List of top Mathematics Questions on Three Dimensional Geometry

If the foot is perpendicular from (1, 2, 3) to the line $\frac{x+1}{2} = \frac{y-2}{5} = \frac{z-1}{1}$ is $( a, \beta, \gamma)$, then find $a + \beta + \gamma$

A straight line drawn from the point P(1,3, 2), parallel to the line $\frac{x-2}{1}=\frac{y-4}{2}=\frac{z-6}{1}$, intersects the plane L₁ : x - y + 3z = 6 at the point Q. Another straight line which passes through Q and is perpendicular to the plane L₁ intersects the plane L₂ : 2x - y + z = -4 at the point R. Then which of the following statements is (are) TRUE ?

Let R³ denote the three-dimensional space. Take two points P = (1, 2, 3) and Q = (4, 2 ,7). Let
dist(X, Y) denote the distance between two points X and Y in R3. Let
$S=\left\{X\in\R^3:(dist(X,P)^2)-(dist(X,Q))^2=50\right\}\ and$
$T=\left\{Y\in \R^3:(dist(Y,Q))^2-(dist(Y,P))^2=50\right\}$
Then which of the following statements is (are) TRUE ?

Let y ∈ R be such that the lines $L_1:\frac{x+11}{1}=\frac{y+21}{2}=\frac{z+29}{3}$ and $L_2:\frac{x+16}{3}=\frac{y+11}{2}=\frac{z+4}{\gamma}$ intersect. Let R₁ be the point of intersection of L₁ and L₂. Let O = (0, 0 ,0), and $\hat{n}$ denote a unit normal vector to the plane containing both the lines L₁ and L₂.
Match each entry in List-I to the correct entry in List-II.

List - I		List - II
(P)	γ equals	(1)	$-\hat{i}-\hat{j}+\hat{k}$
(Q)	A possible choice for $\hat{n}$ is	(2)	$\sqrt{\frac{3}{2}}$
(R)	$\overrightarrow{OR_1}$ equals	(3)	1
(S)	A possible value of $\overrightarrow{OR_1}.\hat{n}$ is	(4)	$\frac{1}{\sqrt6}\hat{i}-\frac{2}{\sqrt6}\hat{j}+\frac{1}{\sqrt6}\hat{k}$
		(5)	$\sqrt{\frac{2}{3}}$

The correct option is

The length of the perpendicular drawn from the point $(1, 2, 3)$ to the line $(X - 6)/3 = (y - 7)/2 = (Z - 7)/-2$ is:

The angle between the lines, whose direction cosines $ l, m, n $ satisfy the equations: \[ l + m + n = 0 \quad \text{and} \quad 2l^2 + 2m^2 - n^2 = 0, \] is:

If $A(1, 1, 1), B(0, λ, 0), C(λ + 1, 0, 1), D(2, -2, 2)$ are co-planer then $\sum (\lambda_i + 2)^2$ is equal to

Let the line $L: \frac{x-1}{2}=\frac{y+1}{-1}=\frac{z-3}{1}$ intersect the plane $2 x+y+3 z=16$ at the point $P$ Let the point $Q$ be the foot of perpendicular from the point $R(1,-1,-3)$ on the line $L$ If $\alpha$ is the area of triangle $P Q R$, then $\alpha^2$ is equal to

Let $\alpha x+\beta y+y z=1$ be the equation of a plane passing through the point$(3,-2,5)$and perpendicular to the line joining the points $(1,2,3)$ and $(-2,3,5)$ Then the value of $\alpha \beta y$is equal to ____

A plane $E$ is perpendicular to the two planes $2 x-2 y+z=0$ and $x-y+2 z=4$, and passes through the point $P (1,-1,1)$ If the distance of the plane $E$ from the point $Q(a, a, 2)$ is $3 \sqrt{2}$, then $( PQ )^2$ is equal to

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 :

Assertion (A): Quadrilateral formed by vertices A(0, 0, 0), B(3, 4, 5), C(8, 8, 8) and D(5, 4, 3) is a rhombus.
Reason (R): ABCD is a rhombus if $AB = BC = CD = DA$, and $AC \ne BD$.

Find the shortest distance between the pair of lines:
Line $ L_1 $ : \[ \frac{x - 1}{2} = \frac{y + 1}{3} = z, \] Line $ L_2 $ : \[ \frac{x + 1}{5} = \frac{y - 2}{1} = z = 2. \]

Show that the lines
\[ \frac{x + 1}{3} = \frac{y + 3}{5} = \frac{z + 5}{7} \quad \text{and} \quad \frac{x - 2}{1} = \frac{y - 4}{3} = \frac{z - 6}{5} \] intersect and find their point of intersection.

If a line makes angles $ \alpha, \beta, \gamma $ with the x-axis, y-axis, and z-axis respectively,
then prove that: \[ \sin^2 \alpha + \sin^2 \beta + \sin^2 \gamma = 2 \]

The vector equation of a line which passes through the point (2, -4, 5) and is parallel to the line $\displaystyle \frac{x+3}{3} = \frac{4-y}{2} = \frac{z+8}{6}$ is:

Let the image of the point $P(2,-1,3)$ in the plane $x+2 y-z=0$ be $Q$ Then the distance of the plane $3 x+2 y+z+29=0$ from the point $Q$ is

Shortest distance between lines $\frac{(x-5)}{4}$=$\frac{(y-3)}{6}$=$\frac{(z-2)}{4}$ and $\frac{(x-3)}{7}=\frac{(y-2)}{5}=\frac{(z-9)}{6}$ is ?

In a triangle BC, if the mid points of sides AB, BC, CA are (3,0,0), (0,4,0),(0,0,5) respectively, then AB²+ BC² + CA² =

The orthocenter of the triangle whose sides are given by x + y + 10 = 0, x - y - 2 = 0 and 2x + y - 7 = 0 is

For l ∈ R, the equation (2l - 3) x² + 2lxy - y² = 0 represents a pair of distinct lines

In △ABC, if a : b : c = 4 : 5 : 6, then the ratio of the circumference to its in radius is

The number of diagonals of a polygon is 35. If A, B are two distinct vertices of this polygon, then the number of all those triangles formed by joining three vertices of the polygon having AB as one of its sides is:

Two lines $2x=3y=-z$ and $6z=-y=-4z$ intersect at a point, then find the angle between them.

ABC is an isosceles triangle with an inscribed circle with center O. Let P be the midpoint of BC. If AB=AC=15 and BC=10,then OP equals