out put of the cross product

i am trying to get the 'cross product' of two vectors. these two vectors represent two planes. so, my vectors are as a1,b1,-1 and a2,b2,-1. (I used, my plane equation as ax+by-z+d=0).

this was my defined function to get the cross product;

vector<double> cross_vector(vector<double> plane1,vector<double> plane2){
vector<double> cross_product;
double a1=plane1.at(0); double a2=plane2.at(0);
double b1=plane1.at(1); double b2=plane2.at(1);
int c1,c2=-1;
double cross_a=(b1*c2)-(b2*c1);
double cross_b=(a2*c1)-(a1*c2);
double cross_c=(a1*b2)-(a2*b1);
cross_pro开发者_StackOverflow中文版duct.push_back(cross_a);
cross_product.push_back(cross_;
cross_product.push_back(cross_c);

return cross_product;
}

for the result i got as below result for different plane combinations;

 523554   -1.3713e+006  -0.00160687

 556340   -1.43908e+006  0.00027957

-568368    1.46225e+006 -0.00034963

 143455   -380017       -0.00027957 

i can't understand the values like 1.46225e+006? is there any wrong with my function? i know, my resultant cross vector should be directed exactly horizontal. So, could you also tell me how can i check whether my cross-vector is horizontal or not? hope your advices.

int c1,c2=-1;

This leaves c1 uninitialized. Use:

int c1=-1, c2=-1;

The math looks correct. Placing a quick A = <1,0,0> and B = <0, 1, 0> gave a reasonable result on the backside of <0, 0, 1>. The e notatin represent the number times 10 to the power after the e. So those might be reasonable as well, but it's hard to say as from your example I can't tell what your input values were. I wouldn't personnaly return the value directly though - I'd prefer to return as a reference or pointer to prevent needless copying. Also, as the above poster mentioned, you do have an initialized var.