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If the position vectors of the vertices $A, B, C$ of a tringle $A B C$ are $4 \hat{\imath}+7 j+8 \hat{k}, 2 \hat{\imath}+3 \hat{\jmath}+4 \hat{k}$ and $2 \hat{\imath}+5 \hat{\jmath}+7 \hat{k}$ respectively, then the position vector of the point where bisector of angle A meets $B C$ is
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The correct answer is:
$\frac{1}{3}(6 \hat{\imath}+13 \hat{\mathbf{j}}+18 \hat{\mathrm{k}})$
Suppose the bisector of angle $\mathrm{A}$ meets $\mathrm{BC}$ at $\mathrm{D}$.
Then $A D$ divides $B C$ in the ratio $A B$ : $A C$
$\therefore$ Position vector of $\mathrm{D}=\frac{(\mathrm{AC})(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+4 \hat{\mathrm{k}})+(\mathrm{AB})(2 \hat{\mathrm{i}}+5 \hat{\mathrm{j}}+7 \hat{\mathrm{k}})}{(\mathrm{AB})+(\mathrm{AC})}$
$\overline{\mathrm{AB}}=-2 \hat{\mathrm{i}}-4 \hat{\mathrm{j}}-4 \hat{\mathrm{k}}$ and $\overline{\mathrm{AC}}=-2 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}-\hat{\mathrm{k}}$
$|\overline{\mathrm{AB}}|=\mathrm{AB}=6$ and $|\overline{\mathrm{AC}}|=\mathrm{AC}=\sqrt{4+4+1}=3$
Position vector of $D$
$=\frac{6(2 \hat{i}+3 \hat{j}+4 \hat{k})+3(2 \hat{i}+5 \hat{j}+7 \hat{k})}{6+3}=\frac{18 \hat{i}+33 \hat{j}+45 \hat{k}}{9}=\frac{1}{3}(6 \hat{i}+11 \hat{j}+15 \hat{k})$
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