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Q.

The shortest distance between the straight line passing through the point A = (6, 2, 2) and parallel to the vector (1, –2, 2) and the straight line passing through $A$ (–4, 0, –1) and parallel to the vector (3, –2, –2) is

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a

12

b

6

c

9

d

3

answer is C.

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Detailed Solution

$\begin{array}{rcl} a & = & (6, 2, 2) b = (1, - 2, 2) \\ c & = & (- 4, 0, - 1) d = (3, 2, - 2) \\ a - c & = & (10, 2, 3) b \times d = |\begin{matrix} i & j & k \\ 1 & - 2 & 2 \\ 3 & - 2 & - 2 \end{matrix}| \\ = & i (4 + 4) - j (- 2 - 6) + k (- 2 + 6) \\ = & 8 i + 8 j + 4 k \\ = & 4 (2 i + 2 j + k) \\ S . D & = & |\frac{a - c \cdot (b \times d)}{| b \times d |}| = |\frac{4 (20 + 4 + 3)}{4 \sqrt{4 + 4 + 1}}| \\ = & \frac{27}{3} = 9 \end{array}$

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