Question

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. The position vector of the point where the bisector of angle A meets BC is

• A. $\frac{2}{3}(-6\hat{i}-8\hat{j}-6\hat{k})$
• B. $\frac{2}{3}(6\hat{i}+8\hat{j}+6\hat{k})$
• C. $\frac{1}{3}(6\hat{i}+13\hat{j}+18\hat{k})$
• D. $\frac{1}{3}(5\hat{i}+12\hat{k})$

Answer 1

The correct option is C $\frac{1}{3}(6\hat{i}+13\hat{j}+18\hat{k})$

Suppose the bisector of angle A meets BC at D. Then, AD divides BC in the ratio AB : AC.
So, PV of D
$=\frac{|\overrightarrow{AB}|(2\hat{i}+5\hat{j}+7\hat{k})+|\overrightarrow{AC}|(2\hat{i}+3\hat{j}+4\hat{k})}{|\overrightarrow{AB}|+\overrightarrow{AC}}$
But, $\overrightarrow{AB}=-2\hat{i}-4\hat{j}-4\hat{k}$
and $\overrightarrow{AC}=-2\hat{i}-2\hat{j}-\hat{k}\Rightarrow |\overrightarrow{AB}|=6$ and $|\overrightarrow{AC}|=3$
$\therefore$ pv of $D=\frac{6(2\hat{i}+5\hat{j}+7\hat{k})+3(2\hat{i}+3\hat{j}+4\hat{k})}{6+3}=\frac{1}{3}(6\hat{i}+13\hat{j}+18\hat{k})$