# Projection of a vector on a cone

In order to proceed with a problem I wish to find the projection of a vector on a conical surface.

For eg., projection of any vector $F = (f_1, f_2, f_3)$ on a cone with half angle say $\alpha$ with vertex at origin itself. I want this because I don't want my vector $F$ to be making an angle greater than $\alpha$ with $z$ axis. So, the least I can do is find its orthogonal projection on the cone and obtain a best possible solution. I am assuming $F$ makes angle greater than $\alpha$. $F$ is a physical force, actually.

I would like to obtain a neat expression for the projection. I believed it would be trivial, but have been unable to proceed.

I have done it this way:-

The desired unit vector of ray of cone would have: $$\lambda = \tan^{-1}{f_2/f_1}, \quad \mu = \pi/2 - \alpha$$ such that the vector becomes $$(\cos\mu\cos\lambda, \cos\mu\sin\lambda, \sin\mu)$$. Hence, the projection of $F$ becomes:

$$f_1 \sin\alpha \cos(\tan^{-1}{f_2/f_1}), f_2\sin\alpha\sin(\tan^{-1}{f_2/f_1}), f_3\cos\alpha$$ after simplification gives, $$\frac{f_1^2\sin\alpha}{\sqrt{(f_1^2 + f_2^2)}}, \frac{f_2^2\sin\alpha}{\sqrt{(f_1^2 + f_2^2)}}, f_3\cos\alpha$$

• Can you define what you mean by the projection of a vector on a surface? Jun 17, 2018 at 9:54
• Component of $F$ along the cone $x^2 + y^2 = \tan^2{\alpha}z^2$. No particular use of surface. Jun 17, 2018 at 9:58

• The projection of $F$ on the $0xy$ plane is $F^\prime= (f_1,f_2,0)$. From it you can get the angle $\theta$ in cylindrical coordinates of the vector. $F^\prime= (f \cos \theta,f \sin \theta,0)$ with $f >0$.
• Define the "projection" on the cone as $p(F) = (f \cos \theta,f \sin \theta,\frac{f}{\tan \alpha})$