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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?

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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.

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Ok, thanks, but if I have more than an equation describing the plane and replacing in one of them gives invalid equation... but other replacing gives correct equation? –  Surfer on the fall Jun 21 '12 at 16:36
    
@user1294101 In that case, your set is defined by something like $S = \{x \in S : f_1(x) = f_2(x) = \cdots = f_n(x) = 0 \}$ (which could be a plane). In order for an element to belong to $S$, it would have to satisfy all the equations. Otherwise (if there's an equation it doesn't satisfy), it doesn't belong to the set. –  talmid Jun 21 '12 at 16:42
    
Thank you very much, put probably I didn't explain well... Let's have a plane described by 3x+4y-2z+q=6 and 2x+3y+1z-q=0.. Verify that point (1,1,1,1) belongs to the plane. What would you say? thanks again. –  Surfer on the fall Jun 21 '12 at 16:44
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@ talmid : is $a_1x_1+a_2x_2+a_3x_3+a_4x_4=d$ an equation of one plane ? (palne is not hyperplane) –  Mohamed Jun 21 '12 at 16:49
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@user1294101 We see that $(1,1,1,1)$ satisfies the first equation because $3\cdot 1 + 4\cdot 1 - 2 \cdot 1+ 1 = 6$, but it doesn't satisfy the second one because $2\cdot 1 + 3 \cdot 1 + 1\cdot - 1 = 5 \neq 0$. So $(1,1,1,1)$ doesn't belong to the set described by both equations, because it doesn't satisfy both of them. –  talmid Jun 21 '12 at 16:51

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