# Tangent Vector to a Space Circle

• January 14th 2011, 12:49 PM
mathematicalbagpiper
Tangent Vector to a Space Circle
Find a vector tangent to the space circle:

$x^2+y^2+z^2=1$
$x+y+z=0$

At the point $(1/\sqrt{14}, 2/\sqrt{14}, -3/\sqrt{14})$

I should know how to do this, but it's been 4 years since I had multivariable calc and I don't remember a darn thing from that class. :-(
• January 14th 2011, 01:39 PM
emakarov
Since the circle belongs to the plane x + y + z = 0, the tangent vector belongs to it as well. Also, the vector is perpendicular to the radius-vector $(1/\sqrt{14}, 2/\sqrt{14}, -3/\sqrt{14})$. So you have two equations: x + y + z = 0 and x + 2y - 3z = 0. I get (-5, 4, 1) up to proportionality.
• January 14th 2011, 01:49 PM
zzzoak
$
r_0=(1/\sqrt{14}, 2/\sqrt{14}, -3/\sqrt{14})
$

The normal vector to the tangent plane to the sphere is $r_0$.

The normal vector of the given plane is p=(1,1,1).

The vector perpendicular to $r_0$ and $p$ is (cross product)

$
n=p\; \times \; r_0.
$