Is there a geometric meaning associated with the condition "dot product equals $1$?"

Question

Consider $x,y \in \mathbb{R}^n$.

Then the condition $x \bullet y = 0$ is easy to understand; it just means that $x$ and $y$ are orthogonal.

Question. Does the condition $x \bullet y = 1$ have an easy-to-grasp geometric meaning?

In the $n=1$ case, this is just the statement that $x$ and $y$ are multiplicative inverses.

Answer 1

Of course we can always (if it is non-zero) chance the value of $x \bullet y$ by multiplying one of the vectors by a scalar. To remove this problem, let me fix $y \bullet y=1$, i.e. $y$ is a unit vector.

As @Ranc mentioned in the comments, the map

$$P: x \mapsto (x \bullet y)y$$

is the orthogonal projection on $y$, hence $(x \bullet y)=1$ - which is equivalent to $Px=y$ - has the geometric meaning, that $x$ lies on the hyperplane, which contains $y$ and is orthogonal to $y$.

More generally, $|(x \bullet y)-1|$ is the shortest distance from $x$ to that hyperplane.

Answer 2

A problem with assigning a geometric meaning to any equation including the value $1$ is that unity depends on the unit. To understand "one" you need to choose a line segment as your standard unit. Then that line segment is your unit length, a square with that side is your unit area, and so on. (You could also start with the unit area and work back to the unit length, of course.)

So let's assume that the unit area has already been chosen. Then we have

$$x\bullet y=|x|\cdot (|y|\cos\theta)$$

where $|x|$ and $|y|$ are vector lengths and $\theta$ is the angle between the vectors. Then we also get $|y|\cos\theta$ as the length of the projection of the vector $y$ onto the vector $x$.

So $x\bullet y=1$ means that the rectangle formed by $x$ and the projection of $y$ onto $x$ has unit area.

That is not as simple as we might like, since in a diagram the vectors $x$ and the projection of $y$ onto $x$ are parallel, not perpendicular as they would be in a rectangle. But it does work.

Answer 3

Okay, I think I've got it.

Fix $x \in \mathbb{R}^n$. Write $x^\perp$ for the collection of all $y \in \mathbb{R}^n$ satisfying $x \bullet y = 0$, and $x^{-1}$ for the collection of all $y \in \mathbb{R}^n$ satisfying $x \bullet y = 1$.

Proposition. For all non-zero $x \in \mathbb{R}^n$, the set $x^{-1}$ is the hyperplane obtained by shifting the linear subspace $x^\perp$ in the direction $x$ a distance of $1/|x|$. In symbols:

$$x^{-1} = \frac{x}{x \bullet x}+ x^\perp$$

To see this, note that it suffices the prove the following two statements.

$$\frac{x}{x \bullet x}+ x^\perp \subseteq x^{-1}\qquad \qquad \frac{-x}{x \bullet x}+ x^{-1} \subseteq x^\perp$$

Both follow easily by assuming that $y$ is an element of the LHS and computing $x \bullet y$, thereby verifying that $y$ is an element of the RHS.

Hence:

Geometric interpretation. The equation $x \bullet y = 1$ is equivalent to the statement that $y$ is an element of the hyperplane obtained by shifting the linear subspace $x^\perp$ in the direction $x$ a distance of $1/|x|$.