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 1

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 $\overrightarrow b$.