Match List-I with List-II, List-I, List-II, Lines, Direction Ratios, (A) (x - 1)/2 = (2 - y)/1 = z, (I) 1, 3, -1, (B) (2x - 1)/2 = (y + 1)/3 = (1 - z)/1, (II) 2, -2, 0, (C) (x + 1)/2 = (3 - y)/2, z = 2, (III) 2, -1, 1, (D) (2x - 3)/4 = (1 - 2y)/2 = z/5, (IV) 2, -1, 5, Choose the correct answer from the options given below: Option 1: (A) - (III), (B) - (IV), (C) - (II), (D) - (I) Option 2: (A) - (III), (B) - (I), (C) - (II), (D) - (IV) Option 3: (A) - (III), (B) - (II), (C) - (I), (D) - (IV) Option 4: (A) - (II), (B) - (III), (C) - (I), (D) - (IV)

Answer: (A) - (III), (B) - (I), (C) - (II), (D) - (IV). Solution: Direction ratios for a line in the form (x - x_0)/a = (y - y_0)/b = (z - z_0)/c are (a, b, c). Line (A): (x - 1)/2 = (2 - y)/1 = z Rewrite (2 - y)/1: (2 - y)/1 = (-(y - 2))/1 = (y - 2)/(-1) Rewrite z as (z - 0)/1 The line becomes: (x - 1)/2 = (y - 2)/(-1) = (z - 0)/1 Direction Ratios: (2, -1, 1) which is (III) Line (B): (2x - 1)/2 = (y + 1)/3 = (1 - z)/1 Simplify (2x - 1)/2: (2x - 1)/2 = (2(x - 1/2))/2 = (x - 1/2)/1 Rewrite (1 - z)/1: (1 - z)/1 = (-(z - 1))/1 = (z - 1)/(-1) The line becomes: (x - 1/2)/1 = (y + 1)/3 = (z - 1)/(-1) Direction Ratios: (1, 3, -1) which is (I) Line (C): (x + 1)/2 = (3 - y)/2, z = 2 Rewrite (3 - y)/2: (3 - y)/2 = (-(y - 3))/2 = (y - 3)/(-2) Since z = 2 is constant, the line doesn't move in the z-direction. Write as: (z - 2)/0 The line becomes: (x + 1)/2 = (y - 3)/(-2) = (z - 2)/0 Direction Ratios: (2, -2, 0) which is (II) Line (D): (2x - 3)/4 = (1 - 2y)/2 = z/5 Simplify (2x - 3)/4: (2x - 3)/4 = (2(x - 3/2))/4 = (x - 3/2)/2 Simplify (1 - 2y)/2: (1 - 2y)/2 = (-2(y - 1/2))/2 = (-(y - 1/2))/1 = (y - 1/2)/(-1) Rewrite z/5 as (z - 0)/5 The line becomes: (x - 3/2)/2 = (y - 1/2)/(-1) = (z - 0)/5 Direction Ratios: (2, -1, 5) which is (IV) Final Matching: (A) → (III) (B) → (I) (C) → (II) (D) → (IV)

Geometry

3D Geometry

Medium

core

Match List-I with List-II

List-I List-II
Lines 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:

List-I	List-II
Lines	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

(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 Option: 2

Direction ratios for a line in the form $\frac{x - x_0}{a} = \frac{y - y_0}{b} = \frac{z - z_0}{c}$ are $(a, b, c)$ .

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

Rewrite $\frac{2 - y}{1}$ :

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

$= \frac{y - 2}{-1}$

Rewrite $z$ as $\frac{z - 0}{1}$

The line becomes:

$\frac{x - 1}{2} = \frac{y - 2}{-1} = \frac{z - 0}{1}$

Direction Ratios: $(2, -1, 1)$ which is (III)

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

Simplify $\frac{2x - 1}{2}$ :

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

$= \frac{x - \frac{1}{2}}{1}$

Rewrite $\frac{1 - z}{1}$ :

$\frac{1 - z}{1} = \frac{-(z - 1)}{1}$

$= \frac{z - 1}{-1}$

The line becomes:

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

Direction Ratios: $(1, 3, -1)$ which is (I)

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

Rewrite $\frac{3 - y}{2}$ :

$\frac{3 - y}{2} = \frac{-(y - 3)}{2}$

$= \frac{y - 3}{-2}$

Since $z = 2$ is constant, the line doesn't move in the z-direction.

Write as: $\frac{z - 2}{0}$

The line becomes:

$\frac{x + 1}{2} = \frac{y - 3}{-2} = \frac{z - 2}{0}$

Direction Ratios: $(2, -2, 0)$ which is (II)

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

Simplify $\frac{2x - 3}{4}$ :

$\frac{2x - 3}{4} = \frac{2(x - \frac{3}{2})}{4}$

$= \frac{x - \frac{3}{2}}{2}$

Simplify $\frac{1 - 2y}{2}$ :

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

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

$= \frac{y - \frac{1}{2}}{-1}$

Rewrite $\frac{z}{5}$ as $\frac{z - 0}{5}$

The line becomes:

$\frac{x - \frac{3}{2}}{2} = \frac{y - \frac{1}{2}}{-1} = \frac{z - 0}{5}$

Direction Ratios: $(2, -1, 5)$ which is (IV)

Final Matching:

(A) → (III)

(B) → (I)

(D) → (IV)

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

The angle between two lines whose direction ratio are proportional to $(-\theta, -2\theta, \theta)$ and $(-2\theta, \theta, \theta)$ , where $\theta \in \mathbb{R} \sim \{0\}$ , is

2026: 19 May Shift 1

If $\theta$ is the angle between the lines $L_1$ and $L_2$ , whose direction cosines are $l_1, m_1, n_1$ and $l_2, m_2, n_2$ respectively, then

Match List-I with List-II

List-I	List-II
(A) $\cos\theta$	(I) $l_1l_2+m_1m_2+n_1n_2$
(B) $\sin\theta$	(II) $l_1l_2+m_1m_2+n_1n_2=0$
(C) $L_1$ and $L_2$ are perpendicular	(III) $\frac{l_1}{l_2}=\frac{m_1}{m_2}=\frac{n_1}{n_2}$
(D) $L_1$ and $L_2$ are parallel	(IV) $\sqrt{\sum(m_1n_2-m_2n_1)^2}$

Choose the correct answer from the options given below:

2026: 22 May Shift 2

The shortest distance (in units) between the lines $L_1:\frac{x-1}{2}=\frac{y-2}{3}=\frac{z+4}{6}$ and $L_2:\frac{x-3}{2}=\frac{3-y}{-3}=\frac{z+5}{6}$ is:

2023: 15 June Shift 2

If the straight lines $x = 1 + s$ , $y = -3 - \lambda s$ , $z = 1 + \lambda s$ and $x = \frac{t}{2}$ , $y = 1 + t$ , $z = 2 - t$ with parameters $s$ and $t$ respectively, are coplanar, then $\lambda$ is equal to :

2026: 29 May Shift 2

The perpendicular distance of point (1, 1, 3) on the line $\frac{x}{2}=\frac{y+1}{3}=\frac{-z-3}{-5}$ is

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