$\mathrm{A}(3,2,0), \mathrm{B}(5,3,2), \mathrm{C}(-9,6,-3)$ are three points forming a triangle. AD , the bisector of angle BAC meets BC in D. Find the co-ordinates of D.

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Aicharya · Accepted Answer

Since $A D$ is the bisector of $\angle B A C$. $\Rightarrow \frac{\mathrm{BD}}{\mathrm{DC}}=\frac{\mathrm{AB}}{\mathrm{AC}}$ ....(i) Now, $\mathrm{AB}=\sqrt{(5-3)^2+(3-2)^2+(2-0)^2}$ $=\sqrt{4+1+4}=\sqrt{9}=3$ $A C=\sqrt{(-9-3)^2+(6-2)^2+(-3-0)^2}$ $=\sqrt{144+16+9}=13$ $	herefore \quad \frac{\mathrm{BD}}{\mathrm{DC}}=\frac{3}{13}$ Hence, D divides BC in the ratio $3: 13$ The co-ordinates of D are $\left(\frac{3(-9)+13(5)}{3+13}, \frac{3(6)+13(3)}{3+13}, \frac{3(-3)+13(2)}{3+13}ight)$ $=\left(\frac{19}{8}, \frac{57}{16}, \frac{17}{16}ight)$

$\mathrm{A}(3,2,0), \mathrm{B}(5,3,2), \mathrm{C}(-9,6,-3)$ are three points forming a triangle. AD , the…

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