Parametric equation reorganization

Question

I'm not very familiar with parametric equations and have come across this textbook problem: Find a system of two equations in three variables $x_1$, $x_2$, and $x_3$, that has the solution set given by the parametric representation

$x_1 = t$, $x_2 = s$, and $x_3 = 3 + s - t$

where s and t are any real numbers. Then show that the solutions to your system can also be written as

$x_1 = 3 + s - t$, $x_2 = s$, and $x_3 = t$.

I've gotten to the fact that $x_1 = x_2$ which leaves me with $x_3$ being 3. I'm not sure if that's correct.

Answer 1

No, $x_1 = x_2$ is not a fact at all.

The way the question is written is slightly confusing, because the $t$ in the first parametric representation is not the same as the $t$ in the second (the $s$ can be the same, because it's $x_2$ in both cases). I think it would be better to use another name in the second parametric representation: I'll call it $u$.

Now you want $u = x_3 = 3 + s - t$. All you need to check is that $t = s_1 = 3 + s - u$.

Answer 2

I feel a slightly wrong assumption. The name of parameters should not matter.

Let's rewrite equation as

$$ S1: \begin{cases} x_1=t\\ x_2=s\\ x_3=3+s-t \end{cases} $$

$$ S_2: \begin{cases} x_1=3+s'-t'\\ x_2=s'\\ x_3=t' \end{cases} $$

Now, is there any variable change which convert $S_1$ to $S_2$?

$$ \begin{cases} t=3+s'-t'\\ s=s'\\ 3+s-t=t' \end{cases} $$

or

$$ \begin{cases} s'=s\\ t'=3+s-t \end{cases} $$

By this variable change, $S1$ is converted into $S2$. Hence, they are equivalent.

Answer 3

If one notices that we are taking about the plane

$$x_1-x_2+x_3=3\tag 1$$

then it will be clear that both of the following parametric representations describe this very plane:

$$\begin{matrix}x_1(s,t)=t\\ x_2(s,t)=s\\ x_3(s,t)=3+s-t \end{matrix}\tag 2$$ and $$\begin{matrix} x_1(s,t)=3+s-t&\ \\ x_2(s,t)=s\\ x_3(s,t)=t \end{matrix}.$$

But why is $(1)$ a representation of $(2) $?

Let's take three points on $(2)$.

• Let $s=t=0$ then our point is $(0,0,3).$
• Let $s=t=1$ then our point is $(1,1,3)$.
• Let $s=1$ and $t=0$ then we have $(0,1,4)$.

Then let the three vectors pointing to the three points given above be denoted by $v_1,v_2,v_3$ . The following difference vectors are parallel to the plane:

$$v_2-v_1=\begin{bmatrix}1\\ 1\\ 0\end{bmatrix}$$

$$v_3-v_1=\ \begin{bmatrix}0\\ 1\\ 1\end{bmatrix}.$$

The vector product of the two vectors above will be a normal vector to our plane:

$$n=\begin{bmatrix}\ 1\\ -1\\ \ 1 \end{bmatrix}.$$

Let $[x_1\ x_2\ x_3]^T$ be a general point of the plane and take the vector $[0\ 0\ 3]^T$ pointing to a specific point of the plane. The vector below is parallel to the plane

$$\begin{bmatrix}x_1\\ x_2\\ x_3\end{bmatrix}-\begin{bmatrix}0\\ 0\\ 3\end{bmatrix}=\begin{bmatrix}x_1\\ x_2\\ x_3-3\end{bmatrix}.$$

The following scalar product has to be zero

$$n^T\begin{bmatrix}x_1\\ x_2\\ x_3-3\end{bmatrix} = [1 \ -1 \ \ 1]\begin{bmatrix}\ x_1\\x_2\\\ x_3-3\end{bmatrix}=0.$$

So,

$$x_1-x_2+x_3=3.$$

This proves that $(1)$ is a true equation of the plane given by $(2)$.