You have two line segments and want to know if they are crossing each other:

This is more complicated than it seems, but you can probably ignore some steps depending on which application you are making. For example, it's often unnecessary to test if the line segments are parallel if you are making a game, because the probability that they are exactly parallel is low.

using UnityEngine; using System.Collections; namespace LinearAlgebra { //How to figure out if two lines are intersecting public class LineLineIntersection : MonoBehaviour { //The start end of each line public Transform L1_start; public Transform L1_end; public Transform L2_start; public Transform L2_end; //Just attach a line renderer to an empty game object public LineRenderer lineRenderer1; public LineRenderer lineRenderer2; void Start() { //Give the line renderer a material so we can change the color if they intersect lineRenderer1.material = new Material(Shader.Find("Unlit/Color")); lineRenderer2.material = new Material(Shader.Find("Unlit/Color")); } void Update() { //Connect the points with line renderers lineRenderer1.SetPosition(0, L1_start.position); lineRenderer1.SetPosition(1, L1_end.position); lineRenderer2.SetPosition(0, L2_start.position); lineRenderer2.SetPosition(1, L2_end.position); //Set the default color lineRenderer1.material.color = Color.red; lineRenderer2.material.color = Color.red; //Change color if they intersect if (IsIntersecting()) { lineRenderer1.material.color = Color.blue; lineRenderer2.material.color = Color.blue; } } //Check if the lines are interesecting in 2d space bool IsIntersecting() { bool isIntersecting = false; //3d -> 2d Vector2 l1_start = new Vector2(L1_start.position.x, L1_start.position.z); Vector2 l1_end = new Vector2(L1_end.position.x, L1_end.position.z); Vector2 l2_start = new Vector2(L2_start.position.x, L2_start.position.z); Vector2 l2_end = new Vector2(L2_end.position.x, L2_end.position.z); //Direction of the lines Vector2 l1_dir = (l1_end - l1_start).normalized; Vector2 l2_dir = (l2_end - l2_start).normalized; //If we know the direction we can get the normal vector to each line Vector2 l1_normal = new Vector2(-l1_dir.y, l1_dir.x); Vector2 l2_normal = new Vector2(-l2_dir.y, l2_dir.x); //Step 1: Rewrite the lines to a general form: Ax + By = k1 and Cx + Dy = k2 //The normal vector is the A, B float A = l1_normal.x; float B = l1_normal.y; float C = l2_normal.x; float D = l2_normal.y; //To get k we just use one point on the line float k1 = (A * l1_start.x) + (B * l1_start.y); float k2 = (C * l2_start.x) + (D * l2_start.y); //Step 2: are the lines parallel? -> no solutions if (IsParallel(l1_normal, l2_normal)) { Debug.Log("The lines are parallel so no solutions!"); return isIntersecting; } //Step 3: are the lines the same line? -> infinite amount of solutions //Pick one point on each line and test if the vector between the points is orthogonal to one of the normals if (IsOrthogonal(l1_start - l2_start, l1_normal)) { Debug.Log("Same line so infinite amount of solutions!"); //Return false anyway return isIntersecting; } //Step 4: calculate the intersection point -> one solution float x_intersect = (D * k1 - B * k2) / (A * D - B * C); float y_intersect = (-C * k1 + A * k2) / (A * D - B * C); Vector2 intersectPoint = new Vector2(x_intersect, y_intersect); //Step 5: but we have line segments so we have to check if the intersection point is within the segment if (IsBetween(l1_start, l1_end, intersectPoint) && IsBetween(l2_start, l2_end, intersectPoint)) { Debug.Log("We have an intersection point!"); isIntersecting = true; } return isIntersecting; } //Are 2 vectors parallel? bool IsParallel(Vector2 v1, Vector2 v2) { //2 vectors are parallel if the angle between the vectors are 0 or 180 degrees if (Vector2.Angle(v1, v2) == 0f || Vector2.Angle(v1, v2) == 180f) { return true; } return false; } //Are 2 vectors orthogonal? bool IsOrthogonal(Vector2 v1, Vector2 v2) { //2 vectors are orthogonal is the dot product is 0 //We have to check if close to 0 because of floating numbers if (Mathf.Abs(Vector2.Dot(v1, v2)) < 0.000001f) { return true; } return false; } //Is a point c between 2 other points a and b? bool IsBetween(Vector2 a, Vector2 b, Vector2 c) { bool isBetween = false; //Entire line segment Vector2 ab = b - a; //The intersection and the first point Vector2 ac = c - a; //Need to check 2 things: //1. If the vectors are pointing in the same direction = if the dot product is positive //2. If the length of the vector between the intersection and the first point is smaller than the entire line if (Vector2.Dot(ab, ac) > 0f && ab.sqrMagnitude >= ac.sqrMagnitude) { isBetween = true; } return isBetween; } } }

