To find a vector-valued function $\mathbf{r}(u, v)$ whose graph represents the plane $x + y + z = 2$, we parameterize the plane in terms of two variables $u$ and $v$. Here's how:

1. The equation $x + y + z = 2$ can be rewritten to express one variable (typically $z$) in terms of $x$ and $y$: $z = 2 - x - y$

2. Assign $x = u$ and $y = v$, where $u$ and $v$ are parameters that can take on any real values. Then, substitute these into the expression for $z$: $z = 2 - u - v$

3. The vector-valued function $\mathbf{r}(u, v)$ can now be written as: $\mathbf{r}(u, v) = \langle u, v, 2 - u - v \rangle$ or equivalently: $\mathbf{r}(u, v) = u \mathbf{i} + v \mathbf{j} + (2 - u - v) \mathbf{k}$

Thus, the answer is: $\mathbf{r}(u, v) = \langle u, v, 2 - u - v \rangle$

Would you like additional details or have any specific questions?
Here are 5 related questions to deepen your understanding:

1. How do you verify that this parameterization lies on the given plane?
2. What happens if you choose a different variable (e.g., $x$ or $y$) to solve for in the equation of the plane?
3. How can you interpret this parameterization geometrically?
4. How would you adapt the parameterization for a different plane, such as $2x + y - z = 5$?
5. How does this parameterization change if $u$ and $v$ are restricted to specific intervals?

Tip: Always double-check your parameterization by substituting it back into the original equation to confirm it satisfies the given surface equation.