New chat PYQ Library Doubts Rank report

Sign in to see Recents

Sign in to save progress →

3D Geometry question

2025 · 23 Jan · Shift 2 · Q31

Guest · try one question or challengeGuest · try one question or challenge

PYQ Library
JEE Main
Mathematics
3D Geometry
2025 · 23 Jan · Shift 2 · Q31

3D Geometry question

2025 · 23 Jan · Shift 2 · Q31

JEE MainMathematics3D GeometryMCQ+4 / −1

If the square of the shortest distance between the lines

\frac{x-2}{1}=\frac{y-1}{2}=\frac{z+3}{-3}

and

\frac{x+1}{2}=\frac{y+3}{4}=\frac{z+5}{-5}

is

\frac{m}{n}

, where

m

,

n

are coprime numbers, then

m+n

is equal to :

A
14
B
6
C
21
D
9

More from 3D Geometry

The distance of the line $\frac{x - 2}{2} = \frac{y - 6}{3} = \frac{z - 3}{4}$ from the point $(1, 4, 0)$ along the line $\frac{x}{1} = \frac{y - 2}{2} = \frac{z + 3}{3}$ is :2025 · MCQ
Let in a $△ A B C$ , the length of the side $(1, 2, 3)$ be 6 , the vertex $B$ be $(1, 2, 3)$ 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 $△ A B C$ is:2025 · MCQ
Let the line passing through the points $(- 1, 2, 1)$ and parallel to the line $\frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z}{4}$ intersect the line $\frac{x + 2}{3} = \frac{y - 3}{2} = \frac{z - 4}{1}$ at the point $P$ . Then the distance of $P$ from the point $Q (4, - 5, 1)$ …2025 · MCQ
Let P be the image of the point $Q (7, - 2, 5)$ in the line $L : \frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z}{4}$ and $R (5, p, q)$ be a point on $L$ . Then the square of the area of $△ P QR$ is …2025 · Numerical
Let $A (x, y, z)$ be a point in $x y$ -plane, which is equidistant from three points $(0, 3, 2), (2, 0, 3)$ and $(0, 0, 1)$ . Let $B = (1, 4, - 1)$ and $C = (2, 0, - 2)$ . Then among the statements (S1) : $△ ABC$ is…2025 · MCQ
If the image of the point $(4, 4, 3)$ in the line $\frac{x - 1}{2} = \frac{y - 2}{1} = \frac{z - 1}{3}$ is $(α, β, γ)$ , then $α + β + γ$ is equal to2025 · MCQ
The square of the distance of the point $(\frac{15}{7}, \frac{32}{7}, 7)$ from the line $\frac{x + 1}{3} = \frac{y + 3}{5} = \frac{z + 5}{7}$ in the direction of the vector $\hat{i} + 4 \hat{j} + 7 \hat{k}$ is:2025 · MCQ
Let $L_{1} : \frac{x - 1}{1} = \frac{y - 2}{- 1} = \frac{z - 1}{2}$ and $L_{2} : \frac{x + 1}{- 1} = \frac{y - 2}{2} = \frac{z}{1}$ be two lines. Let $L_{3}$ be a line passing through the point $(α, β, γ)$ and be perpendicular to both $L_{1}$ …2025 · MCQ