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$If \bar{OA} = \bar{i} + 2 \bar{j} + 3 \bar{k} and \bar{OB} = 4 \bar{I} + \bar{K} are the position vectors of the points A and B, then the position vector of a point on the line passing through B and parallel to the vector \bar{OA} \times \bar{OB} which is at a distance of \sqrt{189} units from B is$

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$4 \bar{i} + 11 \bar{j} - 8 \bar{k}$

$- 2 \bar{i} - 11 \bar{j} + 8 \bar{k}$

$2 \bar{i} - 11 \bar{j} + 8 \bar{k}$

$6 \bar{i} + 11 \bar{j} - 7 \bar{k}$

answer is A.

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Detailed Solution

$\bar{OA} \times \bar{OB} = |\begin{matrix} i & j & k \\ 1 & 2 & 3 \\ 4 & 0 & 1 \end{matrix}| = i (2) - j (11) + k (- 8) = 2 i + 11 j - 8 k$

Eq. of line $\bar{r} = 4 i + k + t (2 i + 11 j - 8 k)$

Point on line $P (4 + 2 t, 11 t, 1 - 8 t)$

$\begin{array}{rcl} |\bar{PB}| & = & \sqrt{189} \Rightarrow {|\bar{PB}|}^{2} = 189 \\ 4 t^{2} + 121 t^{2} + 64 t^{2} & = & 189 \Rightarrow t = \pm 1 \\ ∴ Point P (6, 11, 7) (or) (2, - 11, 9) \end{array}$

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