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I am studying a chapter on vector subspaces and projections. I came across this definition:

We can split the set of all vectors $\mathbb{R}^3$ into two disjoint sets: vectors entirely contained in $S$ and vectors perpendicular to $S$. We say $\mathbb{R}^3$ decomposes into the direct sum of the subspaces $S$ and $S^\bot$:

$$\mathbb{R}^3 = S \oplus S^\bot$$

And then about projection onto a line:

Let $l$ be a line passing through the origin and is in $\mathbb{R}^3$ with direction vector $\vec{v}$:

$$ l: \{(x,y,z) \in \mathbb{R}^3 \mid (x,y,z) = t\vec{v}, t \in \mathbb{R}^3 \} $$

The orthogonal space to the line $l$ consists of all vectors perpendicular to the direction vector $\vec{v}$:

$$l^\bot : \{(x,y,z) \in \mathbb{R}^3 \mid (x,y,z).\vec{v} = 0 \}$$ ...

The orthogonal space for a line $l$ with direction vector $\vec{v}$ is a plane with normal vector $\vec{v}$.

All makes sense except the last sentence which says the orthogonal space is a plane with normal vector $\vec{v}$. What my intuition says is that the orthogonal space should be the set of all planes (not a single plane) in $\mathbb{R}^3$ with normal vector $\vec{v}$ because:

$$\mathbb{R}^3 = S \oplus S^\bot$$

I hope someone can tell me what is it that I am getting wrong here.

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    $\begingroup$ The orthogonal subspace contains the origin though, which fixes one plane. $\endgroup$
    – Paul
    Dec 26, 2020 at 19:47

2 Answers 2

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"We can split the set of all vector in $\mathbb{R}^{3}$ into two disjoint sets: vectors entirely contained in $S$ and vectors perpendicular to $S$."

I don't like this line for several reasons, the two main reasons being that

  1. These two sets $S$ and $S^{\perp}$ are not disjoint, they share the $0$ vector;
  2. it seems to suggest that every vector in $\mathbb{R}^{3}$ is either in $S$ or $S^{\perp}$, which is not the case. There are lots of vectors that are neither contained in $S$, nor are orthogonal to $S$. What is true is that every vector in $\mathbb{R}^{3}$ is a sum of a vector in $S$ and a vector in $S^{\perp}$.

The orthogonal space for a line $\ell$ with direction $v$ is a plane with normal vector $v$.

This is absolutely correct. The orthogonal space is a collection of vectors, not a collection of planes. And all vectors are rooted at the origin. So the orthogonal space consists of all vectors in the plane containing the origin which has normal vector $v$ (this is essentially the definition of the normal vector of a plane).

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  • $\begingroup$ Thank you very much. I think the key point I was missing here was that the vectors in orthogonal space are rooted at the origin. This way, there is only one plane that its vectors are perpendicular to the line. $\endgroup$ Dec 27, 2020 at 6:32
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Actually you're confusing with vectorial spaces and affines spaces. With vectorial spaces there exists only one subspace that has a given normal vector, because a plane is define by two non colinear vectors.

But then if you work with affines spaces, you also need a point : that's what we are used to.

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  • $\begingroup$ what is wrong with my answer? $\endgroup$
    – math
    Dec 26, 2020 at 19:48
  • $\begingroup$ Thanks mate. I don't know about affine space yet. btw I didn't down vote :) $\endgroup$ Dec 27, 2020 at 6:30

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