The above version is more robust and will even work in more dimensions, but if speed is the key and you are working in 2d space, I've found another solution here. So you can replace the fat version with this:

//Check if the lines are interescting in 2d space //Alternative version from http://thirdpartyninjas.com/blog/2008/10/07/line-segment-intersection/ bool IsIntersectingAlternative() { bool isIntersecting = false; //3d -> 2d Vector2 p1 = new Vector2(L1_start.position.x, L1_start.position.z); Vector2 p2 = new Vector2(L1_end.position.x, L1_end.position.z); Vector2 p3 = new Vector2(L2_start.position.x, L2_start.position.z); Vector2 p4 = new Vector2(L2_end.position.x, L2_end.position.z); float denominator = (p4.y - p3.y) * (p2.x - p1.x) - (p4.x - p3.x) * (p2.y - p1.y); //Make sure the denominator is > 0, if so the lines are parallel if (denominator != 0) { float u_a = ((p4.x - p3.x) * (p1.y - p3.y) - (p4.y - p3.y) * (p1.x - p3.x)) / denominator; float u_b = ((p2.x - p1.x) * (p1.y - p3.y) - (p2.y - p1.y) * (p1.x - p3.x)) / denominator; //Is intersecting if u_a and u_b are between 0 and 1 if (u_a >= 0 && u_a <= 1 && u_b >= 0 && u_b <= 1) { isIntersecting = true; } } return isIntersecting; }

There's actually a third way to figure out if two line segments in 2d space are intersecting. The problem with this way is that you can't find the point of intersection, but that doesn't matter if you don't need to. To help us we are going to use the dot product! The basic idea is this (we want to see if the blue line and the black line is intersecting):

We first have to find the normal to a line. Then we calculate the dot product between this normal and the direction between the normal and the two points on the other line. If one of these dot products is negative and the other is positive, then we know the points on the other line are on different sides of the first line. Then we repeat the process for the other line. If you want a more detailed explanation you should watch this video: TypeScript Game Development - Basic 2D Collision Detection. The code looks like this:

//Line segment-line segment intersection in 2d space by using the dot product //p1 and p2 belongs to line 1, and p3 and p4 belongs to line 2 public static bool AreLineSegmentsIntersectingDotProduct(Vector3 p1, Vector3 p2, Vector3 p3, Vector3 p4) { bool isIntersecting = false; if (IsPointsOnDifferentSides(p1, p2, p3, p4) && IsPointsOnDifferentSides(p3, p4, p1, p2)) { isIntersecting = true; } return isIntersecting; } //Are the points on different sides of a line? private static bool IsPointsOnDifferentSides(Vector3 p1, Vector3 p2, Vector3 p3, Vector3 p4) { bool isOnDifferentSides = false; //The direction of the line Vector3 lineDir = p2 - p1; //The normal to a line is just flipping x and z and making z negative Vector3 lineNormal = new Vector3(-lineDir.z, lineDir.y, lineDir.x); //Now we need to take the dot product between the normal and the points on the other line float dot1 = Vector3.Dot(lineNormal, p3 - p1); float dot2 = Vector3.Dot(lineNormal, p4 - p1); //If you multiply them and get a negative value then p3 and p4 are on different sides of the line if (dot1 * dot2 < 0f) { isOnDifferentSides = true; } return isOnDifferentSides; }

So which one is the fastest? I've used C#'s stopwatch timer to measure the time of each line segment-line segment algorithm, and after 1000 updates, this was the total time:

- Algorithm 1: 06.9025942 seconds
- Algorithm 2: 00.0034113 seconds
- Algorithm 3: 00.0096906 seconds