For a linear system, why is direction "stored" in the variables when considering it as linear equations, but in vectors when its as a vector equation?

Question

Given an arbitrary system of equations, why is direction in space "stored" in the variables when considering the system as linear equations, but "stored" in vectors when considering the system as a vector equation? For example suppose we have a system of three equations in three variables where each equation is of the form (a_i)x + (b_i)y + (c_i)z = d_i. Lets also suppose they represent three distinct planes in R^3. In the context of the system representing planes in space, it seems to me that dimension/direction is sort of "stored" in the variables x, y and z. Considering the system as a linear combination of vectors, the coefficients associated with any one variable make a column vector; for example, for variable x lets call the vector v_1 = . The vector equation associated with the system would then be xv_1 + yv_2 + zv_3 = . In this context it seems as though dimension/direction is stored in the vectors, and the variables x,y, and z now just scale them. I realize that both contexts have the same solution set, and both take place in R^3, but is there a more intuitive explanation for this relatedness?

Answer 1

You have stumbled upon the principle of duality in projective geometry, and Dual Theorems. Basically You define a plane as $$\mathbf{P}\cdot \mathbf{x} = 0$$

where $\mathbf{P}=(a,b,c,d)$ are the plane coordinates and $\mathbf{x}=(x,y,z,1)$ are point coordinates. The equation is then $ax+by+cz+d=0$.

It can interpreted as either

1. All the points $\mathbf{x}$ that pass through a fixed plane defined by the coordinates $\mathbf{P}$.
2. All the planes $\mathbf{P}$ that pass through a fixed point defined by the coordinates $\mathbf{x}$.

So for this construction, either $\mathbf{P}$ are coordinates and $\mathbf{x}$ are arbitrary, or the other way around. The caveat is the which ever is the coordinates in cannot contain all zeros. So either $\mathbf{P} \neq (0,0,0,0)$ or $\mathbf{x} \neq (0,0,0,0)$. It is said than in $\rm R^3$ points are dual to planes and vise versa.

What this allows to define is that

• for every point in $\rm R^3$ there is corresponding unique plane (through the point, with the normal away from the origin)
• and for every plane there is a corresponding unique point (closest to the origin)

So I can define a plane using the point coordinates $\mathbf{x}=(\vec{r},1)$ such as $\mathbf{P} = \left( \frac{\vec{r}}{\| \vec{r} \|}, -\vec{r}\cdot\vec{r} \right)$

Or I can define a point using the plane coordinates $\mathbf{P}=(\vec{a},b)$ such as $\mathbf{x}=(-\vec{a} b,\vec{a}\cdot\vec{a})$

Example

A plane has coordinates $\mathbf{P}=(\vec{a},b)=((2,1,3),-5)$ has equation $2x+y+3z-5=0$. The unique point closest to the origin is $\mathbf{x}=\left((-(-5)(2,1,3),(2,1,3)\cdot(2,1,3)\right)=(10,5,15,14)$.

The Cartesian coordinates of $\mathbf{x}$ are $x=\frac{10}{14}=\frac{5}{7}$, $y=\frac{5}{14}$ and $z=\frac{15}{14}$.

Lets prove these coordinates satisfy the plane equation $$ 2 \frac{5}{7} + 1 \frac{5}{14} + 3\frac{15}{14} - 5 =0\;\checkmark$$

So either $(2,1,3,-5)$ are the plane coordinates of this construction, or $(10,5,15,14)$ are the point coordinates of this construction.