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Let $p \in \mathcal{P}^2$ be a point in projective 2-space coincident with a line $l\in\mathcal{P}^2$ such that $l^\top p = 0$. What does $l \times p$ mean?

For example, $p = \left(x,y,1\right)^\top$ and $l=\left(-1, 0, x\right)^\top$, the line is coincident with the point, i.e. $(l^\top p = 0)$. The cross product is $v = l \times p = \left(-xy, 1+x^2, -y\right)^\top$. Wondering, what is the physical meaning of $v$?

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  • $\begingroup$ Hi bendervader: how exactly are you modeling a projective line with a single homogeneous point $(-1,0,x)^\top$? (I'm just not very familiar with the model. The model I knows relies on specifying two points on the line.) $\endgroup$
    – rschwieb
    Jan 13, 2014 at 19:51

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This is not a natural operation between lines and points. The cross product of two different lines is a point (intersection) and the cross product of two different points is a line (connecting the points). In this case you have to take the dual of either the point or the line. In the first case the cross product is the point on $l$ at maximum distance from $p$. In the second case it is the line through $p$ orthogonal to $l$.

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