Perpendicular vector in 6th dimensional space

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SUMMARY

This discussion focuses on finding a vector perpendicular to a unit vector in 6-dimensional Euclidean space, specifically for applications in brain connectivity analysis using MRI. The method involves constructing an orthonormal basis from two unit vectors, u and v, and utilizing orthogonal projections to derive the desired perpendicular vector. The final expression for the perpendicular vector w is defined as w=bf-ag, where f and g are components of the orthonormal basis.

PREREQUISITES
  • Understanding of 6-dimensional Euclidean space
  • Familiarity with unit vectors and orthonormal bases
  • Knowledge of orthogonal projections in vector spaces
  • Basic concepts of vector operations and linear algebra
NEXT STEPS
  • Study the construction of orthonormal bases in higher-dimensional spaces
  • Learn about orthogonal projections and their applications in vector analysis
  • Explore the mathematical properties of unit vectors in multi-dimensional contexts
  • Investigate the implications of vector operations in brain connectivity analysis
USEFUL FOR

Mathematicians, software developers working on brain imaging analysis, and researchers in neuroscience who require a solid understanding of vector mathematics in higher-dimensional spaces.

Sorento7
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I am working on a software for analysis of brain connections using MRI.

Please suggest the simplest way to find a vector which is:

perpendicular to a unit vector that is positioned in the coordinate center,

it should be in the 2D plane containing the given vector and a given point on the unit 5-sphere.

Everything takes place in 6 dimensional Euclidean space.
 
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This should work:

Denote the two given unit vectors by u and v. First we find an orthonormal basis {f,g} for the 2-dimensional subspace spanned by {u,v}. Define g=v. The orthogonal projection of u onto the 1-dimensional subspace spanned by g is <g,u>g. Let's denote this vector by p. The vector u-p will be orthogonal to g. So we define ##f=(u-p)/\|u-p\|##.

Now we can write u=af+bg, where a=<f,u> and b=<g,u>. Define w=bf-ag. This w is orthogonal to u.
 

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