# Find intersection of two planes

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.

• Please take also a look to mathjax tutorial to present your derivation in a proper way. Thanks – user Sep 28 '19 at 22:43
• Sorry man, first question ever posted on Mathematics. – Hector Sep 28 '19 at 22:44
• That's fine, your work looks very good and well presented. Using mathjax is also very important for the quality of the site. – user Sep 28 '19 at 22:46
• 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. – tommsch Feb 6 at 9:23

## 1 Answer

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.

• 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? – Hector Sep 28 '19 at 22:46
• @Hector A vector parallel to the $x-y$ plane would be in the form $(a,b,0)$. – user Sep 28 '19 at 22:47