Question

The position vectors of the vertices $A,B,C$ of a triangle are $\hat{i}-\hat{j}-3\hat{k}$, $2\hat{i}+\hat{j}-2\hat{k}$ and $-5\hat{i}+2\hat{j}-6\hat{k}$ respectively. The length of the bisector $AD$ of the $\angle BAC$ where $D$ is on the line segment $BC$, is

• A. $\frac{15}{2}$
• B. $\frac{1}{4}$
• C. $\frac{11}{2}$
• D. None of these

Answer 1

Answer: (D)

None of these

$A=\hat{i}-\hat{j}-3\hat{k}$,

$\overrightarrow{B}=2\hat{i}+\hat{j}-2\hat{k}$

$\overrightarrow{C}=-5\hat{i}+2\hat{j}-6\hat{k}$

position vector of

$\overrightarrow{D}=\frac{\overrightarrow{B}+\overrightarrow{C}}{2}$

$\overrightarrow{D}=\frac{2\hat{i}+\hat{j}-2\hat{k}-5\hat{i}+2\hat{j}-6\hat{k}}{2}$

$\overrightarrow{D}=\frac{-3}{2}\hat{i}+\frac{3}{2}\hat{j}-4\hat{k}$

$\overrightarrow{AD}=\overrightarrow{D}-\overrightarrow{A}$

$\overrightarrow{AD}=\frac{-3}{2}\hat{i}+\frac{3}{2}\hat{j}-4\hat{k}-\hat{i}+\hat{j}+3\hat{k}$

$\overrightarrow{AD}=\frac{-5}{2}\hat{i}+\frac{5}{2}\hat{j}-\hat{k}$

Length =magnitude of AD

$\left|\overrightarrow{AD}\right|=\sqrt{(\frac{-5}{2}{)}^2+(\frac{5}{2}{)}^2+1^2}$

$\left|\overrightarrow{AD}\right|=\sqrt{\left(\frac{25}{4}\right)+\left(\frac{25}{2}\right)+1}$

$\left|\overrightarrow{AD}\right|=\sqrt{\frac{54}{4}}$