# I Finding a linear combination to enter a sphere

#### johann1301h

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?

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#### StoneTemplePython

Science Advisor
Gold Member
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'?

#### johann1301h

orthogonality: 90 degrees, orthonormal: 90 degrees and length = 1unit ?

#### StoneTemplePython

Science Advisor
Gold Member
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
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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.

Last edited:

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