Question

Let $A(5,4,6),B(1,-1,3)$ and $C(4,3,2)$ form $\Delta ABC.$ If the internal bisector of angle $A$ meets $BC$ in $D$, then the length of $AD$ 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 $\frac{3}{8}\sqrt{170}$

Given:
$A(5,4,6),B=(1,-1,3)$ and $C(4,3,2)$
Using distance formula:
$AB=\sqrt{(1-5{)}^2+(-1-4{)}^2+(3-6{)}^2}$
$=5\sqrt{2}$
$AC=\sqrt{(4-5{)}^2+(3-4{)}^2+(2-6{)}^2}$
$=3\sqrt{2}$
Clearly, $AB:AC=5:3$
Since bisector of $\angle BAC$ meets $BC$ in $D$
$AD$ is the bisector of $\angle BAC$ we have
$\Rightarrow BD:DC=AB:AC=5:3$
$D$ divides $BC$ in the ratio $5:3$
$\Rightarrow D=\frac{5C+3B}{8}=\frac{1}{8}(23,12,19)$
$\Rightarrow AD=\sqrt{{(\frac{23}{8}-5)}^2+{(\frac{12}{8}-4)}^2+{(\frac{19}{8}-6)}^2}=\sqrt{\frac{1530}{64}}=\frac{3}{8}\sqrt{170}$