When are three points of a projective plane collinear?

Question

I have the following points in a projective plane ($P(\mathbb{R}^{3\times 1})$:

$B=\mathbb{R}(1,1,1)^T$

$C=\mathbb{R}(1,0,1)^T$

and

$D=\mathbb{R}(0,2,0)^T$

To-Do:

1) prove that these points are collinear

2) Find the point $A$ of intersection of $Bv C$ (line going through $B$ and $C$) with $4x_0-x_1=0$

3) Find the cross-ratio $(A,B;C,D)$

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I have read the chapter few times. Nevertheless I am not sure how to prove that three points in a projective pane are collinear. Has anyone an idea how to solve this problem?

Answer 1

a) $B,C,D$ are projectively aligned because they appear as belonging to a same $\mathbb{R^3}$ plane. Indeed, their resp coordinates:

$$\pmatrix{x_0\\x_1\\x_2}=\pmatrix{1\\1\\1}, \pmatrix{1\\0\\1}, \pmatrix{0\\2\\0}$$

obey the same linear constraint: $x_0-x_2=0$.

Another explanation: there is a null linear combination: $$2B-2C-D=0.$$

b) A parametric representation of line $BC$ is: $$\lambda\pmatrix{1\\1\\1}+(1-\lambda)\pmatrix{1\\0\\1}=\pmatrix{1\\ \lambda\\1}=\pmatrix{x_0\\ x_1\\ x_2}$$

as we must also have $4x_0=x_1$, the intersection point $A$ must be such that : $\lambda=4.$ Thus $$A:=\pmatrix{1\\ 4\\ 1}.$$

c) $(A,B;C,D)$ can be computed through one of its projections on one of the coordinates' axes. Let us take it on component $x_1$:

$$ (A,B;C,D) = (4,1;0,2) = ((0-4)/(0-1))/((2-4)/(2-1))=4/(-2) = -2$$

Explanation: projection on one of the axes is a linear transform, thus is a particular case of projective transform, and it is a fundamental property that projective transforms preserve cross ratios.

Answer 2

Three points in $\mathbf P^2_K$ ($K$ a field): $[x:y:t]$, $[x':y':t']$, $[x'':y'':t'']$ are collinear if and only if the determinant of their projective coordinates: $$\begin{bmatrix} x&x'&x''\\y&y'&y''\\t&t'&t'' \end{bmatrix}=0.$$ It is the case here.

Without determinants: just show the rank of the matrix is $\le 2$, with row reduction: $$\begin{bmatrix} 1&1&0\\1&0&2\\1&1&0 \end{bmatrix}\rightsquigarrow\begin{bmatrix} 1&1&0\\0&-1&2\\0&0&0 \end{bmatrix}$$ (Of course, for this matrix, it isn't even necessary.)