Question

A(3,2,0) , B(5,3,2) and C(-9,6,-3) are three points joining a triangle and AD is bisector of the angle $\angle$ BAC. AD meets BC at the point

• A. $(\frac{19}{8},\frac{-57}{16},\frac{17}{16})$
• B. $(\frac{-19}{8},\frac{57}{16},\frac{17}{16})$
• C. $(\frac{19}{8},\frac{57}{16},\frac{17}{16})$
• D. None of these

Answer 1

The correct option is C

$(\frac{19}{8},\frac{57}{16},\frac{17}{16})$

From properties of triangle

$\frac{AB}{AC}=\frac{BD}{DC}AB=\sqrt{(5-3{)}^2+(3-2{)}^2+(2-0{)}^2}=3AC=\sqrt{(-9-3{)}^2+(6-2{)}^2+(3-0{)}^2}=13$
$\therefore \frac{AB}{AC}=\frac{3}{13}=\frac{BD}{DC}$ i.e. D divided BC in ratio 3:13
Let D be $(x_1,y_1,z_1)$
Using section formula
$x_1=\frac{3(-9)+13\left(5\right)}{3+13}=\frac{19}{8}{y}_1=\frac{3\left(6\right)+13\left(3\right)}{3+13}=\frac{57}{16}{z}_1=\frac{3(-3)+13\left(2\right)}{3+13}=\frac{17}{16}\therefore \text{D}is(\frac{19}{8},\frac{57}{16},\frac{17}{16})$