If the position vectors of A,B,C are  4i^+7j^+8k^,2i^+3j^+4k^ and 2i^+5j^+7k^   respectively, of triangle ABC, then the position vector of the point where the bisector of angle A meets BC is

Suppose the bisector of angle A meets BC at D. Then AD divides BC in the ratio AB : AC. So, P.V. of D is given by|AB→|(2i^+5j^+7k^)+|AC→|(2i^+3j^+4k^)|AB→|+|AC→| but  AB→=−2i^−4j^−4k^and  AC→=−2i^−2j^−k^⇒|AB→|=6 and |AC→|=3Therefore, P.V. of D is given by  6(2i^+5j^+7k^)+3(2i^+3j^+4k^)6+3=13(6i^+13j^+18k^)