91Ó°ÊÓ

The vector's various coordinates are presented here. We know that in coordinates, the x component of a vector comes first, followed by the y component. To calculate the angle of the vector with respect to the x-axis, just multiply the y component by the x component and then take the inverse of that result.

This will tell you the vector's angle with the x-axis.

It can be represented as

$\mathbf{θ}\mathbf{=}{\mathbf{tan}}^{\mathbf{-}\mathbf{1}}\left(\frac{y}{x}\right)$…â¶Ä¦â¶Ä¦â¶Ä¦â¶Ä¦â¶Ä¦â¶Ä¦(1)

The vector A is$\left(-8.60\mathrm{cm},5.20\mathrm{cm}\right)$

The magnitude of the vector can be calculated by the sum of the square of the magnitude of the individual coordinates

$$\begin{gathered} \mathrm{A}=\sqrt{{\left({\mathrm{A}}_{\mathrm{x}}\right)}^2+{\left({\mathrm{B}}_{\mathrm{x}}\right)}^2} \\ =\sqrt{{\left(-8.6\mathrm{cm}\right)}^2+{\left(5.2\mathrm{cm}\right)}^2} \\ =10.04\mathrm{cm} \end{gathered}$$

Hence the magnitude of vector A is 10.04 cm

The counterclockwise angle taken from the positive x-axis is given by equation (1), such that,

$$\begin{gathered} \mathrm{θ}={\tan }^{-1}\left(\frac{{\mathrm{A}}_{\mathrm{x}}}{{\mathrm{A}}_{\mathrm{y}}}\right) \\ ={\tan }^{-1}\left(\frac{5.2\mathrm{cm}}{-8.6\mathrm{cm}}\right) \\ =\mathrm{ττ}-{\tan }^{-1}\left(\frac{5.2\mathrm{cm}}{-8.6\mathrm{cm}}\right) \\ =149° \end{gathered}$$

Hence vector A makes the angle of $149°$and magnitude is 10.04 cm .

The vector A is$\left(-9.70\mathrm{m},-2.45\mathrm{m}\right)$

The magnitude of the vector can be calculated by the sum of the square of the magnitude of the individual coordinates

$$\begin{gathered} \mathrm{A}=\sqrt{{\left({\mathrm{A}}_{\mathrm{x}}\right)}^2+{\left({\mathrm{B}}_{\mathrm{x}}\right)}^2} \\ =\sqrt{{\left(-9.70\mathrm{m}\right)}^2+{\left(-2.45\mathrm{m}\right)}^2} \\ =10.0\mathrm{cm} \end{gathered}$$

Hence the magnitude of vector A is 10.04 cm

The counterclockwise angle taken from the positive x-axis is given by equation (1), such that,

$$\begin{gathered} \mathrm{θ}={\tan }^{-1}\left(\frac{{\mathrm{A}}_{\mathrm{y}}}{{\mathrm{A}}_{\mathrm{x}}}\right) \\ ={\tan }^{-1}\left(\frac{-2.45\mathrm{m}}{-9.70\mathrm{m}}\right) \\ =14.2° \end{gathered}$$

Hence vector A makes the angle of $194.2°$and magnitude is 10.0 m.

The vector A is$\left(-9.70\mathrm{m},-2.45\mathrm{m}\right)$

The magnitude of the vector can be calculated by the sum of the square of the magnitude of the individual coordinates

$$\begin{gathered} \mathrm{A}=\sqrt{{\left({\mathrm{A}}_{\mathrm{x}}\right)}^2+{\left({\mathrm{B}}_{\mathrm{x}}\right)}^2} \\ =\sqrt{{\left(7.75\mathrm{km}\right)}^2+{\left(-2.70\mathrm{km}\right)}^2} \\ =8.21\mathrm{km} \end{gathered}$$

Hence the magnitude of vector A is 8.21 km

The counterclockwise angle taken from the positive x-axis is given by equation (1), such that,

$$\begin{gathered} \mathrm{θ}={\tan }^{-1}\left(\frac{{\mathrm{A}}_{\mathrm{y}}}{{\mathrm{A}}_{\mathrm{x}}}\right) \\ ={\tan }^{-1}\left(\frac{7.75\mathrm{m}}{-2.70\mathrm{m}}\right) \\ =\mathrm{ττ}-70.79° \\ =109.2° \end{gathered}$$

Hence vector A makes the angle of$109.2°$and magnitude is 8.21 km .

Therefore for case (a) the angle of the vector is and magnitude is , for case (b) the angle of the vector is $14.2°$ and magnitude is 10.0 m , and in case (c) the angle of the vector is $109.2°$ and magnitude is 8.21 km .