A unit vector perpendicular to the lines r¯=(−i¯−2j¯−k¯)+t(3i¯+j¯+2k¯) , r=(2i¯−2j¯+3k¯)+s(i¯+2j¯+3k¯) where s∈R,t∈R , is

− i ¯ − 7 j ¯ + 5 k ¯ 5 3

A unit vector perpendicular to the lines r¯=(−i¯−2j¯−k¯)+t(3i¯+j¯+2k¯) , r=(2i¯−2j¯+3k¯)+s(i¯+2j¯+3k¯) where s∈R,t∈R , is

− i ¯ − 7 j ¯ + 5 k ¯ 5 3

7 i ¯ − 7 j ¯ − k ¯ 3 11

− i ¯ − 7 j ¯ + 5 k ¯ 5 3

− i ¯ + 7 j ¯ + 7 k ¯ 3 11

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$A unit vector perpendicular to the lines \bar{r} = (- \bar{i} - 2 \bar{j} - \bar{k}) + t (3 \bar{i} + \bar{j} + 2 \bar{k}), r = (2 \bar{i} - 2 \bar{j} + 3 \bar{k}) + s (\bar{i} + 2 \bar{j} + 3 \bar{k}) where s \in R, t \in R, is$

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$\frac{- \bar{i} + 7 \bar{j} + 5 \bar{k}}{5 \sqrt{3}}$

$\frac{7 \bar{i} - 7 \bar{j} - \bar{k}}{3 \sqrt{11}}$

$\frac{- \bar{i} - 7 \bar{j} + 5 \bar{k}}{5 \sqrt{3}}$

$\frac{- \bar{i} + 7 \bar{j} + 7 \bar{k}}{3 \sqrt{11}}$

answer is C.

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Detailed Solution

$\begin{array}{l} |\begin{array}{lll} i & j & k \\ 3 & 1 & 2 \\ 1 & 2 & 3 \end{array}| = \bar{i} (- 1) - \bar{j} (7) + \bar{k} (5) = - \bar{i} - 7 \bar{j} + 5 \bar{k} \\ Unit vector = \frac{- \bar{i} - 7 \bar{j} + 5 \bar{k}}{\sqrt{1 + 49 + 25}} \end{array}$