \right\}$
D.
$\left\{ {\sqrt 3 , - \sqrt 3 } \right\}$
2019
Q328
JEE Mains
MCQ
14 Mar 2026
If an angle between the line, ${{x + 1} \over 2} = {{y - 2} \over 1} = {{z - 3} \over { - 2}}$ and the plane, $x - 2y - kz = 3$ is ${\cos ^{ - 1}}\left( {{{2\sqrt 2 } \over 3}} \right),$ then a value of k is :
A.
$\sqrt {{3 \over 5}} $
B.
$ - {5 \over 2}$
C.
$ - {3 \over 2}$
D.
$\sqrt {{5 \over 3}} $
2019
Q329
JEE Mains
MCQ
14 Mar 2026
The perpendicular distance from the origin to the plane containing the two lines,
${{x + 2} \over 3} = {{y - 2} \over 5} = {{z + 5} \over 7}$ and
${{x - 1} \over 1} = {{y - 4} \over 4} = {{z + 4} \over 7},$ is :
${{x + 2} \over 3} = {{y - 2} \over 5} = {{z + 5} \over 7}$ and
${{x - 1} \over 1} = {{y - 4} \over 4} = {{z + 4} \over 7},$ is :
A.
$6\sqrt {11} $
B.
${{11} \over {\sqrt 6 }}$
C.
11
D.
11$\sqrt 6 $
2019
Q330
JEE Mains
MCQ
14 Mar 2026
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 :
A.
cos$-$1$\left( {{{17} \over {31}}} \right)$
B.
cos$-$1$\left( {{{9} \over {35}}} \right)$
C.
cos$-$1$\left( {{{19} \over {35}}} \right)$
D.
cos$-$1$\left( {{7 \over {31}}} \right)$
2019
Q331
JEE Mains
MCQ
14 Mar 2026
Two lines ${{x - 3} \over 1} = {{y + 1} \over 3} = {{z - 6} \over { - 1}}$ and ${{x + 5} \over 7} = {{y - 2} \over { - 6}} = {{z - 3} \over 4}$ intersect at the point R. The reflection of R in the xy-plane has coordinates :
A.
(2, 4, 7)
B.
(2, $-$ 4, $-$7)
C.
(2, $-$ 4, 7)
D.
($-$ 2, 4, 7)
2019
Q332
JEE Mains
MCQ
14 Mar 2026
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
A.
12
B.
7
C.
17
D.
5
2019
Q333
JEE Mains
MCQ
14 Mar 2026
The plane containing the line ${{x - 3} \over 2} = {{y + 2} \over { - 1}} = {{z - 1} \over 3}$ and also containing its projection on the plane 2x + 3y $-$ z = 5, contains which one of the following points ?
A.
($-$ 2, 2, 2)
B.
(2, 2, 0)
C.
(2, 0, $-$ 2)
D.
(0, $-$ 2, 2)
2019
Q334
JEE Mains
MCQ
14 Mar 2026
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 :
A.
2, $-$1, 1
B.
$2\sqrt 3 ,1, - 1$
C.
$\sqrt 2 ,1, - 1$
D.
$\sqrt 2 , - \sqrt 2 $
2019
Q335
JEE Mains
MCQ
14 Mar 2026
On which of the following lines lies the point of intersection of the line, ${{x - 4} \over 2} = {{y - 5} \over 2} = {{z - 3} \over 1}$ and the plane,
x + y + z = 2 ?
A.
${{x - 4} \over 1} = {{y - 5} \over 1} = {{z - 5} \over { - 1}}$
B.
${{x - 2} \over 2} = {{y - 3} \over 2} = {{z + 3} \over 3}$
C.
${{x - 1} \over 1} = {{y - 3} \over 2} = {{z + 4} \over { - 5}}$
D.
${{x + 3} \over 3} = {{4 - y} \over 3} = {{z + 1} \over { - 2}}$
2019
Q336
JEE Mains
MCQ
14 Mar 2026
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 ?
A.
(2, 1, 3)
B.
(4, $-$ 1, 2)
C.
(4, 1, $-$ 2)
D.
($-$ 2, 3, 5)
2019
Q337
JEE Mains
MCQ
14 Mar 2026
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 -
A.
(1, 1, $-$ 1)
B.
