What is the simplest and most efficient ways in numpy to generate two orthonormal vectors a and b such that the cross product of the two vectors equals another unit vector k, which is already known?
I know there are infinitely many such pairs, and it doesn't matter to me which pairs I get as long as the conditions axb=k and a.b=0 are satisfied.
这将做：
>>> k # an arbitrary unit vector k is not array. k is must be numpy class. np.array
np.array([ 0.59500984, 0.09655469, -0.79789754])
要获得第一个：
>>> x = np.random.randn(3) # take a random vector
>>> x -= x.dot(k) * k # make it orthogonal to k
>>> x /= np.linalg.norm(x) # normalize it
要获得第二个：
>>> y = np.cross(k, x) # cross product with k
并验证：
>>> np.linalg.norm(x), np.linalg.norm(y)
(1.0, 1.0)
>>> np.cross(x, y) # same as k
array([ 0.59500984, 0.09655469, -0.79789754])
>>> np.dot(x, y) # and they are orthogonal
-1.3877787807814457e-17
>>> np.dot(x, k)
-1.1102230246251565e-16
>>> np.dot(y, k)
0.0
抱歉，由于缺乏声誉，我无法将其作为评论。
关于@behzad.nouri 的回答，请注意，如果k不是单位向量，则代码将不再给出正交向量！
这样做的正确和通用方法是减去随机向量的纵向部分。通用公式在 这里
因此，您只需在原始代码中替换它：
>>> x -= x.dot(k) * k / np.linalg.norm(k)**2
假设支持正交基的向量是u。
b1 = np.cross(u, [1, 0, 0]) # [1, 0, 0] can be replaced by other vectors, just get a vector orthogonal to u
b2 = np.cross(u, b1)
b1, b2 = b1 / np.linalg.norm(b1), b2 / np.linalg.norm(b2)
如果您愿意，可以使用更短的答案。
获取变换矩阵
B = np.array([b1, b2])
TransB = np.dot(B.T, B)
u2b = TransB.dot(u) # should be like [0, 0, 0]