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Dot product: normal and tangent

  1. Mar 17, 2009 #1
    1. The problem statement, all variables and given/known data
    Tangent plane goes through point P=(a,b,f(a,b)). Any point on the plane is then
    Q=(x,y,z)=(x,y,f(a,b)+fx(a,b)(x-a)+fy(a,b)(y-b)) (fx and fy are partial derivatives)
    and the vector [tex]\overline{PQ}[/tex] is on tangent plane.
    Calculate dot product n.[tex]\overline{PQ}[/tex] and show that the normal vector is perpendicular to the tangent plane.

    2. Relevant equations

    n=fx(a,b)[tex]\hat{i}[/tex] + fy(a,b)[tex]\hat{j}[/tex] - [tex]\hat{k}[/tex]

    3. The attempt at a solution
    Ok, I should get the dot product = 0.

    I don't know how to do that, because I don't know how to get the vector [tex]\overline{PQ}[/tex] into a right form. So far I've tried

    [tex]\overline{PQ}[/tex] = [tex]\overline{Q}[/tex] - [tex]\overline{P}[/tex]
    = xi + yj +f(x,y)k - ai - bj - f(a,b)k
    = (x-a)i + (y-b)j + (f(x,y)-f(a,b))k

    How am I supposed to get zero dot product from THAT and n? :yuck:
    n has partial derivatives, can I write them some other way? I'm stuck.
     
  2. jcsd
  3. Mar 17, 2009 #2

    Mark44

    Staff: Mentor

    So start with actually doing the dot product of PQ and n. What do you get?
     
  4. Mar 17, 2009 #3
    I get

    n dot PQ = fx(a,b)(x-a)+fy(a,b)(y-b)-f(x,y)+f(a,b)

    I still have those partial derivatives, and I don't know what to do with them.
     
  5. Mar 17, 2009 #4

    Mark44

    Staff: Mentor

    That's not what I get. You have an error in what you got for PQ.
     
  6. Mar 17, 2009 #5
    To get PQ I think I should start with Q-P. But it is scalar and I don't see how it turns into a vector.

    Q-P= x,y,f(a,b) + fx(a,b)(x-a) + fy(a,b)(y-b) - a,b,f(a,b)
     
  7. Mar 17, 2009 #6

    Mark44

    Staff: Mentor

    Q is a vector and P is a vector, and Q - P is a vector, and is equal to PQ. When you add or subtract vectors, you do so component-wise.
     
  8. Mar 17, 2009 #7
    Q and P are given as scalars. P=(a,b,f(a,b)) is, I think, relatively easy to vectorize
    ai + bj + f(a,b)k
    Q is a bit ugly
    xi + yj + f(a,b)k + fx(a,b)xi - fx(a,b)ai + fy(a,b)yj - fy(a,b)bj
    and so Q - P is
    xi + yj + fx(a,b)xi - fx(a,b)ai + fy(a,b)yj - fy(a,b)bj - ai - bj
    and as k is gone, can be put as i and j
    (x + fx(a,b)x - fx(a,b)a -a)i + (y + fy(a,b)y - fy(a,b)b -b)j

    Taking dot product with n :

    (x + fx(a,b)x - fx(a,b)a -a)*(fx(a,b)) + (y + fy(a,b)y - fy(a,b)b -b)*(fy(a,b)) = ..

    The product doesn't cancel out, it is not zero.
     
  9. Mar 17, 2009 #8

    Tom Mattson

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    I don't know how you're getting that [itex]\vec{PQ}[/itex], but the dot product with the normal vector is obviously zero.

    ???

    How do you get terms such as [itex]f_x(a,b)x\hat{i}[/itex]?

    [itex]f(a,b)+f_x(a,b)(x-a)+f_y(a,b)(y-b)[/itex] is the z coordinate of Q.
     
  10. Mar 17, 2009 #9

    Mark44

    Staff: Mentor

    The expression above is incorrect. Your x and y components are correct, but all the rest is the multiplier for the unit vector k.
     
  11. Mar 17, 2009 #10
    Thank you for correcting my somewhat unusual (and wrong!) approach..

    I didn't see the z-part of Q as it is.

    Dot product is (of course):
    [tex]\bar{PQ}[/tex].n = fx(a,b)(x-a) + fy(a,b)(y-b) - fx(a,b)(x-a) - fy(a,b)(y-b) = 0

    Once again, thank you for your help.
     
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