(1, 1, 1)
C.
($-$ 1, $-$ 1, $-$1)
D.
($-$ 1, $-$ 1, 1)
2019
Q338
JEE Mains
MCQ
14 Mar 2026
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 -
A.
${1 \over 8}$
B.
${1 \over 2}$
C.
${1 \over 4}$
D.
$-$ ${1 \over 4}$
2019
Q339
JEE Mains
MCQ
14 Mar 2026
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 :
A.
x $-$ 2y + z = 0
B.
3x + 2y $-$ 3z = 0
C.
x + 2y $-$ 2z = 0
D.
5x + 2y $-$ 4z = 0
2019
Q340
JEE Mains
MCQ
14 Mar 2026
If the lines x = ay + b, z = cy + d and x = a'z + b', y = c'z + d' are perpendicular, then :
A.
ab' + bc' + 1 = 0
B.
cc' + a + a' = 0
C.
bb' + cc' + 1 = 0
D.
aa' + c + c' = 0
2019
Q341
JEE Mains
MCQ
14 Mar 2026
The plane through the intersection of the planes x + y + z = 1 and 2x + 3y – z + 4 = 0 and parallel to y-axis
also passes through the point :
A.
(–3, 0, -1)
B.
(3, 2, 1)
C.
(3, 3, -1)
D.
(–3, 1, 1)
2019
Q342
JEE Mains
MCQ
14 Mar 2026
The equation of the line passing through (–4, 3, 1), parallel
to the plane x + 2y – z – 5 = 0 and intersecting
the line ${{x + 1} \over { - 3}} = {{y - 3} \over 2} = {{z - 2} \over { - 1}}$ is :
to the plane x + 2y – z – 5 = 0 and intersecting
the line ${{x + 1} \over { - 3}} = {{y - 3} \over 2} = {{z - 2} \over { - 1}}$ is :
A.
${{x + 4} \over 3} = {{y - 3} \over {-1}} = {{z - 1} \over 1}$
B.
${{x + 4} \over 1} = {{y - 3} \over {1}} = {{z - 1} \over 3}$
C.
${{x + 4} \over -1} = {{y - 3} \over {1}} = {{z - 1} \over 1}$
D.
${{x - 4} \over 2} = {{y + 3} \over {1}} = {{z + 1} \over 4}$
2019
Q343
JEE Advanced
MSQ
14 Mar 2026
Three lines ${L_1}:r = \lambda \widehat i$, $\lambda $ $ \in $ R,
${L_2}:r = \widehat k + \mu \widehat j$, $\mu $ $ \in $ R and
${L_3}:r = \widehat i + \widehat j + v\widehat k$, v $ \in $ R are given.
For which point(s) Q on L2 can we find a point P on L1 and a point R on L3 so that P, Q and R are collinear?
${L_2}:r = \widehat k + \mu \widehat j$, $\mu $ $ \in $ R and
${L_3}:r = \widehat i + \widehat j + v\widehat k$, v $ \in $ R are given.
For which point(s) Q on L2 can we find a point P on L1 and a point R on L3 so that P, Q and R are collinear?
A.
$\widehat k$
B.
$\widehat k$ + $\widehat j$
C.
$\widehat k$ + ${1 \over 2}$$\widehat j$
D.
$\widehat k$ $-$ ${1 \over 2}$$\widehat j$
2019
Q344
JEE Advanced
MSQ
14 Mar 2026
Let L1 and L2 denote the lines
$r = \widehat i + \lambda ( - \widehat i + 2\widehat j + 2\widehat k)$, $\lambda $$ \in $ R
and $r = \mu (2\widehat i - \widehat j + 2\widehat k),\,\mu \in R$
respectively. If L3 is a line which is perpendicular to both L1 and L2 and cuts both of them, then which of the following options describe(s) L3?
$r = \widehat i + \lambda ( - \widehat i + 2\widehat j + 2\widehat k)$, $\lambda $$ \in $ R
and $r = \mu (2\widehat i - \widehat j + 2\widehat k),\,\mu \in R$
respectively. If L3 is a line which is perpendicular to both L1 and L2 and cuts both of them, then which of the following options describe(s) L3?
A.
$r = {2 \over 9}(2\widehat i - \widehat j + 2\widehat k) + t(2\widehat i + 2\widehat j - \widehat k),\,t \in R$
B.
