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. $\frac{1}{3}(6\hat{i}+11\hat{j}+15\hat{k})$
• C. $\frac{1}{4}(8\hat{i}+14\hat{j}+19\hat{k})$
• D. $\frac{1}{3}(6\hat{i}+13\hat{j}+18\hat{k})$

Answer 1

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

$AB=\sqrt{(4-2{)}^2+(7-3{)}^2+(8-4{)}^2}=6$
$AC=\sqrt{(4-2{)}^2+(7-5{)}^2+(8-7{)}^2}=3$
$\therefore AB:AC=2:1=BD:DC$
So, $x=\frac{2\cdot 2+1\cdot 1}{2+1}=2$
$y=\frac{2\cdot 5+1\cdot 3}{2+1}=\frac{13}{3}$
$z=\frac{2\cdot 7+1\cdot 4}{2+1}=6$
$\Rightarrow D(x,y,z)\equiv D(2,\frac{13}{3},6)$
Hence, position vector of $D=\frac{1}{3}(6\hat{i}+13\hat{j}+18\hat{k})$