First slide

Vectors

Question

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

Moderate

A

120°
B

150°
C

135°
D

none of these

Solution

Here $(A + B) \cdot A = 0$
$\begin{array}{l} A^{2} + ABcos θ = 0 \\ or ABcos θ = - A^{2} . . . . (1) \\ Bcos θ = - A . . . (2) \end{array}$
Further, $| A + B | = \frac{1}{2} | B |$
$\begin{array}{l} or (A + B) \cdot (A + B) = {(\frac{1}{2} | B |)}^{2} \\ A^{2} + B^{2} + 2 ABcos θ = \frac{1}{4} B^{2} \\ or (A^{2} + ABcos θ) + B^{2} + ABcos θ = \frac{1}{4} B^{2} \\ 0 + B^{2} + ABcos θ = - \frac{1}{4} B^{2} \end{array}$
$or ABcos θ = - \frac{3}{4} B^{2} \dots (3)$
From eqs. (1) and (3)
$A^{2} = \frac{3}{4} B^{2} or A = (\sqrt{3} / 2) B$
From eq. (2), $Bcos θ = - (\sqrt{3} / 2) B$
or $\cos θ = - \sqrt{3} / 2 or θ = 150^{\circ}$

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