Q7P - Introduction to Electrodynamics

Question

Find the separation vector r from the source point (2,8,7) to the field point ( 4,6,8). Determine its magnitude ( r ), and construct the unit vector  $\hat{\mathit{r}}$

Step-by-Step Solution

Answer

The magnitude is 3 and unit separation vector is $\frac{2}{3}i-\frac{2}{3}j+\frac{1}{3}k$ .

Step 1: Explain the concept and write the expression of position vector

The separation vector is obtained by subtracting the source vector ${\overrightarrow{\mathbf{r}}}_{\mathbf{2}}$ from the destination vector ${\overrightarrow{\mathbf{r}}}_{\mathbf{1}}$ . The expression of position vector is as follows:

 

$\overrightarrow{r}=xi+yj+zk$ 

 

Where $i,j,k$ are unit vectors along  $x,y,z$ coordintaes

Step 2: Find the separation vector

The position vector at  $\left(2,8,7\right)$ is $\overrightarrow{r}=2i+8j+7k$ , and position vector at $\left(4,6,8\right)$ is  $\overrightarrow{r}=4i+6j+8k$

 

The separation vector is given as $\overrightarrow{r}={\overrightarrow{r}}_2-{\overrightarrow{r}}_1$ .

 

Substitute $2i+8j+7k$ for ${\overrightarrow{r}}_1$ , and $4i+6j+8k$ for ${\overrightarrow{r}}_2$ into $\overrightarrow{r}={\overrightarrow{r}}_2-{\overrightarrow{r}}_1$

$$\begin{gathered} \overrightarrow{r}=4i+6j+8k-2i-8j-7k \\ \overrightarrow{r}=2i-2j+k \end{gathered}$$              

Step 3: find the magnitude of separation vector

Find the magnitude of separation vector.

$$\begin{gathered} \left|\overrightarrow{r}\right|=\sqrt{{\left(2\right)}^2+{\left(-2\right)}^2+{\left(1\right)}^2} \\ \left|\overrightarrow{r}\right|=3 \end{gathered}$$

Step 4: Find the unit separation vector

Divide the separation vector by its magnitude to find unit separation vector.

 

$$\begin{gathered} \hat{r}=\frac{\overrightarrow{r}}{\left|\overrightarrow{r}\right|} \\ =\frac{2i-2j+k}{3} \\ =\frac{2}{3}i-\frac{2}{3}j+\frac{1}{3}k \end{gathered}$$ 

 

Thus the unit separation vector is  $\frac{2}{3}i-\frac{2}{3}j+\frac{1}{3}k$ .