Answer
Two vectors are said to be orthogonal vectors if and only if their dot product is zero or they make an angle of ${{90}^{\circ }}$.
Work Step by Step
Consider $\mathbf{v}=2\mathbf{i}+\mathbf{j}$, $\mathbf{w}=2\mathbf{k}$, and $\mathbf{v}\cdot \mathbf{w}=0$.
Find the magnitude of vector v as follows:
$\begin{align}
& \left\| \mathbf{v} \right\|=\sqrt{{{\left( 2 \right)}^{2}}+{{\left( 1 \right)}^{2}}} \\
& =\sqrt{4+1} \\
& =\sqrt{5}
\end{align}$
Therefore, $\left\| \mathbf{v} \right\|=\sqrt{5}$.
Similarly, find magnitude of vector w as follows:
$\begin{align}
& \left\| \mathbf{w} \right\|=\sqrt{{{\left( 1 \right)}^{2}}} \\
& =\sqrt{1}
\end{align}$
Therefore, $\left\| \mathbf{w} \right\|=1$.
Now, find the cosine of the angle between their directions as follows:
$\begin{align}
& \cos \theta =\left( \frac{\mathbf{v}\cdot \mathbf{w}}{\left\| \mathbf{v} \right\|\text{ }\left\| \mathbf{w} \right\|} \right) \\
& =\left( \frac{0}{\sqrt{5}\left( 1 \right)} \right) \\
& =\left( \frac{0}{\sqrt{5}} \right) \\
& =0
\end{align}$
So,
$\begin{align}
& \theta ={{\cos }^{-1}}\left( 0 \right) \\
& ={{90}^{\circ }}
\end{align}$
Hence, the angle between two vectors is ${{90}^{\circ }}$.Therefore, it can be concluded that the vectors are perpendicular to each other or the vectors are orthogonal.
