# Find a plane that passes through the given points and is tanget to the graph

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.

• 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. – colormegone Feb 23 '14 at 19:31
• @RecklessReckoner but how do I find a tangent plane that does contain this points? I mean, how do I verify/prove that? – Thums Feb 23 '14 at 20:26

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:

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.