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 $A B C$, then the position vector of the point where the bisector of angle $A$ meets $B C$ 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 $B C$ at $D$. Then $A D$ divides $B C$ in the ratio $A B: A C$.
So, P.V. of $D$ is given by
$\frac{|\overrightarrow{A B}|(2 \hat{i}+5 \hat{j}+7 \hat{k})+|\overrightarrow{A C}|(2 \hat{i}+3 \hat{j}+4 \hat{k})}{|\overrightarrow{A B}|+|\overrightarrow{A C}|}$
But $\overrightarrow{A B}=-2 \hat{i}-4 \hat{j}-4 \hat{k}$
and $\overrightarrow{A C}=-2 \hat{i}-2 \hat{j}-\hat{k}$
$\Rightarrow |\overrightarrow{A B}|=6$ and $|\overrightarrow{A C}|=3$
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} $
$=\frac{1}{3}(6 \hat{i}+13 \hat{j}+18 \hat{k})$