Space/Plane equation/Example for sets/Example

We consider the set

${\displaystyle {}E={\left\{{\begin{pmatrix}x\\y\\z\end{pmatrix}}\in \mathbb {R} ^{3}\mid 5x-y+3z=0\right\}}\,.}$

This is the subset inside ${\displaystyle {}\mathbb {R} ^{3}}$ containing all those points with coordinates ${\displaystyle {}{\begin{pmatrix}x\\y\\z\end{pmatrix}}}$ fulfilling the condition

${\displaystyle {}5x-y+3z=0\,.}$

This condition has a clear meaning for every point ${\displaystyle {}{\begin{pmatrix}x\\y\\z\end{pmatrix}}}$, it can be true or false. Hence, this is a well-defined subset. For example, the points ${\displaystyle {}{\begin{pmatrix}0\\0\\0\end{pmatrix}}}$ and ${\displaystyle {}{\begin{pmatrix}2\\-3\\-{\frac {13}{3}}\end{pmatrix}}}$ belong to the set, the point ${\displaystyle {}{\begin{pmatrix}2\\4\\0\end{pmatrix}}}$ does not belong to the set. If we want to check for a point ${\displaystyle {}{\begin{pmatrix}x\\y\\z\end{pmatrix}}}$ whether it belongs to ${\displaystyle {}E}$, we just have to check the condition. In this respect, the given description of ${\displaystyle {}E}$ is very good. If instead we would like to have a good overview of ${\displaystyle {}E}$ as a whole, then this description is not so convincing. We claim that ${\displaystyle {}E}$ coincides with the set

${\displaystyle {}E'={\left\{r{\begin{pmatrix}3\\0\\-5\end{pmatrix}}+s{\begin{pmatrix}0\\3\\1\end{pmatrix}}\mid r,s\in \mathbb {R} \right\}}\,.}$

This second description presents the set as the set of all elements that can be built in a certain way, namely as a linear combination of the points ${\displaystyle {}{\begin{pmatrix}3\\0\\-5\end{pmatrix}}}$ and ${\displaystyle {}{\begin{pmatrix}0\\3\\1\end{pmatrix}}}$ with arbitrary real coefficients. The advantage of this description is that one gets immediately an overview of all its elements. For example, it is clear that it contains infinitely many elements. However, in this description, it is more difficult to decide whether a given element belongs to the set or not.

In order to show that both sets are identical, we have to show ${\displaystyle {}E\subseteq E'}$ and ${\displaystyle {}E'\subseteq E}$. For the first inclusion, let ${\displaystyle {}P={\begin{pmatrix}x\\y\\z\end{pmatrix}}\in E}$. Then

${\displaystyle {}{\begin{pmatrix}x\\y\\z\end{pmatrix}}={\frac {x}{3}}{\begin{pmatrix}3\\0\\-5\end{pmatrix}}+{\frac {y}{3}}{\begin{pmatrix}0\\3\\1\end{pmatrix}}\,.}$

Here, the equality in the first and in the second component is clear, and the equality in the third component is a reformulation of the starting equation

${\displaystyle {}5x-y+3z=0\,.}$

Taking ${\displaystyle {}r={\frac {x}{3}}}$ and ${\displaystyle {}s={\frac {y}{3}}}$, we see that ${\displaystyle {}P\in E'}$. Now suppose that ${\displaystyle {}P={\begin{pmatrix}x\\y\\z\end{pmatrix}}\in E'}$. This means that there is a representation

${\displaystyle {}P={\begin{pmatrix}x\\y\\z\end{pmatrix}}=r{\begin{pmatrix}3\\0\\-5\end{pmatrix}}+s{\begin{pmatrix}0\\3\\1\end{pmatrix}}={\begin{pmatrix}3r\\3s\\-5r+s\end{pmatrix}}\,}$

with some real numbers ${\displaystyle {}r,s\in \mathbb {R} }$. In order to show that this point belongs to ${\displaystyle {}E}$, we have to show that it fulfills the defining equation of ${\displaystyle {}E}$. But this is clear because of

${\displaystyle {}5x-y+3z=5(3r)-3s+3(-5r+s)=0\,.}$