The three vectors in Fig. 3-33 have magnitudes a=3.00m,b =4.00 m, and c =10.0 mand angle $\mathit{\theta }\mathbf{=}\mathbf{30}\mathbf{.}\mathbf{0}\mathbf{°}$What are (a) the x component and (b) the y component of$\overrightarrow{\mathbf{a}}$;(c) the x component and (d) the y component of b; and (e) the x component and (f) the y component of$\overrightarrow{\mathbf{c}}$? If$\overrightarrow{\mathbf{c}}\mathbf{=}\mathit{p}\overrightarrow{\mathbf{a}}\mathbf{+}\mathit{q}\overrightarrow{\mathbf{b}}$what is the value of (g) pand q?

1. The magnitude of$\overrightarrow{a}$ is 3.00 m
2. The magnitude of$\overrightarrow{b}$ is 4.00 m
3. The magnitude of$\overrightarrow{c}$ is 10.0 m
4. The angle between$\overrightarrow{a}$ and$\overrightarrow{b}$ 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$\overrightarrow{a}$ is along x –axis.Therefore, the x component of $\overrightarrow{a}$is 3.00 m.

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

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

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

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

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

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

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

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

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

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

Use the vector diagram to find the y component of .

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

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

The vectorcan be written as,

$\overrightarrow{c}=-5\stackrel{⏜}{i}+8.66\stackrel{⏜}{j}$

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

$\overrightarrow{a}=3\stackrel{⏜}{i}$

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

$\overrightarrow{b}=3.46\stackrel{⏜}{i}+2\stackrel{⏜}{j}$

Now, it is given that,

$\overrightarrow{c}=p\overrightarrow{a}+q\overrightarrow{b}$

This can be further elaborated as,

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

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 .