Numpy cross product does not return orthogonal vector
I am using the cross product in numpy to generate a third vector orthogonal to two orthogonal vectors. In the code snippet below, the first operation (cross product) shows my problem, taking the cross product of two vectors gives me just the negation of one of the input vectors, not a third vector orthogonal to both.
The next operation shows that my two vectors are indeed orthogonal, not that this should matter. What's going on here?
np.cross([ 0.36195593, 0.93219521, 0. ],[ 0.65916161, -0.25594151, -0.70710672])
Out[94]: array([-0.6591615 , 0.25594147, -0.70710684])
np.dot([ 0.36195593, 0.93219521, 0. ],[ 0.65916161, -0.25594151, -0.70710672])
Out[95]: 3.905680168170278e-09
python numpy cross-product
asked Nov 21 '18 at 18:28
user7474454user7474454
52
add a comment |
I am using the cross product in numpy to generate a third vector orthogonal to two orthogonal vectors. In the code snippet below, the first operation (cross product) shows my problem, taking the cross product of two vectors gives me just the negation of one of the input vectors, not a third vector orthogonal to both.
The next operation shows that my two vectors are indeed orthogonal, not that this should matter. What's going on here?
np.cross([ 0.36195593, 0.93219521, 0. ],[ 0.65916161, -0.25594151, -0.70710672])
Out[94]: array([-0.6591615 , 0.25594147, -0.70710684])
np.dot([ 0.36195593, 0.93219521, 0. ],[ 0.65916161, -0.25594151, -0.70710672])
Out[95]: 3.905680168170278e-09
python numpy cross-product
asked Nov 21 '18 at 18:28
user7474454user7474454
52
2
It isn't exactly the negation of your last vector. It just happens to be the case for the values you're using: that is the correct cross product.
– Matt Messersmith
Nov 21 '18 at 18:36
"taking the cross product of two vectors gives me just the negation of one of the input vectors" No, it does not. Take a look at the result again. The last term in the result is-0.70710684, which has the same sign as the third term of the second argument ofnp.cross.
– Warren Weckesser
Nov 21 '18 at 18:46
add a comment |
I am using the cross product in numpy to generate a third vector orthogonal to two orthogonal vectors. In the code snippet below, the first operation (cross product) shows my problem, taking the cross product of two vectors gives me just the negation of one of the input vectors, not a third vector orthogonal to both.
The next operation shows that my two vectors are indeed orthogonal, not that this should matter. What's going on here?
np.cross([ 0.36195593, 0.93219521, 0. ],[ 0.65916161, -0.25594151, -0.70710672])
Out[94]: array([-0.6591615 , 0.25594147, -0.70710684])
np.dot([ 0.36195593, 0.93219521, 0. ],[ 0.65916161, -0.25594151, -0.70710672])
Out[95]: 3.905680168170278e-09
python numpy cross-product
asked Nov 21 '18 at 18:28
user7474454user7474454
52
I am using the cross product in numpy to generate a third vector orthogonal to two orthogonal vectors. In the code snippet below, the first operation (cross product) shows my problem, taking the cross product of two vectors gives me just the negation of one of the input vectors, not a third vector orthogonal to both.
The next operation shows that my two vectors are indeed orthogonal, not that this should matter. What's going on here?
np.cross([ 0.36195593, 0.93219521, 0. ],[ 0.65916161, -0.25594151, -0.70710672])
Out[94]: array([-0.6591615 , 0.25594147, -0.70710684])
np.dot([ 0.36195593, 0.93219521, 0. ],[ 0.65916161, -0.25594151, -0.70710672])
Out[95]: 3.905680168170278e-09
python numpy cross-product
python numpy cross-product
asked Nov 21 '18 at 18:28
user7474454user7474454
52
asked Nov 21 '18 at 18:28
user7474454user7474454
52
asked Nov 21 '18 at 18:28
user7474454user7474454
52
asked Nov 21 '18 at 18:28
user7474454user7474454
52
asked Nov 21 '18 at 18:28
user7474454user7474454
52
52
2
It isn't exactly the negation of your last vector. It just happens to be the case for the values you're using: that is the correct cross product.
