suppose I have a normalized vector perpendicular to a plane like
x=(2i+3j+k)√14 , how can I find a rotation matrix A, such that it rotates x into the xy plane like so:
Ax=y=(i,j)√2
According to documentation matrix.ith_to_zero_rotation return a rotation matrix that sends the i-th coordinates of the vector v to zero by doing a rotation with the (i-1)-th coordinate :
sage: v = vector((2,3,1))/sqrt(14)
sage: matrix.ith_to_zero_rotation(v, 2)
[ 1 0 0]
[ 0 3/10*sqrt(14)*sqrt(5/7) 1/10*sqrt(14)*sqrt(5/7)]
[ 0 -1/10*sqrt(14)*sqrt(5/7) 3/10*sqrt(14)*sqrt(5/7)]
sage: matrix.ith_to_zero_rotation(v, 2) * v
(1/7*sqrt(14), sqrt(5/7), 0)See also matrix.vector_on_axis_rotation.
Asked: 12 years ago
Seen: 1,347 times
Last updated: Mar 02 '17
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This is a linear algebra question, not a question about Sage. Try asking on http://math.stackexchange.com/
Look at http://www.hr.shuttle.de:9000/home/pub/105/ There is an example.
why did you rotate the vector around z axis? can't we rotate it around x and y axes instead?
Your original question doesn't have a unique answer. There are infinitely many rotations that take your vector `x` to a vector in the XY-plane.