Q40E - University Physics with Modern Physics

Question

You are given two vectors $\overrightarrow{\mathbf{A}}\mathbf{=}\mathbf{-}\mathbf{3}\mathbf{.00}\widehat{\mathbf{i}}\mathbf{+}\mathbf{6}\mathbf{.00}\widehat{\mathbf{j}}$  and $\overrightarrow{\mathbf{B}}\mathbf{=}\mathbf{7}\mathbf{.00}\widehat{\mathbf{i}}\mathbf{+}\mathbf{2}\mathbf{.00}\widehat{\mathbf{j}}$ . Let counter- clockwise angles be positive. (a) What angle does $\overrightarrow{A}$ make with the +x-axis? (b) What angle does $\overrightarrow{\mathbf{B}}$  make with the +x-axis? (c) Vector $\overrightarrow{\mathbf{C}}$ is the sum of  $\overrightarrow{\mathit{A}}$ and $\overrightarrow{\mathbf{B}}$ , so  $\overrightarrow{\mathbf{C}}\mathbf{=}\overrightarrow{\mathbf{A}}\mathbf{+}\overrightarrow{\mathbf{B}}$ What angle does  $\overrightarrow{\mathbf{C}}$ make with the +x-axis?

Step-by-Step Solution

Answer

Answer

 

(a)  $\overrightarrow{A}$ makes an angle of 116.6^o  with +x-axis.

(b)  $\overrightarrow{B}$ makes an angle of  15.9^o with +x-axis.

(c)  $\overrightarrow{C}$ makes an angle of 63.4^o  with +x-axis.

Step-by-Step Solution Step 1: Identification of given data

The given vectors are $\overrightarrow{A}=-3.00\widehat{i}+6.00\widehat{j} and \overrightarrow{B}=7.00\widehat{i}+2.00\widehat{j}$.

Step 2: Vector Quantities and their magnitudes.

Consider a vector quantity $\overrightarrow{V}=V_x\widehat{i}+V_y\widehat{j}$ , Here V_x  and  V_y are the components along x, and y directions respectively and $\widehat{i},\widehat{j}$ are the unit vectors along x, and y directions respectively. The direction of this vector is expressed as,

$\tan \theta =\frac{V_y}{V_x}$

Step 3: Estimation of angle between x-axis and A →

Part (a)

 

The expression for the direction of a vector is given by, 

$\tan \theta =\frac{A_y}{A_x}$

Here, A_x and  A_y are the components of  $\overrightarrow{A}$.

 

Substitute -3.00  for A_x ,  6.00 for A_y .

$\begin{array}{c}\tan \theta =\frac{6.00}{-3.00}\\ =-2\\ \theta ={\tan }^{-1}\left(-2\right)+180°\\ =116.6°\end{array}$

Thus, $\overrightarrow{A}$ makes an angle of 116.6^o with x-axis.

Step 4: Estimation of angle between x-axis and B →

Part (b)

 

The expression for the direction of a vector is given by, 

 $\tan \theta =\frac{B_y}{B_x}$

Here, B_y  and B_x are the components of $\overrightarrow{B}$.

 

Substitute  7.00 for B_x ,  2.00 for  B_y.

$\begin{array}{c}\tan \theta =\frac{2.00}{7.00}\\ =0.285\\ \theta ={\tan }^{-1}\left(0.285\right)\\ =15.9°\end{array}$

Thus,  $\overrightarrow{B}$ makes an angle of  15.9^o with x-axis.

Step 4: Estimation of angle between x-axis and C →

Part (c)

 

The sum of the given vectors is expressed as,

 $\overrightarrow{C}=\overrightarrow{A}+\overrightarrow{B}$

 

Substitute $-3.00\widehat{i}+6.00\widehat{j}$  for $\overrightarrow{A}$ , $7.00\widehat{i}+2.00\widehat{j}$  for $\overrightarrow{B}$ .

$\begin{array}{c}\overrightarrow{C}=\left(-3.00\widehat{i}+6.00\widehat{j}\right)+\left(7.00\widehat{i}+2.00\widehat{j}\right)\\ =\left(-3.00+7.00\right)\widehat{i}+\left(6.00+2.00\right)\widehat{j}\\ =4.00\widehat{i}+8.00\widehat{j}\end{array}$ 

 

The expression for the direction of a vector is given by, 

 $\tan \theta =\frac{C_y}{C_x}$

Here,  C_x and C_y are the components of  $\overrightarrow{C}$ .

 

Substitute 4.00  for C_x , 8.00  for C_y .

 $\begin{array}{c}\tan \theta =\frac{8.00}{4.00}\\ =2.00\\ \theta ={\tan }^{-1}\left(2.00\right)\\ =63.4°\end{array}$

Thus, $\overrightarrow{C}$  makes an angle of 63.4^o  with x-axis.