Degrees of freedom of a plane if $R^3$ is confuse me. I think ax+by+cz+d=0 has a 4 Dof, but it has a 3 Dof in Multiple View Geometry of Mr. Zisserman . Is the reason of 3 Dof caused by homogeneous coordinates?
- why plane has a 3Dof
- Is the reason of 3Dof caused by homogenous coordinates?
The reason is because you can solve for one of the variables. In your case: $$ax+by+cz+d=0$$ $$\rightarrow cz=-ax-by-d$$ $$\rightarrow z=\frac{-a}{c}x+\frac{-b}{c}y+\frac{-d}{c}$$ From here, we can just chose to rename those coefficients. If we can chose any value for a, b, c, and d, then it's no different than choosing 3 numbers. $$z=Ax+By+C$$ where $A, B, C$ are any real number.