Lines and angles in 3D space

Question

If given the angles between 3 intersecting lines (ie $A, B, C$ pass through $O$, and $\angle AB \angle AC \angle BC$ are given) how can you find the angle between one line and the plane created by the other two lines

Answer 1

Using the convention that $A$ and $B$ are in the plane and $C$ is the line:

We will assume that $A$ is in the direction $(1,0,0)$.

given $\angle AB$ we can then find a direction for $B$: $(\cos(\angle AB), \sin(\angle AB), 0)$.

These are both unit vectors. We will assume $C$ is also a unit vector; this time it will involve all three dimensions. WE can then use the dot product and some linear algebra to find some values for this vector:

$A\cdot C = |A||C|\cos(\angle AC) = \cos(\angle AC)$

$B\cdot C = \cos(\angle BC)$

$A_xC_x + A_yC_y + A_zC_z = \cos(\angle AC)$

$B_xC_x + B_yC_y + B_zC_z = \cos(\angle BC)$

But $A_z$ and $B_z$ are both $0$ so those drop away, so we get:

$$\left[ \begin{array}{cc|c} 1&0&\cos(\angle AC)\\ \cos(\angle AB)&\sin(\angle AB)&\cos(\angle BC)\\ \end{array} \right] $$

Which we can then solve for $C_x$ and $C_y$, which we can square and add to get the square of the cosine of our result angle.

Answer 2

Let $\alpha=\angle{AB}$, $\beta=\angle{AC}$ and $\gamma=\angle{BC}$. The volume of the parallelepiped with these internal angles and unit edge lengths is $$V = \sqrt{1+2\cos\alpha\cos\beta\cos\gamma-\cos^2\alpha-\cos^2\beta-\cos^2\gamma}.$$ On the other hand, if $\theta$ is the angle between line $A$ and the plane defined by $B$ and $C$, we also have $$V=\sin\theta\sin\gamma$$ therefore $$\sin\theta = \csc\gamma\sqrt{1+2\cos\alpha\cos\beta\cos\gamma-\cos^2\alpha-\cos^2\beta-\cos^2\gamma}$$ and similarly for the other line/plane angles.