Question

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 $\frac{1}{2}$, are

• A. $|a|=\frac{1}{2},|b|=1$
• B. $|a|=|b|=1$
• C. $|a|=1,|b|=\frac{1}{2}$
• D. None of these

Answer 1

The correct option is B $|a|=|b|=1$

Given, both vectors have same magnitude i.e., $|a|=|b|$ and scalar product of vectors, $a\cdot b=\frac{1}{2}$ (given).
Let $\theta$ be the angle between two vectors $a$ and $b$. Then,
$\cos \theta =\frac{a\cdot b}{|a||b|}$
$\Rightarrow \cos {60}^{\circ }=\frac{\frac{1}{2}}{|a||a|}[\because |a|=|b|(given\left)\right]$
$\Rightarrow \frac{1}{2}=\frac{1}{2|a{|}^2}$
$\Rightarrow |a|=1$
$\therefore |a|=|a|=1$.