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 =\[\sqrt{4^2 + 5^2 + 3^2} = \sqrt{16 + 25 + 9} = \sqrt{50} = 5\sqrt{2}\]

\[\sqrt{1^2 + 1^2 + 4^2} = \sqrt{18} = 3\sqrt{2}\]

AD is the internal bisector of\[∠BAC\]

\[\therefore \frac{BD}{DC} = \frac{AB}{AC} = \frac{5}{3}\] 

Thus, D divides BC internally in the ratio 5:3.

\[\therefore D \equiv \left( \frac{5 \times 4 + 3 \times 1}{5 + 3}, \frac{5 \times 3 + 3\left( - 1 \right)}{5 + 3}, \frac{5 \times 2 + 3 \times 3}{5 + 3} \right)\]
\[ \Rightarrow D \equiv \left( \frac{23}{8}, \frac{12}{8}, \frac{19}{8} \right)\]
\[ \Rightarrow D \equiv \left( \frac{23}{8}, \frac{3}{2}, \frac{19}{8} \right) \]
\[ \therefore AD = \sqrt{\left( 5 - \frac{23}{8} \right)^2 + \left( 4 - \frac{12}{8} \right)^2 + \left( 6 - \frac{19}{8} \right)^2}\]
\[ = \sqrt{\frac{{17}^2 + {20}^2 + {29}^2}{8^2}}\]
\[ = \sqrt{\frac{289 + 400 + 841}{8^2}}\]
\[ = \frac{\sqrt{1530}}{8}\]