A=−1,2,−3,B=5,0,−6,C=0,4,−1. If l,m,n be the direction cosines of the internal angular bisector of∠BAC, then l−m−n=

# $A=\left(-1,2,-3\right),B=\left(5,0,-6\right),C=\left(0,4,-1\right)$. If $l,m,n$ be the direction cosines of the internal angular bisector of$\angle BAC$, then $l-m-n=$

1. A

$\frac{28}{\sqrt{714}}$

2. B

$\frac{22}{\sqrt{714}}$

3. C

$\frac{38}{\sqrt{714}}$

4. D

$\frac{12}{\sqrt{714}}$

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### Solution:

The vector along the internal angular bisector of $\overline{a},\overline{b}$ is$\left(\frac{\overline{a}}{\left|a\right|}±\frac{\overline{b}}{\left|b\right|}\right)$

Here

It implies that

Hence, the vector along the internal angular bisector of  $\angle BAC$ is $\left(\frac{\overline{a}}{\left|a\right|}+\frac{\overline{b}}{\left|b\right|}\right)=\frac{25i+8j+5k}{21}$

Therefore, the direction cosines are$〈\frac{25}{\sqrt{714}},\frac{8}{\sqrt{714}},\frac{5}{\sqrt{714}}〉$

It implies that  $l-m-n=\overline{)\frac{12}{\sqrt{714}}}$

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