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. `(1)/(3)(6hati+13hatj+18hatk)`
B. `(2)/(3)(6hati+12hatj-8hatk)`
C. `(1)/(3)(-6hati-8hatj-9hatk)`
D. `(2)/(3)(-6hati-12hatj+8hatk)`

Answer

A

Explanation

The the bisector of `angleA` meets BC at D, then AD divides BC in the ratio AB:AC.
`therefore `Position vector of D
`=(|AB|(2hati+5hatj+7hatk)+|AC|(2hati+3hatj+4hatk))/(|AB|+|AC|)`
here, `|AB|=|-2hati-4hatj-4hatk|-6`
and `|AC|=|-2hati-2hatj-hatk|=3`
`therefore`Position vector of D
`=(6(2hati+5hatj+7hatk)+3(2hati+3hatj+4hatk))/(6+3)`
`=(18hati+39hatj+54hatk)/(9)`
`=(1)/(3)(6hati+13hatj+18hatk)`.