Two vectors $\overrightarrow{\mathbf{a}}$and $\overrightarrow{\mathbf{b}}$have the components, in meters ${\mathit{a}}_{\mathbf{x}}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{2}$,role="math" localid="1657003775216" ${\mathit{a}}_{\mathit{y}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{6}$,${\mathit{b}}_{\mathbf{x}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{50}$,${\mathit{b}}_{\mathbf{y}}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{5}$,(a) Find the angle between the directions of $\overrightarrow{\mathbf{a}}$and $\overrightarrow{\mathbf{b}}$.There are two vectors in the xy plane that are perpendicular to $\overrightarrow{\mathbf{a}}$and have a magnitude of 5.0 m. One, vector $\overrightarrow{\mathbf{c}}$, has a positive x component and the other, vector $\overrightarrow{\mathbf{d}}$, a negative x component. What are (b) the x component and (c) the y component of vector, and (d) the x component and (e) the y component of vector$\overrightarrow{\mathbf{d}}$?

The components of$\overrightarrow{a}\mathrm{are},a_x=3.2\mathrm{m},a_y=1.6\mathrm{m}$

The components of$\overrightarrow{b},b_x=0.5\mathrm{m},b_y=4.5\mathrm{m}$$\overrightarrow{b},b_x=0.5\mathrm{m},b_y=4.5\mathrm{m}$

The magnitude of $\overrightarrow{c}$is 5.0 m

The magnitude of $\overrightarrow{d}$is5.0 m

A scalar product is the product of the magnitude of one vector and the scalar component of the second vector along the direction of the first vector.Using the formula for scalar product and vector product we can find the angle between two vectors.

Formula:

$\cos \theta =\frac{\overrightarrow{a}.\overrightarrow{b}}{ab}$ …(i)

Write the vectors in the unit vector notations first.

$$\begin{gathered} \overrightarrow{a}=a_x\stackrel{⏜}{i}+a_y\stackrel{⏜}{j}+a_z\stackrel{⏜}{k} \\ =\left(3.2\stackrel{⏜}{i}+1.6\stackrel{⏜}{j}\right)\mathrm{m} \\ \overrightarrow{b}=b_x\stackrel{⏜}{i}+b_y\stackrel{⏜}{j}+b_z\stackrel{⏜}{k} \\ =\left(0.5\stackrel{⏜}{i}+4.5\stackrel{⏜}{j}\right)\mathrm{m} \end{gathered}$$

Now, use equation (i) to find the angle between the two vectors.

$$\begin{gathered} \cos \theta =\frac{\left(3.2\stackrel{⏜}{i}+1.6\stackrel{⏜}{j}\right)m.\left(0.5\stackrel{⏜}{i}+4.5\stackrel{⏜}{j}\right)m}{\left[\left(\sqrt{3.2^2+1.6^2}\right)m.\left(\sqrt{0.5^2+4.5^2}\right)m\right]} \\ =0.5426 \\ \theta ={\cos }^{-1}\left(0.5426\right) \\ =57° \end{gathered}$$

Therefore, the angle between the two vectors is $57°$.

The angle made bywith x-axis is,

$$\begin{gathered} \theta ={\tan }^{-1}\left(\frac{1.6\mathrm{m}}{3.2\mathrm{m}}\right) \\ =26.6° \end{gathered}$$

As $\overrightarrow{c}$is perpendicular to $\overrightarrow{a}$, the angle made by $\overrightarrow{c}$with x-axis is $26.6°-90°=-63.4°$

The x component of

$$\begin{gathered} c_x=\left(5m\right).\cos \left(-63.4°\right) \\ =2.2\mathrm{m} \end{gathered}$$

Therefore, the x component of $\overrightarrow{c}$is 2.2m.

The y component of $\overrightarrow{c}$is calculated as,

$$\begin{gathered} c_y=\left(5m\right).\sin \left(-63.4°\right) \\ =-4.5m \end{gathered}$$

As $\overrightarrow{d}$is perpendicular to $\overrightarrow{a}$, the angle made by $\overrightarrow{d}$with x-axis is$26.6°+90°=116.6°$

Therefore, the x component of$\overrightarrow{d}$

role="math" localid="1657006734048" $$\begin{gathered} d_x=\left(5m\right).\cos \left(116.6°\right) \\ =-2.2\mathrm{m} \end{gathered}$$

Therefore, the x component of$\overrightarrow{d}$ is -2.2m

The y component ofis calculated as follows

$$\begin{gathered} d_y=\left(5m\right).\sin \left(116.6°\right) \\ =4.5\mathrm{m} \end{gathered}$$

Therefore, the y component of$\overrightarrow{d}$ is 4.5m.