Question

$A(5,4,6),B=(1,-1,3)$ and $C(4,3,2)$ form $\Delta$ABC. If the internal bisector of angle A meets BC in D, then the length of $\overline{AD}$ is?

• A. $\frac{1}{8}\sqrt{170}$
• B. $\frac{3}{8}\sqrt{170}$
• C. $\frac{5}{8}\sqrt{170}$
• D. $\frac{7}{8}\sqrt{170}$

Answer 1

The correct option is B $\frac{3}{8}\sqrt{170}$

ANSWER

$A(5,4,6),B(1.-1,3)andC(4,3,2)$ are the vertices of triangle ABC

Using distance formula

$AB=\sqrt{(1-5{)}^2+(-1-4{)}^2+(3-6{)}^2}$

​$=\sqrt{16+25+9}$

$=\sqrt{50}=5\sqrt{2}$

$AC=\sqrt{(4-5{)}^2+(3-4{)}^2+(2-6{)}^2}$

$=\sqrt{1+1+16}=\sqrt{18}=3\sqrt{2}$

Since bisector of $\angle BAC$ meets $BC$ in $D$

$AD$ is the bisector of $\angle BAC$ we have

$\frac{BD}{DC}=\frac{AB}{AC}=\frac{5\sqrt{2}}{3\sqrt{2}}$

$D$ divides $BC$ in the ratio $5:3$.

Hence the coordinates of $D=(\frac{20+3}{8},\frac{15-3}{8},\frac{10+9}{8})$

$=(\frac{23}{8},\frac{12}{8},\frac{19}{8})$

Distance $AD=\sqrt{(\frac{23}{8}-5{)}^2+(\frac{12}{8}-4{)}^2+(\frac{19}{8}-6{)}^2}$

$⟹AD=\sqrt{\frac{289}{64}+\frac{400}{64}+\frac{841}{64}}$

$⟹AD=\sqrt{\frac{1530}{64}}$

$optionBiscorrect$