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Find the equation for the line of intersection of the planes.

5x - 2y - 2z = 1, 4x + y + z = 6

So I know how to solve this equation:

  1. Cross the normal vectors

  2. Solve for intersection points

  3. plug in to equation.

When solving for part 2, I get the following system:

5x - 2y - 2z = 1,

4x + y + z = 6

set z=0

And voila! I get values through solving the system with the z variable removed by z=0.

Question is, why can I set z=0? It doesnt make sense to me.

What other variables can I "make up" to solve these systems of equations?

2 Answers 2

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multiplying your second equation by $2$ and adding to the first we get $$13x=13$$ thus $$x=1$$ and we get now $$y+z=2$$ and from here we obtain $$y=2-z$$ Setting $$z=t$$ we get the equation of the line as $$x=1,y=2-t,z=t$$ with a real number $t$

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    yeah, but why do you set z=t? – 2017-02-15
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    to have a Parameter in your equation you don't need that – 2017-02-15
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Just solve this system: \begin{cases} 5x - 2y - 2t = 1\\ 4x + y + t = 6 \end{cases} with parameter $t$, or \begin{cases} 5x - 2y = 1+ 2t\\ 4x + y = 6-t \end{cases} then $$x=\frac{13}{13}=1,\,y=2-t$$ and the line equation is $$x=1,\,\dfrac{y-2}{-1}=z$$