If two planes in $\mathbb{R}^3$ pass by the origin, do they necessarily intersect at multiple points?

Question

I had seen an exercise, that asked something like this:

Confirm that for any two planes in $\mathbb{R}^3$ that pass by the origin, they will always intersect at more than one point.

But, unless I don't understand the word "origin", as far as I'm concerned, that statement can't be proven, because there ARE two planes that pass by the origin and their intersection is only one point (the origin), which I illustrate like this:

Ok the drawing sucks but you get the idea: if the red point is the origin, doesn't this drawing demonstrate that the statement is false? Or I am missing something key in this problem?

Answer 1

Your drawing does not show 2 planes. A plane does not have edges, instead planes are infinite in all directions. You can think of them as cutting the entire space over in 2 parts.

This means that they actually will intersect in a line, except for the case where the planes are equal and they intersect in a plane.

Answer 2

The planes are meant to extend infinitely so that they intersect in a line.

Answer 3

A plane passing through the origin is a subspace of dimension two. Call the two planes $P_1$ and $P_2$. From the dimension identiy we have $$ \rm{dim}( P_1 + P_2) = \rm{dim} ( P_1) + \rm{dim} ( P_1) - \rm{dim} ( P_1 \cap P_2) $$ But $\rm{dim}( P_1 + P_2) \leq 3 $ so $$2 + 2 - \rm{dim} ( P_1 \cap P_2) \leq 3 $$ which gives $\rm{dim} ( P_1 \cap P_2) \geq 1$. So the intersection connections at least a line.