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Mathematics
List of top Mathematics Questions on 3D Geometry
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 $(0, 0), (\lambda_1, \lambda_2)$ and $(\lambda_2, \lambda_1)$ is
JEE Main - 2025
JEE Main
Mathematics
3D Geometry
Let a straight line \( L \) pass through the point \(P(2, -1, 3)\) and be perpendicular to the lines
\[ \frac{x - 1}{2} = \frac{y + 1}{1} = \frac{z - 3}{-2} \quad \text{and} \quad \frac{x - 3}{1} = \frac{y - 2}{3} = \frac{z + 2}{4}. \]
If the line \(L\) intersects the yz-plane at the point Q, then the distance between the points P and Q is:
JEE Main - 2025
JEE Main
Mathematics
3D Geometry
Let \( L_1 : \frac{x - 1}{3} = \frac{y - 1}{-1} = \frac{z + 1}{0} \) and \( L_2 : \frac{x - 2}{2} = \frac{y}{0} = \frac{z + 4}{\alpha} \), where \( \alpha \in \mathbb{R} \), be two lines which intersect at the point \( B \). If \( P \) is the foot of the perpendicular from the point \( A(1, 1, -1) \) on \( L_2 \), then the value of \( 26 \alpha (PB)^2 \) is:
JEE Main - 2025
JEE Main
Mathematics
3D Geometry
If the midpoint of a chord of the ellipse \( \frac{x^2}{9} + \frac{y^2}{4} = 1 \) is \( \left( \sqrt{2}, \frac{4}{3} \right) \), and the length of the chord is \( \frac{2\sqrt{\alpha}}{3} \), then \( \alpha \) is:
JEE Main - 2025
JEE Main
Mathematics
3D Geometry
If \( A \) and \( B \) are the points of intersection of the circle \( x^2 + y^2 - 8x = 0 \) and the hyperbola \( \frac{x^2}{9} - \frac{y^2}{4} = 1 \), and a point \( P \) moves on the line \( 2x - 3y + 4 = 0 \), then the centroid of \( \triangle PAB \) lies on the line:
JEE Main - 2025
JEE Main
Mathematics
3D Geometry
The square of the distance of the point \(\left( \frac{15}{7}, \frac{32}{7}, 7 \right)\) from the line \(\frac{x+1}{3} = \frac{y+3}{5} = \frac{z+5}{7}\) in the direction of the vector \(\mathbf{i} + 4\mathbf{j} + 7\mathbf{k}\) is:
JEE Main - 2025
JEE Main
Mathematics
3D Geometry
Let in a \( \triangle ABC \), the length of the side AC is 6, the vertex B is \( (1, 2, 3) \) and the vertices A, C lie on the line
\[ \frac{x - 6}{3} = \frac{y - 7}{2} = \frac{z - 7}{-2}. \]
Then the area (in sq. units) of \( \triangle ABC \) is:
JEE Main - 2025
JEE Main
Mathematics
3D Geometry
Let the line \( x + y = 1 \) meet the circle \( x^2 + y^2 = 4 \) at the points A and B. If the line perpendicular to \( AB \) and passing through the mid point of the chord AB intersects the circle at C and D, then the area of the quadrilateral ABCD is equal to:
JEE Main - 2025
JEE Main
Mathematics
3D Geometry
Find the foot of the perpendicular drawn from point $(2, -1, 5)$ to the line
\[ \frac{x - 11}{10} = \frac{y + 2}{-4} = \frac{z + 8}{-11} \]
Also, find the length of the perpendicular.
CBSE Class XII - 2025
CBSE Class XII
Mathematics
3D Geometry
Let the position vectors of points A, B and C be \( \mathbf{a} = 3\hat{i} - \hat{j} - 2\hat{k} \), \( \mathbf{b} = \hat{i} + 2\hat{j} - \hat{k} \), and \( \mathbf{c} = \hat{i} + 5\hat{j} + 3\hat{k} \), respectively. Find the vector and Cartesian equations of the line passing through \( A \) and parallel to line \( BC \).
