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I'm trying to find the tangent plane to the surface defined by $z^2 =x^2 - y^2$ at the point $P(1, 1, 0)$. It seems trivial, but I hit a roadblock because I end up with a line, not a plane!

I define a function $F(x,y,z) = x^2 - y^2 - z^2$, and the surface is the level at $F=0$. Then, the gradient $\text{grad}(F) = \langle 2x, -2y, -2z\rangle$ which is the normal vector at a point $(x,y,z)$.

It follows that the normal vector at the point $P(1,1,0)$ is $n=\langle 2, -2, 0\rangle$.

The definition of the normal plane should be $2(x - 1) - 2(y - 1) + 0(z - 0) = 0$.

Then, after simplifying, I get $2x -2y =0$, which is not a plane equation! Where did I go wrong?

Thank you,

--Louis

Ben Whitney
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hlouis
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  • I don't know the answer because of bad format, but I know $2x -2y =0$ is a plane equation. – Shuchang Nov 21 '13 at 04:05
  • Yout answer is correct. It represents a plane equation in 3d. – Mhenni Benghorbal Nov 21 '13 at 04:19
  • The set in $\mathbb{R}^3$ defined by $2x-2y$ is indeed a plane: it is precisely ${(x,x,z)\mid x,z\in \mathbb{R}}$ – userCaltech Nov 21 '13 at 04:34
  • oh, I thought a plane equation has to absolutely and explicitly contain all three dimensions... Now that I think about it, on the graph, it is very clear that y=x is a line that is tangent to the surface, and the plane is that line "copied" besides itself for all values of z.

    Thanks a lot to all that answered and to Mr. Singh who edited my question to add proper formatting!

    --Louis

     – hlouis Nov 21 '13 at 04:48

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