91Ó°ÊÓ

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

$\stackrel{→}{r}=xi+yj+zk$

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

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

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

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

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

Find the magnitude of separation vector.

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

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

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

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