Finding Orthogonal Vectors in 4 Space

  • Thread starter Thread starter shane1
  • Start date Start date
  • Tags Tags
    Space Vectors
Click For Summary
SUMMARY

This discussion focuses on finding two orthogonal vectors of norm 1 in four-dimensional space that are perpendicular to the vectors u = (2, 1, -4, 0), v = (-1, -1, 2, 2), and w = (3, 2, 5, 4). The user initially attempted to solve a system of equations but encountered a free variable. Shane suggests using the determinant of a 4x4 matrix as an n-dimensional analog of the cross product to find a vector perpendicular to the given vectors. This method will yield a line of vectors, from which two can be normalized to achieve the desired norm.

PREREQUISITES
  • Understanding of vector norms and orthogonality
  • Familiarity with solving systems of linear equations
  • Knowledge of determinants and their properties in linear algebra
  • Basic concepts of n-dimensional geometry
NEXT STEPS
  • Learn how to calculate the determinant of a 4x4 matrix
  • Study the properties of orthogonal vectors in higher dimensions
  • Explore the n-dimensional analog of the cross product
  • Practice normalizing vectors to achieve a specific norm
USEFUL FOR

Mathematicians, physics students, and anyone working with linear algebra and vector spaces, particularly in higher dimensions.

shane1
Messages
7
Reaction score
0
I have this question that says:
Find two vectors of norm 1 that are orthagonal to the three vectors u = (2, 1, -4, 0), v = (-1, -1, 2, 2), and w = (3, 2, 5, 4).

I've tried setting up a system of equations to solve.
2a + b - 4c = 0
-a - b + 2c + 2d = 0
3a + 2b + 4c + 4d = 0

But when I did that I was left with a free variable. So basically I was wondering if there's another way to do it such as taking the determinate like how you do in 3 space. Except in 4 space.
Eg.
i j k
0 1 0
1 2 5

Shane
 
Physics news on Phys.org
There will be a whole line of vectors perpendicular to those vectors. But only 2 will have norm 1.
 
If you know how to calculate the determinat of an nxn matrix there is an n-dimensional analog of the cross product:
[tex] \vec{v}=\left| \begin{array}{c c c c}<br /> \hat{i} & \hat{j} & \hat{k} & \hat{l} \\<br /> 2 & 1 & -4 & 0 \\<br /> -1 & -1 & 2 & 2 \\<br /> 3 & 2 & 5 & 4 \end{array} \right |[/tex]

Which will give you a vector perpendicular to the n-1 you already have.
 

Similar threads

  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
1
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
9
Views
2K
  • · Replies 9 ·
Replies
9
Views
4K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 5 ·
Replies
5
Views
1K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 16 ·
Replies
16
Views
3K