$\overrightarrow{\mathbf{A}}$has the magnitude$\stackrel{}{\mathbf{12}\mathbf{.}\mathbf{0}\mathbf{m}}$and is angled$\stackrel{}{\mathbf{60}\mathbf{°}}$counterclockwise from the positive direction of the x axis of an$\stackrel{}{\mathbf{xy}}$ coordinate system. Also,$\overrightarrow{\mathbf{B}}\mathbf{=}\mathbf{(}\mathbf{12.0}\mathbf{}\mathbf{\text{m}}\mathbf{)}\widehat{\mathbf{\text{i}}}\mathbf{+}\mathbf{(}\mathbf{8.0}\mathbf{}\mathbf{\text{m}}\mathbf{)}\widehat{\mathbf{\text{j}}}$on that same coordinate system. We now rotate the system counter clockwise about the origin by$\mathbf{20}\mathbf{°}$ to form an${\mathbf{x}}^{\mathbf{\text{'}}}{\mathbf{y}}^{\mathbf{\text{'}}}$ system. On this new system, what are (a)$\overrightarrow{\mathbf{A}}$and (b)$\overrightarrow{\mathbf{B}}$, both in unit-vector notation?

The magnitude of$\overrightarrow{\mathrm{A}}$ is$12.0\text{m}$ and angle$\mathrm{\mu }=60.0°$ counterclockwise.

$\overrightarrow{\mathrm{B}}=12.0\hat{\mathrm{i}}+8.00\hat{\mathrm{j}}$.

The coordinate system rotated through an angle $20.0°$ counterclockwise.

New angles of vectors after rotations can be found by using the original angles and the angle by which the system is rotated. Using these new angels, we can answer the above question.

After rotating the system, new angle,$\mathrm{\mu }\text{'}$is calculated using the old angles as,

$\begin{array}{c}\mathrm{\mu }\text{'}=60.0°-20.0°\\ =40.0°\end{array}$

Hence, the components of $\overrightarrow{\mathrm{A}}$in the new system are calculated as:

The$\mathrm{x}$component is,

$\begin{array}{c}{{\mathrm{A}}^{\text{'}}}_{\mathrm{x}}=\mathrm{Acos}(40°)\\ =12\times 0.7660\\ =9.19\text{\hspace{0.17em}m}\end{array}$

The $\mathrm{y}$component is,

$\begin{array}{c}\mathrm{A}{\text{'}}_{\mathrm{y}}=\mathrm{Asin}(40°)\\ =12\times 0.6428\\ =7.71\text{\hspace{0.17em}m}\end{array}$

Therefore, the vector$\overrightarrow{\mathrm{A}}$ in the unit vector notation is written as $9.19\hat{\mathrm{i}}+7.7\hat{\mathrm{j}}$.

The given vector$\overrightarrow{\mathrm{B}}$is written as,

$\overrightarrow{\mathrm{B}}=12.0\hat{\mathrm{i}}+8.00\hat{\mathrm{j}}$

We first need to magnitude and angle of $\overrightarrow{B}$in the original system.

$\begin{array}{c}\left|\overrightarrow{\mathrm{B}}\right|=\sqrt{{{\mathrm{B}}_{\mathrm{x}}}^2+{{\mathrm{B}}_{\mathrm{y}}}^2}\\ =\sqrt{{12}^2+8^2}\\ =\sqrt{208}\\ =14.42\text{\hspace{0.17em}m}\end{array}$

The angle is found using the tan function as,

$\begin{array}{c}\mathrm{tan\alpha }=\frac{{\mathrm{B}}_{\mathrm{y}}}{{\mathrm{B}}_{\mathrm{x}}}\\ \mathrm{\alpha }={\tan }^{-1}\left(\frac{{\mathrm{B}}_{\mathrm{y}}}{{\mathrm{B}}_{\mathrm{x}}}\right)\\ ={\tan }^{-1}\left(\frac{8.00}{12.0}\right)\\ =33.69°\end{array}$

Now, after rotating the system, new angle $\mathrm{\mu }\text{'}$is,

$\begin{array}{c}\mathrm{\mu }\text{'}=33.69°-20.0°\\ =13.69°\end{array}$

Hence, the components of $\overrightarrow{B}$in the new system are calculated as:

The $x$component is,

$\begin{array}{c}{{\mathrm{B}}^{\text{'}}}_{\mathrm{x}}=\mathrm{Bcos}(13.69°)\\ =14.42\times 0.9716\\ =14.0\text{\hspace{0.17em}m}\end{array}$

And, the $y$component is,

$\begin{array}{c}\mathrm{B}{\text{'}}_{\mathrm{y}}=\mathrm{Bsin}(13.69°)\\ =14.42\times 0.2367\\ =3.41\text{\hspace{0.17em}m}\end{array}$

Therefore, the vector$\overrightarrow{\mathrm{B}}$ in the unit vector notation is written as $14.0\hat{\mathrm{i}}+3.41\hat{\mathrm{j}}$.