Question

Find the magnitude of two vectors a and b having the same magnitude and such that the angle between them is ${60}^{\circ }$ and their scalar product is $\frac{1}{2}$

Answer 1

Both vectors have same magnitude i.e. |a| = |b| and scalar product of vectors, $a.b=\frac{1}{2}$ [Given]

Let $\theta$ be the angle between two vectors a and b, then

$cos\theta =\frac{a.b}{|a||b|}\Rightarrow cos{60}^{\circ }=\frac{\frac{1}{2}}{|a||a|}$ [$\because$ |a| = |b|]

$\Rightarrow \frac{1}{2}=\frac{1}{2|a{|}^2}\Rightarrow |a{|}^2=1\Rightarrow |a|=1$

Thus, $|a|=|b|=1$