`vecR` is the resultant of `vecA` and `vecB` . `vecR` is inclined to `vecA` at angle `theta_(1)` and to `vecB` at angle `theta_(2)`, then :

Options

A. `(3)/(4)(hati -hatj)`
B. `(5)/(2)(hati -hatj)`
C. `hati -hatj`
D. `(1)/(2)(hati +hatj)`

Answer

C

Explanation

Component of `vecA` along `vecB` is
`" "=vecA. vecB=(8hati + 6hatj).((hati-hatj)/(sqrt(2))) = (8-6)/(sqrt(2)) = sqrt(2)`
`therefore`Required component `= sqrt(2)((hati-hatj)/(sqrt2))= (hati - hatj)`