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

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.

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, im working on stokes theorem.

Last edited:
LCKurtz
Homework Helper
Gold Member
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.

LCKurtz