The length of the internal bisector of angle $A$ in $\triangle \mathrm{ABC}$ with vertices $A(4,7,8), B(2,3…
The length of the internal bisector of angle $A$ in $\triangle \mathrm{ABC}$ with vertices $A(4,7,8), B(2,3,4)$ and $C(2,5,7)$ is
$\frac{1}{3} \sqrt{29}$
$\frac{2}{3} \sqrt{29}$
$\frac{2}{3} \sqrt{34}$
$\frac{4}{3} \sqrt{34}$
Solution
Let $A D$ be the bisector of angle $A$ met $B C$ at $D$.
Then $D$ divides $B C$ in ratio $A B: A C$
$\begin{aligned}
& A B=\sqrt{4+16+16}=6, A C=\sqrt{4+4+1}=3 \\
& \therefore D=\left(\frac{6+12}{6+3}, \frac{9+30}{6+3}, \frac{12+42}{6+3}\right)=\left(2, \frac{13}{3}, 6\right) \\
& A D=\sqrt{(-2)^2+\left(-\frac{8}{3}\right)^2+(-2)^2}=\sqrt{\frac{136}{9}}=\frac{2 \sqrt{34}}{3} .
\end{aligned}$