Question

$A(-1,2,3),B(5,0,-6),C(0,4,-1)$ are three points. Show that the direction cosines of the bisectors of $\angle BAC$ are proportional to $(25,8,5)$ and $(-11,20,23)$.

Answer 1

$A(-1,2,-3)B(5,0,-6)$

$C(0,4,-1)$

D.R's of AB are

$=(5-(-1),0-2,-6-(-3\left)\right)$

$=(6,-2,-3)$

D.R's of AC rule

$=(0-(-1),4-2,-1-(-3\left)\right)$

$=(1,2,2)$

DC's of AB are $\frac{6}{7},\frac{-2}{7},\frac{-3}{7}$

$[\because \sqrt{6^2+(-2{)}^2+(-3{)}^2}=7]$

DC's of AC are $\frac{1}{3},\frac{2}{3},\frac{2}{3}$

$[\because \sqrt{1^2+2^2+2^2}=3]$

DC's of Bisector of $\angle$ BAC is proportional to

$\frac{6}{7}\pm \frac{1}{3},\frac{-2}{7}\pm \frac{2}{3},\frac{-3}{7}\pm \frac{2}{3}$

$\frac{18\pm 7}{21},\frac{-6\pm 14}{21},\frac{-9\pm 14}{21}$

$(25,8,5)$ and $(11,-20,-23)$

i.e $(25,8,5)$ and $(-11,20,23)$