CBSE Class XII - 2025
CBSE Class XII
Mathematics
3D Geometry
Find the distance of the point $P(2, 4, -1)$ from the line \[ \frac{x + 5}{1} = \frac{y + 3}{4} = \frac{z - 6}{-9}. \]
CBSE Class XII - 2025
CBSE Class XII
Mathematics
3D Geometry
Determine if the lines $\mathbf{r}_1 = ( \hat{i} + \hat{j} - \hat{k} ) + \lambda ( 3\hat{i} - \hat{j} )$ and $\mathbf{r}_2 = ( 4\hat{i} - \hat{k} ) + \mu ( 2\hat{i} + 3\hat{k} )$ intersect with each other.
CBSE Class XII - 2025
CBSE Class XII
Mathematics
3D Geometry
The straight line \[ \frac{x-3}{3} = \frac{y-2}{1} = \frac{z-1}{0} \] is:
WBJEE - 2025
WBJEE
Mathematics
3D Geometry
Find the vector equation of a straight line passing through the point (5, 2, -4) and parallel to the vector \( 3\hat{i} + 2\hat{j} - 8\hat{k} \).
UP Board XII - 2025
UP Board XII
Mathematics
3D Geometry
Find the shortest distance between the lines whose vector equations are:
\(\vec{r} = (1-t)\hat{i} + (t-2)\hat{j} + (3-2t)\hat{k}\) and \(\vec{r} = (s+1)\hat{i} + (2s-1)\hat{j} - (2s+1)\hat{k}\).
UP Board XII - 2025
UP Board XII
Mathematics
3D Geometry
Find the shortest distance between the lines \(\vec{r} = \hat{i} + \hat{j} + \lambda(2\hat{i} - \hat{j} + \hat{k})\) and \(\vec{r} = 2\hat{i} + \hat{j} - \hat{k} + \mu(3\hat{i} - 5\hat{j} + 2\hat{k})\).
UP Board XII - 2025
UP Board XII
Mathematics
3D Geometry
Find the shortest distance between the lines \( \vec{r} = \hat{i} + 2\hat{j} - 4\hat{k} + \lambda(2\hat{i} + 3\hat{j} + 6\hat{k}) \) and \( \vec{r} = 3\hat{i} + 3\hat{j} - 5\hat{k} + \mu(2\hat{i} + 3\hat{j} + 6\hat{k}) \).
UP Board XII - 2025
UP Board XII
Mathematics
3D Geometry
If the ordered pairs (2x - 3, 5) and (x, y - 1) are equal, then find the numbers x and y.
UP Board XII - 2025
UP Board XII
Mathematics
3D Geometry
Find the direction cosine of Z-axis.
UP Board XII - 2025
UP Board XII
Mathematics
3D Geometry
Find the shortest distance between the lines whose vector equations are \(\vec{r}=(\hat{i}+2\hat{j}+3\hat{k})+\lambda(\hat{i}-3\hat{j}+2\hat{k})\) and \(\vec{r}=(4\hat{i}+5\hat{j}+6\hat{k})+\mu(2\hat{i}+3\hat{j}+\hat{k})\).
UP Board XII - 2025
UP Board XII
Mathematics
3D Geometry
Find the direction-cosines of the y-axis.
UP Board XII - 2025
UP Board XII
Mathematics
3D Geometry
The Cartesian equation of a line is \(\frac{x+3}{2} = \frac{y-5}{4} = \frac{z+6}{2}\). Find its vector equation.
UP Board XII - 2025
UP Board XII
Mathematics
3D Geometry
Let A = (2, 0, -1), B = (1, -2, 0), C = (1, 2, -1), and D = (0, -1, -2) be four points. If \(\theta\) is the acute angle between the plane determined by A, B, C and the plane determined by A, C, D, then \(\tan\theta =\)
AP EAPCET - 2025
AP EAPCET
Mathematics
3D Geometry
If the equation of the plane passing through point $ (3, 2, 5) $ and perpendicular to the planes $$ 2x - 3y + 5z = 7, \quad 5x + 2y - 3z = 11 $$ is $$ x + by + cz + d = 0, $$ then find $ 2b + 3c + d $.
AP EAPCET - 2025
AP EAPCET
Mathematics
3D Geometry
The points $ A(-1, 2, 3), B(2, -3, 1), C(3, 1, -2) $ are:
AP EAPCET - 2025
AP EAPCET
Mathematics
3D Geometry
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