In a class today I provided a method of finding the point of intersection of two planes that goes as follows, however it returned a plane and no one in the class could work out why. Consider the two planes - which I then rearranged:
$$ \newcommand{\tvect}[3]{% {\Bigl(\negthinspace\begin{smallmatrix}#1\\#2\\#3\end{smallmatrix}\Bigr)}} \Pi_1: \vec{r} \cdot \tvect{2}{3}{4} = 4 \rightarrow \vec{r} \cdot \tvect{1/2}{3/4}{1} = 1 $$
$$ \Pi_2: \vec{r} \cdot \tvect{4}{5}{2} = 3 \rightarrow \vec{r} \cdot \tvect{4/3}{5/3}{2/3} = 1$$
I then set both equations to be equal and rearranged:
$$ \vec{r} \cdot \tvect{4/3}{5/3}{2/3} = \vec{r} \cdot \tvect{1/2}{3/4}{1} $$
$$ \vec{r} \cdot ( \tvect{4/3}{5/3}{2/3} - \tvect{1/2}{3/4}{1}) = 0 $$
$$ \vec{r} \cdot \tvect{-5/6}{-11/12}{1/3} = 0 $$
If we rewrite $ \vec{r} $ as $\tvect{x}{y}{z}$, and calculate the dot product, we get:
$$ -\frac{5}{6}x - \frac{11}{12} y + \frac{1}{3}z = 0$$
This is obviously the equation of a plane (which I checked and all three intersect on the same line) but I would have expected it to get me the equation of the line of intersection as that is the only place where $ \vec{r} $ satisfies both equations simultaneously?
Can someone explain where I have gone wrong and an intuitive explanation for why another plane is being produced as the solution?
The intersection of the two planes is the solution of the system of two equations $$ \begin{cases} 2x+3y+4z=4\\ 4x+5y+2z=3 \end{cases} $$
The result of your manipulation is that you can substitute one of the two equation by your new equation $-\frac{5}{6}x - \frac{11}{12} y + \frac{1}{3}z = 0$, but anyway you have to solve a system with this and one of the original equations.
In other words: You have simply found another plane that intersects the given two planes in the same line, but now you have to find this line.
Can you do this?