Consider a A B C where A(1,3,2), B(-2,8,0) and C(3,6,7).

Question

Consider a $	riangle A B C$ where $A(1,3,2), B(-2,8,0)$ and $C(3,6,7)$. If the angle bisector of $\angle B A C$ meets
the line $B C$ at $D$, then the length of the projection of the vector $\overrightarrow{A D}$

Accepted Answer

$D$ divides $B C$ in ratio $1: 1$

$$
D:\left(\frac{1}{2}, 7, \frac{7}{2}\right)
$$

$\begin{aligned} & \overrightarrow{A D}=\left(\frac{1}{2}-1\right) \hat{i}+(7-3) \hat{j}+\left(\frac{7}{2}-2\right) \hat{k} \\ & =-\frac{1}{2} \hat{i}+4 \hat{j}+\frac{3}{2} \hat{k} \\ & \overrightarrow{A C}=2 \hat{i}+3 \hat{j}+5 \hat{k}\end{aligned}$

Projection of $\overrightarrow{A D}$ on $\overrightarrow{A C}$

$$
=\frac{-1+12+\frac{15}{2}}{\sqrt{4+9+25}}=\frac{37}{2 \sqrt{38}}
$$

Solutions · 1