Question

If $A(3,2,0),B(5,3,2)$ and $C(-9,6,-3)$ are the vertices of a $△ABC$ and the bisector of $\angle ABC$ meets $BC$ at $D$ , then coordinates of $D$ are

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

$AB=\sqrt{(5-3{)}^2+(3-2{)}^2+(2-0{)}^2}=\sqrt{4+1+4}=\sqrt{9}=3$

$AC=\sqrt{(3+9{)}^2+(6-2{)}^2+(0+3{)}^2}=\sqrt{144+16+9}=\sqrt{169}=13$

$\frac{BD}{DC}=\frac{AB}{AC}$ (By angle bisector therorm)

D divides BC in the rates $3:13=m:n$

$D=(\frac{m{x}_2+n{x}_1}{m+n},\frac{m{y}_2+ny}{m+n},\frac{m{z}_2+n{z}_1}{m+n})$d

$B(5,3,2)C(-9,6,-3)$

$D=(\frac{13\times 5-3\times 9}{16}\frac{13\times 3+3\times 6}{16},\frac{13\times 2+3(-3)}{16})$

$=(\frac{65-27}{16},\frac{39+18}{16},\frac{26-9}{16})$

$=(\frac{38}{16},\frac{57}{16},\frac{17}{16})=(\frac{19}{8},\frac{57}{16},\frac{17}{16})$