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How to calculate the intersection of two planes ?

These are the planes and the result is gonna be a line in $\Bbb R^3$:

$x + 2y + z - 1 = 0$

$2x + 3y - 2z + 2 = 0$

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What representation of the result do you seek? What have you done so far? –  AlexR Aug 25 '13 at 17:29
    
To calculate an intersection, by definition you must set the equations equal to each other such that the solution will provide the intersection. In short, set $$x + 2y + z - 1 = 2x + 3y - 2z + 2 = 0$$ To get a matrix you must solve. –  Don Larynx Aug 25 '13 at 17:35

2 Answers 2

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You need to solve the two equations $$ x + 2y + z - 1 = 0 \\ 2x + 3y - 2z + 2 = 0. $$

Notice that, these are two equations in three variables, so you have a free variable say $z=t$, then we have

$$ x + 2y = 1-t \\ 2x + 3y = 2t-2. $$

Solving the last system gives

$$ \left\{ x=-7+7\,t,y=4-4\,t \right\} .$$

Then the parametrized equation of the line is given by

$$ (x,y,z)= (-7+7t, 4-4t,t)=(-7,4,0)+(7,-4,1)t . $$

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The first plane has normal vector $\begin{pmatrix}1\\2\\1\end{pmatrix}$ and the second has normal vector $\begin{pmatrix}2\\3\\-2\end{pmatrix}$, so the line of intersection must be orthogonal to both of these. We know that the unique vector orthogonal to two linearly independent vectors $v_1,v_2$ is $v_1\wedge v_2$, so the direction vector of the line of intersection is $$\begin{pmatrix}1\\2\\1\end{pmatrix}\wedge \begin{pmatrix}2\\3\\-2\end{pmatrix}=\begin{pmatrix}-7\\4\\-1\end{pmatrix}$$Next, we need to find a particular point on the line. We can try $y=0$ and solve the resulting system of linear equations:$$\begin{align}x+z-1&=&0\\2x-2z+2&=&0\end{align}$$ giving $x=0, z=1$, thus the line of intersection is $\lbrace{\begin{pmatrix}-7t\\4t\\1-t\end{pmatrix}:t\in \Bbb R\rbrace}$

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