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?
Point belonging to plane
1 Answers
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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0Ok, 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? – 2012-06-21
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0@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. – 2012-06-21
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0Thank 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. – 2012-06-21
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0What a wonderful and clean answer! You are very good at explainig, thanks again! – 2012-06-21
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1@ talmid : is $a_1x_1+a_2x_2+a_3x_3+a_4x_4=d$ an equation of one plane ? (palne is not hyperplane) – 2012-06-21
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1@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. – 2012-06-21
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0@Mohamed Thanks. I've added some sort of disclaimer. – 2012-06-21
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0@user1294101: a single linear equation in $\mathbb R^4$ gives a 3-dimensional subspace called a hyperplane. Two linear equations give the intersection of these two hyperplanes, which will generally be a plane. So to be in the plane, your point must satisfy both equations. The corresponding thing in $\mathbb R^3$ is that a linear equation gives a plane and two give the line that is the intersection of the two planes. If you are asked to verify that a point is on the line defined by two linear equations, it must satisfy them both. – 2012-06-21
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0I'm sorry but I'm just starting studying geometry, so I have a question.... what's the difference in this problem between plane and hyperplane? Anyway, the problem was about a plane. – 2012-06-21
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0@user1294101 A hyperplane is described by only one equation in $\mathbb{R}^4$, and a plane needs two. It's a bigger subspace of $\mathbb{R}^4$ (its elements have to satisfy only one equation, not two). I'd understood your problem to mean that you had a hyperplane, but my general considerations still apply: *if we have one or more equations describing a set, an element has to satisfy all the equations in order to belong to it*. In this case, the point certainly doesn't satisfy both equations, so it doesn't belong to the plane. – 2012-06-21
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0@talmid I have quite understood... but is your first answer wrong for my problem? I'm quite sure it's right. – 2012-06-21
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0@user1294101 I've reworded a bit so that my answer focuses on your particular problem. It's better to understand the general idea of deciding membership of any set, and then apply it to the specific case of a plane in $\mathbb{R}^4$. You'll use the same concept many times. – 2012-06-21
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0I'd never expected a so helpful help... Just the last thing: what do you exactly mean as "set" in "In our case, the set is a plane"? I'm not English... – 2012-06-21
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0@user1294101 A set is any collection of points. Some examples: $A = {1,2,3}$ is the set that contains the numbers $1$, $2$, and $3$; $\mathbb{N}$ is the set that contains all the natural numbers ($1$, $2$, $3$, $4$, etc.); $\mathbb{R}$ is the set of all real numbers, and contains, among other elements, the numbers $6$, $\pi$, $-2$, and $1/2$; $\mathbb{R}^4$ is the set of all lists of four real numbers, one of which might be $(1,3,-2,7)$; and finally, a plane is a set, because it's a collection of points, that together form a particular shape. – 2012-06-21