Here, \(\overset\rightharpoonup a=\begin{pmatrix}6\\1\end{pmatrix}\) and \(\overset\rightharpoonup b=\begin{pmatrix}-1\\6\end{pmatrix}\)
From dot product formula we that, \(\overset\rightharpoonup a.\overset\rightharpoonup b=x_1x_2+y_1y_2\)
= 6x(-1) + 1x6 = 0.
Since the dot product of two vectors is zero they are perpendicular to each other.
To show this, angle between two vectors is \(\cos\left(\theta\right)=\frac{\overset\rightharpoonup a.\overset\rightharpoonup b}{\left|\overset\rightharpoonup a\right|\left|\overset\rightharpoonup b\right|}\)
Or, \(\cos\left(\theta\right)=\frac0{\left|\overset\rightharpoonup a\right|\left|\overset\rightharpoonup b\right|}\)
Or, \(\cos\left(\theta\right)=0\)
Or, \(\theta=\cos^{-1}\left(0\right)\)
∴ \(\theta=90^°\)