First slide

Vectors

Question

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

Difficult

A

$120^{0}$
B

$150^{0}$
C

$135^{o}$
D

None of these

Solution

$\frac{B}{2} = \sqrt{A^{2} + B^{2} + 2 ABcosθ} - - - - (i)$
$∴$ $\tan 90^{0} = \frac{Bsinθ}{A + Bcosθ} \Rightarrow A + Bcosθ = 0$
$∴ cosθ = - \frac{A}{B}$
Hence, from (i) $\frac{B^{2}}{4} = A^{2} + B^{2} - 2 A^{2} \Rightarrow A = \sqrt{3} \frac{B}{2}$
$\Rightarrow cosθ = - \frac{A}{B} = - \frac{\sqrt{3}}{2} ∴ θ = 150^{0}$

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