# Why ax+by+cz=d is a plane

Why is $ax+by+cz=d$ a plane in three dimensions? Because it just is? Not good enough.

There are proofs for this. They involve normals and dot products and other fancy things that you’re not entirely sure are legit (if you’re in high school). I’m going to try a very different approach to convince you that this is a plane, today.

Take $ax+by+cz=d$. Make $z=0$. You have $ax+by=d$. This is a line, as is clearly known (constant slope). Now increase $z$ by an infinitesimal amount: say $\Delta z$. You have $ax+by=d-c\Delta z$, which is again a line. However, what kind of a line is it? It is certainly parallel to the earlier line $ax+by=d$. It just has a different y-axis intercept.

You can continue trying this for different values of $z$. You get an infinite number of parallel lines, all shifted along the y-axis.

Now start raising each straight line $ax+by=d-cz$ by $z$ units along the z-axis. Clearly each point on the raised line satisfies the equation $ax+by+cz=d$. This bunch of raised and shifted parallel lines form a plane.

This might not be a rigorous proof, but it is probably more visually satisfying than the proof with normals and doc products and what not.