Let the angle between two nonzero vector `vecA and vecB` is `120^@` and its resultant be `vecC`.

Options

A. `C=|A-B|`
B. `C lt |A-B|`
C. `C gt |A-B|`
D. `C=|A+B|`

Answer

B

Explanation

If C is the resultant of two vectors A and B, then
`C=sqrt(A^(2)+B^(2)+2AB cos 120^(@))`
`|C|=sqrt(A^(2)+B^(2)-AB)`
`(because cos 120^(@)=-(1)/(2))`
Similarly,
`|A-B|=sqrt(A^(2)+B^(2)+2" AB"cos 60^(@))`
(`because` A-B, direction of B will be reversed, so the angle between A and -B will be `60^(@)`)
`=sqrt(A^(2)+B^(2)+AB)" "(becausecos60^(@)=(1)/(2))`
`implies|A-B|gt|C|" "( :.tan60^(@)=sqrt(3))`
(`:.D" or "A-B`, direction of B will be reversed A and B will be `60^(@)`)