question_answer

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

B) \[\frac{2}{3}(6\hat{i}+12\hat{j}-8\hat{k})\]

C) \[\frac{1}{3}(-6\hat{i}-8\hat{j}-9\hat{k})\]  

D) \[\frac{2}{3}(-6\hat{i}-12\hat{j}+8\hat{k})\]

Correct Answer: A

Solution :

Let O be the origin and the bisector of \[\angle A\]meets BC at D. Then \[\frac{BD}{DC}=\frac{AB}{AC}\] and position vector of D is given by \[\overrightarrow{OD}=\frac{|\overrightarrow{AB}|\overrightarrow{OC}|+|\overrightarrow{AC}|\overrightarrow{OB}}{|\overrightarrow{OC}|+|\overrightarrow{AC}|}\] \[\overrightarrow{AB}=2\hat{i}+3\hat{j}+4\hat{k}-4\hat{i}-7\hat{j}-8\hat{k}\] \[\left| \overrightarrow{AB} \right|=\left| -2i-4j-4\hat{k} \right|\] \[=\sqrt{{{(-2)}^{2}}+{{(-4)}^{2}}+{{(-4)}^{2}}}\] \[=\sqrt{4+16+16}=\sqrt{36}\] \[|\overrightarrow{AB}|=6\] \[\overrightarrow{AC}=2\hat{i}+5\hat{j}+7\hat{k}-4\hat{i}-7\hat{j}-8\hat{k}\] \[\left| \overrightarrow{AC} \right|=\left| -2i-2\hat{j}-\hat{k} \right|\] \[=\sqrt{{{(-2)}^{2}}+{{(-2)}^{2}}+{{(-1)}^{2}}}\] \[=\sqrt{4+4+1}=\sqrt{9}\] \[\left| \overrightarrow{AC} \right|=3\] \[\therefore \] position vector of \[\vec{D}\] \[=\frac{6(2\hat{i}+5\hat{j}+7\hat{k})+3(2\hat{i}+3\hat{j}+4\hat{k})}{6+3}\] \[=\frac{12\hat{i}+30\hat{j}+42\hat{k}+6\hat{i}+9\hat{j}+12\hat{k}}{9}\] \[=\frac{18\hat{i}+39\hat{j}+54\hat{k}}{9}=\frac{(6\hat{i}+13\hat{j}+18\hat{k})}{3}\]