JEE Main · 2023 · 06 Apr 2023 · Shift 1 · Three Dimensional Geometry

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One vertex of a rectangular parallelopiped is at the origin O and the lengths of its edges along x , y and z axes are 3, 4 and 5 units respectively. Let P be the vertex

(3, 4, 5)

. Then the shortest distance between the diagonal OP and an edge parallel to z axis, not passing through O or P is:

(A) $\frac{12}{5}$
(B) $\frac{12}{5 5}$
(C) $125$
(D) $\frac{12}{5}$ ✓ Correct

Answer: (D)

Solution

The diagonal

O P

has direction

(3, 4, 5)

through

O (0, 0, 0)

. Take the vertical edge joining

(3, 0, 0)

(3, 0, 5)

, parallel to the

z

-axis with direction

(0, 0, 1)

and passing through neither

O

nor

P

. For two skew lines,

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

with

a_{1} = (0, 0, 0)

a_{2} = (3, 0, 0)

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

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

. Here

b_{1} \times b_{2} = (4, - 3, 0)

, so

∣ b_{1} \times b_{2} ∣ = 5

and

(a_{2} - a_{1}) \cdot (4, - 3, 0) = 12

. Therefore the shortest distance is

\frac{12}{5}

JEE Main · 2023 · 06 Apr 2023 · Shift 1 · Single correct option

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JEE Main · 2023 · 06 Apr 2023 · Shift 1 · Three Dimensional Geometry

Solution

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