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 $.

Hint: We need to have a basic idea on the vector concept to solve this problem. The angle between two vectors can be found using the dot product of those vectors.

Complete step-by-step answer:

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

Note: The angle between two vectors $$\overrightarrow a $$ and $\overrightarrow b $ is defined by $$\cos \theta = \dfrac{{\overrightarrow a \cdot \overrightarrow b }}{{\left| {\overrightarrow a } \right|\left| {\overrightarrow b } \right|}},$$ $\theta $ is the angle between vectors, $$\left| {\overrightarrow a } \right|,\left| {\overrightarrow b } \right|$$ are the magnitudes of the vectors $\overrightarrow a $ and $\overrightarrow b $.