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 $ABC$, then 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. $\frac{1}{3}(5\hat{j}+12\hat{k})$

Answer 1

Answer: (C)

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
$\frac{|\overrightarrow{AB}|(2\hat{i}+5\hat{j}+7\hat{k})+|\overrightarrow{AC}|(2\hat{i}+3\hat{j}+4\hat{k})}{|\overrightarrow{AB}|+|\overrightarrow{AC}|}$
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, P.V, of $D$ is given by
$\frac{6(2\hat{i}+5\hat{j}+7\hat{k})+3(2\hat{i}+3\hat{j}+4\hat{k})}{6+3}$
$=\frac{1}{3}(6\hat{i}+13\hat{j}+18\hat{k})$