I Finding a linear combination to enter a sphere

Let's say we have n vectors in ℝ3. And say we have defined a subspace inside ℝ3 in the form of a sphere with radius r, and the center of the spheare is at P, where P is a vector in ℝ3.

What methods exists to find any linear combination of the n vectors, so that the sum of all of them, lies within the sphere?

F. ex i have hear of something called least squares. Can that be used for this?
 

StoneTemplePython

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assuming we are using the standard metric / 2norm here, then yes there are nice ways to do this.

First, for now, assume p is the zero vector. Do you know what orthogonality is? And in particular, do you know the term 'orthonormal'?
 
orthogonality: 90 degrees, orthonormal: 90 degrees and length = 1unit ?
 

StoneTemplePython

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orthogonality: 90 degrees, orthonormal: 90 degrees and length = 1unit ?
yes. and the standard case here is we select two vector in ##\{\mathbf x_1, \mathbf x_2, \mathbf x_3\}## and take the dot product (which is the standard inner product in reals). These vectors are mutually orthonormal iff
##\mathbf x_j^T \mathbf x_k= 1## if ##j = k## and ##=0## if ##j \neq k##

- - - -
what does this have to do with your problem? still with ##\mathbf p =\mathbf 0##, you want to select say ##m## of these mutually orthonormal vectors where

##\big \Vert \sum_{i=1}^m \alpha_i \mathbf x\big \Vert_2 \leq r##
or
##\big \Vert \sum_{i=1}^m \alpha_i \mathbf x\big \Vert_2^2 \leq r^2##

this is equivalent to
## \sum_{i=1}^m \alpha_i^2 \leq r^2##

Confirm that this makes sense
- - - -
edit:
for avoidance of doubt, mutually orthonormal implies mutually linearly independent, so in ##\mathbb R^3## it must be the case that ##m\leq 3##, because you cannot have more than 3 linearly independent vectors when your dimension is 3.
 
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