Why are sheafs of planes linear combinations of two other planes In most books, it is simply stated that the sheaf of planes can be described as a linear combination of two planes, but I cannot alight upon a simple explanation as to why.
I can see that a symmetric line equation in 3-D can be expressed as three symmetric 2-D line equations, that when plotted in 3-D describe three planes, and I can see that when you 'tweak' the multiples of these three symmetric equations to create a combined linear plane equation, the combined expression indeed rotates about the line common to all three planes.  (Though one only needs two of these 2-D line equations when plotted in 3-D to produce the same effect.)
Any suggestions as to a simple explanation of why, at, say, the level of transition from school to university ?
 A: Here's one possible proof. Everything is in Euclidean three-space, equipped with a Cartesian coordinate system $(x, y, z)$.
Let $P_{0} = (x_{0}, y_{0}, z_{0})$ be a point, and $\ell$ a line through $P_{0}$. Suppose $n_{1} = (a_{1}, b_{1}, c_{1})$ and $n_{2} = (a_{2}, b_{2}, c_{2})$ are non-proportional vectors orthogonal to $\ell$. The equations
\begin{align*}
  f_{1}(x, y, z) &:= a_{1}(x - x_{0}) + b_{1}(y - y_{0}) + c_{1}(z - z_{0}) = 0, \\
  f_{2}(x, y, z) &:= a_{2}(x - x_{0}) + b_{2}(y - y_{0}) + c_{2}(z - z_{0}) = 0,
\end{align*}
define planes containing $\ell$. If $t_{1}$ and $t_{2}$ are real numbers, then $t_{1}f_{1}(x, y, z) + tf_{2}(x, y, z) = 0$ is the equation of the plane containing $\ell$ and with normal vector $t_{1}n_{1} + t_{2}n_{2}$. Since $n_{1}$ and $n_{2}$ are non-proportional, every non-zero vector orthogonal to $\ell$ may be written as a linear combination of $n_{1}$ and $n_{2}$. Since every plane containing $\ell$ is uniquely specified by a vector orthogonal to $\ell$, every plane containing $\ell$ is a linear combination of the given equations.
