The position vectors of the vertices A,B and C of a triangle are 2i-3j+3k,2i+2j+3k and -i+…

Question

The position vectors of the vertices $A,B$ and $C$ of a triangle are $2\hat{i}-3\hat{j}+3\hat{k},2\hat{i}+2\hat{j}+3\hat{k}$ and $-\hat{i}+\hat{j}+3\hat{k}$ respectively. Let $l$ denotes the length of the angle bisector $AD$ of $\angleBAC$ where $D$ is on the line segment $BC$, then $2{l}^{2}$ equals :

Accepted Answer

$\begin{gathered} AB=5 \ AC=5 \end{gathered}$

$	herefore D$ is midpoint of $BC$

$\begin{gathered} D(\frac{1}{2},\frac{3}{2},3) \ 	herefore l=\sqrt{{(2-\frac{1}{2})}^{2}+{(-3-\frac{3}{2})}^{2}+(3-3{)}^{2}} \ l=\sqrt{\frac{45}{2}} \ 	herefore 2{l}^{2}=45 \end{gathered}$

Solutions · 1