Is there a perp operator for vectors?

Question

I'm coming from a software development background, and working on some geometry related software, that I'm documenting, and was wanting to know what the correct mathematical operator is for the following:

We have a function called $perp$ which when passed a vector $v$ will give a perpendicular vector to $v$.

My question is there a symbol or specific named use in mathematics for such a function - the perp operator or some such?

For example if I wanted to say that point p is the result of the perpendicular vector from points $r$ to $s$ and scaled by some length $d$:

$$v = \overline{s - r}$$ $$p = d\times perp(v)$$

Would that be syntactically/mathematically correct? - on Wikipedia it notes that it may be called a 'rejection', using that syntax it would look like this:

$$p = d\times v_2$$

Answer 1

"$perp(v)$" is an ill-defined concept. Writing something in the form of $f(x)$ implies that there exists exactly one value $y$ such that $y=f(x)$.

In the case of perpendicular vectors, this is not the case. Given a vector $v$, if vector $u$ is perpendicular to $v$, then the vector $2\cdot u$ is also perpendicular to $v$. So, which one of the two is equal to $perp(v)$? Or maybe it's $-u$? Or maybe $-3u$? Or $(\pi + 3e^{sqrt}) u$?

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What you can define is the set of all vectors, perpendicular to $v$. This set would generally be denoted as $v^\top$ or $\{v\}^\top$. This set is actually a vector subspace of the vector space $v$ belongs to, and if $v$ is not $0$, then the dimension of $v^\top$ is one less than the dimension of the total space.

In particular, if $v$ is a $3$-dimensional vector, then all vectors perpendicular to it form a $2$-dimensional vector space, or in other words, a plane.

Answer 2

Remember that a function in software development is not the same thing as a mathematical function. Let's say your function perp(v) is a pure function, ie. it behaves like a mathematical function: it always returns the same value for a given input argument. It turns out there is no canonical way to define a mathematical function this way since (as 5xum wrote) there are many vectors perpendicular to a given vector. You could simply choose one particular perpendicular vector for each input vector, but that's ugly (and it probably won't be a linear operator, in case that matters to you.) However, in certain cases we may be able to do something. Suppose you're working in the plane $\mathbb R^2$ and you want the vector of the same magnitude, but rotated through $90^\circ$ counterclockwise. Then you can simply say "let $\tau$ be the operator which rotates a vector through $90^\circ$" and $\tau(v)$ is your new vector. If you like you can represent $\tau$ as a matrix. Unfortunately (as far as I'm aware) this particular operator has no generally agreed upon name.

Actually, there is one operator that satisfies the letter of your request, but not its spirit: the zero operator (since the zero vector is perpendicular to every vector).

As for the second half of your post, I would avoid using the $\times$ symbol for scalar multiplication of vectors, since that symbol is easy to mistake for the cross product. Better to use the dot $\cdot$ or simply write $\mathbf p = d \operatorname{perp}(\mathbf v)$. Note the difference in typeface for vectors and scalars. Also note that I just said $\mathbf p$ is a vector, not a point! It's simplest to identify points with vectors (although perhaps it's semantically inaccurate.) So no need for the overline in your first equation. If you must make a formal distinction between points and vectors, notation will get much more messy.

Rejection is an interesting term! I've never heard it before. But you don't get to choose the length of a rejection: it is defined as the difference between your vector and its projection onto another specific vector. This doesn't seem to be quite what your looking for (which vector are you projecting onto??)