Question

$A(5,4,6)$, $B(1,-1,3)$ and $C(4,3,2)$ form $△ABC$. 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 $\frac{3}{8}\sqrt{170}$

Consider the problem

The coordinates of $A,B,C$ are $(5,4,6),(1,-1,3),(4,3,2)$ respectively and angle bisector of $A$ meets $BC$ at $D$.

Then,

$\frac{BD}{CD}=\frac{AB}{AC}$

$\frac{BD}{CD}=\frac{\sqrt{(1-5{)}^2+(-1-4{)}^2+(3-6{)}^2}}{\sqrt{(4-5{)}^2+(3-4{)}^2+(2-6{)}^2}}=\frac{\sqrt{16+25+9}}{\sqrt{1+1+16}}=\frac{5}{3}$

$BD:CD=5:3$

Now using section formula coordinates of $D$ are

$(\frac{5\times 4+3\times 1}{5+3},\frac{5\times 3+3\times (-1)}{5+3},\frac{5\times 2+3\times 3}{5+3})$

$=(\frac{23}{8},\frac{3}{2},\frac{19}{8})$

$AD=\sqrt{(\frac{23}{8}-5{)}^2+(\frac{3}{2}-4{)}^2+(\frac{19}{8}-6{)}^2}$

$AD=\sqrt{{\left(\frac{23-40}{8}\right)}^2+{\left(\frac{3-8}{2}\right)}^2+{\left(\frac{19-48}{8}\right)}^2}$

$AD=\sqrt{{\left(\frac{-17}{8}\right)}^2+{\left(\frac{-5}{2}\right)}^2+{\left(\frac{-29}{8}\right)}^2}$

$=\sqrt{\frac{1530}{64}}$

$=\frac{3\sqrt{170}}{8}$ units

Hence, the length of $AD$ is $=\frac{3\sqrt{170}}{8}$ units