Geometry

3D Geometry

Medium

core

If the lines $\frac{x-1}{2} = \frac{y-m}{3} = \frac{z-3}{4}$ and $\frac{x-4}{5} = \frac{y-1}{2} = z$ are interesecting, then the value of $m$ , is

-1

-2

✅ Correct Option: 4

The lines meet when the determinant with rows $(3,\,1-m,\,-3)$ , $(2,3,4)$ and $(5,2,1)$ vanishes, where the first row joins $(1,m,3)$ to $(4,1,0)$ . Expanding gives $36-18m=0$ , so $m=2$ .

2025: 3 June Shift 2

The angle at which the line, $\frac{x-1}{0} = \frac{2-y}{-1} = \frac{2z-3}{-2}$ is inclined with the positive direction of z-axis is

2026: 19 May Shift 2

Points on the line $\frac{x+2}{3} = \frac{y+1}{2} = \frac{z-3}{2}$ at a distance of 5 units from the point $(1, 3, 3)$ , are

2022: 16 July Shift 2

The correct order of steps from A to E, for finding value of p, so that the lines $\frac{1-x}{3} = \frac{7y-14}{2p} = \frac{z-3}{2}$ and $\frac{7-7x}{3p} = \frac{y-5}{1} = \frac{6-z}{5}$ are at right angle is:

A. $p = \frac{70}{11}$

B. $(-3) \times \left(\frac{-3p}{7}\right) + 1 \times \left(\frac{2p}{7}\right) + 2 \times (-5) = 0$

C. $\frac{x-1}{-3} = \frac{y-2}{2\frac{p}{7}} = \frac{z-3}{2}$ , $\frac{x-1}{-\frac{3p}{7}} = \frac{y-5}{1} = \frac{z-6}{-5}$

D. $\frac{9p}{7} + \frac{2p}{7} - 10 = 0$

E. $\frac{11p}{7} = 10$

Choose the correct answer from the options given below:

2025: 2 June Shift 1

The foot of the perpendicular drawn from the point $(1,6,3)$ to the line $\frac{x}{1} = \frac{y-1}{2} = \frac{z-2}{3}$ is

2026: 22 May Shift 2

Match List-I with List-II

List-I	List-II
(A) The distance of a point $P(a,b,c)$ from x-axis	(I) $\sqrt{a^2+b^2+c^2}$
(B) The distance of a point $P(a,b,c)$ from y-axis	(II) $\sqrt{b^2+c^2}$
(C) The distance of a point $P(a,b,c)$ from Z-axis	(III) $\sqrt{a^2+c^2}$
(D) The distance of a point $P(a,b,c)$ from origin	(IV) $\sqrt{a^2+b^2}$

Choose the correct answer from the options given below:

2026: 30 May Shift 1

Given a line $L:\frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3}$ , then

Match the LIST-I with LIST-II

LIST-I	LIST-II
A. The foot of perpendicular from point (1, 6, 3) on line L	I. $2\sqrt{13}$
B. Perpendicular distance from point (1, 6, 3) to line L	II. (1, 0, 7)
C. Image of point (1, 6, 3) with respect to line L	III. (1, 3, 5)
D. Distance between points (1, 6, 3) and (1, 0, 7)	IV. $\sqrt{13}$

Choose the correct answer from the options given below:

If the lines x−12=y−m3=z−34\frac{x-1}{2} = \frac{y-m}{3} = \frac{z-3}{4}2x−1​=3y−m​=4z−3​ and x−45=y−12=z\frac{x-4}{5} = \frac{y-1}{2} = z5x−4​=2y−1​=z are interesecting, then the value of mmm, is

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CUET Mathematics Past Year Questions (Topic-Wise)

If the lines $\frac{x-1}{2} = \frac{y-m}{3} = \frac{z-3}{4}$ and $\frac{x-4}{5} = \frac{y-1}{2} = z$ are interesecting, then the value of $m$ , is