Question

Let $\overline{a}=4\hat{i}+7\hat{j}+8\hat{k},\overline{b}=2\hat{i}+3\hat{j}+4\hat{k}and\overline{c}=2\hat{i}+5\hat{j}+7\hat{k}$ are the position vectors of triangle $ABC.$ The position vectors of the point where the bisector of angle $A$ meets at $BC$, 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. $\frac{1}{3}(5\hat{i}+12\hat{k})$

Answer 1

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

$AB=6,AC=3$
$P$ divides $BC$ in the ratio 6 : 3
$\therefore P.VofP=\frac{6(2\hat{i}+5\hat{j}+7\hat{k})+3(2\hat{i}+3\hat{j}+4\hat{k})}{6+3}$
$\frac{18\hat{i}+39\hat{j}+54\hat{k}}{9}=\frac{3(6\hat{i}+13\hat{j}+18\hat{k})}{9}$
$=\frac{1}{3}(6\hat{i}+13\hat{j}+18\hat{k})$
Hence choice (C) is correct answer.