The resultant of two vectors A and B is perpendicular to the vector A and its magnitude is equal to half the magnitude of vector B. The angle between A and B is -

Options

A. `120^(@)`
B. `150^(@)`
C. `135^(@)`
D. None of these

Answer

B

Explanation

`(B)/(2)= sqrt(A^(2) + B^(2) + 2AB cos theta)` direction of resultant vector with respect to `vec(A)`
`tan alpha= (B sin theta)/(A+ B cos theta)`
`rArr tan 90^(@)= (B sin theta)/(A +B cos theta)`
`therefore A + B cos theta= 0 rArr cos theta= (-A)/(B)`
`(B^(2))/(2 xx 2)= A^(2) + B^(2) + 2AB ((-A)/(B))= A^(2) + B^(2) - 2AB^(2)`
`rArr A^(2)= B^(2)- (B^(2))/(4)= (3B^(2))/(4)`
`rArr A= (sqrt3)/(2) B`
`cos theta= (- A)/(B)= (- sqrt3)/(2) rArr theta= 150^(@)`