JEE Main · 2020 · 08 Jan 2020 · Shift 1 · Three Dimensional Geometry

The shortest distance between the lines

\frac{x - 3}{3} = \frac{y - 8}{- 1} = \frac{z - 3}{1}

and

\frac{x + 3}{- 3} = \frac{y + 7}{2} = \frac{z - 6}{4}

is:

(A) $\frac{7}{2} 30$
(B) $330$ ✓ Correct
(C)3
(D) $230$

Answer: (B)

Solution

The two lines have direction vectors

p = 3 \hat{i} - \hat{j} + \hat{k}

and

q = - 3 \hat{i} + 2 \hat{j} + 4 \hat{k}

and pass through

A (3, 8, 3)

and

B (- 3, - 7, 6)

. Their cross product is

p \times q = - 6 \hat{i} - 15 \hat{j} + 3 \hat{k}

, and the joining vector is

A B = - 6 \hat{i} - 15 \hat{j} + 3 \hat{k}

. The shortest distance is

d = \frac{∣ A B \cdot ( p \times q ) ∣}{∣ p \times q ∣} = \frac{∣36 + 225 + 9∣}{36 + 225 + 9} = \frac{270}{270} = 270 = 330

JEE Main · 2020 · 08 Jan 2020 · Shift 1 · Single correct option

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