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Let $\mathrm{A}(3,2,0), \mathrm{B}(5,3,2) \mathrm{C}(-9,6,-3)$ be three points forming a triangle. $\mathrm{AD}$, the bisector of $\angle \mathrm{BAC}$, meets $\mathrm{BC}$ in D. Find the coordinates of the point D.
Solution:
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Verified Answer
Note that
$$
\begin{aligned}
\mathrm{AB} &=\sqrt{(5-3)^2+(3-2)^2+(2-0)^2} \\
&=\sqrt{4+1+4}=3 \\
\mathrm{AC} &=\sqrt{(-9-3)^2+(6-2)^2+(-3-0)^2} \\
&=\sqrt{144+16+9}=13
\end{aligned}
$$
Since $\mathrm{AD}$ is the bisector of $\angle \mathrm{BAC}$, we have $\frac{\mathrm{BD}}{\mathrm{DC}}=\frac{\mathrm{AB}}{\mathrm{AC}}=\frac{3}{13}$
i.e., D divides $\mathrm{BC}$ in the ratio $3: 13$. Hence, the coordinates of $\mathrm{D}$ are
$$
\begin{array}{r}
\left(\frac{3(-9)+13(5)}{3+13}, \frac{3(6)+13(3)}{3+13}, \frac{3(-3)+13(2)}{3+13}\right) \\
=\left(\frac{19}{8}, \frac{57}{16}, \frac{17}{16}\right)
\end{array}
$$
$$
\begin{aligned}
\mathrm{AB} &=\sqrt{(5-3)^2+(3-2)^2+(2-0)^2} \\
&=\sqrt{4+1+4}=3 \\
\mathrm{AC} &=\sqrt{(-9-3)^2+(6-2)^2+(-3-0)^2} \\
&=\sqrt{144+16+9}=13
\end{aligned}
$$
Since $\mathrm{AD}$ is the bisector of $\angle \mathrm{BAC}$, we have $\frac{\mathrm{BD}}{\mathrm{DC}}=\frac{\mathrm{AB}}{\mathrm{AC}}=\frac{3}{13}$
i.e., D divides $\mathrm{BC}$ in the ratio $3: 13$. Hence, the coordinates of $\mathrm{D}$ are
$$
\begin{array}{r}
\left(\frac{3(-9)+13(5)}{3+13}, \frac{3(6)+13(3)}{3+13}, \frac{3(-3)+13(2)}{3+13}\right) \\
=\left(\frac{19}{8}, \frac{57}{16}, \frac{17}{16}\right)
\end{array}
$$
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