Planar equation with missing variable 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.
 A: 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.
A: 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:
