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If the coordinates of the vertices of a $\triangle A B C$ are $A(7,6,4), B(5,4,6), C(3,2,0)$ and the bisector of $\angle B A C$ meets the side $B C$ at $D$, then the coordinates of $D$ are
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The correct answer is:
$\left(\frac{13}{3}, \frac{10}{3}, 4\right)$
$\because A D$ is the angle bisector of $\angle A$.
$\Rightarrow \quad \frac{A B}{A C}=\frac{B D}{C D}$
Now,
$\begin{aligned} & A B=\sqrt{4+4+4}=2 \sqrt{3} \\ & A C=\sqrt{16+16+16}=4 \sqrt{3}\end{aligned}$
$\Rightarrow \quad \frac{B D}{C D}=\frac{2 \sqrt{3}}{4 \sqrt{3}}=\frac{1}{2}$
$\Rightarrow D$ divides $B C$ in the ratio $1: 2$ internally.
$\Rightarrow \quad D \equiv\left(\frac{3+10}{3}, \frac{8+2}{3}, \frac{12+0}{3}\right)$
$\equiv\left(\frac{13}{3}, \frac{10}{3}, 4\right)$
$\Rightarrow \quad \frac{A B}{A C}=\frac{B D}{C D}$
Now,
$\begin{aligned} & A B=\sqrt{4+4+4}=2 \sqrt{3} \\ & A C=\sqrt{16+16+16}=4 \sqrt{3}\end{aligned}$
$\Rightarrow \quad \frac{B D}{C D}=\frac{2 \sqrt{3}}{4 \sqrt{3}}=\frac{1}{2}$
$\Rightarrow D$ divides $B C$ in the ratio $1: 2$ internally.
$\Rightarrow \quad D \equiv\left(\frac{3+10}{3}, \frac{8+2}{3}, \frac{12+0}{3}\right)$
$\equiv\left(\frac{13}{3}, \frac{10}{3}, 4\right)$
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