## Precalculus (6th Edition) Blitzer

Assume the two vectors A and B subtend at angle $\theta$ such that; $\mathbf{A}=\mathbf{i}-\mathbf{j}+\mathbf{k}$ And, $\mathbf{B}=-\mathbf{i}+\mathbf{j}-\mathbf{k}$ Now consider the dot product of two vectors A and B. \begin{align} & \mathbf{A}\cdot \mathbf{B}=\left( \mathbf{i}-\mathbf{j}+\mathbf{k} \right)\cdot \left( -\mathbf{i}+\mathbf{j}-\mathbf{k} \right) \\ & AB\cos \theta =-1-1-1 \\ & \sqrt{{{\left( 1 \right)}^{2}}+{{\left( -1 \right)}^{2}}+{{\left( 1 \right)}^{2}}}\times \sqrt{{{\left( 1 \right)}^{2}}+{{\left( 1 \right)}^{2}}+{{\left( -1 \right)}^{2}}}\cos \theta =-3 \\ & \sqrt{3}\times \sqrt{3}\cos \theta =-3 \end{align} Simplify further, \begin{align} & 3\cos \theta =-3 \\ & \cos \theta =-\frac{3}{3} \\ & \cos \theta =-1 \\ & \theta =\pi \end{align} Since, the angle between vectors A and B is $\theta =\pi$, hence when the dot product of two vectors is negative, the angle between them is obtuse. Therefore the given statement makes sense.