# Solving for Line Intersecting Two Cones

I am looking for a closed form and recursive form (an approximate form would be okay) solution to evaluate a fixed vector when I know the angle between it and two other known vectors. Any suggestions on how to go about solving this would be helpful.

Given vectors $\vec{M}$, $\vec{V}_{1}$, and $\vec{V}_{2}$ (all in $\mathbb{R ^{3} }$), I know the angle between $M$ and $\vec{V}_{1}$ is $\theta_{1}$ and the angle between $\vec{M}$ and $\vec{V}_{2}$ is $\theta_{2}$.

I can take the dot product between the vectors to get

$$\vec{M}\cdot\vec{V}_{1}=|\vec{M}||\vec{V}_{1}|\cos(\theta_{1})$$

$$\vec{M}\cdot\vec{V}_{2}=|\vec{M}||\vec{V}_{2}|\cos(\theta_{2})$$

I'll simplify the right side and assume the magnitude of all of these vectors is 1.

Using subscripts to break each vector down into its x, y, and z components, I have these simultaneous equations:

$$M_{x}V_{1x}+M_{y}V_{1y}+M_{z}V_{1z}=\cos(\theta_{1})$$ $$M_{x}V_{2x}+M_{y}V_{2y}+M_{z}V_{2z}=\cos(\theta_{2})$$

The only unknown here is vector $\vec{M}$, which is assumed to have unit length.

For what it's worth, each of these describes a cone with a half angle of $\theta$. I'm looking for the vector(s) $\vec{M}$ where the cones intersect.

Is this how I should start? Where do I go from here?

I want to be able to solve this recursively in a Kalman filter (or extended Kalman filter). I'll have a continuous set of vectors $\vec{V}$, and I'd like to update my estimate of $\vec{M}$ recursively. I'm interested in any suggestions on how to approach that.

• I am going to take the word "tangent" out. After looking at it, I see there is an ambiguity where there can be two lines of intersection and two corresponding solutions pointing in the opposite direction (anti-parallel). So I see there can be 1, 2 or 4 solutions. I'll have to determine some way to resolve that ambiguity. – Jim Sep 23 '15 at 20:35

Suppose $V_1$ and $V_2$ point towards the north and south poles, and $\theta_1 = \theta_2 = \frac{\pi}{2}$. Then every unit vector in the equatorial plane is a solution to your problem. You have infinitely many solutions. And "closed form equations" don't tend to produce that kind of thing.
If you're willing to say "the $V$ vectors are not collinear," then perhaps we can take their average, start working with a plane orthogonal to that average, etc., and eventually get to a quadratic. But as it stands: it seems hopeless.