MHB Find Unit Vector Orthogonal to Vector

  • Thread starter Thread starter annie122
  • Start date Start date
  • Tags Tags
    Plane
annie122
Messages
51
Reaction score
0
#q1
how do i find the normal form of a plane given a point thru the plane and a vector parallel to it?

first of all i don't understand what a vector parallel to a plane is :/

#q2

how do i find a unit vector that's orthogonal to a certain vector?
 
Physics news on Phys.org
Re: equation of a plane

Q1: Well, what is the normal form for a plane? What does each symbol represent? How could you find an applicable value for each of those symbols from the given data?

Q2: Can you write down an equation that says "vector $\vec{x}$ is orthogonal to vector $\vec{y}$"?
 
Re: equation of a plane

You may want to take a look at THIS.

If, after reading that, you still have questions, be more specific and we'll try to help :)
 
Re: equation of a plane

Yuuki said:
#q1
how do i find the normal form of a plane given a point thru the plane and a vector parallel to it?
You can't. A vector parallel to a plane defines a line (through the given point) in the plane. But there are an infinite number of planes containing a given line.

first of all i don't understand what a vector parallel to a plane is :/
Well, since a vector is "movable" this is the same a vector lying in the plane.

#q2

how do i find a unit vector that's orthogonal to a certain vector?
There are an infinite number of vectors orthogonal to a given vector. One way find one of them is to find the equation of the plane containing (0, 0, 0) perpendicular to the given vector. Choose any point in that plane and construct the vector from (0, 0, 0) that point. Finally, calculate the length of that vector and divide the vector by its length to get a unit vector.
 
Re: equation of a plane

HallsofIvy said:
You can't. A vector parallel to a plane defines a line (through the given point) in the plane. But there are an infinite number of planes containing a given line.

Right, I thought of that. But it may be that the OP is actually given a vector that defines a line (there's another point on the line). A line and a point not on the line will uniquely determine a plane, at least in Euclidean geometry.
 
##\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