A(4,3,5), B (0,–2,2) and C (3,2,1) are three points. The coordinates of the point in which the bisector of $∠ BAC$ meets the side $BC$ is

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$(\frac{15}{8}, \frac{4}{8}, \frac{11}{8})$

$(\frac{12}{7}, \frac{2}{7}, \frac{10}{7})$

$(\frac{9}{5}, \frac{2}{5}, \frac{7}{5})$

$(\frac{3}{2}, 0, \frac{3}{2})$

answer is A.

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Detailed Solution

In a triangle $A B C$ , the internal angular bisector of $∠ A$ intersect the line $B C$ in the ratio $A B : A C$ internally.

The vertices of triangle $A B C$ are $A (4, 3, 5)$ , $B (0, - 2, 2)$ , $C (3, 2, 1)$

The lengths of the sides $A B, A C$ are as below

$A B = \sqrt{16 + 25 + 9} = \sqrt{50} = 5 \sqrt{2} A C = \sqrt{1 + 1 + 16} = \sqrt{18} = 3 \sqrt{2}$

The ratio

$\begin{array}{rcl} A B : A C & = & 5 \sqrt{2} : 3 \sqrt{2} \\ = & 5 : 3 \end{array}$

Hence, the coordinates of the point in which the bisector of $∠ BAC$ meets the side $BC$ is $\begin{array}{rcl} D (\frac{5 (3) + 3 (0)}{8}, \frac{5 (2) + 3 (- 2)}{8}, \frac{5 (1) + 3 (2)}{8}) & = & (\frac{15}{8}, \frac{4}{8}, \frac{11}{8}) \\ = & (\frac{15}{8}, \frac{1}{2}, \frac{11}{8}) \end{array}$