Q38E - University Physics with Modern Physics

Question

Given two vectors  $\overrightarrow{A}=4.00\widehat{i}+7.00\widehat{j}$ and $\overrightarrow{B}=5.00\widehat{i}-7.00\widehat{j}$ , (a) find the magnitude of each vector; (b) use unit vectors to write an expression for the vector difference $\overrightarrow{A}-\overrightarrow{B}$ ; and (c) find the magnitude and direction of the vector difference $\overrightarrow{A}-\overrightarrow{B}$ . (d) In a vector diagram show $\overrightarrow{A},\overrightarrow{B}$  and $\overrightarrow{A}-\overrightarrow{B}$, and show that your diagram agrees qualitatively with your answer to part (c).

Step-by-Step Solution

Answer

Answer

 

a) The magnitude of  $\overrightarrow{A}$ is 8.06, and the magnitude of $\overrightarrow{B}$  is 5.39.

b) The vector difference  $\overrightarrow{A}-\overrightarrow{B}$ can be expressed as, $\overrightarrow{C}=-1.00\widehat{i}+9.00\widehat{j}$ .

c) The magnitude of $\overrightarrow{A}-\overrightarrow{B}$  is 9.06 and it makes an angle of 96.3^o  with x-axis.

d)The vector diagram agrees with part (c) results.

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

The vector  $\overrightarrow{A}$ is given as  $\overrightarrow{A}=4.00\widehat{i}+7.00\widehat{j}$ and the vector $\overrightarrow{B}$  is given as, $\overrightarrow{B}=5.00\widehat{i}-2.00\widehat{j}$.

Step-2: Magnitude of a vector

The magnitude of vector  $\overrightarrow{G}=G_x\widehat{i}+G_y\widehat{j}$ can be expressed as,

 $\left|\overrightarrow{G}\right|=\sqrt{G_x^2+G_y^2}$

Here  $G_x,G_y$ are the components in x and y direction and $\widehat{i},\widehat{j}$  are the unit vectors in x and y directions receptively.

Step-3: Estimation of magnitudes of given vectors

Part (a)

 

The components of $\overrightarrow{A}$  can be represented as,

 $\begin{array}{l}\overrightarrow{A}=4.00\widehat{i}+7.00\widehat{j}\\ A_x=4.00,A_y=7.00\end{array}$

 

The magnitude of  $\overrightarrow{A}$ can be calculated as,

 $\left|\overrightarrow{A}\right|=\sqrt{A_x^2+A_y^2}$

 

Substitute 4.00 for A_X , and 7.00 for  A_Y.

 $\begin{array}{c}\left|\overrightarrow{A}\right|=\sqrt{{\left(4.00\right)}^2+{\left(7.00\right)}^2}\\ =\sqrt{65}\\ =8.06\end{array}$

 

The components of  $\overrightarrow{B}$can be represented as,

 $\begin{array}{l}\overrightarrow{B}=5.00\widehat{i}-2.00\widehat{j}\\ B_x=5.00,B_y=-2.00\end{array}$

 

The magnitude of  $\overrightarrow{B}$can be expressed as,

 $\left|\overrightarrow{B}\right|=\sqrt{B_x^2+B_y^2}$

 

Substitute 5.00 for  B_x and -2.00 for B_y,  

 $\begin{array}{c}\left|\overrightarrow{B}\right|=\sqrt{{\left(5.00\right)}^2+{\left(-2.00\right)}^2}\\ =\sqrt{29}\\ =5.39\end{array}$

 

Thus, the magnitude $\overrightarrow{A}$ of is 8.06 and magnitude of  $\overrightarrow{B}$ is 5.39.

Step-4: Estimation of vector difference

Part (b)

 

The vector $\overrightarrow{A}$is given as $\overrightarrow{A}=4.00\widehat{i}+7.00\widehat{j}$   and the vector $\overrightarrow{B}$ is given as,

$\overrightarrow{B}=5.00\widehat{i}-2.00\widehat{j}$

Consider vector $\overrightarrow{C}$  is the resultant vector $\overrightarrow{A}-\overrightarrow{B}$of  and is expressed as, $\overrightarrow{C}=\overrightarrow{A}-\overrightarrow{B}$

 $$\begin{gathered} Substitute 4.00\widehat{i}+7.00\widehat{j} for \overrightarrow{A} , and 5.00\widehat{i}-2.00\widehat{j} for \overrightarrow{B} . \\ \begin{array}{c}\overrightarrow{C}=\left(4.00\widehat{i}+7.00\widehat{j}\right)-\left(5.00\widehat{i}-2.00\widehat{j}\right)\\ =\left(4.00-5.00\right)\widehat{i}+\left(7.00-\left(-2.00\right)\right)\widehat{j}\\ =-1.00\widehat{i}+9.00\widehat{j}\end{array} \end{gathered}$$

Thus, the vector difference  $\overrightarrow{A}-\overrightarrow{B}$ can be expressed as,  $\overrightarrow{C}=-1.00\widehat{i}+9.00\widehat{j}$

Step-5: Estimation of magnitude and direction of vector difference

Part (c)

The components of $\overrightarrow{C}$  can be represented as,

$\begin{array}{l}\overrightarrow{C}=-1.00\widehat{i}+9.00\widehat{j}\\ C_x=-1.00,C_y=9.00\end{array}$

Thus, the magnitude of  $\overrightarrow{C}$can be expressed as,

$\left|\overrightarrow{C}\right|=\sqrt{C_x^2+C_y^2}$

Substitute -1.00 for C_x and 9.00 for, C_y ,

$\begin{array}{c}\left|\overrightarrow{C}\right|=\sqrt{{\left(-1.00\right)}^2+{\left(9.00\right)}^2}\\ =\sqrt{82}\\ =9.06\end{array}$

Thus, the magnitude of $\overrightarrow{A}-\overrightarrow{B}$  is 9.06.

 

The direction of a vector quantity can be expressed as, 

 $\tan \theta =\frac{C_y}{C_x}$

 

Substitute -1.00 for C_x and 9.00 for, C_y ,

 $\begin{array}{c}an\theta =\frac{9.00}{-1.00}\\ =-9.00\\ \theta ={\tan }^{-1}\left(-9.00\right)+180°\\ =96.3°\end{array}$

 

Thus, the $\overrightarrow{A}-\overrightarrow{B}$  makes an angle of 96.3^o  with x-axis.

Step-6: Representation of vector diagram

Part (d)

The vector diagram of $\overrightarrow{C}=\overrightarrow{A}-\overrightarrow{B}$  is plotted above. Which agrees with the results obtained in part (c).  C_x is negative and C_y  is positive. Thus, the resultant vector lies in the second quadrant.