Question

$A(-1,2,-3),B(5,0,-6)$ and $C(0,4,-1)$ are the vertices of a triangle. The d.r's of the internal bisector of $\angle BAC$ are:

• A. $(-\frac{5}{2},-\frac{4}{5},-\frac{1}{2})$
• B. $(\frac{5}{2},-\frac{4}{5},-\frac{1}{2})$
• C. $(-\frac{5}{2},\frac{4}{5},\frac{1}{2})$
• D. $(\frac{5}{2},\frac{4}{5},\frac{1}{2})$

Answer 1

The correct option is D $(\frac{5}{2},\frac{4}{5},\frac{1}{2})$

$\text{d.r's of}AB=(5+1,0-2,-6+3)=(6,-2,-3)$
$\therefore AB=\sqrt{6^2+(-2{)}^2+(-3{)}^2}=\sqrt{49}=7$
$\text{d.r's of CA are}(0+1,4-2,-1+3)\equiv (1,2,2)$
Hence $AC=\sqrt{1^2+2^2+2^2}=\sqrt{9}=3$
The bisector of $\angle BAC$ will divide the side BC in the ratio $AB:AC$ i.e.,in the ratio $7:3$ internally.
Let the bisector of $\angle BAC$ , meets the side $BC$ at point $D$.
Therefore, $D$ divides $BC$ in the ratio $7:3$
Coordinates of $D$ are:
$(\frac{7\times 0+3\times 5}{7+3},\frac{7\times 4+3\times 0}{7+3},\frac{7\times (-1)+3\times (-6)}{7+3})$
$D\equiv (\frac{3}{2},\frac{14}{5},-\frac{5}{2})$
Therfore, d.r's of bisector $AD$ are:
$(\frac{3}{2}+1,\frac{14}{5}-2,-\frac{5}{2}+3)$
d.r's of bisector $AD\equiv (\frac{5}{2},\frac{4}{5},\frac{1}{2})$