Question

The angle between the vectors $\overrightarrow{a}$ and $\overrightarrow{b}$ such that $\left|\overrightarrow{a}\right|=\left|\overrightarrow{b}\right|=\sqrt{2}$ and $\overrightarrow{a}\cdot \overrightarrow{b}=1$ is

• A. $\frac{\pi }{2}$
• B. $\frac{\pi }{3}$
• C. $\frac{\pi }{4}$
• D. $0$

Answer 1

The correct option is B $\frac{\pi }{3}$

We know that, $\overrightarrow{a}.\overrightarrow{b}=$ $\left|\overrightarrow{a}\right|.\left|\overrightarrow{b}\right|cos\theta$

$⟹1=\sqrt{2}\ast \sqrt{2}\ast cos\theta$

$\therefore$ $cos\theta =\frac{1}{2}$

$⟹\theta =co{s}^{-1}\left(\frac{1}{2}\right)$

$⟹\theta =\frac{\pi }{3}$

Hence, option B is correct.