Solution (a) We begin by finding several coordinate curves. We choose to use $v=0,$ $v=1,$ $u=0,$ and $u=1,$ but any numbers will do.

When $v=0,$ the $u$-coordinate curve is $\mathbf{r} ( u,0) = (1+2u) \mathbf{i}+u\mathbf{j}+3\mathbf{k}$.

When $v=1,$ the $u$-coordinate curve is $\mathbf{r} ( u,1) = (2+2u) \mathbf{i}+ ( u-1) \mathbf{j}+4\mathbf{k}$.

Figure 41

When $u=0,$ the $v$-coordinate curve is $\mathbf{r} ( 0,v) = ( 1+v) \mathbf{i}-v\mathbf{j}+(3+v) \mathbf{k}$.

When $u=1,$ the $v$-coordinate curve is $\mathbf{r} ( 1,v) =(3+v) \mathbf{i}+ ( 1-v) \mathbf{j}+(3+v) \mathbf{k}$.

The graph of the parametric surface is a plane as shown in Figure 41.

(b) To find a rectangular equation for the surface, we write the components of $\mathbf{r=r}(u,v)$ as a system of equations. Since $\mathbf{r}=x\mathbf{i}+y\mathbf{j}+z\mathbf{k}=x(u,v) \mathbf{i}+y(u,v) \mathbf{j}+z(u,v) \mathbf{k},$ we have $$\left\{ \begin{array}{ll@{\qquad}l} x=2u+v+1 & &{\color{#0066A7}{(1)}}\\ y=u-v & & {\color{#0066A7}{(2)}}\\ z= v+3 & & {\color{#0066A7}{(3)}} \end{array} \right.$$

If we eliminate $u$ and $v$ from this system, we will have an expression involving $x,$ $y,$ and $z.$ We eliminate $v$ by adding equations $(1)$ and $(2)$ and equations $(2)$ and $(3) .$ $$\begin{equation*} \begin{array}{ll@{\qquad\qquad}l} x+y=3u+1 & & \color{#0066A7}{{\hbox{(4) Add \(1\) and \(2\).}}}\\ y+z= \hspace{6pt}u+3 & &\color{#0066A7}{{\hbox{(5) Add \(2\) and \(3\).}}} \end{array} \end{equation*}$$

Now we eliminate $u$: $$\begin{equation*} \begin{array}{ll@{\qquad}l} \hspace{48pt}x+ y = 3u+1 & & \color{#0066A7}{(4)} \\[5pt] \hspace{27pt}\underline{\,\ -3y-3z=-3u-9} & \,\ \ \ \ \ \ \ \ \ \ & \color{#0066A7}{{\hbox{Multiply (5) by \(-3\).}}} \\ \hspace{21pt}x-2y-3z=-8 & & \color{#0066A7}{{\hbox{Add.}}} \end{array} \end{equation*}$$

This is the equation of a plane whose normal vector is $\mathbf{i}-2\mathbf{j}-3\mathbf{k}.$