Answer
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 $325$.
Work Step by Step
(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 $325$.
