Question

If $4i+7j+8k,2i+7j+7k$ and $3i+5j+7k$ are the position vectors of the vertices $A,B$ and $C$ respectively of triangle $ABC$. The position vector of the point where the bisector of angle $A$ meets $BC.$

• A. $\frac{1}{3}(5j+12k)$
• B. $\frac{1}{3}(6i+13j+18k)$
• C. $\frac{2}{3}(6i+8j+6k)$
• D. $\frac{2}{3}(-6i-8j-6k)$

Answer 1

The correct option is B $\frac{1}{3}(6i+13j+18k)$

Suppose the bisector of angle $A$ meets $BC$ at $D$.

Then, $AD$ divides $BC$ in the ratio $AB:AC$

$\therefore$ position vector of $D$ $=\frac{\left(AC\right)(2i+3j+4k)+\left(AB\right)(2i+5j+7k)}{\left(AB\right)+\left(AC\right)}$

But $AB+-2i-4j-4k$ and $AC=-2i-2j-k$

$\therefore AB=\sqrt{4+16+11}=6$ and $AC=\sqrt{4+4+1}=3$

$\therefore$ P.V. of $D$ is $\frac{6(2i+5j+7k)+3(2i+3j+4k)}{6+3}$

$=\frac{18i+39j+54k}{9}=\frac{1}{3}(6i+13j+18k)$