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

Info icon This preview shows pages 18–21. Sign up to view the full content.

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
Image of page 18

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

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.
Image of page 19
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 .
Image of page 20

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

Image of page 21
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern