What is so special about $\underline{i}-2\underline{j}+\underline{k}$?

Question

If you take two 3-vectors whose components are in arithmetic progression, then their cross-product (as long as it is not the zero vector) is always parallel to $$\underline{i}-2\underline{j}+\underline{k}$$

For example, $$\left(\begin{matrix}1\\2\\3\end{matrix}\right)\times\left(\begin{matrix}7\\4\\1\end{matrix}\right)=\left(\begin{matrix}-10\\20\\-10\end{matrix}\right)$$

More generally, provided that $$a_1d_2-a_2d_1\neq0$$ $$\left(\begin{matrix}a_1\\a_1+d_1\\a_1+2d_1\end{matrix}\right)\times\left(\begin{matrix}a_2\\a_2+d_2\\a_2+2d_2\end{matrix}\right)=\left(\begin{matrix}(a_1d_2-a_2d_1)\\-2(a_1d_2-a_2d_1)\\(a_1d_2-a_2d_1)\end{matrix}\right)$$

What is the geometrical reason for this?why this vector perpendicular to the two arithmetic progressions always points in the same direction?

Answer 1

The fact seems that a vector with components in arithmetic progression $(x,x+d,x+2d)$ belongs to the plane $ax+by+cz=0$

$$ax+b(x+d)+c(x+2d)=0\iff (a+b+c)x+bd+2cd=0$$

which is true for $a=1,b=-2$ and $c=1$.

Then the vector $\underline{i}-2\underline{j}+\underline{k}$ represents a normal vector to the plane which contains all the vectors with components in arthmetic progression.

Answer 2

Thanks to @gimusi for the "obvious" insight I missed is that all such points must lie on a plane which includes the origin.

Then if $(x,y,z)$ is a point on the plane and these coordinates are in arithmetic progression, then$$z-y=y-x$$ $$\implies x-2y+z=0$$