# Why is a linear equation with three variables a plane?

In linear algebra, why is the graph of a three variable equation of the form $ax+by+cz+d=0$ a plane? With two variables, it is easy to convince oneself that the graph is a line (using similar triangles, for example). However with three variables, this same technique does not seem to work: setting one of the variables to be a constant yields a line in 3-D space (I think), and one could repeat this process for each constant value of that variable, but in the end there seems not to be an easy way to check that all these lines are coplanar.

I don't remember seeing why this is in a book, and Khan Academy's video, for example, simply states that this is the case.

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Can't you take a point, and look at the lines in each direction from that point which satisfy the equation? –  Alex Becker Jan 22 '12 at 6:21

Fix a point $(x_0,y_0,z_0)$ satisfying the equation, so $ax_0+by_0+cz_0=-d$.

Let $(x,y,z)$ be a point in space. Then $(x,y,z)$ satisfies the equation iff $ax+by+cz=-d=ax_0+by_0+cz_0$, i.e., iff $a(x-x_0)+b(y-y_0)+c(z-z_0)=0$.

Let $\vec r=(a,b,c)$. Then we are saying that $(x,y,z)$ satisfies the equation iff $(x-x_0,y-y_0,z-z_0)$ is a vector orthogonal to $\vec r$. For any vector $\vec r$, the set of such vectors is a plane through the origin, call it $\Pi$.

Then $(x,y,z)$ satisfies the equation iff it belongs to the translation of the plane $\Pi$ by the vector $(x_0,y_0,z_0)$. But, of course, the translation of a plane is again a plane.

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Look at the equation as a dot product or inner product:

$$\left[ \begin{array}{ccc} a & b & c \end{array} \right] \left[ \begin{array}{c} x \\ y \\ z \end{array} \right] = -d.$$

Then it is clear to see that the point $(x, y, z)$ that satisfies the equation is any point in the plane that is perpendicular to the vector $\left[ \begin{array}{ccc} a & b & c \end{array} \right]$ (this fixes its orientation) and is the right distance from the origin to yield a dot product of $-d$.

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