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I have no idea about how to solve this. I know how to find a tanget plane to a surface, but I'm not sure if I understand what "passes through the points" mean.

What is asked: Find a plane that passes though the points (1, 1, 2) and (-1, 1, 1) and that is tangent to the graph of $$f(x,y) = xy$$

I'm afraid I do not have much to share about what I've tried so far since I really couldn't see a way through it, even though it doesn't look that hard. Any help will be appreciated.

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  • $\begingroup$ The tangent plane you seek must contain those two points. (Clearly, neither is on the surface $ \ z \ = xy \ . $ ) The third point determining the plane is the (unknown) tangent point on that surface. $\endgroup$ – colormegone Feb 23 '14 at 19:31
  • $\begingroup$ @RecklessReckoner but how do I find a tangent plane that does contain this points? I mean, how do I verify/prove that? $\endgroup$ – Thums Feb 23 '14 at 20:26
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The normal vector to a point $ \ (X, Y , XY) \ $ on this surface, $ \ xy - z = 0 \ , $ is given by $ \ \nabla f \ = \ \langle y, x , -1 \rangle \vert_{(X,Y, XY)} \ = \langle Y, X , -1\rangle \ . \ $ This normal vector must also be perpendicular to the vectors from $ \ (X, Y , XY) \ $ to $ \ (1,1, 2) \ $ and to $ \ (-1, 1 , 1) \ . \ $ We can construct an equation for the tangent plane after solving for $ \ X \ $ and $ \ Y \ . $

I won't give the result, but here's a picture of the situation:

enter image description here

EDIT (3/15) -- Since this recently got its first vote and has had some time to "cool off", I'll post a result (which I had to reconstruct, since I long ago tossed my notes).

A cross-product calculation with the two vectors in the tangent plane tells us that $ \ \langle Y - 1 \ , \ 2XY - X - 3 \ , \ 2 - 2Y \rangle \ = \ k \ \langle Y, X , -1 \rangle \ . $ We can use the $ \ x-$ and $ \ z-$ components to resolve that $ \ Y = \frac{1}{2} \ $ and $ \ k = -1 \ . $ Comparison of the $ \ y-$ components then yields $ \ X = 3 \ . $

So the tangent point to the surface is $ \ (3 \ , \ \frac{1}{2} \ , \ \frac{3}{2} ) \ $ and an equation for the tangent plane is

$$ \frac{1}{2} x \ + \ 3y \ - \ z \ = \ \frac{3}{2} \ \ . $$

This checks against the coordinates of the two given points and the tangent point we've found.

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