## Precalculus (6th Edition) Blitzer

a) Unit vector is $\frac{\sqrt{3}}{2}\mathbf{i}+\frac{1}{2}\mathbf{j}$. b) Vector projection is $-\frac{325\sqrt{3}}{2}\mathbf{i}-\frac{325}{2}\mathbf{j}$. c) Magnitude is $325$.
(a). The boat is inclined on the ramp at an angle of ${{30}^{\circ }}$. Hence, in the upward direction, the unit vector can be resolved having a unit magnitude. Therefore, unit vector $\mathbf{u}$ is \begin{align} & \mathbf{u}=\cos {{30}^{\circ }}\mathbf{i}+\sin {{30}^{\circ }}\mathbf{j} \\ & =\frac{\sqrt{3}}{2}\mathbf{i}+\frac{1}{2}\mathbf{j} \end{align} (b). It is known that the vector projection of F onto u is $\text{pro}{{\text{j}}_{\mathbf{u}}}\mathbf{F}=\frac{\mathbf{F}\cdot \mathbf{u}}{{{\left\| \mathbf{u} \right\|}^{2}}}\mathbf{u}$ …… (3) Substituting equations (1) and (2) in equation (3), we get \begin{align} & \text{pro}{{\text{j}}_{\mathbf{u}}}\mathbf{F}=\frac{\left( 0,-650 \right)\left( \frac{\sqrt{3}}{2},\frac{1}{2} \right)}{{{\left\| \mathbf{u} \right\|}^{2}}}\left( \frac{\sqrt{3}}{2}\mathbf{i}+\frac{1}{2}\mathbf{j} \right) \\ & =\left( 0\times \frac{\sqrt{3}}{2},-650\times \frac{1}{2} \right)\left( \frac{\sqrt{3}}{2}\mathbf{i}+\frac{1}{2}\mathbf{j} \right) \\ & =-\frac{325\sqrt{3}}{2}\mathbf{i}-\frac{325}{2}\mathbf{j} \end{align} Hence, the vector projection of F onto a unit vector is $-\frac{325\sqrt{3}}{2}\mathbf{i}-\frac{325}{2}\mathbf{j}$. (c). It is known that the magnitude of a vector is $\left\| \mathbf{A} \right\|=\sqrt{{{\left( \text{component of }\mathbf{i} \right)}^{2}}+{{\left( \text{component of }\mathbf{j} \right)}^{2}}}$ Hence, from (1), we get \begin{align} & \left\| \text{pro}{{\text{j}}_{\mathbf{u}}}\mathbf{F} \right\|=\sqrt{{{\left( -\frac{325\sqrt{3}}{2} \right)}^{2}}+{{\left( \frac{325}{2} \right)}^{2}}} \\ & =325 \end{align} This means that a force of $325$ pounds is required to keep the boat from rolling down the ramp. Hence, the magnitude of vector projection is $325$.