I need to prove this:
If all the normal lines to a regular surface pass through a fixed point, then the surface is a portion of the sphere.
I haven't really tried much since I don't know what to do.
Thanks
[Math] Normal lines of a regular surface
surfaces
Related Solutions
Sounds like you have a good strategy. You are looking at the cross section in some plane $z=z_0$, and you want to prove that the cross section here is a circle.
You are given that any normal line will intersect the $z$ axis. Let's write this line as $\vec{p}+\vec{n}t$ where $\vec{n}$ is the normal vector at the point $\vec{p}$ we are discussing. You can then show that a new line $\vec{p} + (\vec{n}-\hbox{proj}_{\hat{z}}\vec{n})t$ will also intersect the $z$ axis (taking away the z component doesn't change the fact that the line is in a radial direction).
Moreover, this new line will be normal to the cross section at $\vec{p}$: the tangent plane projects to a tangent line of the cross section, so must a normal vector of the surface project to a normal vector of the cross section.
This gives you the criteria you wanted -- all horizontal lines orthogonal to cross sections will intersect at the origin. This tells you roughly that the cross sections are circular, and that the overall shape will be a surface of revolution.
Very nice questions indeed!
Change of notation: the surface will be denoted by $\Sigma$ instead of $S$, and the plane by $\Pi$ instead $P$; this done since I had already written up a large portion of my answer employing this altered symbolism when I re-read the problem statement and re-discovered the OP's original use of $S$, $P$, etc. Apologies for any confusion caused .
Let's handle the questions in the order asked:
i.) Suppose $\Sigma$ is a connected surface in $\Bbb R^3$; then $\Sigma$, being a $2$-manifold, is path connected. (See this web page and the citings therein.) Let $p, q \in \Sigma$ be any two points, and let $\gamma:[0, 1] \to \Sigma$ be a smooth path joining $p$ and $q$ in the sense that $\gamma(0) = p$, $\gamma(1) = q$. Then by the given hypothesis, the normal line at every point of $\gamma(t)$ contains the same point $Q$; thus the vector $\gamma(t) - Q$ is normal to $\Sigma$ at $\gamma(t)$ for all $t \in [0, 1]$. Since $\gamma(t)$ is a curve in $\Sigma$, its tangent vectors $\dot \gamma(t)$ are each orthogonal to $\gamma(t) - Q$; i.e.
$\dot \gamma(t) \cdot (\gamma(t) - Q) = 0 \tag{1}$
for all $t \in [0, 1]$. $Q$ being fixed, we have
$\dot Q = \dfrac{dQ}{dt} = 0, \tag{2}$
whence
$\dot \gamma(t) = (\gamma(t) - Q)'; \tag{3}$
here I switch freely 'twixt the "dot" (as in $\dot \gamma$) and "prime" (as in $\gamma'$) notation for derivatives, depending on which is easiest to write in a given expression. In any event, if (3) is substituted into (1), we have
$(\gamma(t) - Q)' \cdot (\gamma(t) - Q) = 0 \tag{4}$
for all $t$, which in turn implies
$(\Vert \gamma(t) - Q \Vert^2)' = ((\gamma(t) - Q ) \cdot (\gamma(t) - Q))' = 2(\gamma(t) - Q)' \cdot (\gamma(t) - Q) = 0; \tag{5}$
i.e., $\Vert \gamma(t) - Q \Vert$ must be a constant independent of $t$. Thus in fact we have
$\Vert p - Q \Vert = \Vert \gamma(0) - Q \Vert = \Vert \gamma(1) - Q \Vert = \Vert q - Q \Vert = \Vert \gamma(t) - Q \Vert \tag{6}$
for all $t \in [0, 1]$. Since $p, q \in \Sigma$ are arbitrary, we see that every point of $\Sigma$ lies at a distance $\Vert p - Q \Vert$ from $Q$; $\Sigma$ thus lies in the sphere if radius $\Vert p - Q \Vert$ centered at $Q$. QED.
