Question

In Problems 14–18, $P_1=\left(3\sqrt{2},7\sqrt{2}\right)$and $P_2=\left(8\sqrt{2},2\sqrt{2}\right)$.

Find the direction angle of $\mathbf{v}$.

Short answer

The directional angles are$$\begin{gathered} \alpha =45° \\ \beta =315° \end{gathered}$$

Step-by-step solution

Step 1. Given information

The points are

$$\begin{gathered} P_1=\left(3\sqrt{2},7\sqrt{2}\right) \\ {P}_2=\left(8\sqrt{2},2\sqrt{2}\right) \end{gathered}$$

Step 2. Find the direction angle of v.

Position vector $\mathbf{v}$is equal to $\overrightarrow{P_1{P}_2}$

$$\begin{gathered} \overrightarrow{P_1{P}_2}=\overrightarrow{O{P}_2}-\overrightarrow{O{P}_1} \\ =\left(8\sqrt{2}\hat{i}+2\sqrt{2}\hat{j}\right)-\left(3\sqrt{2}\hat{i}+7\sqrt{2}\hat{j}\right) \\ =5\sqrt{2}\hat{i}-5\sqrt{2}\hat{j} \end{gathered}$$

The direction angles $\alpha ,\beta$of $\mathrm{v}=a\hat{i}+b\hat{j}$are given by

localid="1647012506438" $$\begin{gathered} \cos \alpha =\frac{a}{\sqrt{a^2+b^2}} \\ =\frac{5\sqrt{2}}{\sqrt{{\left(5\sqrt{2}\right)}^2+{\left(-5\sqrt{2}\right)}^2}} \\ =\frac{1}{\sqrt{2}} \\ \alpha =45° \\ \cos \beta =\frac{b}{\sqrt{a^2+b^2}} \\ =\frac{-5\sqrt{2}}{\sqrt{{\left(5\sqrt{2}\right)}^2+{\left(-5\sqrt{2}\right)}^2}} \\ =-\frac{1}{\sqrt{2}} \\ \beta =315° \end{gathered}$$