– Matt Messersmith
Nov 21 '18 at 18:36
"taking the cross product of two vectors gives me just the negation of one of the input vectors" No, it does not. Take a look at the result again. The last term in the result is-0.70710684, which has the same sign as the third term of the second argument ofnp.cross.
– Warren Weckesser
Nov 21 '18 at 18:46
add a comment |
2
It isn't exactly the negation of your last vector. It just happens to be the case for the values you're using: that is the correct cross product.
– Matt Messersmith
Nov 21 '18 at 18:36
"taking the cross product of two vectors gives me just the negation of one of the input vectors" No, it does not. Take a look at the result again. The last term in the result is-0.70710684, which has the same sign as the third term of the second argument ofnp.cross.
– Warren Weckesser
Nov 21 '18 at 18:46
2
2
It isn't exactly the negation of your last vector. It just happens to be the case for the values you're using: that is the correct cross product.
– Matt Messersmith
Nov 21 '18 at 18:36
It isn't exactly the negation of your last vector. It just happens to be the case for the values you're using: that is the correct cross product.
– Matt Messersmith
Nov 21 '18 at 18:36
"taking the cross product of two vectors gives me just the negation of one of the input vectors" No, it does not. Take a look at the result again. The last term in the result is
-0.70710684, which has the same sign as the third term of the second argument of np.cross.– Warren Weckesser
Nov 21 '18 at 18:46
"taking the cross product of two vectors gives me just the negation of one of the input vectors" No, it does not. Take a look at the result again. The last term in the result is
-0.70710684, which has the same sign as the third term of the second argument of np.cross.– Warren Weckesser
Nov 21 '18 at 18:46
add a comment |
1 Answer
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First, it's isn't exactly the negation. The last term has the same sign. It just so happens, perfectly coincidentally, that it is close to the negation of one of the original vectors.
Secondly, it is the correct cross product. Instead of doing it by hand, I'll appeal to the fact that the cross product is defined geometrically as a vector that must be orthogonal to its two original inputs. The fact that the two inputs are orthogonal is (largely) irrelevant.
In [11]: first = [ 0.36195593, 0.93219521, 0.]
In [12]: second = [ 0.65916161, -0.25594151, -0.70710672]
In [13]: third = np.cross(first, second)
In [14]: third
Out[14]: array([-0.6591615 , 0.25594147, -0.70710684])
In [15]: np.dot(first, third)
Out[15]: 0.0
In [17]: np.dot(second, third)
Out[17]: 1.1102230246251565e-16
In [18]: np.isclose( np.dot(second, third), 0)
Out[18]: True
HTH.
answered Nov 21 '18 at 18:54
Matt MessersmithMatt Messersmith
6,30921933
add a comment |
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First, it's isn't exactly the negation. The last term has the same sign. It just so happens, perfectly coincidentally, that it is close to the negation of one of the original vectors.
Secondly, it is the correct cross product. Instead of doing it by hand, I'll appeal to the fact that the cross product is defined geometrically as a vector that must be orthogonal to its two original inputs. The fact that the two inputs are orthogonal is (largely) irrelevant.
In [11]: first = [ 0.36195593, 0.93219521, 0.]
In [12]: second = [ 0.65916161, -0.25594151, -0.70710672]
In [13]: third = np.cross(first, second)
In [14]: third
Out[14]: array([-0.6591615 , 0.25594147, -0.70710684])
In [15]: np.dot(first, third)
Out[15]: 0.0
In [17]: np.dot(second, third)
Out[17]: 1.1102230246251565e-16
In [18]: np.isclose( np.dot(second, third), 0)
Out[18]: True
HTH.
answered Nov 21 '18 at 18:54
Matt MessersmithMatt Messersmith
6,30921933
add a comment |
First, it's isn't exactly the negation. The last term has the same sign. It just so happens, perfectly coincidentally, that it is close to the negation of one of the original vectors.
