Mathematics - Pair of Lines Question with Solution | TestHub

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