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
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\(\frac{B}{2} = \sqrt{A^{2} + \left(\frac{B}{2}\right)^{2} + 2A\frac{B}{2} \cos \theta}\) …(i)
\ \ \(\tan 90^\circ = \frac{B \sin \theta}{A + B \cos \theta} \Rightarrow A + B \cos \theta = 0\)
\ \ \(\cos \theta = -\frac{A}{B}\)
Hence, from (i) \(\frac{B^{2}}{4} = A^{2} + B^{2} - 2AB \Rightarrow A = \sqrt{3} \frac{B}{2}\)
⇒ ⇒ \(\cos \theta = -\frac{A}{B} = -\frac{\sqrt{3}}{2}\) \ \ \(\theta = 150^\circ\)
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