Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It only takes a minute to sign up.
Sign up to join this community
I want to calculate the angle between the plane with a normal $N = [N_x,N_y,N_z]$ and the vector $V = [V_x,V_y,V_z]$ and I used this formula for angle $$\alpha = \arccos \frac{V \circ N} {|V|\; |N|}$$ My question is does it affect the result the direction of the normal vector (If it is directed to the opposite direction as the incident line on the plane)
Thanks
Suppose the base point, intersection of line and plane, is the origin $(0,0,0)$. The cosine of the angle between the line and the normal to the plane is the dot product of normalized (unit) vectors $N$ and $V$. Then the angle between the line and the plane itself would be the complement of that first angle.
If the base point is not the origin, then we find the intersection between the line and plane and translate. However that seems to be accounted for in your notation already.
Geometrically a ray emerging from the plane makes the same angle with the plane if its direction is reversed. However the reversal of direction of the normal line without a corresponding reversal of the incident line (or vice versa) introduces a minus sign into the dot product. Typically a negative cosine gets mapped to an angle in the second quadrant, so the step above where the angle of the first quadrant is complemented to get the angle between incident line and plane has to be modified, if your purpose to get the same angle between line and plane whichever way the rays are directed.
The easiest way to achieve that is by throwing in an extra absolute value:
$$ \alpha = \cos^{-1} \frac{|V \circ N|}{|V| |N|} $$
and then the acute (or right) angle the line makes with the plane is $\frac{\pi}{2} - \alpha$.
