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!
 A: 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.
A: 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 $t$
