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

Question

Consider $	riangle ABC$ where $A(1,2,3)$, $B(-2,8,0)$ and $C(3,6,7)$. If the angle bisector of $\angle BAC$ meets line $BC$ at $D$, then the length of the projection of $\vec{AD}$ on $\vec{AC}$ is:

Accepted Answer

$\vec{AB}=(-3,6,-3)$, $|AB|=3\sqrt6$; $\vec{AC}=(2,4,4)$, $|AC|=6$. By the angle bisector theorem, $D$ divides $BC$ in ratio $BD:DC = AB:AC$.

The projection of $\vec{AD}$ on $\vec{AC}$ can be computed via $\vec{AD} = \vec{AB}+\dfrac{AB}{AB+AC}(\vec{BC})$ and taking the dot product with $\hat{AC}$.

*(Note: independent recomputation of this projection from the given coordinates did not exactly reproduce any listed option, suggesting a coordinate was altered by OCR corruption in the source; the source's stated answer is retained here.)*

Answer: A ($\frac{37}{2\sqrt{38}}$)

Solutions · 1