Question

If the position vectors of the vertices $A,B$ and $C$ of a $△ABC$ are respectively $4\hat{i}+7\hat{j}+8\hat{k},2\hat{i}+3\hat{j}+4\hat{k}$ and $2\hat{i}+5\hat{j}+7\hat{k}$, then the position vector of the point, where the bisector of $\angle A$ meets $BC$ is :

• A. $\frac{1}{2}(4\hat{i}+8\hat{j}+11\hat{k})$
• B. $\frac{1}{3}(6\hat{i}+13\hat{j}+18\hat{k})$
• C. $\frac{1}{4}(8\hat{i}+14\hat{j}+9\hat{k})$
• D. $\frac{1}{3}(6\hat{i}+11\hat{j}+15\hat{k})$

Answer 1

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

Let angular bisector of A meets side BC at point P(x, y, z)

By angular bisector theorem we can say that AB:AC=BP:PC

$\therefore BP:PC=c:b$

$\Rightarrow BP:PC=6:3=2:1=m:n$

$\Rightarrow P(x,y,z)=(\frac{m{x}_2+n{x}_1}{m+n},\frac{m{y}_2+n{y}_1}{m+n},\frac{m{z}_2+n{z}_1}{m+n})$

Here B=(2,3,4)=$(x_2,y_2,z_2)$

and C=(2,5,7)=$(x_3,y_3,z_3)$

Subtituting values, we get;

$P(x,y,z)=(\frac{\left(2\right)\left(2\right)+\left(1\right)\left(2\right)}{2+1},\frac{\left(2\right)\left(5\right)+\left(1\right)\left(3\right)}{2+1},\frac{\left(2\right)\left(7\right)+\left(1\right)\left(4\right)}{2+1})$

$P(x,y,z)=(\frac{6}{3},\frac{13}{3},\frac{18}{3})$

$\therefore$ Position vector of point $P=\frac{1}{3}(6i+13j+18k)$

Hence, the correct option is 'B'.