Question

A(1, 2, 3), B(0, 4, 1), C(-1, -1, -3) are the vertices of a triangle ABC.

Find the point in which the bisector of the angle $\angle BAC$ meets BC.

Answer 1

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

$=\sqrt{1+4+4}=3=\sqrt{9}=3$

AC = $\sqrt{(-2{)}^2+(-3{)}^2+(-6{)}^2}=\sqrt{4+9+36}=7$

=$\sqrt{49}$=7

AD is the internal bisector of $\angle BAC$

$\therefore \frac{BC}{DC}=\frac{AB}{AC}=\frac{3}{7}$

$\therefore D=(\frac{3(-1)+3\times 0}{3+7},\frac{3(-1)+7\times \left(4\right)}{3+7},\frac{3(-3)+7\times 1}{3+7})$

$\Rightarrow D\equiv (\frac{-3}{10},\frac{25}{10},\frac{-2}{10})$

$\Rightarrow D\equiv (\frac{-3}{10},\frac{5}{2},\frac{-1}{5})$