## Precalculus (6th Edition) Blitzer

The vector is, $\mathbf{v}=a\mathbf{i}+b\mathbf{j}$. The magnitude of the vector $\mathbf{v}$ is given by: $\left\| \mathbf{v} \right\|=\sqrt{{{a}^{2}}+{{b}^{2}}}$ The unit vector $\frac{\mathbf{v}}{\left\| \mathbf{v} \right\|}$ is given by: \begin{align} & \frac{\mathbf{v}}{\left\| \mathbf{v} \right\|}=\frac{a\mathbf{i}+b\mathbf{j}}{\sqrt{{{a}^{2}}+{{b}^{2}}}} \\ & =\frac{a}{\sqrt{{{a}^{2}}+{{b}^{2}}}}\mathbf{i}+\frac{b}{\sqrt{{{a}^{2}}+{{b}^{2}}}}\mathbf{j} \end{align} The dot product of unit vector $\left( \frac{\mathbf{v}}{\left\| \mathbf{v} \right\|} \right)$ with $\mathbf{v}$ is, \begin{align} & \mathbf{v}\cdot \frac{\mathbf{v}}{\left\| \mathbf{v} \right\|}=\left( a\mathbf{i}+b\mathbf{j} \right)\cdot \left( \frac{a}{\sqrt{{{a}^{2}}+{{b}^{2}}}}\mathbf{i}+\frac{b}{\sqrt{{{a}^{2}}+{{b}^{2}}}}\mathbf{j} \right) \\ & =\frac{{{a}^{2}}}{\sqrt{{{a}^{2}}+{{b}^{2}}}}\mathbf{i}\cdot \mathbf{i}+\frac{{{b}^{2}}}{\sqrt{{{a}^{2}}+{{b}^{2}}}}\mathbf{j}\cdot \mathbf{j} \\ & =\frac{{{a}^{2}}}{\sqrt{{{a}^{2}}+{{b}^{2}}}}+\frac{{{b}^{2}}}{\sqrt{{{a}^{2}}+{{b}^{2}}}} \\ & =\frac{{{a}^{2}}+{{b}^{2}}}{\sqrt{{{a}^{2}}+{{b}^{2}}}} \end{align} Thus, the dot product of unit vector $\left( \frac{\mathbf{v}}{\left\| \mathbf{v} \right\|} \right)$ with $\mathbf{v}$ is, $\mathbf{v}\cdot \frac{\mathbf{v}}{\left\| \mathbf{v} \right\|}=\sqrt{{{a}^{2}}+{{b}^{2}}}$ The dot product of unit vector $\left( \frac{\mathbf{v}}{\left\| \mathbf{v} \right\|} \right)$ with $\mathbf{v}$ gives the magnitude of the vector $\mathbf{v}$. That means the $\frac{\mathbf{v}}{\left\| \mathbf{v} \right\|}$ is the unit vector in the direction of $\mathbf{v}$.