Question

Find the unit vector in the direction of vector $\overrightarrow{PQ},$ where P and Q are the points (1, 2, 3) and (4, 5, 6).

Answer 1

Let $\overrightarrow{a}$ and $\overrightarrow{b}$ are the position vectors of the points $P\left(1,2,3\right)$ and $Q\left(4,5,6\right)$
Then,
$$\begin{gathered} \overrightarrow{a}=\hat{i}+2\hat{j}+3\hat{k} \\ \overrightarrow{b}=4\hat{i}+5\hat{j}+6\hat{k} \end{gathered}$$
So,
$$\begin{gathered} \overrightarrow{PQ}=\overrightarrow{b}-\overrightarrow{a} \\ =4\hat{i}+5\hat{j}+6\hat{k}-\hat{i}-2\hat{j}-3\hat{k} \\ =3\hat{i}+3\hat{j}+3\hat{k} \end{gathered}$$

Now, $\left|\overrightarrow{PQ}\right|=\sqrt{3^2+3^2+3^2}=\sqrt{9+9+9}=3\sqrt{3}$
Therefore, Unit vector parallel to $\overrightarrow{PQ}$ = $\frac{\overrightarrow{PQ}}{\left|PQ\right|}=\frac{1}{3\sqrt{3}}\left(3\hat{i}+3\hat{j}+3\hat{k}\right)=\frac{1}{\sqrt{3}}\left(\hat{i}+\hat{j}+\hat{k}\right)$