Match the LIST-I with LIST-II, LIST-I Equation of line, LIST-II Point on the line, direction ratios of the line, A., (x-5)/3 = (y-2)/2 = (z+4)/(-8), I., (-3, -2, 8), 5, 2, -4, B., (x+5)/3 = (y+2)/2 = (z-4)/(-8), II., (3, 2, -8), 5, 2, -4, C., (x-3)/5 = (y-2)/2 = (z+8)/(-4), III., (-5, -2, 4), 3, 2, -8, D., (x+3)/5 = (y+2)/2 = (z-8)/(-4), IV., (5, 2, -4), 3, 2, -8, Choose the correct answer from the options given below: Option 1: A-III, B-IV, C-II, D-I Option 2: A-III, B-IV, C-I, D-II Option 3: A-IV, B-III, C-II, D-I Option 4: A-IV, B-III, C-I, D-II

Answer: A-IV, B-III, C-II, D-I. Solution: In (x-a)/l=(y-b)/m=(z-c)/n the point is (a,b,c) and direction ratios are l,m,n. A gives (5,2,-4) with 3,2,-8 (IV); B gives (-5,-2,4) with 3,2,-8 (III); C gives (3,2,-8) with 5,2,-4 (II); D gives (-3,-2,8) with 5,2,-4 (I).

Geometry

3D Geometry

Easy

core

Match the LIST-I with LIST-II

LIST-I
Equation of line LIST-II
Point on the line, direction ratios of the line
A. $\dfrac{x-5}{3} = \dfrac{y-2}{2} = \dfrac{z+4}{-8}$ I. $(-3, -2, 8), \langle 5, 2, -4 \rangle$
B. $\dfrac{x+5}{3} = \dfrac{y+2}{2} = \dfrac{z-4}{-8}$ II. $(3, 2, -8), \langle 5, 2, -4 \rangle$
C. $\dfrac{x-3}{5} = \dfrac{y-2}{2} = \dfrac{z+8}{-4}$ III. $(-5, -2, 4), \langle 3, 2, -8 \rangle$
D. $\dfrac{x+3}{5} = \dfrac{y+2}{2} = \dfrac{z-8}{-4}$ IV. $(5, 2, -4), \langle 3, 2, -8 \rangle$

Choose the correct answer from the options given below:

	LIST-I Equation of line		LIST-II Point on the line, direction ratios of the line
A.	$\dfrac{x-5}{3} = \dfrac{y-2}{2} = \dfrac{z+4}{-8}$	I.	$(-3, -2, 8), \langle 5, 2, -4 \rangle$
B.	$\dfrac{x+5}{3} = \dfrac{y+2}{2} = \dfrac{z-4}{-8}$	II.	$(3, 2, -8), \langle 5, 2, -4 \rangle$
C.	$\dfrac{x-3}{5} = \dfrac{y-2}{2} = \dfrac{z+8}{-4}$	III.	$(-5, -2, 4), \langle 3, 2, -8 \rangle$
D.	$\dfrac{x+3}{5} = \dfrac{y+2}{2} = \dfrac{z-8}{-4}$	IV.	$(5, 2, -4), \langle 3, 2, -8 \rangle$

A-III, B-IV, C-II, D-I

A-III, B-IV, C-I, D-II

A-IV, B-III, C-II, D-I

A-IV, B-III, C-I, D-II

✅ Correct Option: 3

In $\dfrac{x-a}{l}=\dfrac{y-b}{m}=\dfrac{z-c}{n}$ the point is $(a,b,c)$ and direction ratios are $\langle l,m,n\rangle$ .

A gives $(5,2,-4)$ with $\langle3,2,-8\rangle$ (IV); B gives $(-5,-2,4)$ with $\langle3,2,-8\rangle$ (III); C gives $(3,2,-8)$ with $\langle5,2,-4\rangle$ (II); D gives $(-3,-2,8)$ with $\langle5,2,-4\rangle$ (I).

2023: 30 May Shift 3

The angle between the two planes $x + y - z = 3$ and $3x + 2y + z = 5$ is :

2025: 30 May Shift 1

Consider the line $\vec{r} = -2\hat{i} + 3\hat{j} + \hat{k} + \lambda(5\hat{i} - 3\hat{j} - \hat{k})$ . Match List-I with List-II

List-I	List-II
(A) A point on the given line	(I) $\left(\frac{5}{\sqrt{35}}, \frac{-3}{\sqrt{35}}, \frac{-1}{\sqrt{35}}\right)$
(B) Direction ratios of the given line	(II) (2, 3, 1)
(C) Direction cosines of the given line	(III) (5, -3, -1)
(D) Direction ratios of a line perpendicular to given line	(IV) (-2, 3, 1)

Choose the correct answer from the options given below:

2025: 19 May Shift 1

If the line $\frac{-x+1}{3} = \frac{-y-2}{-2k} = \frac{z+3}{2}$ and $\frac{-1+x}{3k} = \frac{-1 +y}{1} = \frac{-z+6}{5}$ are perpendicular, then the value of k is:

2025: 13 May Shift 2

If the lines $\frac{1-x}{3} = \frac{3y-6}{k} = \frac{3-z}{-2}$ and $\frac{1-x}{2k} = \frac{y-5}{3} = \frac{6-z}{5}$ are perpendicular to each other, then $k$ is equal to

2024: 16 May Shift 1

The direction cosines of the line which is perpendicular to the lines with direction ratios $1,-2,-2$ and $0,2,1$ are :

2026: 18 May Shift 2

The angle between the two lines whose direction cosines are given by the equations $4l+m+n=0$ and $2mn+24nl+4lm=0$ , is

Solution

Related questions:

Keyboard Shortcuts

CUET Mathematics Past Year Questions (Topic-Wise)