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If $4 \hat{i} + 7 \hat{j} + 8 \hat{k}$ , $2 \hat{i} + 3 \hat{j} + 4 \hat{k}$ and $2 \hat{i} + 5 \hat{j} + 7 \hat{k}$ are the position vectors of the vertices $A, B and C$ of triangle $ABC$ , the position vector of the point where the bisector $A$ meets $BC$ is

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$\frac{2}{3} (6 \hat{i} + 8 \hat{j} + 6 \hat{k})$

$\frac{1}{3} (6 \hat{i} + 13 \hat{j} + 18 \hat{k})$

$2 (\hat{i} + \hat{j} + \hat{k})$

$\frac{2}{3} (- 6 \hat{i} - 8 \hat{j} - 6 \hat{k})$

answer is C.

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Detailed Solution

$\bar{B D} : \bar{D C} = \bar{A B} : \bar{A C}$

$= 6 : 3 = 2 : 1$

$\vec{O D} = \frac{(4 \hat{i} + 10 \hat{j} + 14 \hat{k}) + (2 \hat{i} + 3 \hat{j} + 4 \hat{k})}{3}$

$= \frac{6 \hat{i} + 13 \hat{j} + 18 \hat{k}}{3}$

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