Surface described by orthogonality condition for vectors Two vectors $\mathbf{x}=(x_1,x_2)$ and $\mathbf{y}=(y_1,y_2)$ are orthogonal if their scalar product is zero, i.e. $x_1 y_1 + x_2 y_2=0$. What surface does this equation describe in the four-dimensional $(x_1,x_2,y_1,y_2)$ space? How about the orthogonality condition for $3$-dimensional (or higher dimensional) vectors? What is the corresponding surface and how would one visualize it?
Thanks for the answers.
 A: This is only a first approximation to your answer, but I think it's a good start.
Let $x$ and $y$ be vectors in $\mathbb{R}^n$ and assume $x\cdot y = 0$.  Now, I'm going to add an additional condition that neither $x$ nor $y$ is $0$.  To give away the punchline, it will turn out there's a nice description of these points.  The remaining points (where either $x=0$ or $y=0$ or both) will be degenerate in a way, because then the dot product doesn't tell you anything.  It turns out these remaining points destroy the "niceness" of the others (at least, how I usually define nice, i.e., getting a smooth manifold).
So, let $X =\{(x,y)\in \mathbb{R}^{2n}|$ $x\neq 0$, $y\neq 0,$ and $x\cdot y = 0\}$.  The goal is to understand the shape of $X$.
The first thing to notice is that if $(x,y)\in X$, then so is $(\lambda x,y)$ and $(x,\mu y)$ for any $\lambda, \mu > 0$.  Further, if we set $Y =\{(x,y)\in X|\text{  } |x|=|y|=1 \}$, then it's clear that every point in $X$ is of the form $(\lambda x, \mu y)$ for a unique choice of $\lambda, \mu > 0$.
It follows from this that $X = Y\times (0,\infty)\times (0,\infty)$.  That is, $X$ has the shape of $Y$ extended in 2 directions.  But what is the shape of $Y$?
For the moment, let's take an apparent detour.  How do you describe a sphere in $\mathbb{R}^n$?  It's $S^n = \{ x\in\mathbb{R}^{n+1}| \text{  }|x|=1\}$.  (Note that the $n$ in $S^n$ is one less than the $n$ in $\mathbb{R}^{n+1}$)  What is a tangent vector on $S^n$ at a point $x\in S^n$?  More generally, what is the tangent plane through a point $x\in S^n$?
It shouldn't be too hard to convince yourself by looking at this on the circle and usual sphere that the tangent plane/line through a point $x$ in $S^n$ is the plane with normal vector $x$ going through the point $x$ (so $x$ does double duty as both a point and a vector).
Saying the normal vector is $x$ means that for any vector $y$ in the tangent plane, we must have $x\cdot y = 0$.  What if we take a unit vector $y$ in the tangent plane?  Well, we just find that $|x| = |y| = 1$ and $x\cdot y = 0$.  But these were exactly our conditions on belonging to $Y$!
This means that $Y$ is the so called Unit Tangent Bundle of the sphere - it's the collection of all $(p,v)$ where $p$ is a point on the sphere (i.e., $|p| = 1$, $v$ is a tangent vector to there sphere at $p$ (i.e., $p\cdot v = 0$) where $v$ has unit length (i.e., $|v| = 1$).
This means that, in particular, you can think of these spaces as "the collection of points on a sphere paired with a choice of unit length tangent vector".  It turns out this shape is fairly well understood.
For example, when $n=1$, one gets the unit tangent bundle of a circle.  It is know that the tangent bundle of a circle is trivial - that is, is diffeomorphic to $S^1\times \mathbb{R}$.  From here it follows that the unit tangent bundle is diffeomorphic to $S^1\times S^1$.  This shape is called a Torus and is best physically described as a hollow donut.
This, in particular, almost answers your first question about what we get in $\mathbb{R}^4$.
For $n =2$, one gets the unit tangent bundle of $S^2$.  It's known that this is diffeomorphic to $\mathbb{R}P^3$, which can almost be visualized.  For $n=3$, one gets the unit tangent bundle of $S^3$, which is known to be diffeomorphic to $S^3\times S^2$.  I don't know of a suitably nice description of any other unit tangent bundles except for $n=7$, one gets $S^7\times S^6$.
A: Cut your hypersurface $\{(x_1,x_2,x_3,x_4)\in\mathbb R^4:x_1x_2+x_3x_4=0\}$ with the hyperplanes $H_t=\{(x_1,x_2,x_3,x_4)\in\mathbb R^4:x_1=t\}$, for $t\in\mathbb R$, and visualize those intersections. 
