Question

The vertices of a triangle arc $A(5,4,6),B(1,-1,3)$ and $C(4,3,2)$. The internal bisector of $\angle BAC$ meets BC in D. Find $AD$.

Answer 1

Solution -

REF. Image

Vertices of triangle are $A(5,4,6),B(1,-1,3),C(4,3,2)$

$\overrightarrow{AB}=-4\hat{i}-5\hat{j}-3\hat{k}$

$\overrightarrow{AC}=-\hat{i}-\hat{j}-4\hat{k}$

$\left|\overrightarrow{AB}\right|=5\sqrt{2}$

$\left|\overrightarrow{AC}\right|=3\sqrt{2}$

$\left|\overrightarrow{AD}\right|=\left|\overrightarrow{AC}\right|cos\theta$ -(1)

$\overrightarrow{AB}.\overrightarrow{AC}=\left|\overrightarrow{AB}\right|\left|\overrightarrow{AC}\right|cos2\theta$

$4+5+12=\left(5\sqrt{2}\right)\left(3\sqrt{2}\right)cos2\theta$

$cos2\theta =\frac{21}{30}=\frac{7}{10}$

$cos2\theta =1-2co{s}^2\theta$

$cos\theta =\sqrt{\frac{1+cos2\theta }{2}}=\sqrt{\frac{1+\frac{7}{10}}{2}}=\frac{1}{2}\sqrt{\frac{17}{5}}$

$\left|\overrightarrow{AD}\right|=\left|\overrightarrow{AC}\right|cos\theta =\frac{3\sqrt{2}}{2}\sqrt{\frac{17}{5}}=\sqrt{\frac{153}{10}}$

$\overrightarrow{AD}=\sqrt{\frac{153}{10}}\left(\frac{\widehat{AB}+\widehat{AC}}{2}\right)=\sqrt{\frac{153}{40}}(\widehat{AB}+\widehat{AC})$

$\overrightarrow{AD}=\sqrt{\frac{153}{40}}[\frac{(-4\hat{j}-5\hat{j}-3\hat{k})}{5\sqrt{2}}+\frac{C-\hat{i}-\hat{j}-4\hat{k}}{3\sqrt{2}}]$

$\overrightarrow{AD}=\sqrt{\frac{153}{80}}\left[\frac{-17\hat{i}-20\hat{j}-29\hat{k}}{15}\right]$