# Then a ˆ u ˆ w a ˆ u a ˆ w ˆ u ˆ w a is in o 3

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Then A ˆ u ˆ w = A ˆ u A ˆ w = ˆ u ˆ w ( A is in O (3)) = 0 , so A ˆ u is also orthogonal to ˆ w . Thus A ˆ u is a linear combination of ˆ u and ˆ v : A ˆ u = c 1 ˆ u + c 2 ˆ v. Since k A ˆ u k = k ˆ u k = 1, c 2 1 + c 2 2 = 1. Thus there exists an angle θ such that c 1 = cos θ , c 2 = sin θ . That is, A ˆ u = ˆ u cos θ + ˆ v sin θ. We conclude that A ˆ v = A ( ˆ w × ˆ u ) = A ˆ w × A ˆ u (Section D) = ˆ w × u cos θ + ˆ v sin θ ) = ˆ v cos θ - ˆ u sin θ. We now see that A and the rotation R ( ˆ w, θ ) produce the same result on the frame { ˆ u, ˆ v, ˆ w } ! Since every x R 3 is a linear combination of ˆ u , ˆ v , ˆ w , we conclude that Ax = R ( ˆ w, θ ) x for all x R 3 . Thus A = R ( ˆ w, θ ). The final item, that we may require 0 θ π , follows immediately from Problem 7–14. QED

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Cross product 19 DISCUSSION. For a given A SO(3), the representation A = R ( ˆ w, θ ), where ˆ w is a unit vector and 0 θ π , is virtually unique. To see this we start with the result of Problem 7–16, and compute the trace of each matrix: trace A = 3 cos θ + (1 - cos θ )( w 2 1 + w 2 2 + w 2 3 ) = 3 cos θ + (1 - cos θ ) = 2 cos θ + 1 . Thus cos θ = trace A - 1 2 . Therefore, as 0 θ π , the angle is completely determined by the knowledge of trace A . Next notice that Problem 7–16 implies A - A t = 2 sin θ 0 - w 3 w 2 w 3 0 - w 1 - w 2 w 1 0 . We conclude immediately that if 0 < θ < π , the vector w is also uniquely determined. The only cases of nonuniqueness are thus when sin θ = 0, or, equivalently, A = A t . If θ = 0, then A = R ( ˆ w, 0) = I and the “axis” ˆ w is irrelevant. If θ = π , then we have A = - I + 2( w i w j ) . It is easy to check in this case that ˆ w is uniquely determined up to sign. This is because ˆ w 6 = 0 and so some coordinate w j 6 = 0 and we obtain from the above equation 2 w j ˆ w = j th column of ( A + I ) . This determines ˆ w up to a scalar factor, and the fact that k ˆ w k = 1 shows that the only possibilities are ˆ w and - ˆ w . We now have a truly intuitive understanding of SO(3). In particular, the right-hand rule makes wonderful sense in this context, as we now explain. Suppose { ˆ ϕ 1 , ˆ ϕ 2 , ˆ ϕ 3 } is an arbitrary orthonormal frame with standard orientation, and form the usual matrix Φ with those columns, Φ = ( ˆ ϕ 1 ˆ ϕ 2 ˆ ϕ 3 ) . Then Φ SO(3) and Φ - 1 maps the frame { ˆ ϕ 1 , ˆ ϕ 2 , ˆ ϕ 3 } to { ˆ ı, ˆ , ˆ k } . Moreover, Φ is a rotation R ( ˆ w, θ ). Thus if you place your right hand to lie in the correct ˆ k = ˆ ı × ˆ position, then rotating your hand with R ( ˆ w, θ ) produces the ˆ ϕ 3 = ˆ ϕ 1 × ˆ ϕ 2 position.
20 Chapter 7 I would say it is almost amazing that a single rotation can move { ˆ ϕ 1 , ˆ ϕ 2 , ˆ ϕ 3 } to { ˆ ı, ˆ , ˆ k } ! Here is a numerical example. Let A = 0 - 1 0 0 0 - 1 1 0 0 . Clearly, A SO(3). Then we compute trace A = 0, so that cos θ = - 1 2 and we have θ = 2 π/ 3. Then A - A t = 2 sin θ 0 - w 3 w 2 w 3 0 - w 1 - w 2 w 1 0 becomes the equation 0 - 1 - 1 1 0 - 1 1 1 0 = 3 0 - w 3 w 2 w 3 0 - w 1 - w 2 w 1 0 , so we read off immediately that ˆ w = 1 3 1 - 1 1 .

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