$r = {1 \over 3}(2\widehat i + k) + t(2\widehat i + 2\widehat j - \widehat k),\,t \in R$
C.
$r = {2 \over 9}(4\widehat i + \widehat j + \widehat k) + t(2\widehat i + 2\widehat j - \widehat k),\,t \in R$
D.
r = $t(2\widehat i + 2\widehat j - \widehat k)$, $t \in R$
2019
Q345
JEE Advanced
Numerical
14 Mar 2026
Three lines are given by
$r = \lambda \widehat i,\,\lambda \in R$,
$r = \mu (\widehat i + \widehat j),\,\mu \in R$ and
$r = v(\widehat i + \widehat j + \widehat k),\,v\, \in R$
Let the lines cut the plane x + y + z = 1 at the points A, B and C respectively. If the area of the triangle ABC is $\Delta $ then the value of (6$\Delta $)2 equals ..............
$r = \lambda \widehat i,\,\lambda \in R$,
$r = \mu (\widehat i + \widehat j),\,\mu \in R$ and
$r = v(\widehat i + \widehat j + \widehat k),\,v\, \in R$
Let the lines cut the plane x + y + z = 1 at the points A, B and C respectively. If the area of the triangle ABC is $\Delta $ then the value of (6$\Delta $)2 equals ..............
Correct Answer: 0.75
Explanation:
Given three lines
$r = \lambda \widehat i,\,\lambda \in R$, $r = \mu (\widehat i + \widehat j),\,\mu \in R$ and $r = v(\widehat i + \widehat j + \widehat k),\,v\, \in R$
cuts the plane x + y + z = 1 at the points A, B and C, respectively. So, for point A, put ($\lambda $, 0, 0) in the plane, we get $\lambda $ + 0 + 0 = 1 $ \Rightarrow $ $\lambda $ = 1 $ \Rightarrow $ A $ \equiv $ (1, 0, 0).
Similarly, for point B, put ($\mu $, $\mu $, 0) in the plane, we get $\mu $ + $\mu $ + 0 = 1 $ \Rightarrow $ $\mu $ = ${1 \over 2}$
$ \Rightarrow $ B $ \equiv $ $\left( {{1 \over 2},{1 \over 2},0} \right)$.
and for point C, p;ut (v, v, v) in the plane we get
v + v + v = 1 $ \Rightarrow $ v = ${{1 \over 3}}$ $ \Rightarrow $ C $ \equiv $ $\left( {{1 \over 3},{1 \over 3},{1 \over 3}} \right)$
Now, are of $\Delta $ABC = ${{1 \over 2}|AB \times AC| = \Delta }$
$ \because $ AB = ${ - {1 \over 2}\widehat i + {1 \over 2}\widehat j}$,
and AC = ${ - {2 \over 3}\widehat i + {1 \over 3}\widehat j + {1 \over 3}\widehat k}$
$ \therefore $ AB $ \times $ AC = $\left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr { - 1/2} & {1/2} & 0 \cr { - 2/3} & {1/3} & {1/3} \cr } } \right|$
= $\widehat i\left( {{1 \over 6}} \right) - \widehat j\left( { - {1 \over 6}} \right) + \widehat k\left( { - {1 \over 6} + {2 \over 6}} \right)$
$ = {1 \over 6}(\widehat i + \widehat j + \widehat k)$
$ \Rightarrow |AB \times AC| = {1 \over 6}\sqrt 3 = {1 \over {2\sqrt 3 }}$
$ \Rightarrow \Delta = {1 \over {4\sqrt 3 }}$
$ \Rightarrow {(6\Delta )^2} = 36{1 \over {16 \times 3}} = {3 \over 4} = 0.75$
$r = \lambda \widehat i,\,\lambda \in R$, $r = \mu (\widehat i + \widehat j),\,\mu \in R$ and $r = v(\widehat i + \widehat j + \widehat k),\,v\, \in R$
cuts the plane x + y + z = 1 at the points A, B and C, respectively. So, for point A, put ($\lambda $, 0, 0) in the plane, we get $\lambda $ + 0 + 0 = 1 $ \Rightarrow $ $\lambda $ = 1 $ \Rightarrow $ A $ \equiv $ (1, 0, 0).
