Mathematics - Vector Question with Solution | TestHub

Suppose angular bisector of $A$ meets $BC$ at $\mathrm{D}(x,y,z)$ Using angular bisector theorem, $$\begin{array}{rl}\frac{AB}{AC}& =\frac{BD}{DC}\\ \frac{BD}{DC}& =\frac{\sqrt{(4-2{)}^2+(7-3{)}^2+(8-4{)}^2}}{\sqrt{(4-2{)}^2+(7-5{)}^2+(8-7{)}^2}}\\ & =\frac{\sqrt{2^2+4^2+4^2}}{\sqrt{2^2+2^2+1^2}}=\frac{6}{3}=2\end{array}$$
$$D(x,y,z)=\left(\frac{6}{3},\frac{13}{3},\frac{18}{3}\right)$$ Therefore, position vector of point $$P=\frac{1}{3}(6i+13j+18k)$$