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: Show that this set of equations defines a unique vector

  1. Sep 9, 2012 #1
    1. The problem statement, all variables and given/known data

    Show that, given vectors A and B:

    A * X = |a|
    A x X = B

    defines X as a unique vector

    2. Relevant equations

    3. The attempt at a solution

    No idea.

    My only conclusion thus far is that the magnitude of x is equal to 1/cos(theta) where theta is the angle it makes with vector A.

    Any ideas?
  2. jcsd
  3. Sep 9, 2012 #2


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    It appears that you're being a bit careless with upper/lower case.

    If you mean AX = |A|,

    then, yes, the magnitude of vector, X, is given by |X| = 1/cos(θ) .


    So, it looks like you used CD = |C| |D| cos(θ) .

    Now use that the magnitude of the vector product is:
    |C × D| = |C| |D| sin(θ)

  4. Sep 9, 2012 #3
    Sorry! Yes, I tried to make all vectors upper case afterwards to prevent confusion with the cross product.

    I have also noted the magnitude of the cross product to be |A||X|sin(theta) = |B| but I am not sure how this relates to finding anything out about X, since I already know its magnitude based on theta above, and I know nothing about B other than it being perpendicular to both A and X.
  5. Sep 9, 2012 #4


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    And what is |X| ?
  6. Sep 9, 2012 #5

    So then that results in
    Tan(theta) = |B|/|A|

    So it's direction relative to A is

    And it's magnitude is
    Which is of course:


    Geometrically, I figure that cos(arctan(x)) = (x^2 + 1)^(-1/2)
    So 1/cos(arctan(x)) = (x^2 + 1)^(1/2)

    Then letting x = |B|/|A|,

    The magnitude must be
    [itex]\sqrt{\frac{|B|^{2}}{|A|^{2}} + 1}[/itex]

    And its direction relative to A must be

    Hm... but arctan will give the same thing for some angle theta, and 180 - theta. So it's still not unique.. this must be resolved some way?

    Thanks for the help!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook