Need the parametric equation of a circle perpendicular to a vector.

AI Thread Summary
To find a parametric equation of a circle in 3D space that is perpendicular to a vector <a,b,c>, the intersection of the plane ax+by+cz=0 and the sphere x^2+y^2+z^2=constant must be determined. Solving the plane equation for z allows substitution into the sphere's equation, resulting in an ellipse in the xy-plane. This ellipse can be expressed in standard form, enabling parameterization as x = p + a*cos(t) and y = q + b*sin(t). The z-coordinate can then be derived from the plane equation in terms of t. The discussion also highlights the complexity of the problem, particularly in relation to Stoke's Theorem and the potential for simpler solutions depending on the specific context.
okkvlt
Messages
53
Reaction score
0
i need a parametric equation of a circle in 3d space that is perpendicular to a vector <a,b,c>. (as t goes up the circle is traced counterclockwise, as viewed from the head of the vector.)
in the form x[t],y[t],z[t]
i know that x^2+y^2+z^2=constant
and that ax+by+cz=0

But i cannot figure out the parametric equation x[t],y[t],z[t] that describes a circle perpendicular to the vector.
 
Mathematics news on Phys.org
or, phrased in other words, this is the intersection of the plane ax+by+cz=0 and the sphere x^2+y^2+z^2=constant.


in case anybodys wondering, I am working on stokes theorem.
 
Last edited:
There may be a shorter way in some specific cases, but you might try this. Solve the plane for z and put that in the equation of the sphere. This will give you an xy equation which represents the projection of the intersection curve in the xy plane. This will be an ellipse. Complete the square on it and get it in the standard form:

\frac {(x-p)^2}{a^2} + \frac {(y-q)^2}{b^2} = 1

Then you can parameterize it as:

x = p + a\cos(t)\ y=q + b\sin(t)

and use these to get z on the plane in terms of t also.
 
I want to add, after thinking about my reply, that it isn't so simple. The equation in the xy plane will likely be both translated and rotated. And since the OP is working with Stoke's Theorem, my guess is that the circuit integral might be unnecessary and, depending on the specific problem, the surface integral that Stoke's theorem gives may be easy. Hard to say without seeing the specific problem.
 
i want to prove that the magnitude of curl is the line integral around a region perpendicular to the curl vector.
 
Last edited:
Thread 'Video on imaginary numbers and some queries'
Hi, I was watching the following video. I found some points confusing. Could you please help me to understand the gaps? Thanks, in advance! Question 1: Around 4:22, the video says the following. So for those mathematicians, negative numbers didn't exist. You could subtract, that is find the difference between two positive quantities, but you couldn't have a negative answer or negative coefficients. Mathematicians were so averse to negative numbers that there was no single quadratic...
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Thread 'Unit Circle Double Angle Derivations'
Here I made a terrible mistake of assuming this to be an equilateral triangle and set 2sinx=1 => x=pi/6. Although this did derive the double angle formulas it also led into a terrible mess trying to find all the combinations of sides. I must have been tired and just assumed 6x=180 and 2sinx=1. By that time, I was so mindset that I nearly scolded a person for even saying 90-x. I wonder if this is a case of biased observation that seeks to dis credit me like Jesus of Nazareth since in reality...

Similar threads

Replies
25
Views
3K
Replies
4
Views
2K
Replies
3
Views
3K
Replies
1
Views
2K
Replies
11
Views
3K
Replies
20
Views
2K
Back
Top