Question

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

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

Answer 1

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

$|AB|=\sqrt{1+4+1}=\sqrt{6},|BC|=\sqrt{49+1+16}=\sqrt{66}$ and $|AC|=\sqrt{36+9+9}=\sqrt{54}$

Let the angle bisector through $A$ meet $BC$ at point $D.D$ divides $BC$ such that $BD:DC=AB:AC=1:3$

Thus, $|BD|=\frac{\sqrt{66}}{4}$ and $|DC|=3\frac{\sqrt{66}}{4}$

Applying cosine rule, we have $\frac{A{B}^2+A{D}^2-B{D}^2}{2AB.AD}=\frac{A{C}^2+A{D}^2-D{C}^2}{2AC.AD}$

$\Rightarrow 3(A{B}^2+A{D}^2-B{D}^2)=A{C}^2+A{D}^2-D{C}^2$
$\Rightarrow A{D}^2=\frac{A{C}^2-D{C}^2-3A{B}^2+3B{D}^2}{2}$
$A{D}^2=\frac{54-37.125-18+12.375}{2}=5.625=\frac{90}{16}$
$\Rightarrow AD=\frac{3}{4}\sqrt{10}$