Three Dimensional Geometry — JEE Main PYQs

354 previous year questions on Three Dimensional Geometry (Mathematics), newest first. Open any question for its complete step-by-step solution.

A perpendicular is drawn from a point on the line $\frac{x - 1}{2} = \frac{y + 1}{- 1} = \frac{z}{1}$ 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
JEE Main 2019View solution →
If the plane $2 x - y + 2 z + 3 = 0$ has the distances $\frac{1}{3}$ and $\frac{2}{3}$ units from the planes $4 x - 2 y + 4 z + λ = 0$ and $2 x - y + 2 z + μ = 0$ , respectively, then the maximum value of $λ + μ$ us equal to
JEE Main 2019View solution →
The plane containing the line $\frac{x - 3}{2} = \frac{y + 2}{- 1} = \frac{z - 1}{3}$ and also containing its projection on the plane $2 x + 3 y - z = 5$ , contains which one of the following points?
JEE Main 2019View solution →
The direction ratios of normal to the plane through the points $(0, - 1, 0)$ and $(0, 0, 1)$ and making an angle $\frac{π}{4}$ with the plane $y - z + 5 = 0$
JEE Main 2019View solution →
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:
JEE Main 2019View solution →
The perpendicular distance from the origin to the plane containing the two lines, $\frac{x + 2}{3} = \frac{y - 2}{5} = \frac{z + 5}{7}$ and $\frac{x - 1}{1} = \frac{y - 4}{4} = \frac{z + 4}{7}$ , is:
JEE Main 2019View solution →
The length of the perpendicular drawn from the point $(2, 1, 4)$ to the plane containing the lines $r = (\hat{i} + \hat{j}) + λ (\hat{i} + 2 \hat{j} - \hat{k})$ and $r = (\hat{i} + \hat{j}) + μ (- \hat{i} + \hat{j} - 2 \hat{k})$ is:
JEE Main 2019View solution →
A plane which bisects the angle between the two given planes $2 x - y + 2 z - 4 = 0$ and $x + 2 y + 2 z - 2 = 0$ , passes through the point:
JEE Main 2019View solution →
If an angle between the line, $\frac{x + 1}{2} = \frac{y - 2}{1} = \frac{z - 3}{- 2}$ and the plane, $x - 2 y - k z = 3$ is $cos^{- 1} (\frac{2 2}{3})$ , then a value of k is :
JEE Main 2019View solution →
Let S be the set of all real values of $λ$ such that a plane passing through the points $(- λ^{2}, 1, 1), (1, - λ^{2}, 1)$ and $(1, 1, - λ^{2})$ also passes through the point $(- 1, - 1, 1)$ . Then $S$ is equal to:
JEE Main 2019View solution →
The vertices $B$ and $C$ of a $△ A B C$ lie on the line, $\frac{x + 2}{3} = \frac{y - 1}{0} = \frac{z}{4}$ such that $B C = 5$ units. Then the area (in sq. units) of this triangle, given that the point $A (1, - 1, 2)$ is:
JEE Main 2019View solution →
Let P be the plane, which contains the line of intersection of the planes, $x + y + z - 6 = 0$ and $2 x + 3 y + z + 5 = 0$ and it is perpendicular to the $x y$ -plane. Then the distance of the point $(0, 0, 256)$ from $P$ is equal to:
JEE Main 2019View solution →
The equation of the plane containing the straight line $\frac{x}{2} = \frac{y}{3} = \frac{z}{4}$ and perpendicular to the plane containing the straight lines $\frac{x}{3} = \frac{y}{4} = \frac{z}{2}$ and $\frac{x}{4} = \frac{y}{2} = \frac{z}{3}$ is:
JEE Main 2019View solution →
If the lines $x = a y + b, z = cy + d$ and $x = a^{'} z + b^{'}, y = c^{'} z + d^{'}$ are perpendicular, then:
JEE Main 2019View solution →
The equation of line passing through $(- 4, 3, 1)$ , parallel to the plane $x + 2 y - z - 5 = 0$ and intersecting the line $\frac{x + 1}{- 3} = \frac{y - 3}{2} = \frac{z - 2}{- 1}$ is:
JEE Main 2019View solution →
