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Linear algebra with dot product?

  1. Jul 14, 2010 #1
    1. The problem statement, all variables and given/known data

    Let A(theta)x = b for each theta in S. Calculate,
    (x dot b)/(|x||b|)

    A =
    [cos(theta) -sin(theta)
    sin(theta) cos(theta)]

    How is this related to theta?
    Recall that x dot y and |x| are the standard dot-product and magnitude, respectively, from vector-calculus. These operations hold for vectors in Rn
    but now have the following definitions, x dot y = x(transpose)y and |x =sqrt(x(transpose)*x)

    2. Relevant equations



    3. The attempt at a solution

    I'm not sure how to do this because i don't know what x is or what b is, so i'm confused.
     
  2. jcsd
  3. Jul 14, 2010 #2

    Mark44

    Staff: Mentor

    x is a vector in R2, and so is b. The matrix A is a rotation matrix, where the parameter theta indicates how much rotation.
     
  4. Jul 14, 2010 #3
    I don't understand how I would go about calculating it without knowing anything besides that
     
  5. Jul 15, 2010 #4

    Mark44

    Staff: Mentor

    Pick a value for theta, then pick a few values for x. Now calculate Ax. That should give you an idea of what A does, and how A is related to theta.
     
  6. Jul 15, 2010 #5
    don't I need to actually calculate the dot product equation though? I can't just pick something for x and b and actually have it be correct to the equation.
     
  7. Jul 15, 2010 #6

    hunt_mat

    User Avatar
    Homework Helper

    The equation is:[tex]A(\theta )x=b[/tex]. Then [tex]x=A^{-1}(\theta )b[/tex], You can calculate what the inverse of A is. Denote [tex]\mathbf{b}=(b_{1},b_{2})[/tex], you can then calculate x in terms of theta and b, from there compute your dot product.
     
  8. Jul 15, 2010 #7

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Nice to know: if [itex]A(\theta)[/itex] is 'rotation about a given axis by angle [itex]\theta[/itex]' then the inverse rotation is just the rotation about the same axis by angle [itex]-\theta[/itex].

    That is, [itex]A(\theta)^{-1}= A(-\theta)[/itex].
     
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