JEE Main · 2024 · 30 Jan 2024 · Shift 1 · Three Dimensional Geometry

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d_{1}

is the shortest distance between the lines

x + 1 = 2 y = - 12 z

x = y + 2 = 6 z - 6

and

d_{2}

is the shortest distance between the lines

\frac{x - 1}{2} = \frac{y + 8}{- 7} = \frac{z - 4}{5}

\frac{x - 1}{2} = \frac{y - 2}{1} = \frac{z - 6}{- 3}

, then the value of

\frac{32 3 d _{1}}{d _{2}}

is:

Answer: 16

Solution

For

L_{1} : \frac{x + 1}{1} = \frac{y}{1/2} = \frac{z}{- 1/12}

and

L_{2} : \frac{x}{1} = \frac{y + 2}{1} = \frac{z - 1}{1/6}

, the shortest-distance formula gives

d_{1} = 2

.

For

L_{3} : \frac{x - 1}{2} = \frac{y + 8}{- 7} = \frac{z - 4}{5}

and

L_{4} : \frac{x - 1}{2} = \frac{y - 2}{1} = \frac{z - 6}{- 3}

, it gives

d_{2} = \frac{12}{3}

\frac{32 3 d _{1}}{d _{2}} = \frac{32 3 \times 2}{12/ 3} = \frac{64 3 \cdot 3}{12} = \frac{192}{12} = 16

JEE Main · 2024 · 30 Jan 2024 · Shift 1 · Integer answer

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