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.

Let the plane passing through the point $(- 1, 0, - 2)$ and perpendicular to each of the planes $2 x + y - z = 2$ and $x - y - z = 3$ be $a x + b y + cz + 8 = 0$ . Then the value of $a + b + c$ is equal to :
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Let a plane P pass through the point $(3, 7, - 7)$ and contain the line, $\frac{x - 2}{- 3} = \frac{y - 3}{2} = \frac{z + 2}{1}$ . If distance of the plane $P$ from the origin is $d$ , then $d^{2}$ is equal to ____ .
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The equation of the plane which contains the $y$ -axis and passes through the point $(1, 2, 3)$ is:
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If the equation of the plane passing through the line of intersection of the planes $2 x - 7 y + 4 z - 3 = 0, 3 x - 5 y + 4 z + 11 = 0$ and the point $(- 2, 1, 3)$ is $a x + b y + cz - 7 = 0$ , then the value of $2 a + b + c - 7$ is ____
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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 $L$ and point $Q$ . If $α$ is the acute angle between the l …
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If the lines $\frac{x - k}{1} = \frac{y - 2}{2} = \frac{z - 3}{3}$ and $\frac{x + 1}{3} = \frac{y + 2}{2} = \frac{z + 3}{1}$ are co-planar, then the value of $k$ is ____
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For real numbers $α$ and $β \neq = 0$ , if the point of intersection of the straight lines $\frac{x - α}{1} = \frac{y - 1}{2} = \frac{z - 1}{3}$ and $\frac{x - 4}{β} = \frac{y - 6}{3} = \frac{z - 7}{3}$ , lies on the plane $x + 2 y - z = 8$ , then $α - β$ is equal to :
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The distance of the point $P (3, 4, 4)$ from the point of intersection of the line joining the points $Q (3, - 4, - 5)$ and $R (2, - 3, 1)$ and the plane $2 x + y + z = 7$ , is equal to ____
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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:
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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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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 rations of lines PR and QS are $(4, - 1, 2)$ and $(- 2, 1, - 2)$ , respectively. Let lines …
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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 $I x + m y + n z = 0$ are points $C (0, - a, - 1)$ and $D$ respectively, then the length of line segment $C D$ is equal to :
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Let $P$ be a plane $I x + m y + n z = 0$ 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 :
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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 …
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If the foot of the perpendicular from point $(4, 3, 8)$ on the line $L_{1} : \frac{x - a}{ℓ} = \frac{y - 2}{3} = \frac{z - b}{4}, ℓ \neq = 0$ is (3, 5, 7) then the shortest distance between the line $L_{1}$ and line $L_{2} : \frac{x - 2}{3} = \frac{y - 4}{4} = \frac{z - 5}{5}$ is equal to :
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If the distance of the point $(1, - 2, 3)$ from the plane $x + 2 y - 3 z + 10 = 0$ measured parallel to the line, $\frac{x - 1}{3} = \frac{2 - y}{m} = \frac{z + 3}{1}$ is $\frac{7}{2}$ , then the value of $∣ m ∣$ is equal to......
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Let the mirror image of the point $(1, 3, a)$ with respect to the plane $r \cdot (2 \hat{i} - \hat{j} + \hat{k}) - b = 0$ be $(- 3, 5, 2)$ . Then, the value of $∣ a + b ∣$ is equal to
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Let $P$ be a plane containing the line $\frac{x - 1}{3} = \frac{y + 6}{4} = \frac{z + 5}{2}$ and parallel to the line $\frac{x - 3}{4} = \frac{y - 2}{- 3} = \frac{z + 5}{7}$ . If the point $(1, - 1, α)$ lies on the plane $P$ , then the value of $∣5 α ∣$ is equal to....
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If the equation of plane passing through the mirror image of a point $(2, 3, 1)$ with respect to line $\frac{x + 1}{2} = \frac{y - 3}{1} = \frac{z + 2}{- 1}$ and containing the line $\frac{x - 2}{3} = \frac{1 - y}{2} = \frac{z + 1}{1}$ is, $α x + β y + γ z = 24$ then …
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Let $P$ be an arbitrary point having sum of the squares of the distance from the planes $x + y + z = 0$ , $ℓ x - n z = 0$ and $x - 2 y + z = 0$ equal to 9. If the locus of the point $P$ is $x^{2} + y^{2} + z^{2} = 9$ , then the value of $ℓ - n$ is equal to ____.
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Let the plane $a x + b y + cz + d = 0$ bisect the line joining the points $(4, - 3, 1)$ and $(2, 3, - 5)$ at the right angles. If $a, b, c, d$ are integer, then the minimum value of $(a^{2} + b^{2} + c^{2} + d^{2})$ is
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The equation of the planes parallel to the plane $x - 2 y + 2 z - 3 = 0$ which are at unit distance from the point $(1, 2, 3)$ is $a x + b y + cz + d = 0$ . If $(b - d) = K (c - a)$ , then the positive value of $K$ is ____
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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:
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If the foot of the perpendicular drawn from the point $(1, 0, 3)$ on a line passing through $(α, 7, 1)$ is $(\frac{5}{3}, \frac{7}{3}, \frac{17}{3})$ , then $α$ is equal to ____
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The shortest distance between the lines $\frac{x - 3}{3} = \frac{y - 8}{- 1} = \frac{z - 3}{1}$ and $\frac{x + 3}{- 3} = \frac{y + 7}{2} = \frac{z - 6}{4}$ is:
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The mirror image of the point ( $1, 2, 3$ ) in a plane is $(- \frac{7}{3}, - \frac{4}{3}, - \frac{1}{3})$ . Which of the following points lies on this plane?