The plane through the intersection of the plane $x + y + z = 1$ and $2 x + 3 y - z + 4 = 0$ and parallel to $y$ -axis also pass through the point:
JEE Main 2019View solution →
If $Q (0, - 1, - 3)$ is the image of the point $P$ in the plane $3 x - y + 4 z = 2$ and $R$ is the point $(3, - 1, - 2)$ , then the area (in square units) of $△ PQR$ is:
JEE Main 2019View solution →
If the length of the perpendicular from the point $(β, 0, β)$ ( $β \neq = 0$ ) to the line, $\frac{x}{1} = \frac{y - 1}{0} = \frac{z + 1}{- 1}$ is $\frac{3}{2}$ , then $β$ is equal to:
JEE Main 2019View solution →
If $L_{1}$ is the line of intersection of the planes $2 x - 2 y + 3 z - 2 = 0, x - y + z + 1 = 0$ and $L_{2}$ is the line of intersection of the planes $x + 2 y - z - 3 = 0, 3 x - y + 2 z - 1 = 0$ , then the distance of the origin from the plane, containing the lines $L_{1}$ and …
JEE Main 2018View solution →
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 :
JEE Main 2018View solution →
If the image of the point $P (1, - 2, 3)$ in the plane, $2 x + 3 y - 4 z + 22 = 0$ measured parallel to the line, $\frac{x}{1} = \frac{y}{4} = \frac{z}{5}$ is Q , then PQ is equal to :
JEE Main 2017View solution →
The distance of the point $(1, 3, - 7)$ from the plane passing through the point $(1, - 1, - 1)$ , having normal perpendicular to both the lines $\frac{x - 1}{1} = \frac{y + 2}{- 2} = \frac{z - 4}{3}$ and $\frac{x - 2}{2} = \frac{y + 1}{- 1} = \frac{z + 7}{- 1}$ , is :
JEE Main 2017View solution →
If the line, $\frac{x - 3}{2} = \frac{y + 2}{- 1} = \frac{z + 4}{3}$ lies in the plane, $l x + my - z = 9$ , then $l^{2} + m^{2}$ is equal to:
JEE Main 2016View solution →
The distance of the point $(1, - 5, 9)$ from the plane $x - y + z = 5$ measured along the line $x = y = z$ is:
JEE Main 2016View solution →
The distance of the point $(1, 0, 2)$ from the point of intersection of the line $\frac{x - 2}{3} = \frac{y + 1}{4} = \frac{z - 2}{12}$ and the plane $x - y + z = 16$ , is :
JEE Main 2015View solution →
The equation of the plane containing the line $2 x - 5 y + z = 3; x + y + 4 z = 5$ , and parallel to the plane, $x + 3 y + 6 z = 1$ , is :
JEE Main 2015View solution →
The image of the line $\frac{x - 1}{3} = \frac{y - 3}{1} = \frac{z - 4}{- 5}$ in the plane $2 x - y + z + 3 = 0$ is the line
JEE Main 2014View solution →
The angle between the lines whose direction cosines satisfy the equations $l + m + n = 0$ and $l^{2} = m^{2} + n^{2}$ is
JEE Main 2014View solution →
If 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, then $k$ can have
JEE Main 2013View solution →
Distance between two parallel planes $2 x + y + 2 z = 8$ and $4 x + 2 y + 4 z + 5 = 0$ is
JEE Main 2013View solution →
An equation of a plane parallel to the plane $x - 2 y + 2 z - 5 = 0$ and at a unit distance from the origin is
JEE Main 2012View solution →
If the lines $\frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z - 1}{4}$ and $\frac{x - 3}{1} = \frac{y - k}{2} = \frac{z}{1}$ intersect, then $k$ is equal to
JEE Main 2012View solution →
If the angle between the line $x = \frac{y - 1}{2} = \frac{z - 3}{λ}$ and the plane $x + 2 y + 3 z = 4$ is $cos^{- 1} (\frac{5}{14})$ , then $λ$ equals
JEE Main 2011View solution →
Statement-1: The point A(1,0,7) is the mirror image of the point B(1,6,3) in the line
$\frac{x}{1} = \frac{y - 1}{2} = \frac{z - 2}{3} .$

Statement-2: The line: $\frac{x}{1} = \frac{y - 1}{2} = \frac{z - 2}{3}$ bisects the line segment joining $A (1, 0, 7)$ and $B (1, 6, 3)$ .
