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Unit vector tangent to the surface

  1. Feb 23, 2014 #1
    I have the following question:

    Given that ø = (x^2)y + cos(z) find the unit vector n which is both tangent to the surface of constant ø at (1,1,∏/2) and normal to the vector b = x + y - 2z (where x y and z are the unit vectors)

    I have calculated ∇ø = 2x + y - z (again where x y and z are the unit vectors)

    but I am unsure what to do next.

    If I want the unit vector tangent to this surface then its gradient has to be the same as what I calculated above? And if its normal to b then n dotted with b has to = 0? Just don't know how to use this information or anything else to actually calculate the unit vector n.

    Thanks :)
     
  2. jcsd
  3. Feb 23, 2014 #2

    Dick

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    If you call the vector you are looking for ##a=a_x \hat x + a_y \hat y + a_z \hat z##, then ##a \cdot b=0## and ##a \cdot n=0##, write down those equations and see what they tell you about the components of a.
     
  4. Feb 23, 2014 #3

    D H

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    You could follow Dick's advice. Alternatively, you could use some other operation on a pair of vectors that yields a vector that is normal to both.
     
  5. Feb 23, 2014 #4
    Sorry, I'm still a bit confused. I want a vector that is normal to b but tangent to ∇ø, not normal to both?
     
  6. Feb 23, 2014 #5

    D H

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    You don't want a vector tangent to ∇ø. The gradient ∇ø points in the direction along which ø(x,y,z) changes the fastest. That is not along a tangent to the surface of constant ø. The gradient normal to this level surface.
     
  7. Feb 23, 2014 #6
    oh yeah of course! So i do the cross product of ∇ø and b? The vector I get as a result of that, will I then have to divide it by its magnitude to make it into the unit vector?
     
  8. Feb 23, 2014 #7

    D H

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    Yes. You'll get the same vector (possibly multiplied by -1) with Dick's method.
     
  9. Feb 23, 2014 #8
    thanks for your help :)
     
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