If `4hati+ 7hatj+ 8hatk,` `2hati+ 3hatj+ 4hatk` and `2hati+ 5hatj+7hatk` are the position vectors of the vertices A, B and C, respectively, of triangle ABC, then the position vector of the point where the bisector of angle A meets BC is

Options

A. `(2)/(3) (-6hati-8hatj-6hatk)`
B. `(2)/(3)(6hati+8hatj+6hatk)`
C. `(1)/(3) (6hati+ 13hatj+18hatk)`
D. `(1)/(3)(5hatj+12 hatk)`

Answer

C

Explanation

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
`(|vec(AB)| ( 2hati +5hatj + 7hatk)+ |vec(AC)| (2hati + 3hatj +4hatk))/(|vec(AB)| + |vec(AC)|)`
But `vec(AB) = -2hati -4hatj -4hatk`
and `vec(AC) = -2hati -2hatj -hatk`
`rArr |vec(AB)| = 6 and |vec(AC)| =3 `
Therefore, P.V of D is given by
`" "(6(2hati+5hatj+7hatk)+3 (2hati+ 3hatj+4hatk))/(6+3)`
`" " = (1)/(3) ( 6hati+ 13hatj + 18hatk)`