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, respectively, of triangle ABC, then the position vector of the point where the bisector of angle A meets BC is

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$\frac{1}{3} (5 \hat{j} + 12 \hat{k})$

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

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

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

answer is C.

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Detailed Solution

Suppose the bisector of angle A meets BC at D. Then AD divides BC in the ratio AB : AC.

So, P.V. of D is given by

$\frac{| \vec{AB} | (2 \hat{i} + 5 \hat{j} + 7 \hat{k}) + | \vec{AC} | (2 \hat{i} + 3 \hat{j} + 4 \hat{k})}{| \vec{AB} | + | \vec{AC} |}$

$\begin{array}{l} but \vec{AB} = - 2 \hat{i} - 4 \hat{j} - 4 \hat{k} \\ and \vec{AC} = - 2 \hat{i} - 2 \hat{j} - \hat{k} \\ \Rightarrow | \vec{AB} | = 6 and | \vec{AC} | = 3 \end{array}$

Therefore, P.V. of D is given by

$\begin{aligned} \frac{6 (2 \hat{i} + 5 \hat{j} + 7 \hat{k}) + 3 (2 \hat{i} + 3 \hat{j} + 4 \hat{k})}{6 + 3} \\ = \frac{1}{3} (6 \hat{i} + 13 \hat{j} + 18 \hat{k}) \end{aligned}$