91Ó°ÊÓ

1. The magnitude of$\stackrel{→}{a}$ is 3.00 m
2. The magnitude of$\stackrel{→}{b}$ is 4.00 m
3. The magnitude of$\stackrel{→}{c}$ is 10.0 m
4. The angle between$\stackrel{→}{a}$ and$\stackrel{→}{b}$ is$30°$ .

The vector diagram can be used to find the angles and linearity of the vectors. Using the vector diagram, we can find the x and y components of the vector. Also, we can find whether the vectors are linear or nonlinear.

Vector diagram:

From the vector diagram, it can be seen that$\stackrel{→}{a}$ is along x –axis.Therefore, the x component of $\stackrel{→}{a}$is 3.00 m.

From the vector diagram, it can be seen that$\stackrel{→}{a}$ is along x –axis. Therefore, the y component of $\stackrel{→}{a}$is 0m.

Use the vector diagram and the angle between$\stackrel{→}{b}$and x-axis to calculate the x component of $\stackrel{→}{b}$.

role="math" localid="1657948864992" $$\begin{gathered} {\stackrel{\mathit{→}}{b}}_x=\stackrel{\mathit{→}}{b}\cos 30° \\ =4\mathrm{m}\left(0.866\right) \\ =3.46\mathrm{m} \end{gathered}$$

Therefore, the x component of$\stackrel{→}{b}$ is 3.46m.

Use the vector diagram and the angle between $\stackrel{→}{b}$and x-axis to calculate the y component of $\stackrel{→}{b}$.

role="math" localid="1657948986680" $$\begin{gathered} {\stackrel{\mathit{→}}{b}}_y=\stackrel{\mathit{→}}{b}3\sin 30° \\ =4\left(0.5\right) \\ =2.00\mathrm{m} \end{gathered}$$

Therefore, the y component of$\stackrel{→}{b}$ is 2.00 m.

Use the vector diagram to find the x component of $\stackrel{→}{c}$.

role="math" localid="1657949144986" $$\begin{gathered} {\stackrel{→}{c}}_x=\stackrel{→}{c}\sin 30° \\ =10\left(-0.5\right) \\ =-5.00\mathrm{m} \end{gathered}$$

Therefore, the x component$\stackrel{→}{c}$ is -5 m.

Use the vector diagram to find the y component of .

$$\begin{gathered} {\stackrel{→}{c}}_y=\stackrel{→}{c}\sin 30° \\ =10\left(0.866\right) \\ =0.866\mathrm{m} \end{gathered}$$

The y component of$\stackrel{→}{c}$ is 8.66 m.

The vectorcan be written as,

$\stackrel{→}{c}=-5\stackrel{Áåœ}{i}+8.66\stackrel{Áåœ}{j}$

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

$\stackrel{→}{a}=3\stackrel{Áåœ}{i}$

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

$\stackrel{→}{b}=3.46\stackrel{Áåœ}{i}+2\stackrel{Áåœ}{j}$

Now, it is given that,

$\stackrel{→}{c}=p\stackrel{→}{a}+q\stackrel{→}{b}$

This can be further elaborated as,

$$\begin{gathered} \stackrel{→}{c}=c_x\stackrel{Áåœ}{i}+c_y\stackrel{Áåœ}{j} \\ =p\left(a_x\stackrel{Áåœ}{i}\right)+q\left(b_x\stackrel{Áåœ}{i}+b_y\stackrel{Áåœ}{j}\right) \\ =\left(p{a}_x+q{b}_x\right)\stackrel{Áåœ}{i}+q{b}_y\stackrel{Áåœ}{j} \end{gathered}$$

Hence,

$c_x=p{a}_x+q{b}_x$

And,

$c_y=q{b}_y$

Therefore, after substituting the values of the respective quantities, we get

$$\begin{gathered} -5.00\mathrm{m}=\mathrm{p}\left(3.00\mathrm{m}\right)+\mathrm{q}\left(3.46\mathrm{m}\right) \\ 8.66\mathrm{m}=\mathrm{q}\left(2.00\mathrm{m}\right) \end{gathered}$$

Solving above two equations, we can find the values of p as,

$p=-6.67\mathrm{m}$

Therefore, the value ofpis -6.67 m .

Also, from step 9, we get,

$2q=8.66$

The value of q is,

$q=4.33\mathrm{m}$

Therefore, the value of q is 4.33 m .