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 $$\mathbb{R}^3$$?