Finding a Basis for a Linear Subspace Orthogonal to a Given Point P in R^3

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SUMMARY

The discussion focuses on finding a basis for a 2-dimensional linear subspace S in R^3 that is orthogonal to a given point P. The user initially attempted to express points in S using the parameter form x = P + αu + βv but struggled to identify two vectors u and v that are orthogonal to P. The solution involves recognizing that if P = (x₀, y₀, z₀), then vectors of the form <0, y, (y₀/z₀)y> and are orthogonal to P, forming a valid basis for the subspace S. The discussion clarifies the distinction between points and vectors in mathematical contexts.

PREREQUISITES
  • Understanding of linear algebra concepts, specifically orthogonal vectors.
  • Familiarity with R^3 vector space and its properties.
  • Knowledge of parameterization of planes in three-dimensional space.
  • Basic grasp of vector notation and operations, including transposition.
NEXT STEPS
  • Study the concept of orthogonal complements in linear algebra.
  • Learn about the Gram-Schmidt process for orthogonalization of vectors.
  • Explore the implications of vector spaces in higher dimensions, such as R^n.
  • Investigate applications of orthogonal projections in computational geometry.
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Mathematicians, students of linear algebra, and anyone interested in understanding vector spaces and orthogonality in R^3.

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I have a given point (vector) P in R^3 and a 2-dimensional linear subspace S (a plane) which consists of all elements of R^3 orthogonal to P.
The point P itself is element of S.

So I can write

P' ( x - P ) = 0

to characterize all such points x in R^3 orthogonal to P. P' means the transpose of P.

My problem is to find a basis of S. This basis should depend on point P.

I tried to find such a basis (alpha,beta) using the parameter form of the plane

x = P + alpha u + beta v

but I am unable to find two vectors u and v orthogonal to P.

I expect that this problem should be easy but I am nevertheless unable to solve it :(

Please help me a bit.
 
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Technically a "point" is not a "vector" but, yes, given a point, $p= (x_0, y_0, z_0)$ we can assign to to it the vector from (0, 0 0) to p, $<x_0, y_0, z_0>$. If <x, y, z> is orthogonal to that vector then $xx_0+ yy_0+ zz_0= 0$ so, solving for z, $z= \frac{xx_0+ yy_0}{z_0}$ (this is assuming $z_0\ne 0$. If it is choose to solve for x or y instead. If all three of $x_0$, $y_0$, and $z_0$, this is the zero vector and all vectors are orthogonal to it.)

Letting x= 0, $\left<0, y, \frac{y_0}{z_0}y\right>$ and taking y= 0, $\left<x, 0, \frac{x_0}{z_0}x\right>$.

Those two vectors are orthogonal to p. You should be able to show that they are independent and so a basis for the subspace. Choosing a value for y in the first and a value for x in the second gives vectors of specific length. They can be chosen so that the vectors have length one.
 
Thanks for your reply - it helped me out of my blockade.
I was too fixed at the implicit characterisation of orthogonal points to P by the equation

P' ( x - P ) = 0.

You are right, exept a typo in the sign for z in your formula.

Points or vectors, that is an old discussion. "Vectors" are often used in physics and "points" are often used in math.
As you know in math a "vector space" consists of some set with an additional algebraic structure. His elements (for example real numbers
complex numbers or integrable functions) are often called points.
 

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