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= - Sri Chaitanya Infinity Learn Best Online Courses for NCERT Solutions, CBSE, ICSE, JEE, NEET, Olympiad and Class 6 to 12

The vector along the internal angular bisector of $\bar{a}, \bar{b}$ is $(\frac{\bar{a}}{|a|} \pm \frac{\bar{b}}{|b|})$

Here $\bar{a} = \bar{A B} and \bar{b} = \bar{A C}$

It implies that $\bar{a} = 6 i - 2 j - 3 k and \bar{b} = i + 2 j + 2 k$

Hence, the vector along the internal angular bisector of $∠ B A C$ is $(\frac{\bar{a}}{|a|} + \frac{\bar{b}}{|b|}) = \frac{25 i + 8 j + 5 k}{21}$

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

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

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