Question

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

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

Answer 1

The correct option is A $\frac{2}{3}(-6\overline{i}-8\overline{j}-6\overline{k})$

Let AD be the bisector of $\angle A$.

AB touches BC at D.

D divides BC in the ratio $AB:AC$.

Now,

$\overrightarrow{AB}=-2i-4j-4k\Rightarrow AB=\sqrt{4+16+16}=6$

$\overrightarrow{AC}=-2i-2j-k\Rightarrow AC=\sqrt{4+4+1}=3$

position vector of D by section formula :

$=\frac{6(2i+5j+7k)+3(2i+3j+4k)}{6+3}$

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