Let $\vec{AD}$ be the angle bisector of ∠A of ΔABC such that $\vec{AD} = α \vec{AB}+β \vec{AC}$, then
Answer & explanation
Correct answer: option 3
Clearly, AD divides BC in the ratio AB : AC.
$∴\vec{AD}=\frac{|\vec{AB}|\vec{AC}+|\vec{AC}|\vec{AB}}{|\vec{AB}|+|\vec{AC}|}$
$⇒\vec{AD}=α \vec{AB}+β \vec{AC}$, where
$α=\frac{|\vec{AC}|}{|\vec{AB}|+|\vec{AC}|}$ and $β=\frac{|\vec{AB}|}{|\vec{AB}|+|\vec{AC}|}$