Finding a Vector Equation of a Line in Space

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 \[ \mathbf{r}=\mathbf{r}_{0}+t \mathbf{D}=\mathbf{i}+2\mathbf{j}-\mathbf{k} +t (3\mathbf{i}+\mathbf{j}-\mathbf{k})=(1+3t)\mathbf{i}+(2+t)\mathbf{j} +(-1-t)\mathbf{k} \]