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I'm new to this forum but I've done my best to check that my question hasn't been answered before, but I may have missed something, feel free to correct me if that's the case.

Anyway, I'm trying to figure out how to solve this problem:

"Determine the equation on the parameter form for the line that represents the intersection of the two planes" $$x+2y+2z=5$$ $$2x-y+2z=2$$

I am unsure of how to even begin thinking.

Thanks!

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You can simply solve this as an algebraic system of two linear equations in the three unknowns. There will be one free variable, so you can introduce a parameter. Let $z=t$, then solve: $$\left\{ \begin{array}{rcl} x+2y&=&5-2t \\ 2x-y&=&2-2t \end{array}\right.$$ for $(x,y)$. The solution set will be in the form of a parametric representation of the line.


You can solve the system with methods of your choice. For example:

  • add twice the second equation to the first equation to get: $5x=9-6t$;
  • subtract twice the first equation from the second to get: $-5y=-8+2t$;

which gives you: $$\left\{ \begin{array}{rcl} x &=& \tfrac{9}{5}-\tfrac{6}{5}t \\ y &=& \tfrac{8}{5}-\tfrac{2}{5}t \\ z &=& t \end{array}\right. \quad \quad (t \in \mathbb{R})$$ Notice that this has indeed the parametric form of a line.

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  • $\begingroup$ Thank you very much! That really cleared it up for me. $\endgroup$ – SmhConfused Dec 7 '16 at 20:20
  • $\begingroup$ Alright, you're welcome! $\endgroup$ – StackTD Dec 7 '16 at 20:28
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multiplying the second equation by $2$ and adding both we get $$5x+6z=9$$ from here we get $$x=\frac{9}{5}-\frac{6}{5}z$$ setting $$z=5t$$ with a real number $t$ we get for $y$ $$y=\frac{8}{5}-2t$$ and our straight line is given by $$x=\frac{9}{5}-6t$$ $$y=\frac{8}{5}-2t$$ $$z=5t$$ with a real number $t$

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