91Ó°ÊÓ

Following vectors are given

$$\begin{gathered} \stackrel{→}{\mathrm{E}}=(6.00\mathrm{m})\mathrm{at}+(0.900\mathrm{rad}) \\ \stackrel{→}{\mathrm{F}}=(5.00\mathrm{m})\mathrm{at}-75° \\ \stackrel{→}{\mathrm{G}}=(4.00\mathrm{m})\mathrm{at}+(1.20\mathrm{rad}) \\ \stackrel{→}{\mathrm{H}}=(6.00\mathrm{m})\mathrm{at}-210° \end{gathered}$$

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

Required formulae are as follow:

$\stackrel{→}{\mathrm{A}}=\mathrm{x}\hat{\mathrm{i}}+\mathrm{y}\hat{\mathrm{j}}\left(\mathrm{i}\right)$

Formula for magnitude:

$\left|\stackrel{→}{\mathrm{A}}\right|=\sqrt{{\mathrm{x}}^2+{\mathrm{y}}^2}\left(\mathrm{ii}\right)$

Formula for angle:

role="math" localid="1657022472163" $\mathrm{θ}={\tan }^{-1}\frac{\mathrm{y}}{\mathrm{x}}\left(\mathrm{iii}\right)$

Where,x ,y, $\stackrel{→}{A}$ are vectors and $\mathrm{θ}$is the angle between x and y .

Now, convert vectors in the unit vector form,

$$\begin{gathered} \stackrel{→}{\mathrm{E}}=\mathrm{x}\hat{\mathrm{i}}+\mathrm{y}\hat{\mathrm{j}} \\ \mathrm{x}=6.00×\cos \left(0.9^{°}\right) \\ =3.73 \\ \mathrm{y}=5.00×\sin \left(-{75}^{°}\right) \\ =4.70 \end{gathered}$$

Therefore, $\stackrel{→}{E}$ in vector form is,

$\stackrel{→}{\mathrm{E}}=3.73\hat{\mathrm{i}}+4.70\hat{\mathrm{j}}$

Similarly,$\stackrel{→}{F}$ can be written as

$\stackrel{→}{\mathrm{F}}=1.29\hat{\mathrm{i}}-4.83\hat{\mathrm{j}}$

The vector$\stackrel{→}{G}$ can be written as,

$\stackrel{→}{\mathrm{G}}=1.45\hat{\mathrm{i}}+3.73\hat{\mathrm{j}}$

The vector$\stackrel{→}{\mathrm{H}}$ can be written as,

$\stackrel{→}{\mathrm{H}}=-5.20\hat{\mathrm{i}}+3.00\hat{\mathrm{j}}$

To find the sum, add all the above vectors using equation (i). Therefore,

$$\begin{gathered} \stackrel{→}{\mathrm{A}}=\stackrel{→}{\mathrm{E}}+\stackrel{→}{\mathrm{F}}+\stackrel{→}{\mathrm{G}}+\stackrel{→}{\mathrm{H}} \\ =\left(3.73+1.29+1.45+5.20\right)\hat{\mathrm{i}}+\left(4.7-4.83+3.73+3.00\right)\hat{\mathrm{j}} \\ =1.28\stackrel{→}{\mathrm{i}}+6.60\hat{\mathrm{j}} \end{gathered}$$

Therefore, sum of unit vectors is $1.28\mathrm{i}⃗+6.60\hat{\mathrm{j}}$.

Use the equation (ii) to find the magnitude.

$$\begin{gathered} \left|\stackrel{→}{\mathrm{A}}\right|=\sqrt{1.{28}^2+6.{60}^2} \\ =6.72\mathrm{m} \end{gathered}$$

Therefore, magnitude of $\stackrel{→}{A}$is$6.72\mathrm{m}$ .

Use the equation (iii) to find the angle in degrees.

$$\begin{gathered} \mathrm{θ}={\tan }^{-1}\left(\frac{\mathrm{y}}{\mathrm{x}}\right) \\ ={\tan }^{-1}\left(\frac{6.60}{1.28}\right) \\ =79.0^{°} \end{gathered}$$

Therefore, angle in degrees is ${\text{θ=79.0}}^{°}$.

Now, convert the angle into radians.

$$\begin{gathered} \mathrm{θ}=\frac{79.0×\mathrm{π}}{180}\mathrm{red} \\ =1.38\mathrm{rad} \end{gathered}$$

Therefore, angle in radian is $1.38\mathrm{rad}$.