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Q.

The shortest distance between the lines whose equations are $r = t (i + j + k), r = k + s (i - 2 j + 3 k)$ is

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a

$\frac{2}{\sqrt{13}}$

b

3

c

$\frac{3}{\sqrt{38}}$

d

$\sqrt{\frac{3}{14}}$

answer is B.

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Detailed Solution

$\begin{array}{rcl} a & = & 0, b = i + j + k \\ c & = & k, d = i - 2 j + 3 k \\ a - c & = & - k, b \times d = |\begin{matrix} i & j & k \\ 1 & 1 & 1 \\ 1 & - 2 & 3 \end{matrix}| \\ = & i (5) - j (2) + k (- 3) \\ S . D & = & |\frac{(a \cdot c) \cdot (b \times d)}{| b \times d |}| = \frac{3}{\sqrt{25 + 4 + 9}} = \frac{3}{\sqrt{38}} \end{array}$

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