If the position vectors of the vertices $A, B$ and $C$ of a $\triangle \mathrm{ABC}$ are respectively $4…
If the position vectors of the vertices $A, B$ and $C$ of a $\triangle \mathrm{ABC}$ are respectively $4 \hat{i}+7 \hat{j}+8 \hat{k}, 2 \hat{i}+3 \hat{j}+4 \hat{k}$ and $2 \hat{i}+5 \hat{j}+7 \hat{k}$, then the position vector of the point, where the bisector of $\angle A$ meets $B C$ is
$\frac{1}{2}(4 \hat{i}+8 \hat{j}+11 \hat{k})$
$\frac{1}{3}(6 \hat{i}+13 \hat{j}+18 \hat{k})$
$\frac{1}{4}(8 \hat{i}+14 \hat{j}+9 \hat{k})$
$\frac{1}{3}(6 \hat{i}+11 \hat{j}+15 \hat{k})$
Solution
Suppose angular bisector of $A$ meets $B C$ at $\mathrm{D}(x, y, z)$
Using angular bisector theorem,
$
\begin{aligned}
\frac{A B}{A C} &=\frac{B D}{D C} \\
\frac{B D}{D C} &=\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{aligned}
$
$
D(x, y, z)=\left(\frac{6}{3}, \frac{13}{3}, \frac{18}{3}\right)
$
Therefore, position vector of point
$
P=\frac{1}{3}(6 i+13 j+18 k)
$
