If \(\overset\rightharpoonup a=\begin{pmatrix}2\\1\end{pmatrix}\) and \(\overset\rightharpoonup b=\begin{pmatrix}0\\-2\end{pmatrix}\), find the angle between \(\overset\rightharpoonup a\) and \(\overset\rightharpoonup b\)
1 Answer
Here, \(\overset\rightharpoonup a=\begin{pmatrix}2\\1\end{pmatrix}\) and \(\overset\rightharpoonup b=\begin{pmatrix}0\\-2\end{pmatrix}\)
\(\overset\rightharpoonup a.\overset\rightharpoonup b=x_1x_2+y_1y_2\)
\(\left|\overset\rightharpoonup a\right|=\sqrt{x^2+y^2}\) = \(\sqrt{2^2+1^2}\) = \(\sqrt5\)
\(\left|\overset\rightharpoonup b\right|=\sqrt{x^2+y^2}\) = \(\sqrt{0^2+-2^2}\) = \(\sqrt4\) = 2
Now angle between two vectors is given by,
\(\cos\left(\theta\right)=\frac{\overset\rightharpoonup a.\overset\rightharpoonup b}{\left|\overset\rightharpoonup a\right|\left|\overset\rightharpoonup b\right|}\)
\(=\frac{-2}{2\sqrt5}=-\frac1{\sqrt5}\)
∴ \(\theta=\cos^{-1}\left(-\frac1{\sqrt5}\right)\)