# Intersection of a line segment and a paraboloid in 3D

Suppose I have a line segment $L$ in 3D: $$x=a_1(1-t)+b_1t$$ $$y=a_2(1-t)+b_2t$$ $$z=(a_1^2+a_2^2-k_1^2)(1-t)+(b_1^2+b_2^2-k_2^2)t$$

Because $L$ is line segment then $0\leq t\leq 1$.

And defining paraboloid $P$ in 3D: $$P:z=2x^2+2y^2-1$$

Where $a_1,a_2,b_1,b_2,k_1,k_2$ are all variables and $k_1$ and $k_2$ are positive numbers.

I want to put some constraints on these variables such that $L$ and $P$ intersect or do not intersect.

I know by substituting $L$ into $P$ and then solving for $t$ where $0\leq t\leq 1$ is the solution of the intersection. But still i could not figure out about the constraints i am looking for.

Thanks for any suggestions.

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You're almost there. First, you need to figure out how many roots of quadratic equation, that you got from substitution, you'll get based on the discriminant. It'll give you an answer about in how many points the line (not segment) will intersect your paraboloid. But since you have also restriction on $t \in [0, 1]$ even if you have intersection you need to analyze root to check whether it satisfies given inequality. It's not much, but problem looks quite tedious to me, but straight forward. – Kaster Dec 7 '12 at 22:06

The endpoints of your segment are at $(a_1, a_2, a_1^2+a_2^2-k_1^2)$ and $(b_1, b_2, b_1^2+b_2^2-k_2^2)$. If both endpoints are above, or both below the paraboloid, then the segment does not intersect the segment. For instance, the first endpoint is above the paraboloid if $a_1^2+a_2^2-k_1^2 > 2a_1^2+2a_2^2-1$. Working it out, both endpoints will be above the paraboloid if
$$k_1^2<1-a_1^2-a_2^2,$$ $$k_2^2<1-b_1^2-b_2^2,$$