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2 vectors orthogonal to 3 4-d vectors

  1. Nov 18, 2008 #1
    I'm a little stuck on this algebra question

    Find 2 unit vectors orthogonal to v1 = <3,1,1,-1>, v2 = <-1,2,2,0> and v3 = <1,0,2,-1>

    I know that this means that if I let z be the vector orthogonal to these 3 then,

    z.v1 = z.v2= z.v3 = 0

    And that my two vectors is likely +/- z.

    I can do something like this:

    z = <a,b,c,d>

    z.v1 = z.v2

    3a + b + c -d = -a + 2b + 2c

    but now what? Or is this the wrong idea?

    Thanks in advance!
  2. jcsd
  3. Nov 18, 2008 #2
    This is rather tricky. Hm. If you're only using:
    [tex]Z \cdot V1 = Z \cdot V2[/tex]
    then you're only solving for a vector that is orthogonal to V1 and V2, means you're excluding V3. So you need to use the third equality as well.

    Even so, you have 4 unknowns and 3 equations. Something's amiss.
  4. Nov 18, 2008 #3


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    Greetings Damascus Road! :smile:

    Yes, z.v1 = z.v2= z.v3 = 0 …

    so your last line should be 3a + b + c -d = -a + 2b + 2c = 0 (= a + 2c - d) … three equations in four variables (which is ok, since you only need the ratios :wink:)
  5. Nov 18, 2008 #4
    Thanks for the replies!

    I'm confused about the ratios however. Can I set the individual components equal?
    3a = a = -a?
  6. Nov 18, 2008 #5


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    Nooo … these are three separate linear equations …

    if it makes you happier, you can reduce the number of variables to 3 (a/d b/d and c/d) by rewriting the equations as 3a/d + b/d + c/d = 1 etc. :wink:
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