# Equation on parametric form

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!

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.

• Thank you very much! That really cleared it up for me. – SmhConfused Dec 7 '16 at 20:20
• Alright, you're welcome! – StackTD Dec 7 '16 at 20:28

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 $t$