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

AD divides BC in ratio AB : AC
So, P.V. of D $=\frac{|\overline{AB}|(2\hat{i}+5\hat{j}+7\hat{k})+|\overline{AC}|(2\hat{i}+3\hat{j}+4\hat{k})}{|\overline{AB}|+|\overline{AC}|}$