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If the distance between the plane, $23 x - 10 y - 2 z + 48 = 0$ and the plane containing the lines $\frac{x + 1}{2} = \frac{y - 3}{4} = \frac{z + 1}{3}$ and $\frac{x + 3}{2} = \frac{y + 2}{6} = \frac{z - 1}{λ} (λ \in R)$ is equal to $\frac{k}{633}$ , then $k$ is equal to ______.
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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 ______
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If for some $α \in R$ , the lines $L_{1} : \frac{x + 1}{2} = \frac{y - 2}{- 1} = \frac{z - 1}{1}$ and $L_{2} : \frac{x + 2}{α} = \frac{y + 1}{5 - α} = \frac{z + 1}{1}$ are coplanar, then the line $L_{2}$ passes through the point:
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The lines
$r = (\hat{i} - \hat{j}) + ℓ (2 \hat{i} + \hat{k})$ and
$r = (2 \hat{i} - \hat{j}) + m (\hat{i} + \hat{j} - \hat{k})$
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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:
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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:
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Let a plane $P$ contain two lines $r = \hat{i} + λ (\hat{i} + \hat{j}), λ \in R$ and $r = - \hat{j} + μ (\hat{j} - \hat{k}), μ \in R$ . If $Q (α, β, γ)$ is the foot of the perpendicular drawn from the point $M (1, 0, 1)$ to $P$ , then …
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A plane P meets the coordinate axes at $A, B$ and C respectively. The centroid of $△ ABC$ is given to be ( $1, 1, 2$ ). Then the equation of the line through this centroid and perpendicular to the plane P is:
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If $(a, b, c)$ is the image of the point $(1, 2, - 3)$ in the line, $\frac{x + 1}{2} = \frac{y - 3}{- 2} = \frac{z}{- 1}$ , then $a + b + c$ is equal to
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The shortest distance between the lines $\frac{x - 1}{0} = \frac{y + 1}{- 1} = \frac{z}{1}$ and $x + y + z + 1 = 0, 2 x - y + z + 3 = 0$ is:
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If the equation of a plane $P$ , passing through the intersection of the planes, $x + 4 y - z + 7 = 0$ and $3 x + y + 5 z = 8$ is ax + by $+ 6 z = 15$ for some $a, b \in R$ , then the distance of the point $(3, 2, - 1)$ from the plane $P$ is ____ .
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The distance of the point $(1, - 2, 3)$ from the plane $x - y + z = 5$ measured parallel to the line $\frac{x}{2} = \frac{y}{3} = \frac{z}{- 6}$ is:
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The length of the perpendicular from the point $(2, - 1, 4)$ on the straight line, $\frac{x + 3}{10} = \frac{y - 2}{- 7} = \frac{z}{1}$ is:
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The equation of a plane containing the line of intersection of the planes $2 x - y - 4 = 0$ and $y + 2 z - 4 = 0$ and passing through the point $(1, 1, 0)$ is:
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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:
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The vector equation of the plane through the line of intersection of the planes $x + y + z = 1$ and $2 x + 3 y + 4 z = 5$ which is perpendicular to the plane $x - y + z = 0$ is:
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A plane passing through the points $(0, - 1, 0)$ and $(0, 0, 1)$ and making an angle $\frac{π}{4}$ with plane $y - z + 5 = 0$ , also passes through the point:
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If the line, $\frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z - 2}{4}$ meets the plane, $x + 2 y + 3 z = 15$ at a point $P$ , then the distance of $P$ from the origin is:
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The plane passing through the point $(4, - 1, 2)$ and parallel to the lines $\frac{x + 2}{3} = \frac{y - 2}{- 1} = \frac{z + 1}{2}$ and $\frac{x - 2}{1} = \frac{y - 3}{2} = \frac{z - 4}{3}$ also passes through the point:
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Let A be a point on the line $r = (1 - 3 μ) \hat{i} + (μ - 1) \hat{j} + (2 + 5 μ) \hat{k}$ and $B (3, 2, 6)$ be a point in the space. Then the value of $μ$ for which the vector $AB$ is parallel to th …
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Two lines $\frac{x - 3}{1} = \frac{y + 1}{3} = \frac{z - 6}{- 1}$ and $\frac{x + 5}{7} = \frac{y - 2}{- 6} = \frac{z - 3}{4}$ intersect at the point $R$ . The reflection of $R$ in the $x y$ -plane has coordinates:
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If the point $(2, α, β)$ lies on the plane which passes through the points $(3, 4, 2)$ and $(7, 0, 6)$ and is perpendicular to the plane $2 x - 5 y = 15$ , then $2 α - 3 β$ is equal to:
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If the line $\frac{x - 2}{3} = \frac{y + 1}{2} = \frac{z - 1}{- 1}$ intersects the plane $2 x + 3 y - z + 13 = 0$ at a point $P$ and the plane $3 x + y + 4 z = 16$ at a point $Q$ , then $P Q$ is equal to :
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The distance of the point having position vector $- \hat{i} + 2 \hat{j} + 6 \hat{k}$ from the straight line passing through the point ( $2, 3, - 4$ ) and parallel to the vector $6 \hat{i} + 3 \hat{j} - 4 \hat{k}$ is
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