Vector Algebra
Angle Bisector & Projection
nta_pyq_2024_jan
Grade 12
Question:
Consider a $\triangle ABC$ where $A(1,2,3)$, $B(-2,8,0)$ and $C(3,6,7)$. If the angle bisector of $\angle BAC$ meets the line $BC$ at $D$, then the length of the projection of the vector $\overrightarrow{AD}$ on the vector $\overrightarrow{AC}$ is:
$\dfrac{37}{2\sqrt{38}}$
$\dfrac{\sqrt{38}}{2}$
$\dfrac{39}{2\sqrt{38}}$
$\sqrt{19}$
Step-by-Step Solution
Key Concept: Compute $AB=\sqrt{9+25+4}=\sqrt{38}$ and $AC=\sqrt{4+9+25}=\sqrt{38}$. Since $AB=AC$, the angle bisector from $A$ meets $BC$ at its midpoint $D$. Then $\overrightarrow{AD}=\frac{1}{2}(\overrightarrow{AB}+\overrightarrow{AC})$. Projection of $\overrightarrow{AD}$ on $\overrightarrow{AC}=\dfrac{\overrightarrow{AD}\cdot\overrightarrow{AC}}{|\overrightarrow{AC}|}$.
$\overrightarrow{AB}=(-3,6,-3)$, $\overrightarrow{AC}=(2,3,5)$, $|AB|=|AC|=\sqrt{38}$. $D$ is midpoint of $BC$: $D=\left(\frac{1}{2},7,\frac{7}{2}\right)$. $\overrightarrow{AD}=\left(-\frac{1}{2},5,\frac{1}{2}\right)=\frac{1}{2}(-1,10,1)$. Projection $=\dfrac{\overrightarrow{AD}\cdot\overrightarrow{AC}}{|\overrightarrow{AC}|}=\dfrac{\frac{1}{2}(-2+30+5)}{\sqrt{38}}=\dfrac{37}{2\sqrt{38}}$.
Correct Answer: 1