# finding rotation matrix in 3d

 0 suppose I have a normalized vector perpendicular to a plane like $x = \frac{(2i+3j+k)}{\sqrt{14}}$ , how can I find a rotation matrix A, such that it rotates x into the xy plane like so: $Ax = y = \frac{(i, j)} {\sqrt{2}}$ asked Jun 19 '12 ebs 129 ● 1 ● 10 ● 15 This is a linear algebra question, not a question about Sage. Try asking on http://math.stackexchange.com/benjaminfjones (Jun 19 '12)Look at http://www.hr.shuttle.de:9000/home/pub/105/ There is an example.ndomes (Jun 19 '12)why did you rotate the vector around z axis? can't we rotate it around x and y axes instead?ebs (Jun 21 '12)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.benjaminfjones (Jun 21 '12)

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