Question

$A(1,-1,-3),B(2,1,-2),C(-5,2,-6)$ are the position vectors of the vertices of a triangle ABC. The length of the bisector of its internal angle at A is :

• A. $\frac{3\sqrt{10}}{2}$
• B. $\frac{3\sqrt{10}}{4}$
• C. $\frac{3\sqrt{10}}{5}$
• D. $\frac{3\sqrt{10}}{7}$

Answer 1

The correct option is B $\frac{3\sqrt{10}}{4}$

Given: $A(1,-1,-3),B(2,1,-2),C(-5,2,-6)$
$\overrightarrow{AB}=\overrightarrow{B}-\overrightarrow{A}=(2\hat{i}+\hat{j}-2\hat{k})-(\hat{i}-\hat{j}-3\hat{k})=\hat{i}+2\hat{j}+\hat{k}$
$\overrightarrow{AC}=\overrightarrow{C}-\overrightarrow{A}=(-5\hat{i}+2\hat{j}-6\hat{k})-(\hat{i}-\hat{j}-3\hat{k})=-6\hat{i}+3\hat{j}-3\hat{k}$
$|\overrightarrow{AB}|=\sqrt{1+4+1}=\sqrt{6}$
$|\overrightarrow{AC}|=\sqrt{36+9+9}=\sqrt{54}=3\sqrt{6}$
Suppose the bisector of angle $A$ meets $\overrightarrow{BC}$ at $D$.
Then $\overrightarrow{AD}$ divides $\overrightarrow{BC}$ in the ratio $\overrightarrow{AB}:\overrightarrow{AC}$
$\therefore P.V.ofD=\frac{|\overrightarrow{AC}|(2\hat{i}+\hat{j}-2\hat{k})+|\overrightarrow{AB}|(-5\hat{i}+2\hat{j}-6\hat{k})}{|\overrightarrow{AB}|+|\overrightarrow{AC}|}$
$\therefore P.V.ofD=\frac{3\sqrt{6}(2\hat{i}+\hat{j}-2\hat{k})+\sqrt{6}(-5\hat{i}+2\hat{j}-6\hat{k})}{\sqrt{6}+3\sqrt{6}}$
$\therefore P.V.ofD=\frac{3(2\hat{i}+\hat{j}-2\hat{k})+(-5\hat{i}+2\hat{j}-6\hat{k})}{4}$
$\therefore P.V.ofD=\frac{\hat{i}+5\hat{j}-12\hat{k}}{4}$
Hence $\overrightarrow{AD}=\overrightarrow{OD}-\overrightarrow{OA}=\frac{\hat{i}+5\hat{j}-12\hat{k}}{4}-(\hat{i}-\hat{j}-3\hat{k})$
$\overrightarrow{AD}=\frac{1}{4}(-3\hat{i}+9\hat{j})$
Length of Angle Bisector will be: $\frac{1}{4}\sqrt{9+81+0}=\frac{\sqrt{90}}{4}=\frac{3\sqrt{10}}{4}$