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.