Question

If the vertices of a $\Delta ABC$ are $A=(1,-1,-3)$ , $B=(2,1,-2)$ and $C=(-5,2,-6)$ then the length of the internal bisector of angle $A$ is

• A. $\frac{3\sqrt{10}}{2}$
• B. $\frac{3\sqrt{10}}{5}$
• C. $\frac{3\sqrt{10}}{7}$
• D. $\frac{3\sqrt{10}}{4}$

Answer 1

Answer: (D)

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

$AD$ bisects the $BC$ in ratio of $\frac{|AB|}{|AC|}=\frac{|BD|}{|CD|}=\frac{\sqrt{6}}{\sqrt{54}}=\frac{1}{3}$

$\frac{BD}{DC}=\frac{1}{3}=\frac{c}{b}$

Coordinates of $D$ is $=\frac{1}{4}(3B+C)$ (section formula)

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

Length of $AD=\sqrt{(1-\frac{1}{4}{)}^2+(-1-\frac{5}{4}{)}^2+(-3+3{)}^2}$

$=\frac{\sqrt{90}}{4}$

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