Question

A (1,2,3) , B (0,4,1) ,C (-1,-1,-3) are the vertices of $\Delta$ABC and the bisector of angle A meets BC in D then D point is

• A. $(-3,5,-1)$
• B. $\left(10,\frac{2}{5},\frac{1}{2}\right)$
• C. $(\frac{-7}{10},\frac{5}{10},-\frac{18}{10})$
• D. $(3,-5,1)$

Answer 1

Answer: (C)

$(\frac{-7}{10},\frac{5}{10},-\frac{18}{10})$

Given $AD$ is the angular bisector of $\angle BAC$

$⟹\frac{BD}{DC}=\frac{AB}{AC}$

$AB=\sqrt{(1-0{)}^2+(2-4{)}^2+(3-1{)}^2}=\sqrt{1+4+4}=3$

$AC=\sqrt{(1+1{)}^2+(2+1{)}^2+(3+3{)}^2}=\sqrt{4+9+36}=7$

$⟹BD:DC=3:7$

$D=(\frac{3\left(0\right)+7(-1)}{3+7},\frac{3\left(4\right)+7(-1)}{3+7},\frac{3\left(1\right)+7(-3)}{3+7})=(\frac{-7}{10},\frac{5}{10},-\frac{18}{10})$