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Tangent Plane

  1. Sep 8, 2009 #1
    1. The problem statement, all variables and given/known data

    Find the tangent plane to the surface x2 + 2y2 + 3z2 = 21, which is parallel to the plane x+4y+6z=0

    2. Relevant equations

    Gradients.

    3. The attempt at a solution
    Here is my solution:
    So gradient of the surface is (2x, 4y, 6z). The normal of the given plane is (1,4,6). Equating both gives us x=0.5, y=1, z=1.
    So this is the point where the plane meets the surface.
    Finding the equation:
    1(x-0.5)+4(y-1)+6(z-1) = x - 0.5 + 4y - 4 + 6z - 6
    Equation for the plane is: x + 4y + 6z = 10.5

    The answer given is x + 4y + 6z = ±21. Where did I go wrong? What did I forget to multiply? And why the two solutions (positive and negative)?

    Thanks!
     
  2. jcsd
  3. Sep 8, 2009 #2

    Mark44

    Staff: Mentor

    The point (.5, 1, 1) is not a point on the quadric surface, so can't be the point where the plane intersects the surface.
    The reason for the two equations for planes is that there are two points on the quadric surface that satisfy the given conditions. The surface is an ellipsoid.
     
  4. Sep 8, 2009 #3
    So how should I do it? I tried finding the point - but no success/
     
  5. Sep 8, 2009 #4

    Hurkyl

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    In your calculation, you used the following step:
    If two vectors are parallel, then they are equal.​
    Do you believe this to be a theorem? If not, then what statement is a theorem? And how does your calculation change if you use that statement instead?
     
  6. Sep 8, 2009 #5
    If two vectors are parallel, then they have the same direction (unit vector). Or maybe, one is a linear combination of the other.
    I still can't see though how this relates to my problem.
     
  7. Sep 8, 2009 #6

    Mark44

    Staff: Mentor

    If two vectors are parallel, then one is a scalar multiple of the other.
     
  8. Sep 8, 2009 #7
    ok got it!
    so the gradient was (2x, 4y, 6z). The normal to the plane is (1,4,6).
    both lines are parallel, so:
    a(2x,4y,6z)=(1,4,6)
    from here we get:
    x=0.5a y=a z=a
    we need to find the a which satisfies the ellipsoid equation:
    x2 + 2y2 + 3z2 = 21
    (0.5a)2 + 2a2 +3a2 = 21
    21a2 = 84
    a2 = 4
    a = ±2
    so we got 2 sets of points where the required plane will intersect:
    one is (1, 2, 2) the other is -(1, 2, 2) - makes sense - two opposite sides of the ellipsoid.
    finding the equation of the plane is easy now:
    1(x-1)+4(x-2)+6(z-2) and 1(x+1)+4(x+2)+6(z+2)
    which gives the two required planes.

    thanks everyone!
     
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