JEE Main 2011View solution →
Statement-1: The point A(3,1,6) is the mirror image of the point B(1,3,4) in the plane x-y+z=5.

Statement-2: The plane x-y+z=5 bisects the line segment joining A(3,1,6) and B(1,3,4).
JEE Main 2010View solution →
A line AB in three-dimensional space makes angles $4 5^{\circ}$ and $12 0^{\circ}$ with the positive x-axis and the positive y-axis respectively. If AB makes an acute angle $θ$ with the positive z-axis, then $θ$ equals
JEE Main 2010View solution →
Let the line $\frac{x - 2}{3} = \frac{y - 1}{- 5} = \frac{z + 2}{2}$ lies in the plane $x + 3 y - α z + β = 0$ . Then $(α, β)$ equals
JEE Main 2009View solution →
The projections of a vector on the three coordinate axis are $6, - 3, 2$ respectively. The direction cosines of the vector are
JEE Main 2009View solution →
The line passing through the points $(5, 1, a)$ and $(3, b, 1)$ crosses the yz-plane at the point $(0, \frac{17}{2}, \frac{- 13}{2})$ . Then
JEE Main 2008View solution →
If the straight lines $\frac{x - 1}{k} = \frac{y - 2}{2} = \frac{z - 3}{3}$ and $\frac{x - 2}{3} = \frac{y - 3}{k} = \frac{z - 1}{2}$ intersect at a point, then the integer k is equal to
JEE Main 2008View solution →
Let $L$ be the line of intersection of the planes $2 x + 3 y + z = 1$ and $x + 3 y + 2 z = 2$ . If $L$ makes an angles $α$ with the positive $x$ -axis, then $cos α$ equals
JEE Main 2007View solution →
If a line makes an angle of $\frac{π}{4}$ with the positive directions of each of $x$ -axis and $y$ -axis, then the angle that the line makes with the positive direction of the $z$ -axis is
JEE Main 2007View solution →
If $(2, 3, 5)$ is one end of a diameter of the sphere $x^{2} + y^{2} + z^{2} - 6 x - 12 y - 2 z + 20 = 0$ , then the coordinates of the other end of the diameter are
JEE Main 2007View solution →
The two lines $x = a y + b, z = cy + d$ ; and $x = a^{'} y + b^{'}, z = c^{'} y + d^{'}$ are perpendicular to each other if
JEE Main 2006View solution →
If the angle $θ$ between the line $\frac{x + 1}{1} = \frac{y - 1}{2} = \frac{z - 2}{2}$ and the plane $2 x - y + λ z + 4 = 0$ is such that $sin θ = \frac{1}{3}$ , the value of $λ$ is
JEE Main 2005View solution →
The angle between the lines $2 x = 3 y = - z$ and $6 x = - y = - 4 z$ is
JEE Main 2005View solution →
If the plane $2 a x - 3 a y + 4 a z + 6 = 0$ passes through the midpoint of the line joining the centres of the spheres
$x^{2} + y^{2} + z^{2} + 6 x - 8 y - 2 z = 13$ and
$x^{2} + y^{2} + z^{2} - 10 x + 4 y - 2 z = 8$ , then $a$ equals
JEE Main 2005View solution →
The distance between the line $r = 2 \hat{i} - 2 \hat{j} + 3 \hat{k} + λ (\hat{i} - \hat{j} + 4 \hat{k})$ and the plane $r \cdot (\hat{i} + 5 \hat{j} + \hat{k}) = 5$ is
JEE Main 2005View solution →
The plane $x + 2 y - z = 4$ cuts the sphere $x^{2} + y^{2} + z^{2} - x + z - 2 = 0$ in a circle of radius
JEE Main 2005View solution →
A line makes the same angle $θ$ , with each of the $x$ and $z$ axis. If the angle $β$ , which it makes with $y$ -axis, is such that $sin^{2} β = 3 sin^{2} θ$ , then $cos^{2} θ$ equals
JEE Main 2004View solution →