JEE Main · 2023 · 10 Apr 2023 · Shift 1 · Three Dimensional Geometry

The shortest distance between the lines

\frac{x + 2}{1} = \frac{y}{- 2} = \frac{z - 5}{2}

and

\frac{x - 4}{1} = \frac{y - 1}{2} = \frac{z + 3}{0}

(A)6
(B)9✓ Correct
(C)7
(D)8

Answer: (B)

Solution

The shortest distance between two skew lines is given by

d = \frac{∣ A B \cdot ( b _{1} \times b _{2} ) ∣}{∣ b _{1} \times b _{2} ∣}

, where

A B

connects points on the two lines and

b_{1}, b_{2}

are direction vectors.

Here

A B = (6, 1, - 8)

b_{1} = (1, - 2, 2)

b_{2} = (1, 2, 0)

b_{1} \times b_{2} = (- 4, 2, 4)

with magnitude

6

.

The numerator is

∣6 (- 4) + 1 (2) + (- 8) (4) ∣ = ∣ - 24 + 2 - 32∣ = 54

.

Thus

d = \frac{54}{6} = 9

JEE Main · 2023 · 10 Apr 2023 · Shift 1 · Single correct option

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