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Find vector parallel to yz plane and perpendicular to other given vector

  1. Jan 21, 2010 #1
    1. The problem statement, all variables and given/known data
    Find a vector u that is parallel to the yz-plane and perpendicular to the vector v=<5,0,4>.

    2. The attempt at a solution

    I first tried to find a unit vector parallel to the yz-plane, I then crossed this vector with vector v, but then I remembered that the resulting vector would be perpendicular to both, so that wouldn't work.

    Let me be more precise:

    1. I found a unit vector to (0,1,1)

    I got (0,1/sqrt(2), 1/sqrt(2))

    2. I was going to cross this vector with v (5,0,4) but then I realized that the resulting vector would be perpendicular both the yz plane and vector v so I didn't follow through.

    So then :

    1. I crossed u(a,b,c) with the yz plane (0,1,1) and got:
    (b-c)i -(a)j +(a)k and set it equal to zero because I want them to be parallel.

    Clarifying that I'm using a,b, and c as my unknowns.

    2.I tried dotting this vector I found with vector v(5,0,4) and I set them equal to zero but now I have an equation with three unknowns but I don't know how to solve them.

    Here's what I got:

    5b-5c +4a=0
  2. jcsd
  3. Jan 21, 2010 #2


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    Hi Raziel2701! :wink:
    oooh, you're making this so complicated :cry:

    forget cross products, forget unit vectors …

    what's a typical vector parallel to the yz-plane? …

    now when is it perpendicular to <5,0,4> ? :smile:
  4. Jan 21, 2010 #3
    I don't think I'm following. Apparently any vector of the form <0,y,z,> as long as y and z are equal should be parallel to the yz plane correct?

    So if that is right then that I get, but as to when it is perpendicular to vector <5,0,4>, I honestly don't know. I'm rather stubborn about using the dot product to find what those vector components would be, but I can't get the right answer yet.

    Any further hints would be great.
  5. Jan 21, 2010 #4


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    No, you are following, you just think you aren't. :biggrin:

    Yes, any (0,b,c) will do it. :smile:
    Why this aversion to the dot product … it's much simpler than the cross product anyway! :smile:

    So, what is (0,b,c).(5,0,4) ? And what does it need to be? :wink:
  6. Jan 21, 2010 #5
    Ok, so I got a valid answer, but there's still something that seems wrong.

    So I got <0,2,0> for my answer, the middle component can be any number apparently, and this vector is indeed perpendicular with the vector <5,0,4> but unless my math is wrong, the dot product of my answer and <0,1,1> does not give me an angle of zero to indicate that the two vectors are parallel but rather, I get an angle of 45 degrees.

    So what's up with that?

    And thank you for your help, I like the puzzling, yet helpful advice rather than straight out giving me an answer and an explanation.
  7. Jan 21, 2010 #6


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    Yes, <0,b,0> for any value of b will do.

    It lies in the yz-plane, and it's perpendicular to <5,0,4> because the dot product is 5*0 + 0*b + 4*0.

    (0,1,1) is a vector you invented in your first post … i did say to forget it! :wink:

    ok, now just go back to your fist post and check why you didn't need a unit vector, and why the cross product wouldn't work …

    you can use cross product to make a new vector perpendicular to a given vector, but not to check that two given vectors are perpendicular … for that, you need the dot product. :smile:
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