Graphing a Vector Function

Graph the curve \(C\) traced out by the vector function \begin{equation*} \mathbf{r}(t)=(2+3t)\mathbf{i}+(3-t)\mathbf{j}+2t\mathbf{k} \end{equation*}

Solution The components of \(\mathbf{r=r}(t)\) are the parametric equations \begin{equation*} x(t)=2+3t\qquad y(t)=3-t\qquad z(t)=2t \end{equation*}

These are the parametric equations of a line in space containing the point \((2,3,0)\) (corresponding to \(t=0\)) and in the direction of the vector \(3\mathbf{i}-\mathbf{j}+2\mathbf{k}\). Solving for \(t\), we obtain the symmetric equations of this line \begin{equation*} \frac{x-2}{3}=\frac{y-3}{-1}=\frac{z}{2} \end{equation*}

Since \(\mathbf{r}(0)=2\mathbf{i}\,{+}\,3\mathbf{j}\) and \(\mathbf{r}(1)=5\mathbf{i}\,{+}\,2\mathbf{j}\,{+}\,2\mathbf{k}\), the positive direction of \(C\) is given by \(\mathbf{r}(1)-\mathbf{r}(0)=3\mathbf{i}-\mathbf{j}\,{+}\,2\mathbf{k}\). See Figure 5.