The position vectors of the vertices A, B, C of a ΔABC are $\hat i-\hat j-3\hat k, 2\hat i+\hat j-2\hat k$ and $-5\hat i+2\hat j-6\hat k$ respectively. The length of the bisector AD of the angle ∠BAC where D is on the line segment BC, is

none of these

We have,

$\vec{AB}=\hat i+2\hat j+\hat k,\vec{AC}=-6\hat i+3\hat j-3\hat k$

$⇒|\vec{AB}|=\sqrt{6}$ and $|\vec{AC}|=3\sqrt{6}$

Clearly, point D divides BC in the ratio AB : AC i.e. 1:3.

∴ Position vector of D $=\frac{(-5\hat i+2\hat j-6\hat k)+3(2\hat i+\hat j-2\hat k)}{1+3}$

⇒ Position vector of D = $\frac{1}{4}(\hat i+5\hat j-12\hat k)$

$∴\vec{AD}=\frac{1}{4}(\hat i+5\hat j-12\hat k)-(\hat i-\hat j-3\hat k)=\frac{3}{4}(-\hat i+3\hat j)$

$⇒AD=|\vec{AD}|=\frac{3}{4}\sqrt{10}$