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

• A

  $\frac{19}{8}, \frac{57}{15}, \frac{57}{15}$
• Correct answer
  B

  $\frac{19}{8}, \frac{57}{16}, \frac{17}{16}$
• C

  $(2,3,0)$
• D

  $(4,5,6)$

Solution

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

Hence, the correct answer is option: 'B'.