question_answer

If the position vectors of the vertices A, B and C of a \[\Delta ABC\] are respectively \[4\widehat{i}+7\widehat{j}+8\widehat{k},2\widehat{i}+3\widehat{j}+4\widehat{k}\] and \[2\widehat{i}+5\widehat{j}+7\widehat{k}\], then the position vector of the point, where the bisector of \[\angle A\] meets BC is                                                         [JEE Online 15-04-2018 (II)]

A)                     \[\frac{1}{2}(4\widehat{i}+8\widehat{j}+11\widehat{k})\]          

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

C) \[\frac{1}{4}(8\widehat{i}+14\widehat{j}+9\widehat{k})\]          

D) \[\frac{1}{3}(6\widehat{i}+11\widehat{j}+15\widehat{k})\]

Correct Answer: B

Solution :

Let angular bisector of A meets side BC at point P(x, y, z) By angular bisector theorem we can say that AB:AC=BP:PC \[\therefore BP:PC=c:b\] \[\Rightarrow BP:PC=6:3=2:1=m:n\] \[\Rightarrow P(x,y,z)=\left( \frac{m{{x}_{2}}+n{{x}_{1}}}{m+n},\frac{m{{y}_{2}}+n{{y}_{1}}}{m+n},\frac{m{{z}_{2}}+n{{z}_{1}}}{m+n} \right)\] Here \[B=\left( 2,3,4 \right)=\left( {{x}_{2}},{{y}_{2}},{{z}_{2}} \right)\] and \[C=\left( 2,5,7 \right)=\left( {{x}_{3}},{{y}_{3}},{{z}_{3}} \right)\] Subtituting values, we get; \[P(x,y,z)=\left( \frac{(2)(2)+(1)(2)}{2+1},\frac{(2)(5)+(1)(3)}{2+1},\frac{(2)(7)+(1)(4)}{2+1} \right)\] \[P(x,y,z)=\left( \frac{6}{3},\frac{13}{3},\frac{18}{3} \right)\] \[\therefore \] Position vector of point \[P=\frac{1}{3}(6i+13j+18k)\] Hence, the correct option is 'B'.