ii.) We will prove this one by contraposition; that is, we will show that, if $\Pi$ is any plane intersecting $\Sigma$ at $p$ non-tangentially, then in any neighborhood $U \subset \Sigma$ of $p$ there is a regular curve $\gamma:(-\epsilon, \epsilon) \to \Sigma \cap \Pi$, for some $0 \ne \epsilon \in \Bbb R$ with $\gamma(0) = p$; the existence such $\gamma(t)$ then implies not merely that there are more than one points of intersection of $\Sigma$ and $\Pi$, but that there are in fact an uncoutable infinity of them, all within $U$. The proof proceeds as follows: let $p \in U \subset \Sigma$, with $U$ a sufficiently small open set that a smooth unit normal field $N_\Sigma$ to $\Sigma$ is defined everywhere on $U$. Clearly, if $\Sigma$ is orientable, we can take $U$ to be all of $\Sigma$, but in any event, $U$ may always be taken to be diffeomophic to an open disk in $\Bbb R^2$. Next, we observe that $\Pi$, being a plane, is itself possessed of a constant unit normal field $N_\Pi$, and that indeed the equation of the plane $\Pi$ may be written
$(r - \pi) \cdot N_\Pi = 0, \tag{7}$
where $\pi \in \Pi$ is any fixed point, and $r =(x, y, z)$ represents the coordinates of a variable point in $\Pi$. As is well-known, (7) presents the equation of a plane passing through point $\pi$ with normal vector $N_\Pi$; taking $\pi$ to be $p$, we obtain the equation of the plane with normal $N_\Pi$ passing through the point $p \in \Sigma$:
$(r - p) \cdot N_\Pi = 0. \tag{8}$
If $\Pi$ is not tangent to $\Sigma$ at $p$, then $N_\Pi$ is not collinear with $N_\Sigma(p)$, which implies that
$N_\Pi \times N_\Sigma(p) \ne 0. \tag{9}$
Now consider the smooth vector field $X$ on $U$ defined by
$X(q) = N_\Pi \times N_\Sigma (q) \tag{10}$
for $q \in U$. In defining $X(q)$ via (10), we simply extend $N_\Pi$ to all of $\Bbb R^3$; in so doing, it is defined at every point of $\Sigma$ to be the same, constant vector. Inspecting the definition (10), we see that
$X(q) \cdot N_\Sigma (q) = (N_\Pi \times N_\Sigma(q)) \cdot N_\Sigma(q) = 0 \tag{11}$
and
$X(q) \cdot N_\Pi = (N_\Pi \times N_\Sigma (q)) \cdot N_\Pi = 0, \tag{12}$
since the vector cross product is always orthogonal to each of its arguments. (11) shows that $X(q)$ is in fact tangent to $\Sigma$ at all points $q$ at which it is defined; furthermore, since $X(q)$ is smooth and nonzero at $q = p$, we can if necessary, by taking $U$ to be an even smaller neighborhood of $p$, insure that $X(q)$ is nonzero for all $q \in U$. This being the case, and since $X(q)$ being smooth implies it is (locally, at least) Lipschitz continuous, for any $q \in U$ there is an $\epsilon > 0$, an open interval $I _\epsilon = (-\epsilon, \epsilon) \subset \Bbb R$, and a smooth curve $\gamma(q, t): I_\epsilon \to U$ with $\gamma(q, 0) = q$ and
$\dot \gamma(q, t) = X(\gamma(q, t)); \tag{13}$
that is, $\gamma(q, t)$ is the unique solution to the differential equation (13) with $\gamma(q, 0) = q$. Taking $q = p$, we have $\gamma(p, t) \subset U$ and $\gamma(p, 0) = p$. By (12) and (13), we see that
$\dot \gamma(p, t) \cdot N_\Pi = X(\gamma(p, t)) \cdot N_\Pi = 0, \tag{14}$
and this implies that
$((\gamma(p, t) - p) \cdot N_\Pi)' = \dot \gamma(p, t) \cdot N_\Pi =0, \tag{15}$
since $p$ and $N_\Pi$ are constants. (15) shows that $(\gamma(p, t) - p) \cdot N_\Pi$ is constant, and since
$(\gamma(p, 0) - p) \cdot N_\Pi = (p - p) \cdot N_\Pi = 0, \tag{16}$
we have that
$(\gamma(p, t) - p) \cdot N_\Pi = 0 \tag{17}$
for all $t$ in $(-\epsilon, \epsilon)$ for some $\epsilon > 0$. (17) implies the curve $\gamma(p, t)$ lies in the plane $\Pi$; thus it lies in the intersection $\Sigma \cap \Pi$ of the surface $\Sigma$ and the plane $\Pi$; since as we have seen $\dot \gamma(p, t) = X(\gamma(p, t)) \ne 0$ in $U$, $\gamma(p, t)$ is regular, the image of $\gamma(p, t)$ in $\Sigma \cap \Pi$ contains an uncountable infinity of distinct points.
We have just shown that if $\Pi$ is not tangent to $\Sigma$ at $p$, then $\Pi \cap \Sigma$ contains the regular curve $\gamma(p, t)$, hence is uncountable. So if we assume that $\Pi \cap \Sigma = \{ p \}$ a singleton, then we must have $\Pi$ tangent to $\Sigma$ at $p$. QED.
Note: It may also be possible to prove this, that is, the existence of $\gamma(p, t)$ (or perhaps even $\gamma(q, t)$, $q \ne p$), using the implicit function theorem, but I haven't got it completely figured out as of this writing. End of Note.
