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Finding orthogonal vector

  1. Mar 6, 2013 #1
    1. The problem statement, all variables and given/known data

    Find a unit vector that is orthogonal to both [itex]u=(1,1,0)[/itex] and [itex]v=(-1,0,1)[/itex]

    Any help appreciated thanks!
     
  2. jcsd
  3. Mar 6, 2013 #2

    haruspex

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    What do you know about the direction of the cross product of two vectors?
     
  4. Mar 6, 2013 #3
    Yeah, I see where you are going with that but I need to answer the question without using the cross product
     
  5. Mar 6, 2013 #4

    haruspex

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    Ok, so how about supposing the vector is (x, y, z) and obtaining some equations based on dot products?
     
  6. Mar 6, 2013 #5
    I could try that, how do I do it?
     
  7. Mar 6, 2013 #6

    haruspex

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    If you take the dot product with each given vector, what should the result be?
     
  8. Mar 6, 2013 #7
    [itex]u\cdot v=(1,1,0)\cdot (-1,0,1)=-1+0+0=-1[/itex]

    [itex]u\cdot v=\left \| u \right \|\left \| v \right \|cos\theta[/itex]

    the dot product represents the angle between the vectors, is that correct?
     
  9. Mar 6, 2013 #8

    haruspex

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    No, I meant take the dot product of (x, y, z) (a vector intended to be orthogonal to the two given vectors) with each of them.
     
  10. Mar 6, 2013 #9
    let [itex]x=(x,y,z)[/itex]

    [itex]x\cdot u=(x,y,z)\cdot (1,1,0)[/itex]
    [itex]x+y=\left \| x \right \|\left \| u \right \|cos\theta=0[/itex]

    [itex]x\cdot v=(x,y,z)\cdot (-1,0,1)[/itex]
    [itex]z-x=\left \| x \right \|\left \| v \right \|cos\theta=0[/itex]

    [itex]x+y=z-x[/itex]
    [itex]y=z[/itex]

    something like this?
     
    Last edited: Mar 6, 2013
  11. Mar 6, 2013 #10

    haruspex

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    I don't understand what u and v represent in there. looks like in the first line they represent (x,y,z) and (1,1,0), and in the second line (x,y,z) and (-1,0,1). But then you deduce an equation based on u and v representing the same in both cases??
    If two vectors are orthogonal, what is their dot product?
    Edit: It's late here ... and so to bed.
     
  12. Mar 6, 2013 #11
    Whoops, I fixed up the code. Does it look better now?

    The dot product of two orthogonal vectors is equal to zero.
     
  13. Mar 6, 2013 #12
    Ah, I figured it out... thanks for your help!
     
  14. Mar 6, 2013 #13

    Fredrik

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    It's confusing to have one symbol mean two different things, so I recommend that you don't use this notation. I'll use ##p=(x,y,z)##.

    You should write this as
    $$0 =p\cdot u=(x,y,z)\cdot (1,1,0) =x+y.$$ No need to mention the angle between the vectors. Also, you're confusing your readers when you put the zero at the end, instead of at the beginning next to ##p\cdot u## which is equal to 0 by definition of p. (Edit: OK, I see that the thing at the end is also obviously equal to zero because the angle between the vectors is ##\pi/2##, but I still prefer to put the zero at the beginning, where no thought is required). You also left out an equality sign.

    Similarly,
    $$0=p\cdot v =(x,y,z)\cdot (-1,0,1) =-x+z.$$

    Yes, something like that. But not that. First you threw away useful information from the equalities ##p\cdot u=0## and ##p\cdot v=0##, and then you incorrectly cancelled x and -x from what you had left.
     
    Last edited: Mar 6, 2013
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