Using determinant for handiness

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2 comments, last by Sneftel 14 years ago
I've been reading how the determinant of 3 basis vectors can be taken to determine handiness of the coordinate system. (u x v) . z > 0 == right hand (u x v) . z < 0 == left hand I'm not clear on how this works though. Say my three vectors are: [1,0,0], [0,1,0], [0,0,1]. The determinant is always 1, yet I can still use these basis vectors for either a right handed or left handed system. What am I missing?
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Order matters.

For instance, the determinant of the identity matrix is one, but the determinant of the matrix

  0  1  0  1  0  0  0  0  1


(identity with first two columns swapped) is -1.

So for the purpose of handedness, the basis {(1,0,0),(0,1,0),(0,0,1)} is not the same as the basis {(0,1,0),(1,0,0),(0,0,1)}.
Quote:Original post by Emergent
So for the purpose of handedness, the basis {(1,0,0),(0,1,0),(0,0,1)} is not the same as the basis {(0,1,0),(1,0,0),(0,0,1)}.


But if someone comes up to me and says is {(1,0,0),(0,1,0),(0,0,1)} left handed or right, there's no way I can determine that, correct? It would be an arbitrary decision based on if I use my left or right hand to visualize the cross of (1,0,0) and (0,1,0).
That's correct. The handedness of a coordinate system does not affect the calculation of the cross product or the determinant, and a particular set of basis vectors with a triple product of 1 may refer either to a right-handed or left-handed coordinate system.

HOWEVER, you can use this to determine whether a set of basis vectors, expressed in a coordinate system of known handedness, refers to a basis which is of the same handedness (>0) or the opposite handedness (<0).

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