JEE Main · 2023 · 08 Apr 2023 · Shift 1 · Three Dimensional Geometry

The shortest distance between the lines

\frac{x - 4}{4} = \frac{y + 2}{5} = \frac{z + 3}{3}

and

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

Answer: (A)

Solution

The lines pass through

a_{1} = (4, - 2, - 3)

with direction

b_{1} = (4, 5, 3)

, and through

a_{2} = (1, 3, 4)

with direction

b_{2} = (3, 4, 2)

. For skew lines the shortest distance is

d = \frac{( a _{2} - a _{1} ) \cdot ( b _{1} \times b _{2} )}{b _{1} \times b _{2}}

.

Here

b_{1} \times b_{2} = \hat{i} 43 \hat{j} 54 \hat{k} 32 = (- 2, 1, 1), ∣ b_{1} \times b_{2} ∣ = 4 + 1 + 1 = 6 .

With

a_{2} - a_{1} = (- 3, 5, 7)

, the numerator is

(- 3) (- 2) + 5 (1) + 7 (1) = 18

.

Therefore

d = \frac{18}{6} = 36

, so the shortest distance between the two lines is

36 .

JEE Main · 2023 · 08 Apr 2023 · Shift 1 · Single correct option