# Proving that for any point on an intersection of two surfaces, their respective tangent planes at the point are orthogonal.

Consider the surfaces $$x^2+y^2-2z^2=2$$ and $$(x^2+y^2)z=4$$

Prove that, for any point $p_0$, on the intersection of these two surfaces, the two tangent planes at $p_0$ are orthogonal.

Would it be satisfactory to find where the surfaces intersect and then find their respective gradient vectors? I know the tangent planes will be orthogonal if the normal line from one plane is on the tangent plane of the other which occurs if: $$\nabla f(x,y,1)\cdot \nabla g(x,y,1)=0.$$

Any help would be appreciated.

• Yes, that'll be enough. Commented Mar 31, 2018 at 16:39

first the intersection points are: $$r^{2}=2+2z^{2}$$ $$r^{2}=4/z$$ $$2+2z^{2}=4/z$$ $$z=1$$ for $z=1$ we have: $$r=2$$ so the intersection point is: $$x=2cos\theta ,y=2sin\theta, z=1$$

The normal of the tangent plane of the first surface: $$n=(2x,2y,-4z)$$

the normal of the tangent plane of the second one is:

$$n=(2xz,2yz,x^{2}+y^{2})$$

the inner product of these normal vectors are: $$<n_{1}.n_{2}>=(4x^{2}z+4y^{2}z-4z(x^{2}+y^{2})$$

so in z=1 we have:$$<n_{1}.n_{2}>=0$$ Notice that the inner product is always zero. I just wanted to show that these surfaces have intersection.

• You get zero everywhere, not only at $\;z=1\;$ ... Commented Mar 31, 2018 at 16:40
• @DonAntonio it was more exact to write about intersection point Commented Mar 31, 2018 at 16:47
• Perhaps, yet getting that the inner product is always zero makes things way easier, as we don't need to worry about any intersection point... Commented Mar 31, 2018 at 16:51
• @DonAntonio yes but I liked to write intersection points. it shows that they have intersection. but you are right too. I will edit my post Commented Mar 31, 2018 at 16:56
• Good point indeed: to show there's at least one intersection point. +1 Commented Mar 31, 2018 at 17:09

The gradients to the surfaces are

\begin{align*}&\nabla_1=\left(\,2x,\,2y,\,-4z\,\right)\\{}\\ &\nabla_2=\left(\,2xz,\,2yz,\,x^2+y^2\,\right)\end{align*}

It is trivial now to check $\;\nabla_1\cdot\nabla_2=0\;$ everywhere, and thus also in any common point to the surfaces