[Math] Check angle between two lines greater than 90
geometry
I want to determine whether the measure of angle Q is greater than 90 degrees.
Best Answer
Calculate as follows (in C# or any similar language):
ux = x2 - x1 ; uy = y2 - y1; // u = (ux,uy) = vector from p1 to p2
vx = x3 - x1 ; vy = y3 - y1; // v = (vx,vy) = vector from p1 to p3
d = ux*vx + uy*vy; // Dot product of u and v
So, the number $d$ is the dot product of the vectors $ \mathbf{u} =\mathbf{p}_2 - \mathbf{p}_1$ and $\mathbf{v} = \mathbf{p}_3 - \mathbf{p}_1$. If $ d<0$, then the angle between the vectors $\mathbf{u}$ and $\mathbf{v}$ is greater than 90 degrees. If you actually want to find the angle, it's a bit more work (but that's not what you asked).
The angle of line $2$ from the horizontal is $\text{Atan2}(y_2-C_y,x_2-C_x)$, a function supplied in most programming languages. Similarly for line $1$, the angle from the horizontal is $\text{Atan2}(y_1-C_y,x_1-C_x)$. The angle between is just the difference, but you have to worry adding or subtracting $2\pi$ as required. I am accustomed to Atan2 returning radians, but you could check your documentation.
Take a point $A$ on the first line and a point $B$ on the second line. For instance:
$$
A=(0,0.384,1);\quad B=(1,-0.3864,0).
$$
To find the angle $\theta=\angle AOB$ between lines, you can choose among several different ways. If you are comfortable with the cosine rule, you can for instance apply it to triangle $ABO$ (where $O=(0,0,0)$):
$$
AB^2=OA^2+OB^2-2OA\cdot OB\cos\theta
$$
and solve for $\cos\theta$.
Or you may exploit the definition of dot product:
$$
\vec{OA}\cdot\vec{OB}=OA\cdot OB\cos\theta.
$$
Best Answer
Calculate as follows (in C# or any similar language):
So, the number $d$ is the dot product of the vectors $ \mathbf{u} =\mathbf{p}_2 - \mathbf{p}_1$ and $\mathbf{v} = \mathbf{p}_3 - \mathbf{p}_1$. If $ d<0$, then the angle between the vectors $\mathbf{u}$ and $\mathbf{v}$ is greater than 90 degrees. If you actually want to find the angle, it's a bit more work (but that's not what you asked).