USU Math 4610
Routine Name: cross_product
Author: Philip Nelson
Language: C++. The code can be compiled using the GNU C++ compiler (gcc). A make file is included to compile an example program
For example,
make
will produce an executable ./vectorOps.out that can be executed.
Description/Purpose: The routine calculates the cross product of two vectors. In mathematics and vector algebra, the cross product or vector product is a binary operation on two vectors in three-dimensional space. Given two linearly independent vectors a and b, the cross_product(a, b) is a vector that is perpendicular to both a and b and thus normal to the plane containing them.
Input: The routine takes two inputs which are two vectors.
@tparam T Type of the elements in the first vector
@tparam U Type of the elements in the second vector
@tparam R Type of the elements in the result vector
@param a The first vector
@param b The second vector
Output: The routine returns the resulting vector of the cross product of the two input vectors
Usage/Example:
int main()
{
std::vector<double> a = {1.1, 2.3, 3.5};
std::vector<double> b = {4.2, 5.4, 6.6};
std::cout << "a\t" << a << '\n';
std::cout << "b\t" << b << '\n';
std::cout << "a x b\t" << cross_product(a, b) << "\n";
}
Output from the lines above
a [ 1.1 2.3 3.5 ]
b [ 4.2 5.4 6.6 ]
a x b [ -3.72 7.44 -3.72 ]
explanation of output:
The first two lines display two vectors a and b.
The third line is the result of a x b.
Implementation/Code: The following is the code for cross_product
This code implements the cross product in 3 dimensions. It ensures the sizes of the vectors are the same then returns a vector with the cross product.
template <typename T, typename U, typename R = decltype(T() * U())>
std::vector<R> cross_product(std::vector<T> const& a, std::vector<U> const& b)
{
// check sizes are 3
if (a.size() != 3 || b.size() != 3)
{
std::cerr << "ERROR: bad size in vector cross product\n";
exit(EXIT_FAILURE);
}
// return the cross product
return {a[1] * b[2] - a[2] * b[1],
a[2] * b[0] - a[0] * b[2],
a[0] * b[1] - a[1] * b[0]};
}
Last Modified: September 2018