Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Three-dimensional Geometry

Question:

Match List-I with List-II

List-I Lines

List-II Direction Ratios

(A) $\frac{x-1}{2} = \frac{2-y}{1}=z$

(I) 1, 3, -1

(B) $\frac{2x-1}{2}=\frac{y+1}{3}=\frac{1-z}{1}$

(II) 2, -2, 0

(C) $\frac{x+1}{2}=\frac{3-y}{2},z=2$

(III) 2, -1, 1

(D) $\frac{2x-3}{4}=\frac{1-2y}{2}=\frac{z}{5}$

(IV) 2, -1, 5

Choose the correct answer from the options given below:

Options:

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

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

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

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

Correct Answer:

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

Explanation:

The correct answer is Option (2) → (A)-(III), (B)-(I), (C)-(II), (D)-(IV)

List-I Lines

List-II Direction Ratios

(A) $\frac{x-1}{2} = \frac{2-y}{1}=z$

(III) 2, -1, 1

(B) $\frac{2x-1}{2}=\frac{y+1}{3}=\frac{1-z}{1}$

(I) 1, 3, -1

(C) $\frac{x+1}{2}=\frac{3-y}{2},z=2$

(II) 2, -2, 0

(D) $\frac{2x-3}{4}=\frac{1-2y}{2}=\frac{z}{5}$

(IV) 2, -1, 5

Extract direction ratios from each line.

(A) $\displaystyle \frac{x-1}{2}=\frac{2-y}{1}=z=t$

$x=1+2t,\; y=2-t,\; z=t$

Direction ratios: $(2,-1,1)$ → matches (III)

(B) $\displaystyle \frac{2x-1}{2}=\frac{y+1}{3}=\frac{1-z}{1}=t$

$x=\frac{1+2t}{2},\; y=-1+3t,\; z=1-t$

Direction ratios: $(1,3,-1)$ → matches (I)

(C) $\displaystyle \frac{x+1}{2}=\frac{3-y}{2}=t,\; z=2$

$x=-1+2t,\; y=3-2t,\; z$ constant

Direction ratios: $(2,-2,0)$ → matches (II)

(D) $\displaystyle \frac{2x-3}{4}=\frac{1-2y}{2}=\frac{z}{5}=t$

$x=\frac{3+4t}{2},\; y=\frac{1-2t}{2},\; z=5t$

Direction ratios: $(2,-1,5)$ → matches (IV)

Final matching:

(A)–(III), (B)–(I), (C)–(II), (D)–(IV)