91影视

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

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${}^{\stackrel{\mathbf{鈫�}}{\mathbf{A}}}$ along the X and Y axis are,

$$\begin{gathered} {\mathbf{A}}_{\mathbf{x}}\mathbf{=}\stackrel{\mathbf{鈫�}}{\mathbf{A}\mathbf{}}\mathbf{肠辞蝉胃} \\ {\mathbf{A}}_{\mathbf{y}}\mathbf{=}\stackrel{\mathbf{鈫�}}{\mathbf{A}}\mathbf{}\mathbf{蝉颈苍胃} \end{gathered}$$

(where${}^{\mathbf{胃}}$is the angle made by a vector${}^{\stackrel{\mathbf{鈫�}}{\mathbf{A}}}$with the x-axis.)

From above vector diagram,

• The x-component of ${}^{\stackrel{鈫�}{a}}$is${}^{{\mathrm{a}}_{\mathrm{x}}=5.0\mathrm{m},\mathrm{as}\mathrm{a}\mathrm{along}\mathrm{x}鈥�\mathrm{axis}.}$

• The y-component of ${}^{\stackrel{鈫�}{a}}$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 ${}^{\stackrel{鈫�}{\mathrm{b}}}$is,

$$\begin{gathered} b_x=\stackrel{鈫�}{b}\cos {35}^{掳} \\ =-2.29\mathrm{m} \end{gathered}$$

• The y-component of ${}^{\stackrel{鈫�}{b}}$is,

$$\begin{gathered} b_y=\stackrel{鈫�}{b}\sin {35}^{掳} \\ =3.28\mathrm{m} \end{gathered}$$

Let鈥檚 assume that ${}^{\stackrel{鈫�}{c}=\stackrel{鈫�}{b}+\stackrel{鈫�}{a}}$. Therefore, the x-component of ${}^{\stackrel{鈫�}{c}}$is,

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

The y-component of ${}^{\stackrel{鈫�}{c}}$is,

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

Then, the magnitude of ${}^{\stackrel{鈫�}{c}}$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 ${}^{\stackrel{鈫�}{a}+\stackrel{鈫�}{b}}$is ${}^{4.2m}$.

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

$$\begin{gathered} \tan 胃=\frac{c_y}{c_x} \\ =\frac{3.28}{2.71} \\ 胃={\tan }^{-1}\left(1.21\right) \\ =50.{44}^{掳} \\ 鈮�{50}^{掳} \end{gathered}$$

Therefore, the direction of ${}^{\stackrel{鈫�}{a}+\stackrel{鈫�}{b}}$is at ${}^{{50}^{掳}}$.

Now, let鈥檚 assume that ${}^{\stackrel{鈫�}{d}=\stackrel{鈫�}{b}-\stackrel{鈫�}{a}}$. The x-component of ${}^{\stackrel{鈫�}{d}}$is,

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

The y-component of ${}^{\stackrel{鈫�}{d}}$is,

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

Then, the magnitude of ${}^{\stackrel{鈫�}{d}}$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${}^{\stackrel{鈫�}{b}-\stackrel{鈫�}{a}}$is${}^{8.0m}$.

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

$$\begin{gathered} \tan 胃=\frac{d_y}{d_x} \\ =\frac{3.28}{-7.29} \\ 胃={\tan }^{-1}\left(-4.50\right) \\ =24.2^{掳} \\ 鈮�{24}^{掳} \end{gathered}$$

Therefore, the direction of ${}^{\stackrel{鈫�}{b}-\stackrel{鈫�}{a}}$is at${}^{{24}^{掳}}$.

For addition

For subtraction