Question

$A(1,-1,-3),B(2,1,-2)\&C(-5,2,-6)$ are the position vectors of the vertices of a triangle ABC The
length of the bisector of its internal angle at A is:

• A. $\frac{\sqrt{10}}{4}$
• B. $\frac{3\sqrt{10}}{4}$
• C. $\sqrt{10}$
• D. None

Answer 1

The correct option is B $\frac{3\sqrt{10}}{4}$

Let BC be the base of $△ABC$ and E be the mid point of BC.

Given, $A(1,-1,3),B(2,1,-2),C(-5,2,-6)$

We use the distance formula: $d=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2+(z_2-z_1{)}^2}$

$AB=\sqrt{1+4+1}=\sqrt{6}$

$AC=\sqrt{36+5+9}=\sqrt{54}=3\sqrt{6}$

$\frac{AB}{AC}=\frac{\sqrt{6}}{3\sqrt{6}}=\frac{1}{3}$

$\frac{AB}{AC}=\frac{BE}{EC}\Rightarrow \frac{BE}{EC}=\frac{1}{3}$

$E=(\frac{-5+6}{4},\frac{2+3}{4},\frac{-6-6}{4})=(\frac{1}{4},\frac{5}{4},-3)$

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

$AE=\sqrt{\frac{9}{16}+\frac{81}{16}}=\sqrt{\frac{90}{16}}$

$\therefore AE=\frac{3\sqrt{10}}{4}$