Point belonging to plane In $\Bbb R^4$, I have a plane (given by its cartesian equation) and a point (given by its coordinates).
How can I check if it belongs to the plane?
 A: Some related cases:


*

*If the equation of a hyperplane is in the form $a_1x_1 + a_2x_2 +
   a_3x_3 + a_4x_4 = b$, to check whether a point $(p_1,p_2,p_3,p_4)$ is
in the hyperplane, we have to see if the point satisfies the
equation, i.e., if when we replace $(x_1,x_2,x_3,x_4)$ by
$(p_1,p_2,p_3,p_4)$ we get a valid equation, that is, if $a_1p_1 +
   a_2p_2 + a_3p_3 + a_4p_4 = b$.

*An example in $\mathbb{R}^3$: to see whether the point $(1,3,2)$
belongs to the (hyper)plane $x-y+2z = 3$, we see that $1-3+2\cdot 2 =
   2 \neq 3$, so it doesn't belong to the plane; however, $(1,0,1)$
does, because $1-0+2\cdot 1= 3$.
In general, if we have a Cartesian equation for some subset of $\mathbb{R}^n$ (or any set $S$) given by $f(x) = 0$, this actually means that the subset is $\{x \in S : f(x) = 0\}$, and what we have to do to decide membership of any $x_0 \in S$ is be to check whether $f(x_0) =0$. If we have more than one equation describing a set, an element has to satisfy all the equations in order to belong to it.
In our case, the set is a plane and is described by two equations, so when we want to decide whether an element belongs to it, we check if the element satisfies both equations. Otherwise, it doesn't belong to the plane.
