The resultant of two vector A and B is at right angles to A and its magnitude is half of B. Find the angle between A and B.

Options

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

Answer

B

Explanation

`B/2= sqrt(A^(2)+B^(2)+2AB cos theta)`…..(i)
`:. Tan90^(@)=(B sin theta)/(A+B cos theta)implies A+B cos theta=0`
`:. cos theta= -(A)/(B)`
Hence, from (i), `(B^(2))/4= A^(2)+B^(2)-2A^(2)implies A= sqrt(3)B/2`
`implies cos theta = -(A)/(B)= -(sqrt(3))/2, :. theta= 150^(@)`