Question

Let vertices of a $△ABC$ are $A(-3,2,0),B(-2,0,2)$ and $C(1,-1,0).$ If $D$ is a point on $BC$ and $AD$ bisects the angle $\angle BAC,$ then point $D$ is

• A. $(-\frac{7}{8},-\frac{3}{8},-\frac{10}{8})$
• B. $(\frac{7}{8},\frac{3}{8},-\frac{10}{8})$
• C. $(-\frac{7}{8},-\frac{3}{8},\frac{10}{8})$
• D. $(\frac{7}{8},\frac{3}{8},\frac{10}{8})$

Answer 1

The correct option is C $(-\frac{7}{8},-\frac{3}{8},\frac{10}{8})$

Given : vertices $A(-3,2,0),B(-2,0,2)$ and $C(1,-1,0)$
$\therefore AB=\sqrt{(-3+2{)}^2+(2-0{)}^2+(0-2{)}^2}=3$
and $AC=\sqrt{(-3-1{)}^2+(2+1{)}^2}=5$
As, $AD$ is the angle bisector : $AB:AC=BD:DC=3:5$
So, $D\equiv (\frac{3-10}{3+5},\frac{-3+0}{3+5},\frac{0+10}{3+5})$
$\Rightarrow D\equiv (-\frac{7}{8},-\frac{3}{8},\frac{10}{8})$