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To calculate the intersection of two straight lines the cross product of the line vectors can be used, i.e. when the lines start in points p and q, and have direction vectors r and s, then if the cross product r x s is nonzero, the intersection point is q+us, and can be found from

[itex]p+t\cdot r = q+u\cdot s[/itex].

using

[itex]t=\frac{(q-p)\times s}{r \times s}[/itex]

I was wondering how to derive such a relationship for the intersection between a straight line and a second order polynomial.

Specifically, I'm interested in second order Lagrange (and 3rd order Hermite) polynomials:

[itex]x=\Psi_1x_1+\Psi_2x_2+\Psi_3x_3[/itex],

with

[itex]\Psi_i=\prod_{M=1,M ≠ N}^{n}\frac{\xi-\xi_M}{\xi_N-\xi_M}[/itex]

where [itex]\xi=0..1[/itex] and [itex]x_1[/itex] is the starting point, [itex]x_2[/itex] the midpoint and [itex]x_3[/itex] the endpoint

My guess is that standard techniques to find the intersection first transform the second order polynomial to the unit plane where the polynomial reduces to a line, then find the intersection and then transform back, but a (quick) search didn't give me anything.

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# Intersection of straight line with (lagrange) polynomial

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