Hyperplane is a set of points satisfying a linear equation: How does a 3-dimension vector fill out a plane? A hyperplane H in $ R^n$ is the set of points $ (x_1, x_2, ..., x_n)$ that satisfy a linear equation:
$$ a_1 x_1 + a_2 x_2 + ... + a_n x_n = b$$
where the vector $ u = [a_1, a_2, ..., a_n]$ of coefficients is not zero. Thus a hyperplane in H in $R^2$ is a line, and a hyperplane H in $R^3$ is a plane.
I'm trying to conceptually understand the definition. But having issue with the picture in 3-dimensions.  So the set of x's that satisfy the equation in $R^3$ will be a vectors with 3 corresponding elements. Unless there's more than one solution, how would a 3 element vector (made from the X's solving the system) fill out a plane in 3 dimensions?
 A: The definition of a plane in $R^3$ is the infinite set of points $(x, y, z)$ that are solutions to the equation $ax + by + cz = d$. So, your assumption / statement of a single solution is not correct. A single $(x_0, y_0, z_0)$ represents a point on the plane.
The real number line has an infinite number of points on it.
So, you can choose $x$ and $y$ independently and they can take an infinite number of values and you can solve for $z$ for those chosen $x, y$ to give you the infinite set of $(x,y,z)$ that satisfy the planar equation.
A: Maybe it's easier to conceptualize if we drop the $b$ for a moment and assume it is zero. Then the equation reads
$$a_1x_1+a_2x_2+a_3x_3=0.$$
The left side is essentially just the scalar product of the vectors $\vec a=(a_1,a_2,a_3)$ and $\vec x=(x_1,x_2,x_3)$. But the scalar product of two vectors is zero if and only if they are perpendicular. So the solution set consists of all the vectors $\vec x$ which are perpendicular to the vector $\vec a$. And that's a plane containing the origin.
Now if we take $b\neq0$ and let's say $\vec y$ is a vector satisfying $\vec a\cdot\vec y=b$, while $\vec x_0$ is a vector from the plane above, so $\vec a\cdot\vec x_0=0$. Then the vector $\vec x_0+\vec y$ is a solution of the original equation, because
$$\vec a\cdot(\vec x_0+\vec y)=\vec a\cdot\vec x_0+\vec a\cdot \vec y=0+b=b.$$
So if we take all the vectors in the plane from the start and then shift each of them by $\vec y$, we get the solutions of the original equation. And that's still a plane, but shifted away from the origin.
