Q. Let $A(3,2,0)$, $B(5,3,2)$, $C(-9,6, -3)$ be three points forming a triangle. $AD$, the bisector of $∠BAC$, meets $BC$ at $D$. Find the coordinates of the point $D$.

Introduction to Three Dimensional Geometry Report Error

A

$\left(\frac{19}{8}, \frac{57}{16}, \frac{-17}{16}\right)$

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B

$\left(\frac{-19}{8}, \frac{-57}{16}, \frac{17}{16}\right)$

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C

$\left(\frac{19}{8}, \frac{57}{16}, \frac{17}{16}\right)$

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D

None of these

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Solution:

$AB = \sqrt{\left(5-3\right)^{2}+\left(3-2\right)^{2}+\left(2-0\right)^{2}}$
$= \sqrt{4+1+4}=3$
$AC = \sqrt{\left(-9-3\right)^{2}+\left(6-2\right)^{2}+\left(-3-0\right)^{2}}$
$=\sqrt{144+16+9}=13$
Since $AD$ is the bisector of $∠BAC$, we have
$\frac{BD}{DC} = \frac{AB}{AC}=\frac{3}{13}$

($\because$ Internal bisector of angle of a triangle divides the opposite side in the ratio of sides containing the angle).
i.e., $D$ divides $BC$ in the ratio $3:13$.
Hence, the coordinates of $D$ are
$\left(\frac{3\left(-9\right)+13\left(5\right)}{3+13}, \frac{3\left(6\right)+13\left(3\right)}{3+13}, \frac{3\left(-3\right)+13\left(2\right)}{3+13}\right)$
$= \left(\frac{19}{8}, \frac{57}{16}, \frac{17}{16}\right)$.

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