Question

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.

Short answer

1. The magnitude of${}^{\overrightarrow{\mathbf{a}}\mathbf{+}\overrightarrow{\mathbf{b}}}$is${}^{\mathbf{4}\mathbf{.}\mathbf{2}\mathbf{}\mathbf{m}}$.
2. The direction of${}^{\overrightarrow{\mathbf{a}}\mathbf{+}\overrightarrow{\mathbf{b}}}$is given by the angle subtended by it with x axis which is${}^{{\mathbf{50}}^{\mathbf{o}}}$.
3. The magnitude${}^{\overrightarrow{\mathbf{b}}\mathbf{-}\overrightarrow{\mathbf{a}}}$of is 8 m.
4. The direction of${}^{\overrightarrow{\mathbf{b}}\mathbf{-}\overrightarrow{\mathbf{a}}}$is given by the angle subtended by it with x axis which is${}^{\mathbf{-}{\mathbf{24}}^{\mathbf{o}}}$.

Step-by-step solution

Given information

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${}^{{\mathbf{35}}^{\mathbf{°}}\mathbf{}\mathbf{west}\mathbf{}\mathbf{of}\mathbf{}\mathbf{north}}$

Understanding the concept of vector diagram

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.)

Draw the vector diagram

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,

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

(a) Calculate the magnitude of a→+b→

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 ${}^{4.2m}$.

(b) Calculate the direction ofa→+b→

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 ${}^{{50}^{°}}$.

(c) Calculate the magnitude of b→-a→

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${}^{8.0m}$.

(d) Calculate the direction of b→+a→ 

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${}^{{24}^{°}}$.

(d) Draw a vector diagram for each combination

For addition

For subtraction