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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.

russell11
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3 Answers3

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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.

Andrés E. Caicedo
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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 $-d$.

Tpofofn
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Here's how I visualised this: Let the linear equation be $Ax+By+Cz=D$. Now, one thing we can definitely conclude is that for every $x$ and $y$, there exists a $z$ that can satisfy the equation. Hence, if imagine an x-y plane in 3-d space, for every point on the plane, there certainly exists a solution somewhere above, below or on the plane.

Now coming to your questions, what's left is to visualise that all those solutions are co-planar. Well, how do we know for 2 dimensions, that a given equation represents a straight line? We know that because a linear function has a constant rate of change.

Similarly, for three dimensions, consider $x$ and $y$ as independent variables and $z$ as the dependent variable. If we start from the origin, as the value of $x$ increases by $1$ unit, $z$ decreases by $\frac{A}{C}$. As $y$ increases by $1$ unit, $z$ decreases by $\frac{B}{C}$. This constant rate of change implies that the equation represents a plane.

Arham Jain
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