# Find the dimension of the sum of three planes

I have the following problem:

Let $$\pi_1$$, $$\pi_2$$, $$\pi_3$$ be three planes in a projective space such that $$\pi_1 \cap \pi_2 \cap \pi_3$$ = $$\emptyset$$, $$\dim\,(\pi_1 \cap \pi_2) = 1$$, $$\dim\,(\pi_1 \cap \pi_3) = \dim (\pi_2 \cap \pi_3) =0.$$ Find the dimension of $$\pi_1 \lor \pi_2\lor \pi_3$$.

I have almost solved it, but I'm struggling at the end. So far, I have already done the following:

Using Grassmann Formula, I get: $$\dim\,(\pi_1\lor \pi_2\lor \pi_3)=5-\dim\,((\pi_1\lor \pi_2)\cap \pi_3)$$. Since $$\pi_1\cap \pi_3 \subseteq (\pi_1\lor \pi_2)\cap \pi_3 \subseteq \pi_3$$, we know that $$0 \le \dim\,((\pi_1 \lor \pi_2)\cap \pi_3)\le2$$.

If $$\dim\,((\pi_1\lor \pi_2)\cap \pi_3)=0$$, then $$(\pi_1\lor \pi_2)\cap \pi_3= \pi_1\cap\pi_3= p$$, $$(\pi_1\lor \pi_2)\cap \pi_3=\pi_2\cap\pi_3={p}$$. Thus, $$\pi_1\cap\pi_2\cap\pi_3={p}$$, which leads to a contradiction.

Therefore, we know that $$\dim\,((\pi_1\lor\pi_2)\cap\pi_3)$$ can only be either 1 or 2.

I don't know how to solve this and I would appreciate a lot if you could help me out.

A projective space of dimension $$n$$ is the space of lines through the origin of a linear space of dimension $$n+1$$, and the projective planes correspond to 3 dimensional subspaces, and, generally, the dimension of the corresponding subspace $$\hat H$$ in the linear space is increased by one, compared to the original object projective subspace $$H$$, for any $$H$$.
So, $$\dim(\pi_1\cap\pi_2)=1$$ means that $$\dim(\hat\pi_1\cap\hat\pi_2)=2$$, thus we can choose a basis $$u_1,u_2\in\hat\pi_1\cap\hat\pi_2$$.
Now, $$\dim(\pi_1\cap\pi_3)=0$$ means $$\dim(\hat\pi_1\cap\hat\pi_3)=1$$, choose any nonzero element of it, $$a$$,
and similarly choose a nonzero $$b$$ in $$\hat\pi_2\cap\hat\pi_3$$.
Clearly, $$a$$ and $$b$$ are not scalar multiple of one another, so they are linearly independent. Extend them to a basis $$a,b,c$$ of $$\hat\pi_3$$.
Using $$\pi_1\cap\pi_2\cap\pi_3=\emptyset\text{,}\$$ that is, $$\ \hat\pi_1\cap\hat\pi_2\cap\hat\pi_3=\{0\}$$, prove that $$u_1,u_2,a,b,c$$ are linearly independent, yielding in particular that $$u_1,u_2,a$$ is a basis for $$\hat\pi_1$$ and $$u_1,u_2,b$$ is a basis for $$\hat\pi_2$$.
Consequently, $$u_1,u_2,a,b,c$$ is a basis for $$\hat\pi_1+\hat\pi_2+\hat\pi_3=(\pi_1\lor\pi_2\lor\pi_3)\hat{}$$,
so its projective dimension is $$4$$.