Question

$A(1,-1,-3)$, $B(2,1,-2)$ & $C(-5,2,-6)$ are the position vectors of the vertices of a triangle ABC. The length of the bisector of its internal angle at A is?

• A. $\sqrt{10}/4$
• B. $3\sqrt{10}/4$
• C. $\sqrt{10}$
• D. None

Answer 1

The correct option is B $3\sqrt{10}/4$

$AB=\sqrt{{(2-1)}^2+{(1+1)}^2+{(-2+3)}^2}$

$=\sqrt{1+4+1}$

$=\sqrt{6}$

$AC=\sqrt{{(-5-1)}^2+{(2+1)}^2+{(-6+3)}^2}$

$=\sqrt{36+9+9}$

$=\sqrt{54}$

$=3\sqrt{6}$

$\frac{AB}{AC}=\frac{BE}{EC}(\because △AEB\sim △AEC\therefore \angle BAE=\angle EAC)$

$⟹\frac{BE}{EC}=\frac{AB}{AC}=\frac{\sqrt{6}}{3\sqrt{6}}=\frac{1}{3}$

Using section formula,

$E=(\frac{-5+2\left(3\right)}{1+3},\frac{2+3\left(1\right)}{1+3},\frac{-6+3(-2)}{4})$

$⟹E=(\frac{-5+6}{4},\frac{2+3}{4},\frac{-6-6}{4})$

$=(\frac{1}{4},\frac{5}{4},-3)$

$\therefore LengthAE=\sqrt{{(1-\frac{1}{4})}^2+{(-1-\frac{5}{4})}^2+{(-3+3)}^2}$

$=\sqrt{\frac{9}{16}+\frac{81}{16}+0}$

$=\sqrt{\frac{90}{16}}$

$=\sqrt{\frac{9\times 10}{16}}$

$=\frac{3\sqrt{10}}{4}$