So yes, this is homework, but for the class that I'm TA'ing. The question is something like:

Given two lines in 2D homogeneous representation ($\mathbb{R}^{2+1}$): $\bf a$ and $\bf b$, their intersection is $\bf p = a \times b$ (so far so good)

"The sine of the angle of their intersection is the homogeneous weight of this point." Prove this.

I assumed that by homogeneous weight, what is meant is the third value ($\lambda$) of the point $\bf p$, where $\mathbf{p} = (\lambda p_1, \lambda p_2, \lambda) $, then what needs to be proven is that $p_3 = a_1b_2 - a_2b_1 = \sin \theta$ with $\theta$ the angle between those lines. But I can't figure this out.

Aside: This would have been an easier question if in stead of homogeneous weight they asked for the magnitude of the intersection, then it is solved by assuming that $\bf a$ and $\bf b$ are normalized, and using $\mathbf{a} \times \mathbf{b} = \left\| \mathbf{a} \right\| \left\| \mathbf{b} \right\| \sin \theta \ \mathbf{n}$, but I'm not sure if that's what is meant.

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    $\begingroup$ Do you have any description of the particular homogenous coordinates chosen for each line? Because in general, you may scale these coordinates arbitrarily without changing the line they describe, and that scale factor directly affects the homogenous weight. So I guess your lines are normalized to $\lVert a\rVert=\lVert b\rVert=1$ and by “homogenous weight” they actually mean the magnitude $\lVert p\rVert$ just like you suggested. But this is pure guesswork. $\endgroup$
    – MvG
    Dec 10 '12 at 9:06
  • $\begingroup$ Apparently the lines are normalized such that $a_1^2 + a_2^2 = 1$ and idem for $\bf b$. I.e. the direction part of the line is a unit vector. Then $a_1b_2 - a_2b_1$ is the determinant of a matrix formed by the direction vectors of the lines. $\endgroup$
    – noio
    Dec 10 '12 at 10:53
  • $\begingroup$ Perhaps, this math.stackexchange.com/q/4067348 is useful. $\endgroup$
    – MathArt
    Mar 18 '21 at 20:32

You can interpret $$p_3 = \begin{vmatrix}a_1 & b_1 \\ a_2 & b_2\end{vmatrix}$$ as the area of the parallelogram described by these two planar (Euclidean) vectors. That area is equal to the length of one vector times the height measured perpendicular to that vector. Which in turn is equal to the product of the lengths times the sine of the intersection angle.

If your vectors $(a_1,a_2)^T$ and $(b_1,b_2)^T$ have unit length due to a normalization to $a_1^2+a_2^2 = b_1^2+b_2^2 = 1$, as you wrote, then the lengths in the above computation become one, and the sine is all that remains.


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