Step III, Q8:
Equation of normal to
π through
x is
x+λn, distance is
∣(x−r)⋅n∣.
Consider a single circle C with center
r lying in the plane
x⋅n=p, and suppose C lies on the surface of a sphere S with center
s, radius R.
Claim:
s=r+λn for some
λ. In other words,
s must lie on the normal to the plane through
r.
\I think it's arguable that this is "obvious" and doesn't need to be proved. I will prove it, but it's going to be a little tedious...
Put
q=s−r. Consider
e=q−(q⋅n)n. Note
Unparseable latex formula:\mathbf{e}\cdot\mathbf{n} = \mathbf{q}\cdot{\mathbf{n}-(\mathbf{q} \cdot \mathbf{n})\mathbf{n}\cdot\mathbf{n} = 0
. (
e is basically what we get if we "remove" the part of
q that's parallel to
n).
Suppose that
e is non-zero. We will (eventually!) show this gives us a contradiction. Let
μ=∣e∣−1, so that
∣μe∣ = 1. Then
r±μe are both points on the circle C, so if they lie on the sphere S, they must both be the same distance from
s.
Now
s−(r+μe)=q−μe.
So
∣s−(r+μe)∣2=(q−μe)⋅(q−μe)=∣q∣2−2μq⋅e+∣μe∣2Similarly,
∣s−(r−μe)∣2=(q+μe)⋅(q+μe)=∣q∣2+2μq⋅e+∣μe∣2So we must have
q.e=0⟹q⋅q−(q⋅n)2, which since
∣n∣=1⟹q=κn for some
κ (either argue geometrically, or use the equality case of Cauchy-Schwartz). But then
e=κn−(κn⋅n)n=0, giving our contradiction.
So
e=0, so
q=(q⋅n)n and so
q is a multiple of
n. Thus
s=r+λn as claimed.
Finally, let
x be any point on C.
x⋅n=r⋅n, and so(x−r)⋅n=0. Then
Unparseable latex formula:|\bf{s}-\bf{x}|^2 = |\bf{s}-\bf{r}-(\bf{x}-\bf{r})|^2 = |\lambda \bf{n} - (\bf{x}-\bf{r})|^2 \\[br]= \lambda^2 |\bf{n}|^2 -2 (\bf{x}-\bf{r})\cdot\bf{n}+|\bf{x}-\bf{r}|^2 = \lambda^2 + 1
(Since
n is a unit vector, and
x lies on a unit circle, center
r so |
x-r| = 1).
But
x lies on the sphere, so
∣s−x∣ is just the radius of the sphere, R. So
λ2+1=RSo for S to contain both circles
C1,C2, we must have
s=r1+λ1n1=r2+λ2n2 for some real
λ1,λ2 satisfying
λ12+1=λ22+1⟹λ1=±λ2.
This proves that for such S, we can find
λ with
r1+λn1=r2±λn2 as desired.
But then
(r1−r2)=−λn1±λ2n2. Since
n1⋅(n1×n2)=n2⋅(n1×n2)=0 we have
(r1−r2)⋅(n1×n2)=0 as desired. Geometrically, this means that the line between the two circle centers has no component perpendicular to both plane normals; equivalently that it lies in the plane through the origin generated by the two plane normals.
Finally, if
r1+λn1=r2+λn2. Dotting with
n1,n2 gives:
r1⋅n1+λ=r2⋅n1+λn2⋅n1⟹p1−r2⋅n1=−λ+λn2⋅n1r1⋅n2+λn2⋅n1=r2⋅n2+λ⟹p2−r1⋅n2=λ−λn2⋅n1If instead
r1+λn1=r2−λn2. Dotting with
n1,n2 gives:
r1⋅n1+λ=r2⋅n1−λn2⋅n1⟹p1−r2⋅n1=−λ−λn2⋅n1r1⋅n2+λn2⋅n1=r2⋅n2−λ⟹p2−r1⋅n2=λ+λn2⋅n1In either case we have
(p1−r2⋅n1)2=(p2−r1⋅n2)2 as desired.
Geometrically, this says that this distance of
r1 from the plane
π2 must equal the distance of
r2 from the plane
π1Comment: Doing questions directly into Latex rather than pen/paper is always tricky; I'd done some roughing out on paper, but there was more work still to be done than I thought, and I hadn't realised just how hard Latex is for questions like this involving vectors. It was
not a good plan! Just about every other character needs to be enclosed in \bf{}, and the source text ends up near illegible. I must admit to losing the will to live, let alone answer the question, towards the end.