MHB How do I Find the Vector Coordinates to Solve for Trihedral Angle?

Julian1
Messages
2
Reaction score
0
Hi everyone,

Here's the problem I have.

Given two unit vectors A, B and angle φ between them. Find the coordinates (in 3D) of a unit vector C so that the angles between C and A,B be α and β respectively.
α + β => φ and α + β + φ <= 360°

It looks trivial to me and yet here I am asking for help:)

I have two equations from the dot products:
A.C = cos(α)
B.C = cos(β)

and I need a third one to solve the problem?

Thanks!Julian.
 
Physics news on Phys.org
Re: find the vector coordinates

I have found the solution.

The third equation is the triple product.

A.BxC = V

where

V2 = 1 + 2*cos(α)*cos(β)*cos(φ) – cos2(α) – cos2(β) – cos2(φ)

Trihedral Angle | OPEN MIND
 
##\textbf{Exercise 10}:## I came across the following solution online: Questions: 1. When the author states in "that ring (not sure if he is referring to ##R## or ##R/\mathfrak{p}##, but I am guessing the later) ##x_n x_{n+1}=0## for all odd $n$ and ##x_{n+1}## is invertible, so that ##x_n=0##" 2. How does ##x_nx_{n+1}=0## implies that ##x_{n+1}## is invertible and ##x_n=0##. I mean if the quotient ring ##R/\mathfrak{p}## is an integral domain, and ##x_{n+1}## is invertible then...
The following are taken from the two sources, 1) from this online page and the book An Introduction to Module Theory by: Ibrahim Assem, Flavio U. Coelho. In the Abelian Categories chapter in the module theory text on page 157, right after presenting IV.2.21 Definition, the authors states "Image and coimage may or may not exist, but if they do, then they are unique up to isomorphism (because so are kernels and cokernels). Also in the reference url page above, the authors present two...
When decomposing a representation ##\rho## of a finite group ##G## into irreducible representations, we can find the number of times the representation contains a particular irrep ##\rho_0## through the character inner product $$ \langle \chi, \chi_0\rangle = \frac{1}{|G|} \sum_{g\in G} \chi(g) \chi_0(g)^*$$ where ##\chi## and ##\chi_0## are the characters of ##\rho## and ##\rho_0##, respectively. Since all group elements in the same conjugacy class have the same characters, this may be...
Back
Top