This is easily proved by the device of using so3 to

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This is easily proved by the device of using SO(3) to move your right hand from the { ˆ ı, ˆ , ˆ k } frame to an orthonormal frame associated with the frame { x, y, x × y } . Here’s the picture, looking “down” at the plane determined by 0, x , y :
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Cross product 13 O y x y x Let x 0 = x/ k x k , and let y 0 be the unit vector shown: y 0 = y - x 0 y x 0 k y - x 0 y x 0 k . Thus y = c 1 x 0 + c 2 y 0 , where c 2 > 0. Then x × y = c 2 x × y 0 = c 2 k x k x 0 × y 0 . Thus x × y is a positive scalar multiple of x 0 × y 0 . Now define the matrix A in terms of its columns, A = ( x 0 y 0 x 0 × y 0 ) . Then A ˆ ı = x 0 , A ˆ = y 0 , A ˆ k = x 0 × y 0 . And we know A SO(3). Thus A takes the right- handed frame { ˆ ı, ˆ , ˆ k } to the right-handed frame { x 0 , y 0 , x 0 × y 0 } . This concludes the proof of the right-hand rule. F. What i s SO ( 3 ) ? This seems to be a good place to discuss the fine details of SO(3), the special orthogonal group on R 3 . The members of this group are of course 3 × 3 orthogonal matrices with determi- nants equal to 1. Let A SO(3). This matrix of course “acts on” R 3 by matrix multiplication, sending column vectors x to the column vectors Ax . We are going to prove that this “action” can be described as an ordinary rotation of R 3 about an axis. We first pause to describe rotations. Suppose that ˆ w is a fixed unit vector in R 3 , and that θ is a fixed angle. We use the standard right-handed orientation for R 3 . Now we take a view of R 3 looking from ˆ w towards 0, and rotate the entire space counter-clockwise about the axis ˆ w through the angle θ . Let us call the resulting matrix R ( ˆ w, θ ). A couple of things should be noticed. One is that a counter-clockwise rotation with a negative angle θ is actually a
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14 Chapter 7 clockwise rotation with the positive angle - θ . The other is that adding any integer multiple of 2 π to θ results in the same rotation, according to our convention here. Now we show how to calculate R ( ˆ w, θ ). To do this use a right-handed orthonormal frame { ˆ u, ˆ v, ˆ w } . This can be computed by choosing any unit vector ˆ u which is orthogonal to ˆ w , and then ˆ v is uniquely determined by the required formula ˆ v = ˆ w × ˆ u . Now think of projecting R 3 orthogonally onto the ˆ u , ˆ v plane, and then rotating through the angle θ . Here’s the view looking down from ˆ w : R(w, ) p θ θ v p u Thus we have R ( ˆ w, θ u = ˆ u cos θ + ˆ v sin θ, R ( ˆ w, θ v = - ˆ u sin θ + ˆ v cos θ. Of course, since ˆ w is the axis of rotation, R ( ˆ w, θ ) ˆ w = ˆ w. The summary of this information is expressed in terms of matrices as R ( ˆ w, θ )(ˆ u ˆ v ˆ w ) = u cos θ + ˆ v sin θ - ˆ u sin θ + ˆ v cos θ ˆ w ) = u ˆ v ˆ w ) cos θ - sin θ 0 sin θ cos θ 0 0 0 1 . Therefore R ( ˆ w, θ ) = (ˆ u ˆ v ˆ w ) cos θ - sin θ 0 sin θ cos θ 0 0 0 1 u ˆ v ˆ w ) - 1 . ( * )
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Cross product 15 Remember: the computation of the inverse of the orthogonal matrix (ˆ u ˆ v ˆ w ) is just a matter of writing down its transpose ˆ u t ˆ v t ˆ w t .
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