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The intersection defined by the two planes $v \bullet \begin{pmatrix} 8 \\ 1 \\ -12 \end{pmatrix} = 35$ and $v \bullet \begin{pmatrix} 6 \\ 7 \\ -9 \end{pmatrix} = 70$ is a line. What is the equation of this line?

This is what I have so far: I set $v = \begin{pmatrix} a \\ b \\ c \end{pmatrix}$. I was able to simplify both LHS of the two given equations to:

$8a+b-12c = 35$

$6a+7b-9c = 70$

I could find the values of $a, b, c$, but I don't know if it will be helpful. How can I find the equation of the line of intersection?

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    $\begingroup$ If you are looking for a parametric equation of the line you will need two things: a point $p$, which is any solution to the system constituted by the planes' equations; and a director vector $v$ which is perpendicular to both planes' normal vectors, so it can be obtained by the cross product of them. After you have done that, your equation is $(x,y,z)=p+\lambda v$ $\endgroup$ – Smurf Jul 23 '15 at 23:14
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There are two basic methods: either

  1. You can obtain the direction of the line by doing the cross-product of the normals $$\left(\begin {matrix}8\\1\\-12\end {matrix}\right)\times\left(\begin {matrix}6\\7\\-9\end {matrix}\right)=\left(\begin {matrix}75\\0\\50\end {matrix}\right) \text{which is parallel to}\left(\begin {matrix}3\\0\\2\end {matrix}\right)$$

And then somehow obtain a point which satisifies both plane equations, perhaps by assigning a value to one of the variables and solving for the other two.

Or, a more systematic way is:

  1. Start with your plane equations (I am using $x$, $y$ and $z$) and eliminate any one of the variables, So, for example, if you eliminate $y$ you get $$2x-3z=7$$ Now separate the letters onto opposite sides of the equation and intoduce $=\lambda$. So now you have $$2x=7+3z=\lambda$$

Now get each of $x$, $y$ and $z$ in terms of $\lambda$ and you end up with the equation of the line in one of many possible vector forms; in this case $$\underline{r}=\left(\begin {matrix}0\\7\\\frac{-7}{3}\end {matrix}\right)+\lambda \left(\begin {matrix}\frac 12\\0\\\frac 13\end {matrix}\right)$$

Of course you can write the direction vector as $$\left(\begin {matrix}3\\0\\2\end {matrix}\right)$$ as before.

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With two equations and three unknowns your solution space will be a one-dimensional set (a line), exactly what you are looking for. You will need to write $(a,b,c)$ in terms of one of the parameters. For instance $a=f_{1}(c),b=f_{2}(c),c=c$.

It will not always be the case that a solution space will be one-dimensional, it could be the case that the column space is one-dimensional and it is possible that no solution would exist.

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Your line is defined by your two equations; you just need to set them up as a system $$\begin{pmatrix} 8&1&-12\\6&7&-9 \end{pmatrix} v = \begin{pmatrix} 35\\70 \end{pmatrix}$$ This is the exact equation you have written, in matrix form. You should be able to row reduce, end up with a single free variable, and then you can parametrically write your solution set.

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