1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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


    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


    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


    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


    User Avatar
    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].
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook