## Precalculus (6th Edition) Blitzer

Assume the two vectors A and B are: $\mathbf{A}=a\mathbf{i}+b\mathbf{j}+c\mathbf{k}$ And, $\mathbf{B}=x\mathbf{i}+y\mathbf{j}+z\mathbf{k}$ Now consider the addition operation of vector A and B. Assume, \begin{align} & \mathbf{C}=\mathbf{A}+\mathbf{B} \\ & =\left( a\mathbf{i}+b\mathbf{j}+c\mathbf{k} \right)+\left( x\mathbf{i}+y\mathbf{j}+z\mathbf{k} \right) \\ & =\left( a+x \right)\mathbf{i}+\left( b+y \right)\mathbf{j}+\left( c+z \right)\mathbf{k} \end{align} The expression $\mathbf{C}=\left( a+x \right)\mathbf{i}+\left( b+y \right)\mathbf{j}+\left( c+z \right)\mathbf{k}$ shows a vector itself. Hence, the vector operations can produce another vector also. Now consider the dot product of two vectors A and B. \begin{align} & \mathbf{A}\cdot \mathbf{B}=\left( a\mathbf{i}+b\mathbf{j}+c\mathbf{k} \right)\cdot \left( x\mathbf{i}+y\mathbf{j}+z\mathbf{k} \right) \\ & =ax+by+cz \end{align} The expression $\mathbf{A}\cdot \mathbf{B}=ax+by+cz$ has no vector sign in the result. Hence, the dot product of two vectors is a real number. Therefore, the given statement makes sense.