Question

Find the angle between the given vectors to the nearest tenth of a degree.

u = <8, 4>, v = <9, -9>

Answer 1

Use the scalar product, which can be calculated in two ways:
\[ \mathbf{u}\cdot\mathbf{v}=|\mathbf{u}||\mathbf{v}|\cos\theta=u_xv_x+u_yv_y \]
Here \(\theta\) is the sought angle between the vectors and the right hand side refers to the components of the vectors. Since \(|\mathbf{u}|=4\sqrt{5}\) and \(|\mathbf{v}|=9\sqrt{2}\) we have
\[(4\sqrt{5})(9\sqrt{2})\cos\theta=8\times9+4\times(-9).\]
Solving for \(\theta\) we obtain
\[ \theta=\cos^{-1}(10^{-1/2}) \]