Quadratic form is well defined in $\mathbb{RP}^2$ I'm learning projective geometry and need help to understand the following statement : 
We give the quadratic form in $\mathbb{R}^3$ (where $\mathbf{x} = (x_1, x_2, x_3)$) :
$$q(\mathbf{x}) = 2x_1^2 - 2x_1x_2 + ax_2^2 + 2x_1x_3 - 4x_2x_3 + 3x_3^2.$$
The quadratic equation $q = 0$ is well defined in the real projective plane.

Given the title of my post I think you can guess my next question: why is $q = 0$ well defined in $\mathbb{RP}^2$? In general, what does it mean for a quadratic equation to be well defined in $\mathbb{RP}^2$? Since the real projective plane is the collection of all lines passing through the origin in $\mathbb{R}^3$, is $q = 0$ a line in $\mathbb{RP}^2$? I think the answer to the last question is no but it's still unclear to me what $q = 0$ represents in $\mathbb{RP}^2$. 
Note : if anyone asks for the parameter $a$, this is part of an exercise where we first need to find the values of $a$ for which $q(\mathbf{x})$ is positive definite. I'm only interested for answers to the questions related to the real projective plane. 
 A: The set in $\mathbb R^3$ at which the quadratic form is zero is not usually a single line. However, for the example where the form (its Hessian matrix) has at least one positive eigenvalue and at least one negative, and none exactly zero, the null set is called the null cone, and is a cone. In some cases, such as $x^2 + y^2 - z^2,$ the cone is a circular cone or cone of revolution, but usually $x^2 + 4 y^2 - z^2$ it is a cone over an ellipse. 
Oh, the cone is a union of lines through the origin. Think about it.
A: The statement "[t]he quadratic equation $q = 0$ is well defined in the real projective plane" means that the set $\{\ell \in \mathbb{RP}^2 \mid q(\ell) = 0\}$ (which is a collection of lines) is a well-defined set; that is, there is an unambiguous definition of when $\ell \in \mathbb{RP}^2$ satisfies $q(\ell) = 0$. 

Definition: Given a point $(x_1, x_2, x_3) \in \ell$ with $(x_1, x_2, x_3) \neq (0, 0, 0)$, we say $q(\ell) = 0$ if $q(x_1, x_2, x_3) = 0$. 

For this to be well-defined, we need to make sure that whether $q(\ell) = 0$ or not depends only upon the line itself, not the point we chose on it; i.e. that the property $q(\ell) = 0$ is a property of the line $\ell$, not a property of the point $(x_1, x_2, x_3)$. In order to do this, we need to show that if $(y_1, y_2, y_3)$ is another point in $\ell$ with $(y_1, y_2, y_3) \neq (0, 0, 0)$, then $q(y_1, y_2, y_3) = 0$ if and only if $q(x_1, x_2, x_3) = 0$ so that it doesn't matter which non-origin point on $\ell$ we chose.
As $\ell$ is a one-dimensional vector space and $(x_1, x_2, x_3) \in \ell$ is not the zero vector, it forms a basis. In particular, for any other vector in $\ell$, it must be a multiple of $(x_1, x_2, x_3)$. In particular, $(y_1, y_2, y_3) = \lambda(x_1, x_2, x_3) = (\lambda x_1, \lambda x_2, \lambda x_3)$ for some real number $\lambda$; as $(y_1, y_2, y_3)$ is not the zero vector, $\lambda$ is non-zero. Now note that
\begin{align*}
q(y_1, y_2, y_3) &= q(\lambda x_1, \lambda x_2, \lambda x_3)\\
&= 2(\lambda x_1)^2 - 2(\lambda x_1)(\lambda x_2) + a(\lambda x_2)^2 + 2(\lambda x_1)(\lambda x_3) - 4(\lambda x_2)(\lambda x_3) + 3(\lambda x_3)^2\\
&= 2\lambda^2x_1^2 - 2\lambda^2x_1x_2 + a\lambda^2x_2^2 + 2\lambda^2x_1x_3 - 4\lambda^2x_2x_3 + 3\lambda^2x_3^2\\
&= \lambda^2(2x_1^2 - 2x_1x_2 + ax_2^2 + 2x_1x_3 - 4x_2x_3 + 3x_3^2)\\
&= \lambda^2q(x_1, x_2, x_3).
\end{align*}
As $\lambda \neq 0$, $\lambda^2 \neq 0$ and hence $q(y_1, y_2, y_3) = 0$ if and only if $q(x_1, x_2, x_3) = 0$. Therefore, the notion $q(\ell) = 0$ is well-defined (i.e. only depends on the line $\ell$), and hence so is the set $\{\ell \in \mathbb{RP}^2 \mid q(\ell) = 0\}$.
More generally, a polynomial $p$ is called homogeneous if there is $k \in \mathbb{R}$ such that $p(\lambda{\bf x}) = \lambda^kf({\bf x})$ for all $\lambda \in \mathbb{R}$. Note, the calculation above shows that the polynomial you asked about is homogeneous with $k = 2$. It's not hard to show that if $p$ is a homogeneous polynomial, then $k = \deg p$ and every monomial in $p$ has top degree (as is the case for your polynomial). If $p$ is a homogeneous polynomial of $n + 1$ variables, then the notion $p(\ell) = 0$ is well-defined in $\mathbb{RP}^n$.