Secondly, it is the correct cross product. Instead of doing it by hand, I'll appeal to the fact that the cross product is defined geometrically as a vector that must be orthogonal to its two original inputs. The fact that the two inputs are orthogonal is (largely) irrelevant.
In [11]: first = [ 0.36195593, 0.93219521, 0.]
In [12]: second = [ 0.65916161, -0.25594151, -0.70710672]
In [13]: third = np.cross(first, second)
In [14]: third
Out[14]: array([-0.6591615 , 0.25594147, -0.70710684])
In [15]: np.dot(first, third)
Out[15]: 0.0
In [17]: np.dot(second, third)
Out[17]: 1.1102230246251565e-16
In [18]: np.isclose( np.dot(second, third), 0)
Out[18]: True
HTH.
answered Nov 21 '18 at 18:54
Matt MessersmithMatt Messersmith
6,30921933
add a comment |
First, it's isn't exactly the negation. The last term has the same sign. It just so happens, perfectly coincidentally, that it is close to the negation of one of the original vectors.
Secondly, it is the correct cross product. Instead of doing it by hand, I'll appeal to the fact that the cross product is defined geometrically as a vector that must be orthogonal to its two original inputs. The fact that the two inputs are orthogonal is (largely) irrelevant.
In [11]: first = [ 0.36195593, 0.93219521, 0.]
In [12]: second = [ 0.65916161, -0.25594151, -0.70710672]
In [13]: third = np.cross(first, second)
In [14]: third
Out[14]: array([-0.6591615 , 0.25594147, -0.70710684])
In [15]: np.dot(first, third)
Out[15]: 0.0
In [17]: np.dot(second, third)
Out[17]: 1.1102230246251565e-16
In [18]: np.isclose( np.dot(second, third), 0)
Out[18]: True
HTH.
answered Nov 21 '18 at 18:54
Matt MessersmithMatt Messersmith
6,30921933
First, it's isn't exactly the negation. The last term has the same sign. It just so happens, perfectly coincidentally, that it is close to the negation of one of the original vectors.
Secondly, it is the correct cross product. Instead of doing it by hand, I'll appeal to the fact that the cross product is defined geometrically as a vector that must be orthogonal to its two original inputs. The fact that the two inputs are orthogonal is (largely) irrelevant.
In [11]: first = [ 0.36195593, 0.93219521, 0.]
In [12]: second = [ 0.65916161, -0.25594151, -0.70710672]
In [13]: third = np.cross(first, second)
In [14]: third
Out[14]: array([-0.6591615 , 0.25594147, -0.70710684])
In [15]: np.dot(first, third)
Out[15]: 0.0
In [17]: np.dot(second, third)
Out[17]: 1.1102230246251565e-16
In [18]: np.isclose( np.dot(second, third), 0)
Out[18]: True
HTH.
answered Nov 21 '18 at 18:54
Matt MessersmithMatt Messersmith
6,30921933
answered Nov 21 '18 at 18:54
Matt MessersmithMatt Messersmith
6,30921933
answered Nov 21 '18 at 18:54
Matt MessersmithMatt Messersmith
6,30921933
answered Nov 21 '18 at 18:54
Matt MessersmithMatt Messersmith
6,30921933
6,30921933
add a comment |
add a comment |
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It isn't exactly the negation of your last vector. It just happens to be the case for the values you're using: that is the correct cross product.
– Matt Messersmith
Nov 21 '18 at 18:36
"taking the cross product of two vectors gives me just the negation of one of the input vectors" No, it does not. Take a look at the result again. The last term in the result is
-0.70710684, which has the same sign as the third term of the second argument ofnp.cross.– Warren Weckesser
Nov 21 '18 at 18:46