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 $\hat{\mathit{r}}$

Short answer

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

Step-by-step solution

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

Find the separation vector

The position vector at $\left(2,8,7\right)$is $\overrightarrow{r}=2i+8j+7k$, and position vector at $\left(4,6,8\right)$isrole="math" localid="1657351115792" $\overrightarrow{r}=4i+6j+8k$

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$\overrightarrow{r}={\overrightarrow{r}}_2-{\overrightarrow{r}}_1$

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

find the magnitude of separation vector

Find the magnitude of separation vector.

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

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 $\frac{2}{3}i-\frac{2}{3}j+\frac{1}{3}k$.