Question

The points $(3,2,0),(5,3,2)$ and $(-9,6,-3)$, are the vertices of a triangle $ABC.AD$ is the internal bisector of $\angle BAC$ which meets $BC$ at $D$. Then the co-ordinates of $D$, are

• A. $\left[\frac{17}{16},\frac{57}{16},\frac{19}{8}\right]$
• B. $\left[\frac{19}{8},\frac{57}{16},\frac{17}{16}\right]$
• C. $[0,0,\frac{17}{16}]$
• D. $[\frac{17}{16},0,0]$

Answer 1

The correct option is B $\left[\frac{19}{8},\frac{57}{16},\frac{17}{16}\right]$

$AD$ is bisector of $\angle BAC$

$\Rightarrow$ Ratio at $D$ is $c:b$ where

$c=AB=\sqrt{(3-5{)}^2+(2-3)+(0-2{)}^2}$

$=\sqrt{4+1+4}=\sqrt{9}=3$

$b=AC=\sqrt{(3+9{)}^2+(2-6{)}^2+(0+3{)}^2}$

$=\sqrt{144+16+9}=\sqrt{169}=13$

for point $D$

$x=\frac{c(-9)+b\left(6\right)}{c+b}=\frac{3(-9)+13\left(5\right)}{3+13}=\frac{38}{16}=\frac{19}{8}$

$y=\frac{c\left(6\right)+b\left(3\right)}{c+b}=\frac{3\left(6\right)+13\left(3\right)}{3+13}=\frac{57}{16}$

$z=\frac{c(-3)+b\left(2\right)}{c+b}=\frac{3(-3)+13\left(2\right)}{3+13}=\frac{17}{16}$

Hence, point $D$ is $[\frac{19}{8},\frac{57}{16},\frac{17}{6}]$