## Precalculus (6th Edition) Blitzer

Dot proct of vectors: For two vectors $\mathbf{v}={{a}_{1}}\mathbf{i}+{{b}_{1}}\mathbf{j}$ and $\mathbf{w}={{a}_{2}}\mathbf{i}+{{b}_{2}}\mathbf{j}$ $\mathbf{v}\cdot \mathbf{w}={{a}_{1}}{{a}_{2}}+{{b}_{1}}{{b}_{2}}$. If the dot product of two nonzero vectors is zero then the vectors are said to be orthogonal vectors. Here, ${{a}_{1}}=1,{{a}_{2}}=-3,{{b}_{1}}=3,{{b}_{2}}=-1$. So, \begin{align} & \mathbf{v}\cdot \mathbf{w}={{a}_{1}}{{a}_{2}}+{{b}_{1}}{{b}_{2}} \\ & =1\left( -3 \right)+\left( 3 \right)\left( -1 \right) \\ & =-6-6 \\ & =-12 \end{align} Since, the dot product is not zero so the vectors are not orthogonal vectors.