Question

The vertices of the triangle are A(5, 4, 6), B(1, –1, 3) and C(4, 3, 2). The internal bisector of angle A meets BC at D. Find the coordinates of D and the length AD.

Answer 1

AB = $\sqrt{4^2+5^2+3^2}=\sqrt{16+25+9}=\sqrt{50}=5\sqrt{2}$
AC = $\sqrt{1^2+1^2+4^2}=\sqrt{18}=3\sqrt{2}$
AD is the internal bisector of $\angle BAC$.
$\therefore \frac{BD}{DC}=\frac{AB}{AC}=\frac{5}{3}$
Thus, D divides BC internally in the ratio 5:3.
$$\begin{gathered} \therefore D\equiv \left(\frac{5\times 4+3\times 1}{5+3},\frac{5\times 3+3\left(-1\right)}{5+3},\frac{5\times 2+3\times 3}{5+3}\right) \\ \Rightarrow D\equiv \left(\frac{23}{8},\frac{12}{8},\frac{19}{8}\right) \\ \Rightarrow D\equiv \left(\frac{23}{8},\frac{3}{2},\frac{19}{8}\right) \\ \therefore AD=\sqrt{{\left(5-\frac{23}{8}\right)}^2+{\left(4-\frac{12}{8}\right)}^2+{\left(6-\frac{19}{8}\right)}^2} \\ =\sqrt{\frac{{17}^2+{20}^2+{29}^2}{8^2}} \\ =\sqrt{\frac{289+400+841}{8^2}} \\ =\frac{\sqrt{1530}}{8} \end{gathered}$$