Find a vector equation of the line containing the points $P_{0}\,\,{=}\,\,(1,2,-1)$ and $P_{1}=(4,3,-2)$.

Solution The vector $\mathbf{D}$ in the direction from $P_{0}=(1,2,-1)$ to $P_{1}=(4,3,-2)$ is $$\begin{equation*} \mathbf{D}=\overrightarrow{P_{0}P_{1}}=3\mathbf{i+j-k} \end{equation*}$$

If we let $\mathbf{r}_{0}=\mathbf{i}+2\mathbf{j-k}$, then a vector equation of the line is $$\mathbf{r}=\mathbf{r}_{0}+t \mathbf{D}=\mathbf{i}+2\mathbf{j}-\mathbf{k} +t (3\mathbf{i}+\mathbf{j}-\mathbf{k})=(1+3t)\mathbf{i}+(2+t)\mathbf{j} +(-1-t)\mathbf{k}$$