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The planes are x+2y+3z=1 and x-y+z=1. My guess would be to set them equal to each other, since they are both equal to 1, we could write that as x+2y+3z=x-y+z. This simplifies to 3y+2z=0, it doesn't seem like this would be our answer though.

Update: I now understand that the cross product of the two normal vectors gives us the direction vector for the line that we're interested in finding but I don't know how to find a point on the line.

ShaneBird
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  • Find the normal vectors for each planes and take the cross product of those vectors. The cross product yields a new vector parallel to the line of intersection. Use this to solve. – 123 Sep 01 '14 at 19:11
  • @mathtastic I now understand how to find the direction vector of the line but how do we find a point on the line so we can create an equation for the line. – ShaneBird Sep 01 '14 at 20:16
  • Ah! Yes, I should have included that information. You can solve the two equations simultaneously. The easiest method would be to use a simple augmented matrix (like those listed below by a few posters)and then apply gaussian elimination to solve your augmented matrix. Here is a link to a wiki than explains gaussian elimination and has a nice example that helps illustrate the method! Don't let the name trick you - it is a very easy process and taking a few minutes to learn the method will help a lot with cal 3. http://en.wikipedia.org/wiki/Gaussian_elimination – 123 Sep 01 '14 at 20:48
  • Excellent, thanks. – ShaneBird Sep 01 '14 at 20:55

3 Answers3

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The method is explained here.

MartinG
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You have $x+2y+3z=1$ and $x-y+z=1$ iff $3y+2z=0$ and $x=1-2y-3z$ iff $y=-{2 \over 3} z$ and $x=1-2y-3z=1-{5 \over 3} z$.

So, the line is given by $\{(1-{5 \over 3} z, -{2 \over 3} z, z) \}_{z \in \mathbb{R}}$.

copper.hat
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Your planes are equivalent to the linear system

$$\left[ \begin{array}{ccc} 1 & 2 & 3 \\ 1 & -1 & 1 \\ \end{array}\right] \left[ \begin{array}{c} x \\ y \\ z \\ \end{array}\right] = \left[ \begin{array}{c} 1 \\ 1 \\ \end{array}\right] $$

If the normal vectors are nonzero and are not parallel then you can always perform row operations so that $\left[\begin{array}{c}1\\0\\\end{array}\right]$ is in one column and $\left[\begin{array}{c}0 \\1\\\end{array}\right]$ is in another one. Then you can pick an arbitrary value for the variable corresponding to the third column and determine the corresponding values for the other two.

$$\left[ \begin{array}{ccc|c} 1 & 2 & 3 & 1\\ 1 & -1 & 1 & 1\\ \end{array}\right]\\ \left[ \begin{array}{ccc|c} 1 & 2 & 3 & 1\\ 0 & -3 & -2 & 0\\ \end{array}\right]\\ \left[\begin{array}{ccc|c} 1 & 0 & \frac53 & 1\\ 0 & -3 & -2 & 0\\ \end{array}\right]\\ \left[\begin{array}{ccc|c} 1 & 0 & \frac53 & 1\\ 0 & 1 & \frac23 & 0\\ \end{array}\right]\\ $$

Picking $z=0$ gives $(1,0,0)$. Picking $z=3$ gives $(-4,-2,3)$.

NovaDenizen
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