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Homework Help: Find a basis for U perp given U = {(3,-1,2),(2,0,-3)}

  1. Apr 8, 2006 #1
    Find a basis for U perp given U = {(3,-1,2),(2,0,-3)}

    to find U perp i have to find a vector such that it perpendicular to BOTH of those vectors up there
    3x - 2y + 2z = 0
    2x - 3z = 0

    i get z = t
    y = 13t/2
    x = 2t
    so the only basis for U perp is
    {2t,13t/2,t}?? ALl values of t are simply multiples of each other anyway...
    AM i right? Only one vector?

    Also
    If [tex] R^n = span\{X_{1},X_{2},...,X_{N}\ [/tex]} and [tex] X \bullet X_{i} = Y \bullet X_{i} [/tex]
    for all i, then show that X = Y


    WEll all the Xi form an orthogonal basis right? So from the given condition couldnt we say
    for n not i
    [tex] X \bullet \left(\frac{X_{i} \bullet X_{1}}{||X_{1}||^2} X_{1} + ... + \frac{X_{i} \bullet X_{n}}{||X_{n}||^2} X_{n}\right) [/tex]
    and [tex] Y \bullet \left(\frac{X_{i} \bullet X_{1}}{||X_{1}||^2} X_{1} + ... + \frac{X_{i} \bullet X_{n}}{||X_{n}||^2} X_{n}\right) [/tex]
    since the two are equal.. and Xi dot X and Xi dot are all equal... then X must be same as Y
     
    Last edited: Apr 8, 2006
  2. jcsd
  3. Apr 8, 2006 #2

    AKG

    User Avatar
    Science Advisor
    Homework Helper

    Strictly speaking, {2t,13t/2,t} is not a basis for U perp, you have to pick some value for t. Also, you need to put your vector in brackets, otherwise this is just a set of 3 numbers, which has nothing to do with anything. So if you pick some non-zero t, then {(2t,13t/2,t)} is a basis for U perp.

    How do you know the Xi form an orthogonal basis? And do you intend N to be the same as n? Anyways, you're overcomplicating things, and what you've written doesn't make sense, so I don't know how to help you. As a general tip, if you want to show two things are equal, a good way to do it is to show that their difference is 0. So as a start, consider the fact that since X.Xi = Y.Xi for all i, (X-Y).Xi = 0 for all i.
     
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