91Ó°ÊÓ

• The vector$\stackrel{→}{\mathrm{A}}$ is 2.80 cm long and makes an angle of$60°$ with x-axis in the first quadrant
• The vector $\stackrel{→}{\mathrm{B}}$is 1.90 cm long and makes an angle of$60°$ with x-axis in the fourth quadrant

Consider a vector quantity $\stackrel{→}{V}=V_x\stackrel{Áåœ}{i}+V_y\stackrel{Áåœ}{j}$, Here$V_x$ and$V_y$ are the components along x, andy directions respectively and$\stackrel{Áåœ}{i},\stackrel{Áåœ}{j}$ are the unit vectors along x, andy directions respectively.

The magnitude of$\stackrel{→}{V}$ can be expressed as,

$\left|\stackrel{→}{V}\right|=\sqrt{V_x^2+V_y^2}$

The direction of this vector is expressed as,

$\tan \mathrm{θ}=\frac{V_y}{V_x}$

Part(a)

The Representation of $\stackrel{→}{A}$in terms of unit vectors is,

$\stackrel{→}{A}=A_x\hat{i}+A_y\hat{j}$.

From the given diagram angle between $\stackrel{→}{A}$and x is,

$\mathrm{θ}=60°$

Thus, the components of vector $\stackrel{→}{A}$is ,

$$\begin{gathered} A_x=A\cos \mathrm{θ} \\ {A}_y=A\sin \mathrm{θ} \end{gathered}$$

Substitute 2.80 cm for A and $60°$for $\mathrm{θ}$in the above equations,

$$\begin{gathered} A_x=(2.80\mathrm{cm})\cos \left(60°\right) \\ =1.40\mathrm{cm} \\ {A}_y=(2.80\mathrm{cm})\sin \left(60°\right) \\ =2.42\mathrm{cm} \end{gathered}$$

Thus, the representation of vector $\stackrel{→}{A}$in terms of unit vectors is ,

$\stackrel{→}{A}=(1.40\mathrm{cm})\hat{i}+(2.42\mathrm{cm})\hat{j}$

The Representation of$\stackrel{→}{B}$in terms of unit vectors is,

$\stackrel{→}{B}=B_x\hat{i}+B_y\hat{j}$.

From the given diagram angle between $\stackrel{→}{B}$and x is,

$\mathrm{θ}=-60°$

Thus, the components of vector$\stackrel{→}{B}$is ,

$$\begin{gathered} B_x=B\cos \mathrm{θ} \\ {B}_y=B\sin \mathrm{θ} \end{gathered}$$

Substitute 1.90 cm for B and$60°$for$\mathrm{θ}$in the above equations,

$$\begin{gathered} B_x=(1.90\mathrm{cm})\cos \left(-60°\right) \\ =0.95\mathrm{cm} \\ {B}_y=(1.90\mathrm{mm})\sin \left(-60°\right) \\ =-1.64\mathrm{cm} \end{gathered}$$

Thus, the representation of vector$\stackrel{→}{B}$in terms of unit vectors is ,

$\stackrel{→}{B}=(0.95\mathrm{cm})\hat{i}+(-1.64\mathrm{cm})\hat{j}$

Thus, the vector sum of$\stackrel{→}{A}+\stackrel{→}{B}$can be expressed as,

$\stackrel{→}{R}=\stackrel{→}{A}+\stackrel{→}{B}$

Substitute for$\stackrel{→}{A}$and $\stackrel{→}{B}$,

$$\begin{gathered} \stackrel{→}{R}=\left(\right(1.40\mathrm{cm})\hat{i}+(2.42\mathrm{cm}\left)\hat{j}\right)+\left(\right(0.95\mathrm{cm})\hat{i}+(-1.64\mathrm{cm}\left)\hat{j}\right) \\ =(1.40\mathrm{cm}+0.95\mathrm{cm})\hat{i}+(2.42\mathrm{cm}+(-1.64\mathrm{cm}\left)\right)\hat{j} \\ =(2.35\mathrm{cm})\hat{i}+(0.78\mathrm{cm})\hat{j} \end{gathered}$$

The vector diagram representation of$\stackrel{→}{R}$is shown below,

The magnitude of$\stackrel{→}{R}$ can be expressed as,

$R=\sqrt{R_x^2+R_y^2}$

Substitute 2.35 cm for$R_x$ and 0.78 cm$R_y$ in the above equation

$$\begin{gathered} R=\sqrt{(2.35\mathrm{cm}{)}^2+(0.78\mathrm{cm}{)}^2} \\ =2.48\mathrm{cm} \end{gathered}$$

The direction of$\stackrel{→}{R}$ can be expressed as,

$tan\mathrm{θ}=\frac{R_y}{R_x}$

Substitute 2.35 cm for$R_x$ and 0.78 cm$R_y$ in the above equation

$$\begin{gathered} \tan \mathrm{θ}=\frac{0.78\mathrm{cm}}{2.35\mathrm{cm}} \\ =0.3319 \\ \mathrm{θ}={\tan }^{-1}(0.3319) \\ =18.4° \end{gathered}$$

Thus, the magnitude of$\stackrel{→}{A}+\stackrel{→}{B}$ is 2.48 cm and its angle with x-axis is$18.4°$ and it is matches with graphical diagram.

Part(b)

The vector difference of$\stackrel{→}{A}-\stackrel{→}{B}$ can be expressed as,

$\stackrel{→}{R}=\stackrel{→}{A}+\stackrel{→}{B}$

Substitute for$\stackrel{→}{A}$ and $\stackrel{→}{B}$,

$$\begin{gathered} \stackrel{→}{R}=\left(\right(1.40\mathrm{cm})\hat{i}+(2.42\mathrm{cm}\left)\hat{j}\right)-\left(\right(0.95\mathrm{cm})\hat{i}+(-1.64\mathrm{cm}\left)\hat{j}\right) \\ =(1.40\mathrm{cm}-0.95\mathrm{cm})\hat{i}+(2.42\mathrm{cm}-(-1.64\mathrm{cm}\left)\right)\hat{j} \\ =(0.45\mathrm{cm})\hat{i}+(4.06\mathrm{cm})\hat{j} \end{gathered}$$

The vector diagram representation of$\stackrel{→}{R}$ is shown below,

The magnitude of$\stackrel{→}{R}$ can be expressed as,

$R=\sqrt{R_x^2+R_y^2}$

Substitute 0.45 cm for$R_x$ and 4.06 cm$R_y$ in the above equation

$$\begin{gathered} R=\sqrt{(0.45\mathrm{cm}{)}^2+(4.06\mathrm{cm}{)}^2} \\ =4.09\mathrm{cm} \end{gathered}$$

The direction of$\stackrel{→}{R}$ can be expressed as,

$\tan \mathrm{θ}=\frac{R_y}{R_x}$

Substitute 0.45 cm for$R_x$ and 4.06 cm$R_y$ in the above equation

$$\begin{gathered} \tan \mathrm{θ}=\frac{4.06\mathrm{cm}}{0.45\mathrm{cm}} \\ =9.02 \\ \mathrm{θ}={\tan }^{-1}(9.02) \\ =83.7° \end{gathered}$$

Thus, the magnitude of$\stackrel{→}{A}+\stackrel{→}{B}$ is 4.09 cm and its angle with x-axis is$83.7°$ and it is matches with graphical diagram

Part(b)

The vector difference of$\stackrel{→}{B}-\stackrel{→}{A}$ can be expressed as,

$\stackrel{→}{R}=\stackrel{→}{B}-\stackrel{→}{A}$

Substitute for$\stackrel{→}{B}$ and $\stackrel{→}{A}$,

$$\begin{gathered} \stackrel{→}{R}=\left(\right(0.95\mathrm{cm})\hat{i}+(-1.64\mathrm{cm}\left)\hat{j}\right)-\left(\right(1.40\mathrm{cm})\hat{i}+(2.42\mathrm{cm}\left)\hat{j}\right) \\ =(0.95\mathrm{cm}-1.40\mathrm{cm})\hat{i}+(-1.64\mathrm{cm}-2.42\mathrm{cm})\hat{j} \\ =-0.45\hat{i}-4.06\hat{j} \end{gathered}$$

The vector diagram representation of$\stackrel{→}{R}$ is shown below,

The magnitude of$\stackrel{→}{R}$ can be expressed as,

$R=\sqrt{R_x^2+R_y^2}$

Substitute -0.45 cm for$R_x$ and -4.06 cm$R_y$ in the above equation

$$\begin{gathered} R=\sqrt{{\left(-0.45cm\right)}^2+{\left(-4.06cm\right)}^2} \\ =4.09cm \end{gathered}$$

The direction of$\stackrel{→}{R}$ can be expressed as,

$\tan \mathrm{θ}=\frac{R_y}{R_x}$

Substitute 0.45 cm for$R_x$ and 4.06 cm$R_y$ in the above equation

$$\begin{gathered} \tan \mathrm{θ}=\frac{4.06cm}{0.45cm} \\ =9.02 \\ \mathrm{θ}={\tan }^{-1}\left(9.02\right) \\ =83.7°+180° \\ =264° \end{gathered}$$

Thus, the magnitude of$\stackrel{→}{B}-\stackrel{→}{A}$ is 4.09 cm and its angle with x-axis is$264°$ and it is matches with graphical diagram