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

Click For Summary
To find a basis for a 2-dimensional linear subspace S orthogonal to a given point P in R^3, one can express the relationship using the equation P' (x - P) = 0. The challenge lies in identifying two vectors u and v that are orthogonal to P. By using parameterization, specific vectors can be derived, such as <0, y, (y_0/z_0)y> and <x, 0, (x_0/z_0)x>, which are confirmed to be independent and form a basis for the subspace. The discussion highlights the distinction between points and vectors, emphasizing their interchangeable use in different contexts. This approach effectively resolves the initial confusion regarding the characterization of orthogonal points.
JWS1
Messages
2
Reaction score
0
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.
 
Physics news on Phys.org
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.
 
Thread 'How to define a vector field?'
Hello! In one book I saw that function ##V## of 3 variables ##V_x, V_y, V_z## (vector field in 3D) can be decomposed in a Taylor series without higher-order terms (partial derivative of second power and higher) at point ##(0,0,0)## such way: I think so: higher-order terms can be neglected because partial derivative of second power and higher are equal to 0. Is this true? And how to define vector field correctly for this case? (In the book I found nothing and my attempt was wrong...

Similar threads

  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 39 ·
2
Replies
39
Views
4K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 5 ·
Replies
5
Views
4K
  • · Replies 52 ·
2
Replies
52
Views
4K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K