Similarly, for point B, put ($\mu $, $\mu $, 0) in the plane, we get $\mu $ + $\mu $ + 0 = 1 $ \Rightarrow $ $\mu $ = ${1 \over 2}$
$ \Rightarrow $ B $ \equiv $ $\left( {{1 \over 2},{1 \over 2},0} \right)$.
and for point C, p;ut (v, v, v) in the plane we get
v + v + v = 1 $ \Rightarrow $ v = ${{1 \over 3}}$ $ \Rightarrow $ C $ \equiv $ $\left( {{1 \over 3},{1 \over 3},{1 \over 3}} \right)$
Now, are of $\Delta $ABC = ${{1 \over 2}|AB \times AC| = \Delta }$
$ \because $ AB = ${ - {1 \over 2}\widehat i + {1 \over 2}\widehat j}$,
and AC = ${ - {2 \over 3}\widehat i + {1 \over 3}\widehat j + {1 \over 3}\widehat k}$
$ \therefore $ AB $ \times $ AC = $\left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr { - 1/2} & {1/2} & 0 \cr { - 2/3} & {1/3} & {1/3} \cr } } \right|$
= $\widehat i\left( {{1 \over 6}} \right) - \widehat j\left( { - {1 \over 6}} \right) + \widehat k\left( { - {1 \over 6} + {2 \over 6}} \right)$
$ = {1 \over 6}(\widehat i + \widehat j + \widehat k)$
$ \Rightarrow |AB \times AC| = {1 \over 6}\sqrt 3 = {1 \over {2\sqrt 3 }}$
$ \Rightarrow \Delta = {1 \over {4\sqrt 3 }}$
$ \Rightarrow {(6\Delta )^2} = 36{1 \over {16 \times 3}} = {3 \over 4} = 0.75$
2018
Q346
JEE Mains
MCQ
14 Mar 2026
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 :
A.
4
B.
$-$ 4
C.
$-$ 8
D.
12
2018
Q347
JEE Mains
MCQ
14 Mar 2026
If the angle between the lines, ${x \over 2} = {y \over 2} = {z \over 1}$
and ${{5 - x} \over { - 2}} = {{7y - 14} \over p} = {{z - 3} \over 4}\,\,$ is ${\cos ^{ - 1}}\left( {{2 \over 3}} \right),$ then p is equal to :
and ${{5 - x} \over { - 2}} = {{7y - 14} \over p} = {{z - 3} \over 4}\,\,$ is ${\cos ^{ - 1}}\left( {{2 \over 3}} \right),$ then p is equal to :
A.
${7 \over 2}$
B.
${2 \over 7}$
C.
$-$ ${7 \over 4}$
D.
$-$ ${4 \over 7}$
2018
Q348
JEE Mains
MCQ
14 Mar 2026
The length of the projection of the line segment joining the points (5, -1, 4) and (4, -1, 3) on the plane,
x + y + z = 7 is :
A.
$\sqrt {{2 \over 3}} $
B.
${2 \over {\sqrt 3 }}$
C.
${2 \over 3}$
D.
${1 \over 3}$
2018
Q349
JEE Mains
MCQ
14 Mar 2026
If L1 is the line of intersection of the planes 2x - 2y + 3z - 2 = 0, x - y + z + 1 = 0 and L2 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 L1 and L2, is :
A.
${1 \over {\sqrt 2 }}$
B.
${1 \over {4\sqrt 2 }}$
C.
${1 \over {3\sqrt 2 }}$
D.
${1 \over {2\sqrt 2 }}$
2018
Q350
JEE Mains
MCQ
14 Mar 2026
An angle between the lines whose direction cosines are gien by the equations,
$l$ + 3m + 5n = 0 and 5$l$m $-$ 2mn + 6n$l$ = 0, is :
$l$ + 3m + 5n = 0 and 5$l$m $-$ 2mn + 6n$l$ = 0, is :
A.
${\cos ^{ - 1}}\left( {{1 \over 3}} \right)$
B.
${\cos ^{ - 1}}\left( {{1 \over 4}} \right)$
C.
${\cos ^{ - 1}}\left( {{1 \over 6}} \right)$
D.
${\cos ^{ - 1}}\left( {{1 \over 8}} \right)$