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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.A. `120^(@)`B. `150^(@)`C. `135^(@)`D. None of these |
Answer» Correct Answer - B `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^(@)` |
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