vector space $\perp$ notation My teacher asked:
given three vectors x, v, w of the same length, find the projection of x on $w^\perp$ parallel to v (v and w are assumed to be 
non-zero and v not orthogonal to w).
My question is, in this context, what is $w^\perp$? Is it a plane? For instance, if $w=(1,0,0)$, what is $w^\perp$?
 A: $w^\bot$ denotes the subspace of all vectors orthogonal to $w$, i.e. $$
  w^\bot = \{v \,:\, v \cdot w = 0 \} \text{.}
$$
More generally, for an arbitrary subspace $U$, $U^\bot$ denotes the orthogonal complement of $U$, which is the subspace of all vectors orthogonal to all vectors in $U$, i.e $$
  U^\bot = \{v \,:\, v \cdot w = 0 \text{ for all $w\in U$}\} \text{.}
$$
If $U = \textrm{span } \{u\}$, then $U^\bot = u^\bot$. If $U = \textrm{span } \{u_1,\ldots,u_n\}$ then $U^\bot = \bigcap_{k=1}^n u_k^\bot$. For finite-dimensional vector space you always have $V = U \oplus U^\bot$. It follows that if $U$ is a $k$-dimensional subspace of an $n$-dimensional space, then $\dim U^\bot = n - k$
So in particuarl, in a $n$-dimensional vector space, $w^\bot$ always has dimension $n-1$. For $\mathbb{R}^2$, $w^\bot$ is therefore line (through the origin), whereas in $\mathbb{R}^3$, $w^\bot$ is a plane (which includes the origin).
A: $w^\perp$ is the subspace of all vectors such that $\langle w,w^\perp\rangle=0$, i.e., $w$ is orthogonal to $w^\perp$. See Wolfram Mathworld.
A: Consider subspaces in $\mathbb{R}^3$ generated by the span of $W$, a set of either one or two linearly independent vectors.  If $Span(W)$ is a line, then $W^\perp$ is a plane.  Alternately, if $Span(W)$ is a plane, then $W^\perp$ is a line.  Can you see why?
In this case, if $W$ consists of the vector $\langle1, 0, 0 \rangle$, then $W^\perp$ is the $y,z$ plane.
