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If you need the intersection point, then the answer by OMG_peanuts is a faster approach. However, if you just want to find whether the lines intersect or not, you can do so by using the line equation (ax + by + c = 0). The approach is as follows: Let's start with two line segments: segment 1 and segment 2.
Since these two numbers are not the same, the lines are not parallel, and they intersect somewhere. Now, if what you are considering is only the line segment between the two points, I would first consider the slopes as we just did.
Suppose the two line segments run from p to p + r and from q to q + s. Then any point on the first line is representable as p + t r (for a scalar parameter t) and any point on the second line as q + u s (for a scalar parameter u). The two lines intersect if we can find t and u such that: p + t r = q + u s. Cross both sides with s, getting.
If there is an intersection, then we must check that the intersection actually lies between both sets of points. If lambda is greater than 1, the intersection is beyond the second point. If lambda is less than 0, the intersection is before the first point. Hence, 0<lambda<1 && 0<gamma<1 indicates that the two lines intersect!
intersection = linePoint1 + (lineVec1 * s); return true; else. intersection = Vector3.zero; return false; So, in your situation, you could use that, then check if the intersection point is between a1 and a2 and b1 and b2: float aSqrMagnitude = aDiff.sqrMagnitude; float bSqrMagnitude = bDiff.sqrMagnitude;
@rayryeng: The above question (already answered) is to find point of intersection (x,y) of two lines. But my question is to check whether two lines intersect or not (just boolean answer yes or no). I tried my best on S.O., but not found a duplicate? Please remove duplicate tag, if you agree. –
The two lines intersect if and only if there is a solution s t to the system of linear equations. a1 + t(b1 −a1) a2 + t(b2 −a2) a3 + t(b3 −a3) = c1 + s(d1 −c1) = c2 + s(d2 −c2) = c3 + s(d3 −c3). If is a solution to this system, then plugging in to the equation for or to the equation for yields thep oint of intersection.
Your two lines intersect if [4,-3,2] + t[1,8,-3] = [1,0,3] + v[4,-5,-9] or. 4 + t = 1 + 4v -3 + 8t = 0 - 5v 2 - 3t = 3 - 9v Thus, you have 3 simultaneous equations with only 2 unknowns, so you are good to go! If you can find a solution for t and v that satisfies these equations, then the lines intersect.
Using the algorithm above it returns true for two coinciding lines which can obviously not return a single point of intersection (since mathematically speaking there are infinite number of intersection points for this case). I think that the algorithms needs to handle this in a separate case since simply intersecting and coinciding lines are ...
0. To find out whether two lines defined by points (p1, p2) and (p3, p4) intersect within the part of the line truncated by the points, first of all check whether any of the points are identical. If not, then compute δ = p2 − p1 and γ = p4 − p3, then compute the determinant d = |δx γx δy γy| If this is zero, the two lines are parallel ...