Question

Find the angle between two vectors $\overrightarrow{a}$ and $\overrightarrow{b}$ with magnitudes $\sqrt{3}$ and $2$, respectively having $\overrightarrow{a}\cdot \overrightarrow{b}=\sqrt{6}$
Answer required

Answer 1

It is given that,
$\left|\overrightarrow{a}\right|=\sqrt{3},\left|\overrightarrow{b}\right|=2$ and, $\overrightarrow{a}\cdot \overrightarrow{b}=\sqrt{6}$
Now, we know that $\overrightarrow{a}\cdot \overrightarrow{b}=|\overrightarrow{a}||\overrightarrow{b}|\cos \theta$
$\therefore \sqrt{6}=\sqrt{3}\times 2\times \cos \theta$
$\Rightarrow \cos \theta =\frac{\sqrt{6}}{\sqrt{3}\times 2}$
$\Rightarrow \cos \theta =\frac{1}{\sqrt{2}}$
$\Rightarrow \theta =\frac{\pi }{4}$
Hence, the angle between the given vectors $\overrightarrow{a}$ and $\overrightarrow{b}$ is $\frac{\pi }{4}$.