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michquelld michquelld
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11 years ago
Supposed s(10, -14) , and s(3, 2, -8) are the directional vectors.

To find the normal of these 2 directional vectors are you suppose to randomly flip one of the positive/negative signs? And if so, is this true for all directional vectors in R^2 and R^3?
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11 years ago
In 2D space, normals to the direction vector (a, b) are just (-b, a) or (b, -a), or in fact any non-zero multiple of this. Why? The dot product works out to zero. Restricting yourself to unit vectors, there are still two different normals.

In 3D space, normals are harder to nail down since they're much more plentiful. Restricting yourself to unit vectors, there are still infinitely many normals. You can pick one by solving the equation (a, b, c) dot (n1, n2, n3) = 0 where (a, b, c) is your direction vector and n1, n2, and n3 are unknowns, but you'll have to make some arbitrary choices.
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