Question

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

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

Answer 1

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

$\overrightarrow{A}B=-2\hat{i}-4\hat{j}-4\hat{k},\overrightarrow{B}C=2\hat{j}+3\hat{k},\overrightarrow{A}C=-2\hat{i}-2\hat{j}-\hat{k}$

So, $|AB|=6,|BC|=\sqrt{1}3$ and $|AC|=3$

Let the angle bisector meet side $BC$ at point $D$.
$AD$ divides $BC$ such that the ratio $BD:DC=AB:AC=2:1$

Thus, $D$ is the internal bisector of side $BC$ in ratio $2:1.$
$\Rightarrow ,\overrightarrow{D}=\frac{2\overrightarrow{C}+\overrightarrow{B}}{2+1}$
$\overrightarrow{D}=\frac{(4\hat{i}+10\hat{j}+14\hat{k})+(2\hat{i}+3\hat{j}+4\hat{k})}{3}$
$\overrightarrow{D}=2\hat{i}+\frac{13}{3}\hat{j}+6\hat{k}$

Hence option $C$ is correct.