Given: The vertices of the triangle are A(5, 4, 6), B(1, -1, 3) and C(4, 3, 2)
To find: the coordinates of D and the length AD
Formula used:
Distance Formula: The distance between any two points (a, b, c) and (m, n, o) is given by,
\(\sqrt{(a-m)^2+(b-n)^2+(c-o)^2}\)
Section Formula:
A line AB is divided by C in m:n where A(x, y, z) and B(a, b, c).
The coordinates of C is given by,
\(\Big(\frac{nx+ma}{m+n},\frac{ny+mb}{m+n},\frac{nz+mc}{m+n}\Big)\)
We know angle bisector divides opposite side in the ratio of the other two sides.
As AD is angle bisector of A and meets BC at D
⇒ BD : DC = AB : BC
Distance between A(5, 4, 6) and B(1, -1, 3) is AB,
The distance between A(5, 4, 6) and C(4, 3, 2) is AC,
⇒ BD: DC = 5:3
Therefore, m = 5 and n = 3
B(1, -1, 3) and C(4, 3, 2)
Coordinates of D using section formula:
= \(\Big(\frac{3(1)+5(4)}{5+3},\frac{3(-1)+5(3)}{5+3},\frac{3(3)+5(2)}{5+3}\Big)\)
= \(\Big(\frac{3+20}{8}, \frac{-3+15}{8},\frac{9+10}{8}\Big)\)
= \(\Big(\frac{23}{8},\frac{12}{8},\frac{19}{8}\Big)\)
= \(\Big(\frac{23}{8},\frac{3}{2},\frac{19}{8}\Big)\)
The distance between A(5, 4, 6) and D \(\Big(\frac{23}{8},\frac{3}{2},\frac{19}{8}\Big)\) is AD,
= \(\sqrt{\frac{1530}{64}}\)
= \(\sqrt{\frac{765}{32}}\) units
Hence, Coordinates of D are \(\Big(\frac{23}{8},\frac{3}{2},\frac{19}{8}\Big)\) and the length of AD is \(\sqrt{\frac{765}{32}}\) units
