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I completed this problem in which I have to find the intersection of two planes. How is it that I can assume the point of intersection is at z=0? May someone please elaborate.

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  • $\begingroup$ Please take also a look to mathjax tutorial to present your derivation in a proper way. Thanks $\endgroup$ – user Sep 28 '19 at 22:43
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    $\begingroup$ Sorry man, first question ever posted on Mathematics. $\endgroup$ – Hector Sep 28 '19 at 22:44
  • $\begingroup$ That's fine, your work looks very good and well presented. Using mathjax is also very important for the quality of the site. $\endgroup$ – user Sep 28 '19 at 22:46
  • $\begingroup$ Actually, you are guessing that there exists a point in the intersection with z=0. If there is no such point, then this method fails, in which case you can make another guess, e.g. x=0, y=0. Note, that any guess is possible, e.g. z=4. $\endgroup$ – tommsch Feb 6 at 9:23
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In that case the intersection is a line not parallel to $x-y$ plane, indeed $\vec \alpha =(3,-13,-5)$, then we can assume any value for $z$ to define the parametric equation.

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  • $\begingroup$ How can you tell from that direction vector? What about it tells you that it is not parallel to the x-y plane? What would it be if it were parallel to the x-y plane? $\endgroup$ – Hector Sep 28 '19 at 22:46
  • $\begingroup$ @Hector A vector parallel to the $x-y$ plane would be in the form $(a,b,0)$. $\endgroup$ – user Sep 28 '19 at 22:47

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