# Geometry: Finding a surface that is orthogonal to a given family of surfaces

Problem Consider a one-parameter family of surfaces ${S_c}$ in $\mathbb{R^3}$ (with $c$ being a real constant) described by

$$S_c = \{(x, y, z)|f(x, y, z) = c\}, c ∈ \mathbb{R}$$

(a) Assume that we seek the surface $T$ described as the graph of a function of $x$ and $y$ (i.e., $z = \phi(x, y)$). Show that the graph $$\{(x, y, \phi(x, y)) | x, y ∈ \mathbb{R}\}$$ is orthogonal to each $S_c$ if and only if the function  satisfies the PDE

$$f_x(x, y, \phi (x, y))\phi_x + f_y(x, y, \phi (x, y))\phi_ y = f_z(x, y, \phi (x, y))$$ (b) Use the method of characteristics to find the surface that intersects each of the surfaces $$z(x + y) = c(3z + 1)$$ orthogonally, and which contains the circle $x^2 + y^2 = 1, z = 1$. Here, $c ∈ \mathbb{R}$ just like above.

My Attempt we know that two planes are orthogonal to each other if the dot product of their normal vectors is zero. I think this is how part (a) should be approached, also because the PDE contains the terms $f_x$,$f_y$, and $f_z$ such that $<f_x,f_y, f_z>$ is one of the normal vectors and $<\phi_x,\phi_y, 1>$ is the other, but I am not sure how to proceed. Help is extremely appreciated. Thanks!!

• I think the other one is $<\phi_x,-\phi_y, 1>$ instead of $<\phi_x,\phi_y, 1>$. – Hoseyn Heydari Dec 2 '13 at 7:48
• Thank you for the input @HoseynHeydari . Why do you think that's the case? – johnsteck Dec 2 '13 at 15:18
• Let's $\mathbf{x}(x,y)=(x,y,\phi(x,y))$. Then we have $N=\mathbf{x}_x \times \mathbf{x}_y=(1,0,\phi_x)\times (0,1,\phi_y)=(\phi_x,-\phi_y, 1)$ – Hoseyn Heydari Dec 2 '13 at 17:46

I can help just with question $(a)$ since I don't know about the "characteristics" method.

First of all notice that, in general, the normal vector at a point $P$ of a surface defined by the equation $g(x,y,z)=0$ is given by the gradient of $g$ evaluated at $P$: $$\nabla g(P) = [\;\partial_x g(P),\,\partial_y g(P),\, \partial_z g(P) \;]$$ (Let me know if the reason behind this statement is not clear for you, and I will elaborate). In our case the surfaces have equations:

$$f(x,y,z)-c=0 \quad\text{ and }\quad \phi(x,y)-z=0$$

$$V = [\; \partial_x\, f,\,\partial_y\, f ,\,\partial_z\, f \;] \quad\text{ and }\quad W = [\; \partial_x \phi,\,\partial_y \phi ,\, -1 \;]$$
Hence the $(a)$ follows, since $V(P)\cdot W(P) = 0 \iff$ your PDE holds, for every point $P$ of $T$.