The shortest distance between lines (-x-3)/4 = (y-6)/3 = z/2 and (-x-2)/4 = y/1 = (z-7)/1 is: Option 1: 1/9 units Option 2: 81 units Option 3: 9 units Option 4: 7 units

Answer: 9 units. Solution: The lines need to be rewritten so that -x is absorbed into the direction ratios. Line 1: (-x-3)/4 = (y-6)/3 = z/2 (x+3)/(-4) = (y-6)/3 = (z-0)/2 Point: a_1 = (-3,\ 6,\ 0) Direction ratios: b_1 = (-4,\ 3,\ 2) Line 2: (-x-2)/4 = y/1 = (z-7)/1 (x+2)/(-4) = (y-0)/1 = (z-7)/1 Point: a_2 = (-2,\ 0,\ 7) Direction ratios: b_2 = (-4,\ 1,\ 1) The shortest distance between two skew lines is given by: d = (, (a_2 - a_1) * (b_1 x b_2), )/(, b_1 x b_2, ) a_2 - a_1 = (-2 - (-3),\ 0 - 6,\ 7 - 0) = (1,\ -6,\ 7) b_1 x b_2 = vmatrix i & j & k \\ -4 & 3 & 2 \\ -4 & 1 & 1 vmatrix i: (3)(1) - (2)(1) = 1 j: -[(-4)(1) - (2)(-4)] = -[-4 + 8] = -4 k: (-4)(1) - (3)(-4) = -4 + 12 = 8 b_1 x b_2 = (1,\ -4,\ 8), b_1 x b_2, = √(1^2 + (-4)^2 + 8^2) = √(1 + 16 + 64) = √81 = 9 (a_2 - a_1) * (b_1 x b_2) = (1)(1) + (-6)(-4) + (7)(8) = 1 + 24 + 56 = 81 d = (, 81, )/9 = 9 Therefore, the shortest distance is 9 units.

Geometry

3D Geometry

Medium

core

The shortest distance between lines $\frac{-x-3}{4} = \frac{y-6}{3} = \frac{z}{2}$ and $\frac{-x-2}{4} = \frac{y}{1} = \frac{z-7}{1}$ is:

$\frac{1}{9}$ units

81 units

9 units

7 units

✅ Correct Option: 3

The lines need to be rewritten so that $-x$ is absorbed into the direction ratios.

Line 1: $\dfrac{-x-3}{4} = \dfrac{y-6}{3} = \dfrac{z}{2}$

$\frac{x+3}{-4} = \frac{y-6}{3} = \frac{z-0}{2}$

Point: $\mathbf{a_1} = (-3,\ 6,\ 0)$

Direction ratios: $\mathbf{b_1} = (-4,\ 3,\ 2)$

Line 2: $\dfrac{-x-2}{4} = \dfrac{y}{1} = \dfrac{z-7}{1}$

$\frac{x+2}{-4} = \frac{y-0}{1} = \frac{z-7}{1}$

Point: $\mathbf{a_2} = (-2,\ 0,\ 7)$

Direction ratios: $\mathbf{b_2} = (-4,\ 1,\ 1)$

The shortest distance between two skew lines is given by:

$d = \dfrac{\left|(\mathbf{a_2} - \mathbf{a_1}) \cdot (\mathbf{b_1} \times \mathbf{b_2})\right|}{|\mathbf{b_1} \times \mathbf{b_2}|}$

$\mathbf{a_2} - \mathbf{a_1} = (-2 - (-3),\ 0 - 6,\ 7 - 0)$

$= (1,\ -6,\ 7)$

$\mathbf{b_1} \times \mathbf{b_2} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -4 & 3 & 2 \\ -4 & 1 & 1 \end{vmatrix}$

$\mathbf{i}$ : $(3)(1) - (2)(1) = 1$

$\mathbf{j}$ : $-[(-4)(1) - (2)(-4)] = -[-4 + 8] = -4$

$\mathbf{k}$ : $(-4)(1) - (3)(-4) = -4 + 12 = 8$

$\mathbf{b_1} \times \mathbf{b_2} = (1,\ -4,\ 8)$

$|\mathbf{b_1} \times \mathbf{b_2}| = \sqrt{1^2 + (-4)^2 + 8^2}$

$= \sqrt{1 + 16 + 64}$

$= \sqrt{81} = 9$

$(\mathbf{a_2} - \mathbf{a_1}) \cdot (\mathbf{b_1} \times \mathbf{b_2}) = (1)(1) + (-6)(-4) + (7)(8)$

$= 1 + 24 + 56 = 81$

$d = \dfrac{|81|}{9} = 9$

Therefore, the shortest distance is $9$ units.

2025: 14 May Shift 1

The shortest distance between the lines $\frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4}$ and $\frac{x-2}{4} = \frac{y-4}{6} = \frac{z-5}{8}$ is equal to

2026: 14 May Shift 2

The shortest distance between the lines whose vector equations are $\vec r=(\hat i+2\hat j+3\hat k)+\lambda(2\hat i+3\hat j+4\hat k)$ and $\vec r=(2\hat i+4\hat j+5\hat k)+\mu(4\hat i+6\hat j+8\hat k)$ is:

2026: 22 May Shift 1

Direction cosines of 2 lines are given by the equations $3l+m+5n=0, 6mn-2nl+5lm=0$ then which of the following statements are correct ?

(A) Direction ratios of two lines are (1, 2, -1) and (-2, 1, 1).

(B) Angle between lines is $\frac{\pi}{3}$ .

(D) Vectors parallel to these lines are $\hat i+2\hat j-\hat k$ and $-2\hat i+\hat j+\hat k$ .

Choose the correct answer from the options given below:

2026: 30 May Shift 2

Direction ratios of the line perpendicular to the lines $\frac{x-3}{2}=\frac{y+7}{-3}=\frac{z-2}{1}$ and $\frac{x+2}{1}=\frac{y+3}{2}=\frac{z-5}{-2}$ are

2025: 16 May Shift 1

Consider the line $\frac{x-2}{2} = \frac{2y-5}{-3}, z = -1$ . Then which of the following is/are true?

(A) It has Direction ratios (2, -3, -1)

(B) It has Direction cosines $\left(\frac{4}{5}, \frac{-3}{5}, \frac{-1}{5}\right)$

(D) It has Direction cosines $\left(\frac{4}{5}, \frac{-3}{5}, 0\right)$

Choose the correct answer from the options given below:

2026: 12 May Shift 2

If angle between the lines $\dfrac{x-1}{\propto} = \dfrac{y+2}{-5} = \dfrac{z+\frac{7}{3}}{\beta}$ and $x=z, y=0$ is $\dfrac{\pi}{4}$ , then

The shortest distance between lines −x−34=y−63=z2\frac{-x-3}{4} = \frac{y-6}{3} = \frac{z}{2}4−x−3​=3y−6​=2z​ and −x−24=y1=z−71\frac{-x-2}{4} = \frac{y}{1} = \frac{z-7}{1}4−x−2​=1y​=1z−7​ is:

Solution

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CUET Mathematics Past Year Questions (Topic-Wise)

The shortest distance between lines $\frac{-x-3}{4} = \frac{y-6}{3} = \frac{z}{2}$ and $\frac{-x-2}{4} = \frac{y}{1} = \frac{z-7}{1}$ is: