What is the sum of the following four vectors in (a) a unitvector notation, and (b) a magnitude and (c) an angle?

$$\begin{gathered} \overrightarrow{\mathbf{A}}\mathbf{=}\left(2.00m\right)\stackrel{\mathbf{⏜}}{\mathbf{i}}\mathbf{+}\left(3.00m\right)\stackrel{\mathbf{⏜}}{\mathbf{j}} \\ \overrightarrow{\mathbf{B}}\mathbf{=}\left(4.00m\right)\mathit{a}\mathit{t}\mathbf{}\mathbf{+}\mathbf{65}\mathbf{.}\mathbf{0} \\ \overrightarrow{\mathbf{C}}\mathbf{=}\left(-4.00m\right)\stackrel{\mathbf{⏜}}{\mathbf{i}}\mathbf{+}\left(-6.00m\right)\stackrel{\mathbf{⏜}}{\mathbf{j}} \\ \overrightarrow{\mathbf{D}}\mathbf{=}\left(5.00m\right)\mathit{a}\mathit{t}\mathbf{}\mathbf{-}\mathbf{235} \end{gathered}$$

The following four vectors are given:

With the help of given magnitude and the angle the conversion of vectors into unit vector is possible.After addition of four vectors, the answer will come out in the form of a unit vector, and the magnitude and angle of that unit vector can be determined.

Formulae are as follows:

$\overrightarrow{E}=\left(A_x+B_x+C_x+D_x\right)\stackrel{⏜}{i}+\left(A_y+B_y+C_y+D_y\right)\stackrel{⏜}{j}$ (i)

The formula for magnitude:

$\left|\overrightarrow{E}\right|=\sqrt{x^2+y^2}$ (ii)

The formula for angle:

$\theta ={\tan }^{-1}\frac{y}{x}$ (iii)

Where $\theta$is the angle between x and y.

The in the unit vector form is written as,

$\overrightarrow{A}=\left(2.00m\right)\stackrel{⏜}{i}+\left(3.00m\right)\stackrel{⏜}{j}$

Now, write vector $\overrightarrow{B}$in the unit vector form.

$$\begin{gathered} B_x=5.00\times \cos \left(-235°\right) \\ =1.69 \\ {B}_y=5.00\times \sin \left(-235°\right) \\ =3.62 \end{gathered}$$

Therefore,

$\overrightarrow{B}=1.69\stackrel{⏜}{i}+3.62\stackrel{⏜}{j}$

Now, vector $\overrightarrow{C}$is written as,

$\overrightarrow{C}=\left(-4.00m\right)\stackrel{⏜}{i}+\left(-6.00m\right)\stackrel{⏜}{j}$

Now, write vector $\overrightarrow{D}$ in the unit vector form.

$$\begin{gathered} D_x=5.00\times \cos \left(-235°\right) \\ =-2.87 \\ {D}_y=4.00\times \sin \left(-235°\right) \\ =4.10 \end{gathered}$$

Therefore,

$\overrightarrow{D}=-2.87\stackrel{⏜}{i}+4.10\stackrel{⏜}{j}$

Now, the sum of four vectors is calculated from equation (i) as follows:

$$\begin{gathered} \overrightarrow{G}=\overrightarrow{A}+\overrightarrow{B}+\overrightarrow{C}+\overrightarrow{D} \\ =\left(2.00-4.00+1.69-2.87\right)\stackrel{⏜}{i}+\left(3.00-6.00+3.63+4.10\right)\stackrel{⏜}{j} \\ =3.18\stackrel{⏜}{i}+4.72\stackrel{⏜}{j} \end{gathered}$$

Therefore, sum of the four vectors is $3.18\stackrel{⏜}{i}+4.72\stackrel{⏜}{j}$.

From equation (ii), find the magnitude of vector$\overrightarrow{G}$.

$$\begin{gathered} \left|\overrightarrow{G}\right|=\sqrt{{\left(-3.18\right)}^2+{\left(4.72\right)}^2} \\ =5.69m \end{gathered}$$

Therefore, magnitude is 5.69m .

Use equation (iii) to determine the angle as below:

$$\begin{gathered} \theta ={\tan }^{-1}\frac{y}{x} \\ ={\tan }^{-1}\left(\frac{4.72}{-3.18}\right) \\ =-56.0° \end{gathered}$$

Therefore, anglemade by the sum of the vector is$-56.0°$ .