Elegant solution to this vector equation?

[monea83]

Given are a plane E and a line l in general position. I need to find a plane that contains l and intersects E at a given angle \alpha. All of this happens in R^3.

The interesting part is to find the normal of the unknown plane, let us call this normal x. I came up with the following equations:

x^Tx = 1

x^Tn = \cos \alpha

x^Td = 0

in which n is the unit normal vector of the given plane, and d is the direction vector of l.

(1) says I want a unit vector, (2) says the planes need to intersect at a given angle, and (3) says the plane needs to be parallel to the given line.

These equations can easily be solved by writing them out in component form with x=(x1, x2, x3) and doing some substitutions, which will yield a quadratic equation in one of the parameters.

However, I am wondering if there is a more elegant way to express the solution - something more matrixy-vectory? The equations look simple enough...?

 

[fzero]

If the line is normal to the plane E, then the plane we're looking for is not uniquely defined. So we may assume that \hat{n}\cdot \hat{d} \neq 1. In this case, \hat{n}, \hat{d} , \hat{n}\cdot \hat{d} span \mathbb{R}^3. We can solve your conditions to find the normals

\hat{x} = \frac{\cos\alpha}{1-(\hat{n}\cdot \hat{d})^2} [ \hat{n} -(\hat{n}\cdot \hat{d}) \hat{d} ] \pm \frac{\sin\alpha}{\sqrt{1-(\hat{n}\cdot \hat{d})^2}}~ \hat{n}\times \hat{d}.

The sign depends on the orientation of positive \alpha.

 

[monea83]

This looks interesting... I tried to reverse engineer your solution, but didn't quite get there. How do you use the fact that the vectors you mention span \mathbb{R}^3?