(iii.) To answer this question, one needs to adopt some definition or other of just what it means for a point to be on one side or the other of a plane. For the present purposes I will deploy a definition based upon equation (7); consider the function
$h(r) = (r - \pi) \cdot N_\Pi, \tag{18}$
where $\pi \in \Pi$ is a specified (non-variable) point and $r \in \Bbb R^3$. It is easy to see that
$\nabla h = N_\Pi, \tag{19}$
everywhere the same. We have also seen that $h(r) = 0$ for $r \in \Pi$, ; this may be interpreted as saying that the line segment or "vector" from $\pi$ to $r$ is in fact perpendicular to $N_\Pi$. If, on the other hand, $h(r) > 0$, then $r - \pi$ has a positive component in the $N_\Pi$ direction, similarly if $h(r) < 0$, then such a component is negative. Since
$h(r)= (r - \pi) \cdot N_\Pi = r \cdot N_\Pi - \pi \cdot N_\Pi, \tag{20}$
and $r \cdot N_\Pi$, $\pi \cdot N_\Pi$ are the respective distances of $r$, $\pi$ from the origin in the direction $N_\Pi$ , we see that $h(r)$ is in fact the excess distance, along $N_\Pi$, of $r$ over $\pi$ from the point $(0, 0, 0)$. I don't want to go into much more detail about this here an now, since these remarks I have made concerning $h(r)$ are really part of elementary $3$-space geometry, almost at the high-school level; I re-iterate them here in order to convey a mathematically formal yet clear understanding of what is meant by being on one side of the plane $\Pi$ or the other. These things being (hopefully adequately) explained, we may proceed. We will say a point $r$ is on the positive side of $\Pi$ provided that $h(r) \ge 0$, and on the negative side if $h(r) \le 0$. We allow $h(r) = 0$ in this terminology to cover the possibility that $\Sigma \cap \Pi \ne \emptyset$, as in the statement of the problem. These matters being settled, suppose then that $\Sigma$ lies on the positive side of $\Pi$. Then $h(\sigma) \ge 0$ for any $\sigma \in \Sigma$. If $\gamma(u)$ is a smooth path in $\Sigma$, then
$(h(\gamma(u)))'_{u = u_0} = (\gamma'(u_0)[h(\gamma(u))])_{u = u_0} = \langle \gamma'(u_0), \nabla H(u_0) \rangle = \langle \gamma'(u_0), N_\Pi \rangle. \tag{21}$
Now if $\sigma \in \Sigma \cap \Pi$, then $h(\sigma) = 0$ so if $\gamma(u_0) = \sigma$, then $h(\gamma(u)) \ge h(\gamma(u_0))$ for all $u$, we must, by the standard result from calculus on derivatives of single variable functions at local extrema, have
$0 = (h(\gamma(u)))'_{u = u_0} = \langle \gamma'(u_0), N_\Pi \rangle. \tag{22}$
But any tangent vector to $\Sigma$ at $\sigma$ is given by $\gamma'(u_0)$ for some path $\gamma(u)$ with $\gamma(u_0) = \sigma$; this shows the entire tangent plane to $\Sigma$ at $\sigma$ is normal to $N_\Pi$; hence it must in fact be $\Pi$. QED.
Note: The above results all generalize nicely to the case of $\Sigma$ an $n$ dimensional hypersurface in $\Bbb R^{n + 1}$; the proofs of (i) and (iii) need very little change to do so; as for (ii), if $n \ge 3$ we of course cannot use (10), since the cross product $\times$ is not defined in this case. But we can in fact find a vector field $X$ tangent to $\Sigma$ in $U$, with $X \cdot N_\Sigma = X \cdot N_\Pi = 0$, simply by taking $X$ to be a non-vanishing section the sub-bundle of $T\Bbb R^{n + 1}$ on $U$ which is normal to both $N_\Sigma$ and $N_\Pi$; then $X$ is a section of $TU$ and the above above argument extends virtually unaltered. By taking $U$ sufficiently small, it is straighforward to present such an $X$ in terms of local coordinates in $U$. End of Note.
Hope this helps! Holiday Cheers,
and as ever,
Fiat Lux!!!
Best Answer
This is only true if the surfaced (let's call it $M$) is connected, I don't know whether your definition of 'regular' would imply this (otherwise the claim is true for each component).
Denote by $p_0$ the joint intersection point of the nomals to $M$. Consider a smooth curve $c(t)$ in $M$ and let $f(t) = ||c(t)- p_0||^2$ Then $$f^\prime(t) = 2 \langle c^{\prime}(t), c(t)-p_0\rangle = 0$$ since $c^\prime$ is tangent to $M$ and, by assumption, $c(t)-p_0$ normal to $M$. This shows that the distance from $M$ to $p_0$ is constant.