If two vectors OA and OB are there, then their resultant OA+OB can be found by completing the parallelogram OACB and OC=OA+OB. Also, if |OA|=|OB|, then the resultant will bisect the angle between them.
Q. A vector C directed along internal bisector of angle between vectors `A=7hati-4hatj-4hatk and B=-2hati-hatj+2hatk` with `|C|=5sqrt(6)` is
a. `(5)/(3)(hati-hatj+hatk)`
b. `(5)/(3)(hati-7hatj+2hatk)`
c. `(5)/(3)(5hati+5hatj+2hatk)`
d. `(5)/(3)(-5hati+5hatj+3hatk)`

Options

A. `(5)/(3)(hati-hatj+hatk)`
B. `(5)/(3)(hati-7hatj+2hatk)`
C. `(5)/(3)(5hati+5hatj+2hatk)`
D. `(5)/(3)(-5hati+5hatj+3hatk)`

Answer

B

Explanation

Here, `c=t(hata+hatb)=t((7hati-4hatj-4hatk)/(9)+((-2hati-hatj+2hatk))/(3))`
`implies c =t((hati-7hatj+2hatk)/(9))`
also, `|c|=5sqrt(6)implies (t)/(9)*sqrt(1+49+4)=5sqrt(6)`
`therefore t=15 implies c=(15)/(9)(hati-7hatj+2hatk)`
or `=(5)/(3)(hati-7hatj+2hatk)`.