2
$\begingroup$

I define a circle $T$ in $\mathbb{R}^3$ with equations $y^2+z^2=3/4$ and $x=-1/2$. This circle lies on the unit sphere $S$. I define the set $B$ to be all points on $T$ with y-values between $\epsilon$ and $-\epsilon$ and positive z-values (that is, a small arc at the top of $T$). For every two points in $B$, consider their corresponding vectors, and find the unit vector orthogonal to both. (Assume we choose the orthogonal vector with positive x-value). Let's call the set of all of these orthogonal vectors $C$. (I will also use $C$ to describe the set of points in $\mathbb{R}^3$ corresponding to these vectors).

My question is, what does $C$ look like (as a set of points)? Is it also an arc of a circle? Perhaps a circle parallel to the xy-plane? For my research problem I'm trying to show that we can choose $\epsilon$ small enough that the points of $C$ will be very close together (this seems obvious to me but I'm having trouble showing it). Instead of defining $T$ as is, is it better to define a larger circle closer to the Prime Meridian (so to speak) of $S$? Thank you.

$\endgroup$
2
  • $\begingroup$ What do you mean by "corresponding vector"? Do you mean the radial vector from the origin to the point? $\endgroup$ Feb 18, 2017 at 23:56
  • $\begingroup$ Yes. So the point $(-1/2, 1/4, \sqrt{11}/4)$ would have corresponding vector $<-1/2, 1/4, \sqrt{11}/4>$ $\endgroup$ Feb 19, 2017 at 0:02

1 Answer 1

0
$\begingroup$

Not a precisely quantitative answer. Just attempting to track the specified geometry. I used $\epsilon=0.1$ radians to delimit the subset $B \subset T$.

My question is, what does $C$ look like (as a set of points)?

It "looks like" as indicated below:


  SphereArcQ
Pardon that I did not

(Assume we choose the orthogonal vector with positive x-value)

but instead showed both $\pm$.

$C$ appears to be bounded by a circular arc connected to a V whose angle is determined by $\epsilon$, in particular, the straight-line boundaries of $C$ are orthogonal to the $\epsilon$ extremes of $B$, as is the circular arc boundary, more clearly seen from $(+\infty,0,0)$:


Smiley
In any case, have a nice day! :-)

$\endgroup$
2
  • $\begingroup$ This is very helpful, thank you! So it seems that by choosing a smaller arc of the circle $T$, I can make that red orthogonal region as small as I need. $\endgroup$ Feb 19, 2017 at 21:50
  • $\begingroup$ @BradElliott: Yes. $\endgroup$ Feb 19, 2017 at 22:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.