- #1
bigfooted
Gold Member
- 685
- 222
Hi,
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
p+t\cdot r = q+u\cdot s.
using
t=\frac{(q-p)\times s}{r \times s}
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:
x=\Psi_1x_1+\Psi_2x_2+\Psi_3x_3,
with
\Psi_i=\prod_{M=1,M ≠ N}^{n}\frac{\xi-\xi_M}{\xi_N-\xi_M}
where \xi=0..1 and x_1 is the starting point, x_2 the midpoint and x_3 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.
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
p+t\cdot r = q+u\cdot s.
using
t=\frac{(q-p)\times s}{r \times s}
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:
x=\Psi_1x_1+\Psi_2x_2+\Psi_3x_3,
with
\Psi_i=\prod_{M=1,M ≠ N}^{n}\frac{\xi-\xi_M}{\xi_N-\xi_M}
where \xi=0..1 and x_1 is the starting point, x_2 the midpoint and x_3 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.