Q41E - University Physics with Modern Physics

Question

Given two vectors  $\overrightarrow{\mathbf{A}}\mathbf{=}\mathbf{-}\mathbf{2}\mathbf{.00}\widehat{\mathbf{i}}\mathbf{+}\mathbf{3}\mathbf{.00}\widehat{\mathbf{j}}\mathbf{+}\mathbf{4}\mathbf{.00}\widehat{\mathbf{k}}$ and $\overrightarrow{\mathbf{B}}\mathbf{=}\mathbf{3}\mathbf{.00}\widehat{\mathbf{i}}\mathbf{+}\mathbf{1}\mathbf{.00}\widehat{\mathbf{j}}\mathbf{-}\mathbf{3}\mathbf{.00}\widehat{\mathbf{k}}$, (a) find the magnitude of each vector; (b) use unit vectors to write an expression for the vector difference $\overrightarrow{\mathbf{A}}\mathbf{-}\overrightarrow{\mathbf{B}}$ ; and (c) find the magnitude of the vector difference $\overrightarrow{\mathbf{A}}\mathbf{-}\overrightarrow{\mathbf{B}}$ . Is this the same as the magnitude of  $\overrightarrow{\mathbf{B}}\mathbf{-}\overrightarrow{\mathbf{A}}$? Explain.

Step-by-Step Solution

Answer

Answer

 

a) The magnitude of  $\overrightarrow{A}$ is 5.38 and magnitude of  $\overrightarrow{A}$ is 4.36.

b) The vector difference is, $\overrightarrow{A}-\overrightarrow{B}=-5.00\widehat{i}+2.00\widehat{j}+7.00\widehat{k}$.

c) The magnitude of  $\overrightarrow{A}-\overrightarrow{B}$ is 8.83 and it is same as the magnitude of $\overrightarrow{B}-\overrightarrow{A}$ .

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

The given vectors are $\overrightarrow{A}=-2.00\widehat{i}+3.00\widehat{j}+4.00\widehat{k}$  and  $\overrightarrow{B}=3.00\widehat{i}+1.00\widehat{j}-3.00\widehat{k}$.

Step-2: Vector quantities and their magnitudes.

The magnitude of a vector   $\overrightarrow{V}=V_x\widehat{i}+V_y\widehat{j}+V_z\widehat{k}$is given by,

 $\left|\overrightarrow{V}\right|=\sqrt{{V_x}^2+{V_y}^2+{V_z}^2}$

Step-3: Estimation of magnitude of given vectors

Part(a)

 

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

$\left|\overrightarrow{A}\right|=\sqrt{A_{{}_x}^2+A_{{}_y}^2+A_{{}_z}^2}$

Substitute -2.00 for A_x  ,3.00 for A_y  and 4.00 for  A_z

 $\begin{array}{c}\overrightarrow{A}=-2.00\widehat{i}+3.00\widehat{j}+4.00\widehat{k}\\ \left|\overrightarrow{A}\right|=\sqrt{{\left(-2.00\right)}^2+{\left(3.00\right)}^2+{\left(4.00\right)}^2}\\ =\sqrt{29}\\ =5.38\end{array}$

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

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

 

Substitute 3.00 for B_x , 1.00 for  B_y and -3.00 for  B_z

 $\begin{array}{c}\left|\overrightarrow{B}\right|=\sqrt{{\left(3.00\right)}^2+{\left(1.00\right)}^2+{\left(-3.00\right)}^2}\\ =\sqrt{19}\\ =4.36\end{array}$

 

Thus, the magnitude of  $\overrightarrow{A}$ is 5.38 and magnitude of  $\overrightarrow{B}$ is 4.36.

Step-4: Estimation of vector difference

Part(b)

 

The vector $\overrightarrow{A}$  is given as,

 $\overrightarrow{A}=-2.00\widehat{i}+3.00\widehat{j}+4.00\widehat{k}$ and,

 

The vector  $\overrightarrow{B}$ is given as,

 $\overrightarrow{B}=3.00\widehat{i}+1.00\widehat{j}-3.00\widehat{k}$

 

The difference $\overrightarrow{A}-\overrightarrow{B}$  can be evaluated as,

 $\begin{array}{c}\overrightarrow{A}-\overrightarrow{B}=\left(-2.00\widehat{i}+3.00\widehat{j}+4.00\widehat{k}\right)-\left(3.00\widehat{i}+1.00\widehat{j}-3.00\widehat{k}\right)\\ =\left(-2.00-3.00\right)\widehat{i}+\left(3.00-1.00\right)\widehat{j}-\left(4.00-\left(-3.00\right)\right)\widehat{k}\\ =-5.00\widehat{i}+2.00\widehat{j}+7.00\widehat{k}\end{array}$

 

Thus, the vector difference $\begin{array}{c}\overrightarrow{A}-\overrightarrow{B}\end{array}$  is  $-5.00\widehat{i}+2.00\widehat{j}+7.00\widehat{k}$.

Step-5: Estimation of magnitude of vector difference

Part(c)

 

Consider $\overrightarrow{A}-\overrightarrow{B}=\overrightarrow{G}$ . The vector $\overrightarrow{G}$  can be expressed as,

 $\overrightarrow{G}=-5.00\widehat{i}+2.00\widehat{j}+7.00\widehat{k}$

 

Hence, the magnitude of vector $\overrightarrow{G}$  can be expressed as,

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

 

Substitute -5.00 for G_x , 2.00 for  G_y and 7.00 for  G_z.

 $\begin{array}{c}\left|\overrightarrow{G}\right|=\sqrt{{\left(-5.00\right)}^2+{\left(2.00\right)}^2+{\left(7.00\right)}^2}\\ =\sqrt{78}\\ =8.83\end{array}$

 

The magnitude of $\overrightarrow{A}-\overrightarrow{B}$ is 8.83.

 

The vector difference  $\overrightarrow{B}-\overrightarrow{A}$ can be evaluated as,

 $\begin{array}{c}\overrightarrow{B}-\overrightarrow{A}=\left(3.00\widehat{i}+1.00\widehat{j}-3.00\widehat{k}\right)-\left(-2.00\widehat{i}+3.00\widehat{j}+4.00\widehat{k}\right)-\\ =\left(-3.00-\left(-2.00\right)\right)\widehat{i}+\left(1.00-3.00\right)\widehat{j}-\left(-3.00-4.00\right)\widehat{k}\\ =5.00\widehat{i}-2.00\widehat{j}-7.00\widehat{k}\end{array}$

 

Thus, the magnitude of  $\overrightarrow{B}-\overrightarrow{A}$ is calculated as,

  $\begin{array}{c}\left|\overrightarrow{B}-\overrightarrow{A}\right|=\sqrt{{\left(5.00\right)}^2+{\left(-2.00\right)}^2+{\left(-7.00\right)}^2}\\ =\sqrt{78}\\ =8.83\end{array}$

 

Thus,  $\overrightarrow{A}-\overrightarrow{B}$ and  $\overrightarrow{B}-\overrightarrow{A}$ have the same magnitude.