Find the sum of the following four vectors in (a) unit-vector notation, and as (b) a magnitude and (c) an angle relative to +x.

$\overrightarrow{\mathbf{P}}\mathbf{:}\mathbf{}\mathbf{10}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{m}\mathbf{,}\mathbf{at}\mathbf{}\mathbf{\hspace{0.33em}}\mathbf{25}\mathbf{.}\mathbf{0}\mathbf{°}$counterclockwise from +x

$\overrightarrow{\mathbf{Q}}\mathbf{:}\mathbf{}\mathbf{12}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{m}\mathbf{,}\mathbf{at}\mathbf{}\mathbf{10}\mathbf{.}\mathbf{0}\mathbf{°}$counterclockwise from +y

$\overrightarrow{\mathbf{R}}\mathbf{:}\mathbf{}\mathbf{8}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{m}\mathbf{,}\mathbf{}\mathbf{at}\mathbf{\hspace{0.33em}}\mathbf{}\mathbf{20}\mathbf{.}\mathbf{0}\mathbf{°}$clockwise from –y

$\mathbf{}\overrightarrow{\mathbf{S}}\mathbf{:}\mathbf{}\mathbf{9}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{m}\mathbf{,}\mathbf{at}\mathbf{\hspace{0.33em}}\mathbf{}\mathbf{40}\mathbf{.}\mathbf{0}\mathbf{°}$counterclockwise from -y

1)$\overrightarrow{\mathrm{P}}=10.0\hspace{0.33em}\mathrm{m}\hspace{0.33em}\mathrm{at}\hspace{0.33em}25.0°\mathrm{counterclockwise}\mathrm{from}\hspace{0.33em}+\mathrm{x}$

2)$\overrightarrow{\mathrm{Q}}=12.0\hspace{0.33em}\mathrm{m}\hspace{0.33em}\mathrm{at}\hspace{0.33em}10.0°\mathrm{counterclockwise}\mathrm{from}\hspace{0.33em}+\mathrm{y}$

3)$\overrightarrow{\mathrm{R}}=8.00\hspace{0.33em}\mathrm{m}\hspace{0.33em}\mathrm{at}\hspace{0.33em}20.0°\mathrm{clockwise}\mathrm{from}\hspace{0.33em}-\mathrm{y}$

4)$\overrightarrow{\mathrm{S}}=9.00\hspace{0.33em}\mathrm{m}\mathrm{at}\hspace{0.33em}40.0°\mathrm{counterclockwise}\mathrm{from}-\mathrm{y}$

We can use vector addition law and find the net displacement of given vectors. By using the trigonometry formula, we find its direction and magnitude.

Formula:

$\overrightarrow{d}=\overrightarrow{P}+\overrightarrow{Q}+\overrightarrow{R}+\overrightarrow{S}$ i)

$d=\sqrt{{\left(d_x\right)}^2+{\left(d_y\right)}^2}$ (ii)$\tan \theta =\frac{d_y}{d_x}$ (iii)

The vector configuration can be drawn as below:

The sum of the four vectors is,

$\overrightarrow{d}=\overrightarrow{P}+\overrightarrow{Q}+\overrightarrow{R}+\overrightarrow{S}$

The x component of four vectors is,

$\overrightarrow{d_x}={\overrightarrow{P}}_x+\overrightarrow{Q_x}+\overrightarrow{R_x}+\overrightarrow{S_x}$

Substitute the value of each component from the figure of vector configuration.

$$\begin{gathered} {\overrightarrow{\mathrm{d}}}_{\mathrm{x}}=\left(\mathrm{Pcos}25-\mathrm{Qsin}10-\mathrm{Rsin}20+\mathrm{Ssin}40\right)\stackrel{⏜}{\mathrm{i}} \\ =\left(10.0\cos 25-12.0\sin 10-8.00\sin 20+9.00\sin 40\right)\stackrel{⏜}{\mathrm{i}} \\ =10.03\stackrel{⏜}{\mathrm{i}} \end{gathered}$$

Similarly, the y component of four vectors is

$\overrightarrow{d_y}={\overrightarrow{P}}_y+\overrightarrow{Q_y}+\overrightarrow{R_y}+\overrightarrow{S_y}$

Substitute the value of each component from the figure of vector configuration.

$$\begin{gathered} {\overrightarrow{\mathrm{d}}}_{\mathrm{y}}=\left(Ps\mathrm{in}25+Q\cos 10-R\cos 20-S\cos 40\right)\stackrel{⏜}{j} \\ =\left(10\sin 25+12\cos 10-8\cos 20-9\cos 40\right)\stackrel{⏜}{j} \\ =1.6319\stackrel{⏜}{j} \end{gathered}$$

In unit vector notation,

$$\begin{gathered} \overrightarrow{d}=\overrightarrow{d_x\stackrel{⏜}{i}}+\overrightarrow{d_y}\stackrel{⏜}{j} \\ =10.03\stackrel{⏜}{i}+1.63\stackrel{⏜}{j} \end{gathered}$$

Therefore, the sum of four vectors,$\overrightarrow{d}$ is equal to $10.03\stackrel{⏜}{i}+1.63\stackrel{⏜}{j}$.

Use the equation (ii) to calculate the magnitude of vector$\overrightarrow{d}$.

$$\begin{gathered} \mathrm{d}=\sqrt{{\left({\mathrm{d}}_{\mathrm{x}}\right)}^2+{\left({\mathrm{d}}_{\mathrm{y}}\right)}^2} \\ =\sqrt{{\left(10.03\right)}^3+{\left(1.6319\right)}^2} \\ =10.2\mathrm{m} \end{gathered}$$

Therefore, the magnitude of$\overrightarrow{d}$ is 10.2m.

Now, calculate the direction using equation (iii) and the x and y components of the vector $\overrightarrow{d}$.Therefore, the direction is,

$$\begin{gathered} \tan \theta =\frac{d_y}{d_x} \\ \theta ={\tan }^{-1}\left(\frac{d_y}{d_x}\right) \\ ={\tan }^{-1}\left(\frac{1.63}{10.03}\right) \\ =9.24° \end{gathered}$$

Therefore, the angle of the vector$\overrightarrow{d}$ is$9.24°$ in counterclockwise direction from the positive x-axis.