Question

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

• A. $(\frac{25}{\sqrt{714}},\frac{-8}{\sqrt{714}},\frac{-5}{\sqrt{714}})$
• B. $(\frac{5}{\sqrt{74}},\frac{6}{\sqrt{74}},\frac{8}{\sqrt{74}})$
• C. $(\frac{25}{\sqrt{714}},\frac{8}{\sqrt{714}},\frac{5}{\sqrt{714}})$
• D. $(\frac{-5}{\sqrt{74}},\frac{6}{\sqrt{74}},\frac{-8}{\sqrt{74}})$

Answer 1

The correct option is C $(\frac{25}{\sqrt{714}},\frac{8}{\sqrt{714}},\frac{5}{\sqrt{714}})$

Here,

$\begin{array}{c}AB=\sqrt{{\left(5+1\right)}^2+{\left(0-2\right)}^2+{\left(-6+3\right)}^2}=\sqrt{49}=7\\ AC=\sqrt{{\left(0+1\right)}^2+{\left(4-2\right)}^2+{\left(1+3\right)}^2}=\sqrt{9}=3\end{array}$

by geometry , the bisector of$\angle BAC$

will divide the side BC in the ration 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

$\begin{array}{c}\left(\frac{7\times 0+3\times 5}{7+3},\frac{7\times 4+3\times 0}{7+3},\frac{7\times \left(-1\right)+3\times \left(-6\right)}{7+3}\right)\\ \left(\frac{3}{2},\frac{14}{5},-\frac{5}{3}\right)\end{array}$

Therefore, direction ratio of the bisector AD are

$\frac{3}{2}-\left(-1\right),\frac{14}{5}-2,\frac{-5}{2}+3i.e.\frac{5}{2},\frac{4}{5},\frac{1}{2}$

Hence , direction cosines of the bisector AD are

$\begin{array}{c}\frac{\frac{5}{2}}{\sqrt{{\left(\frac{5}{2}\right)}^2+{\left(\frac{4}{5}\right)}^2+{\left(\frac{1}{2}\right)}^2}}\\ \frac{\frac{4}{5}}{\sqrt{{\left(\frac{5}{2}\right)}^2+{\left(\frac{4}{5}\right)}^2+{\left(\frac{1}{2}\right)}^2}}\\ \frac{\frac{1}{2}}{\sqrt{{\left(\frac{5}{2}\right)}^2+{\left(\frac{4}{5}\right)}^2+{\left(\frac{1}{2}\right)}^2}}\\ \frac{25}{\sqrt{714}},\frac{8}{\sqrt{714}},\frac{5}{\sqrt{714}}\end{array}$