## Precalculus (6th Edition) Blitzer

Assume that the group member should research the forces acting on the object; the forces are: ${{\mathbf{F}}_{1}}=\left( 3\mathbf{i}+4\mathbf{j} \right)\text{ N}$ and ${{\mathbf{F}}_{2}}=\left( \mathbf{i}+\mathbf{j} \right)\text{ N}$ To find out the angle between these two vectors, we use the dot product: ${{\mathbf{F}}_{1}}\cdot {{\mathbf{F}}_{2}}=\left\| {{\mathbf{F}}_{1}} \right\|\left\| {{\mathbf{F}}_{2}} \right\|\cos \theta$ The magnitude of ${{\mathbf{F}}_{1}}$ is, \begin{align} & {{\mathbf{F}}_{1}}=\sqrt{{{3}^{2}}+{{4}^{2}}} \\ & =5 \end{align} The magnitude of ${{\mathbf{F}}_{2}}$ is, \begin{align} & {{\mathbf{F}}_{2}}=\sqrt{{{1}^{2}}+41} \\ & =\sqrt{2} \end{align} Where $\theta$ is the angle between these two vectors: \begin{align} & \left( 3\mathbf{i}+4\mathbf{j} \right)\cdot \left( \mathbf{i}+\mathbf{j} \right)=5\sqrt{2}\cos \theta \\ & 3\text{ }\mathbf{i}\cdot \mathbf{i}+3\text{ }\mathbf{i}\cdot \mathbf{j}+4\text{ }\mathbf{j}\cdot \mathbf{i}+4\text{ }\mathbf{j}\cdot \mathbf{j}=5\sqrt{2}\cos \theta \\ & \cos \theta =\frac{3+4}{5\sqrt{2}} \\ & \theta =8.13{}^\circ \end{align} Thus, the angle between these two vectors is $8.13{}^\circ$.