# 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

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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.

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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