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A planar equation with one missing variable, i.e.

$ax+bz-d=0$

has shown in my Math exam.

Is this a valid planar equation? It seems to me as $y = 0$, this would produce only a straight line.

Pedro Sobota
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    Could you elaborate why you think $y=0$? – Math Lover Sep 23 '17 at 02:53
  • This is a completely legitimate plane equation. Think of it this way: pretend there's no $y$ involved, then this is a line equation on the $xz$ plane, say the line is $L$; but in fact $y$ can be any value in $\Bbb R$, so what the equation describes is ${(x,y,z)\mid (x,z)\in L, y\in\Bbb R}$. – Vim Sep 23 '17 at 02:53
  • Use the 3D Graphing calculator here: https://www.desmos.com/calculator/nqom2ih05g . It's definitely a plane. – Pritt Balagopal Sep 23 '17 at 02:54
  • @MathLover Because if a term is omitted from a equation it must be because its coefficient is zero? – Pedro Sobota Sep 23 '17 at 03:06
  • Note that the equation $z=0$ is the equation of a plane (a familiar one - the standard two dimensional co-ordinate plane) and that has two variables missing. – Mark Bennet Sep 23 '17 at 06:56

2 Answers2

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The graph in $\mathbb{R}^3$ of the equation $ax+bz-d=0$ is, in fact, a plane, so long as $a,b,d \in \mathbb{R}$, and $a,b$ are not both zero.

For the given equation, the coefficient of the $y$-term is zero, but the equation doesn't force the value of $y$ to be zero.

In fact, since the $y$-term is absent, $y$ is not constrained by the equation (i.e., $y$ is "free"). Thus, for any pair $x,z$ of real numbers satisfying the given equatiion, and any $y \in \mathbb{R}$, the point $(x,y,z)$ satisfies the equation, hence is a point of the graph.

The way to visualize the graph is to first graph the equation in the $xz$-plane, which does, in fact, yield a line (provided $a,b$ are not both zero), but then, since $y$ is free, we can extend that line in the $y$-direction (either forwards or backwards), which thus yields a plane.

Alternatively, you can just use the fact (assuming it's prior knowledge) that the equation form $$Ax + By + Cz + D = 0$$ where $A,B,C,D \in \mathbb{R}$, and $A,B,C$ are not all zero, is the general form for the equation of a plane in $\mathbb{R}^3$. Hence, assuming $a,b$ are not both zero, the given equation $$ax + bz-d=0$$ can be rewritten as $$ax + 0y + bz + (-d) =0$$ which fulfills the requirements of the general form.

quasi
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  • I might be having difficulty understanding how having a term have a coefficient of zero would still permit the end value of the term to be any different from a constant zero. – Pedro Sobota Sep 23 '17 at 03:11
  • For example, let $a = b = 1$ and let $d = 2$. Then, for example, the point $(1,y,1)$ satisfies the equation, for any $y \in \mathbb{R}$. In particular, the points $(1,-5,1),;(1,0,1),;(1,8,1)$ satisfy the equation. – quasi Sep 23 '17 at 03:14
  • Using the example in my previous comment, to verify that the point $(1,8,1)$ satisfies the equation $x + z - 2 = 0$, rewrite the equation as $1x + 0y + 1z - 2 = 0$, then substitute $x = 1,;y=8,;z=1$, and check to see if the equation is satisfied. – quasi Sep 23 '17 at 03:23
  • I think I get it. Precisely because the coefficient is $0$, the variable could be "anything" and not change the result of the equation. But, if I say: $a + b = 4$, and then say, $a = 4$, that might only be because $b = 0$? So any other $b$ in this case would invalidate the equation. – Pedro Sobota Sep 23 '17 at 03:25
  • Right, that's the concept. – quasi Sep 23 '17 at 03:25
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Not every equation of a place has to have all three variables. In your case, there were only two variables present, $x$ and $z$. That means that the value of $y$ does not matter; it can be any real number. Each of the constants $a$, $b$, and $d$ can also any real number, and your equation will still be that of a plane. For example, if $a$, $b$, and $d$ are all equal to $1$, your plane will look like the following:enter image description here

Dibbs
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