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If 4hati+ 7hatj+ 8hatk, 2hati+ 3hatj+ 4h...

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

A

`(1)/(3)(6hati+13hatj+18hatk)`

B

`(2)/(3)(6hati+12hatj-8hatk)`

C

`(1)/(3)(-6hati-8hatj-9hatk)`

D

`(2)/(3)(-6hati-12hatj+8hatk)`

Text Solution

Verified by Experts

The correct Answer is:
A
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)`.
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