向量的點乘與叉乘
阿新 • • 發佈：2019-01-27
//Compute the dot product AB BC
int dot(int[] A, int[] B, int[] C){
AB = new int[2];
BC = new int[2];
AB[0] = B[0]-A[0];
AB[1] = B[1]-A[1];
BC[0] = C[0]-B[0];
BC[1] = C[1]-B[1];
int dot = AB[0] * BC[0] + AB[1] * BC[1];
return dot;
}
//Compute the cross product AB x AC
int cross(int[] A, int[] B, int[] C){
AB = new int[2];
AC = new int[2];
AB[0] = B[0]-A[0];
AB[1] = B[1]-A[1];
AC[0] = C[0]-A[0];
AC[1] = C[1]-A[1];
int cross = AB[0] * AC[1] - AB[1] * AC[0];
return cross;
}
//Compute the distance from A to B
double distance(int[] A, int[] B){
int d1 = A[0] - B[0];
int d2 = A[1] - B[1];
return sqrt(d1*d1+d2*d2);
}
//Compute the distance from AB to C
//if isSegment is true, AB is a segment, not a line.
double linePointDist(int[] A, int[] B, int[] C, boolean isSegment){
double dist = cross(A,B,C) / distance(A,B);
if(isSegment){
int dot1 = dot(A,B,C);
if(dot1 > 0)return distance(B,C);
int dot2 = dot(B,A,C);
if(dot2 > 0)return distance(A,C);
}
return abs(dist);
}
int dot(int[] A, int[] B, int[] C){
AB = new int[2];
BC = new int[2];
AB[0] = B[0]-A[0];
AB[1] = B[1]-A[1];
BC[0] = C[0]-B[0];
BC[1] = C[1]-B[1];
int dot = AB[0] * BC[0] + AB[1] * BC[1];
return dot;
}
//Compute the cross product AB x AC
int cross(int[] A, int[] B, int[] C){
AB = new int[2];
AC = new int[2];
AB[0] = B[0]-A[0];
AB[1] = B[1]-A[1];
AC[0] = C[0]-A[0];
AC[1] = C[1]-A[1];
int cross = AB[0] * AC[1] - AB[1] * AC[0];
return cross;
}
//Compute the distance from A to B
double distance(int[] A, int[] B){
int d1 = A[0] - B[0];
int d2 = A[1] - B[1];
return sqrt(d1*d1+d2*d2);
}
//Compute the distance from AB to C
//if isSegment is true, AB is a segment, not a line.
double linePointDist(int[] A, int[] B, int[] C, boolean isSegment){
double dist = cross(A,B,C) / distance(A,B);
if(isSegment){
int dot1 = dot(A,B,C);
if(dot1 > 0)return distance(B,C);
int dot2 = dot(B,A,C);
if(dot2 > 0)return distance(A,C);
}
return abs(dist);
}