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Consider the planes:

$$\begin{eqnarray}x + y + 2z = 1\\ −x + 2y + z = 5\end{eqnarray}$$

  1. Find the angle between the two planes.
  2. Find the line of intersection of the two planes.

I was able to answer part (1), and I found that the angle was $60$ degrees ($\pi/3$ radians). The part that I'm having trouble with is finding the line of intersection of the two planes. I have no idea where to start, any help would be much appreciated.

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Adding both equations we get $$3y+3z=6$$ or $$y=2-z$$ substituting $z=t$ we obtain $$y=2-t$$ and $$x=2-t+2t=1$$ gives us $$x=-1-t$$ So our line has the equation $$[x,y,z]=[-1,2,0]+t[-1,-1,1]$$ where $t$ is a real number.

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  • $\begingroup$ Thanks for the help. The only part I don't understand is how you came up with the second last line. I'm a bit confused. $\endgroup$ – N_Mathematics_B Sep 25 '19 at 8:35
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    $\begingroup$ @N_Mathematics_B The usual method for this question: let one of the variables be a "parameter" (that you can rename $t$), then solve the system for the other two variables: you get a parametric equation of the line. Here $z$ is the parameter, so the parametric equation is $(x=f(t), y=g(t), z=t)$, with linear functions $f$ and $g$. $\endgroup$ – Jean-Claude Arbaut Sep 25 '19 at 8:38
  • $\begingroup$ I understand now, thanks for the help. $\endgroup$ – N_Mathematics_B Sep 25 '19 at 8:39
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As an alternative, the normal vectors to the planes are:

  • $n_1=(1,1,2)$
  • $n_2=(-1,2,1)$

then a direction vector for the line is

$$n_1 \wedge n_2 = (-3,-3,3)$$

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    $\begingroup$ +1 Interestingly, normal vectors also give very easily an answer to the first question. $\endgroup$ – Jean-Claude Arbaut Sep 25 '19 at 8:36
  • $\begingroup$ @Jean-ClaudeArbaut Yes, that's a very good point! $\endgroup$ – user Sep 25 '19 at 8:50

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