Chapter 11 - Vectors and the Geometry of Space - 11.3 Exercises - Page 773: 9

$\frac{\pi}{2}$ rad $=90^\circ$

$\mathbf{u}=<1.1>$ and $\mathbf{v}=<2,-2>$ $\mathbf{u}\cdot\mathbf{v}=1\times2+1\times(-2)=2-2=0$ $\|\mathbf{u}\|=\sqrt{1^2+1^2}=\sqrt{2}$ and $\|\mathbf{v}\|=\sqrt{2^2+(-2)^2}=\sqrt{8}=2\sqrt{2}$ ${\theta}=\cos^{-1}\frac{\mathbf{u}\cdot\mathbf{v}}{\|\mathbf{u}\|\|\mathbf{v}\|}=\cos^{-1}\frac{0}{2\sqrt{2}\sqrt{2}}=\cos^{-1}{0}=\frac{\pi}{2}$ rad = $90^\circ$