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Find parametric equations for the line
$$x − 2y + 3z = 4, 2x + y − z = −2$$

I know that the first thing you do is cross the vectors perpendicular to the planes, but I'm having a little trouble finding the point of intersection - can I just set $x, y,$ or $z$ to $0$ and take it from there?

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You need a point to form the equation of the line. – Babak S. Dec 8 '12 at 8:15
That's what I'm having trouble finding. Can I just pick an arbitrary point? – Linda Dec 8 '12 at 8:16
Of course, you can not. – Babak S. Dec 8 '12 at 8:18
Wrong question to ask - I'm sleep deprived right now as finals are around the corner. I guess what I was asking in the question was for these two equations, can I just set x, y, or z to 0, and then solving for the two unknowns? I'm terribly confused. – Linda Dec 8 '12 at 8:21

It seems that you are looking for a point lying on the intersection of two planes which is our wanted line. Take $z=t$ into two equations, you get:

$$x-2y+3t=4, 2x+y-t=-2$$ Now eliminate the $y$ by every method you know:


Do the same for $x$:


So the points are as forms: $$\left(\frac{t}{-5},\frac{7}{5}t-2,t\right)$$

Take for example $t=0$....

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Oh man this helps so much.. thanks! I'm gonna do a couple more problems then head to bed, I'm sure I'm just too fatigued to figure things out right now – Linda Dec 8 '12 at 8:37

Points on the line that is the intersection of the two planes will satisfy the system of linear equations $$x-2y+3z=4\\2x+y-z=-2$$ which has as it's solution $$x=-t/5\\ y=-2+7t/5\\ z=t$$ where $t$ is a parameter which maps out the intersection line as it varies over $\mathbb{R}$.

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@GerryMyerson You are correct. – Daryl Dec 8 '12 at 20:25

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