Vector ${}^{\overrightarrow{\mathbf{a}}}$ has a magnitude of localid="1654586257370" ${}^{\mathbf{5}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{m}}$and is directed east. Vector localid="1654586265256" ${}^{\stackrel{\mathbf{\rightharpoonup }}{\mathbf{b}}}$has a magnitude of localid="1654586273265" ${}^{\mathbf{4}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{m}}$and is directedlocalid="1654586281097" ${}^{{\mathbf{35}}^{\mathbf{°}}}$west of due north. Calculate (a) the magnitude and (b) the direction oflocalid="1654586314371" role="math" ${}^{\stackrel{\mathbf{\rightharpoonup }}{\mathbf{a}}\mathbf{+}\stackrel{\mathbf{\rightharpoonup }}{\mathbf{b}}}$? What are (c) the magnitude and (d) the direction oflocalid="1654586297109" ${}^{\stackrel{\mathbf{\rightharpoonup }}{\mathbf{b}}\mathbf{+}\stackrel{\mathbf{\rightharpoonup }}{\mathbf{a}}}$? (e) Draw a vector diagram for each combination.

1. Magnitude of${}^{\overrightarrow{\mathbf{a}}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{m}}$,${}^{\overrightarrow{\mathbf{a}}}$is directed to east.
2. Magnitude of${}^{\overrightarrow{\mathbf{b}}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{m}}$,${}^{\overrightarrow{\mathbf{b}}}$is directed to${}^{{\mathbf{35}}^{\mathbf{°}}\mathbf{}\mathbf{west}\mathbf{}\mathbf{of}\mathbf{}\mathbf{north}}$

Using the vector diagram, we can do the addition and subtraction of vectors. Also, we can find the direction of vectors.

The components of the vector${}^{\overrightarrow{\mathbf{A}}}$ along the X and Y axis are,

$$\begin{gathered} {\mathbf{A}}_{\mathbf{x}}\mathbf{=}\overrightarrow{\mathbf{A}\mathbf{}}\mathbf{cos\theta } \\ {\mathbf{A}}_{\mathbf{y}}\mathbf{=}\overrightarrow{\mathbf{A}}\mathbf{}\mathbf{sin\theta } \end{gathered}$$

(where${}^{\mathbf{\theta }}$is the angle made by a vector${}^{\overrightarrow{\mathbf{A}}}$with the x-axis.)

From above vector diagram,

• The x-component of ${}^{\overrightarrow{a}}$is${}^{{\mathrm{a}}_{\mathrm{x}}=5.0\mathrm{m},\mathrm{as}\mathrm{a}\mathrm{along}\mathrm{x}‐\mathrm{axis}.}$

• The y-component of ${}^{\overrightarrow{a}}$is${}^{a_{y\mathit{=}\mathit{0}\mathit{.}\mathit{0}\mathit{}m\mathit{,}\mathit{}as\mathit{}a\mathit{}along\mathit{}\mathit{-}axis\mathit{.}}}$
  

• The x-component of ${}^{\overrightarrow{\mathrm{b}}}$is,

$$\begin{gathered} b_x=\overrightarrow{b}\cos {35}^{°} \\ =-2.29\mathrm{m} \end{gathered}$$

• The y-component of ${}^{\overrightarrow{b}}$is,

$$\begin{gathered} b_y=\overrightarrow{b}\sin {35}^{°} \\ =3.28\mathrm{m} \end{gathered}$$

Let’s assume that ${}^{\overrightarrow{c}=\overrightarrow{b}+\overrightarrow{a}}$. Therefore, the x-component of ${}^{\overrightarrow{c}}$is,

$$\begin{gathered} c_x=a_x+b_x \\ =5-2.29 \\ =2.71\mathrm{m} \end{gathered}$$

The y-component of ${}^{\overrightarrow{c}}$is,

$$\begin{gathered} c_y=a_y+b_y \\ =0+3.28 \\ =3.28 \end{gathered}$$

Then, the magnitude of ${}^{\overrightarrow{c}}$is,

$$\begin{gathered} c=\sqrt{{c_x}^2+{c_y}^2} \\ =\sqrt{2.{71}^2+3.{28}^2} \\ =4.2m \end{gathered}$$

Therefore, the magnitude of ${}^{\overrightarrow{a}+\overrightarrow{b}}$is ${}^{4.2m}$.

The direction of ${}^{\overrightarrow{\mathbf{a}}\mathbf{+}\overrightarrow{\mathbf{b}}}$is given by the angle subtended by it with x axis which is

$$\begin{gathered} \tan \theta =\frac{c_y}{c_x} \\ =\frac{3.28}{2.71} \\ \theta ={\tan }^{-1}\left(1.21\right) \\ =50.{44}^{°} \\ \approx {50}^{°} \end{gathered}$$

Therefore, the direction of ${}^{\overrightarrow{a}+\overrightarrow{b}}$is at ${}^{{50}^{°}}$.

Now, let’s assume that ${}^{\overrightarrow{d}=\overrightarrow{b}-\overrightarrow{a}}$. The x-component of ${}^{\overrightarrow{d}}$is,

$$\begin{gathered} d_x=b_x-a_x \\ =-2.29-5 \\ =-7.29\mathrm{m} \end{gathered}$$

The y-component of ${}^{\overrightarrow{d}}$is,

$$\begin{gathered} d_y=b_y-a_y \\ =3.28-0 \\ =3.28 \end{gathered}$$

Then, the magnitude of ${}^{\overrightarrow{d}}$is,

$$\begin{gathered} d=\sqrt{{d_x}^2+{d^2}_y} \\ =\sqrt{{\left(-7.29\right)}^3+3.{28}^2} \\ =8.0m \end{gathered}$$

Therefore, the magnitude of${}^{\overrightarrow{b}-\overrightarrow{a}}$is${}^{8.0m}$.

The direction of${}^{\overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{a}}}$is given by the angle subtended by it with x axis which is

$$\begin{gathered} \tan \theta =\frac{d_y}{d_x} \\ =\frac{3.28}{-7.29} \\ \theta ={\tan }^{-1}\left(-4.50\right) \\ =24.2^{°} \\ \approx {24}^{°} \end{gathered}$$

Therefore, the direction of ${}^{\overrightarrow{b}-\overrightarrow{a}}$is at${}^{{24}^{°}}$.

For addition

For subtraction