# 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...?

Homework Helper
Gold Member
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 $$\mathbb{R}^3$$?