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The planes $5x + 3y + 2z = 0$ and $ 2x + 8y - 5z = 0$ intersect. Find the equation of the intersecting line.

I get the parametric equation:

$x = t$

y = $\frac{29}{34}t$

z = $\frac{-121}{170}t$

which I don't think is the right answer. Can someone walk me through the problem?

user127778
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  • What are the normal vectors for each plane? The cross-product of those vectors gives the vector direction for a line in both planes. Using a parametric form for that line (with the constants yet unknown), find a point on that line which must be in both planes. (There will be a "free variable", which you may choose to be zero.) This would then provide the necessary information to write the equation for the line of intersection. – colormegone Feb 14 '14 at 05:51
  • Wolfram Alpha agrees with your solution for $y$. – Daryl Feb 14 '14 at 05:52
  • Please check here for a more elegant solution: http://math.stackexchange.com/questions/475953/how-to-calculate-the-intersection-of-two-planes/1937116#1937116 – Tolga Birdal Sep 22 '16 at 15:21

2 Answers2

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An algebraic solution: solve the equations simultaneously. If you take the first and then $5$ times the first plus twice the second you get $$5x+3y+2z=0\quad\hbox{and}\quad 29x+31y=0\ .$$ Let $x=t$, solve for $y$ from the second equation, solve for $z$ from the first. As long as your arithmetic is accurate the answer will be right.

A geometric solution: find the normal vector to each plane; the line you want is perpendicular to both these vectors; so its direction can be found by using the cross product. You also need a point on the line, but the origin is clearly on both planes and is therefore a point on their line of intersection.

David
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  • Even if the planes didn't pass through the origin, you can always choose one of the coordinates for the point on the line to be zero, since it has to intersect any of the planes eventually (if there is a plane the line doesn't meet, that will be evident from the direction of the line). – colormegone Feb 14 '14 at 05:57
  • Your result does work in the second original equation, $ \ 2x + 8y - 5z = 0 \ $ , so it checks. Unfortunately, the parameterized values for the coordinates in the original post don't check in either equation... – colormegone Feb 14 '14 at 06:13
  • @RecklessReckoner Either the OP or my calculation is wrong. I think it's the OP (surprise) but as no working is given I can't say what happened. Can you see an error in how I found the second equation? If so please tell me ;-) Note that my equations both agree with the answer provided by Semsem. – David Feb 14 '14 at 07:38
  • I am agreeing with your result (and thus also Semsem's). – colormegone Feb 14 '14 at 07:59
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  1. Find a parallel vector to the line of intersection
    $ n_1=(5,3,2),n_2=(2,8,-5)\implies u=n_1×n_2=(-31,29,34)$
  2. Find a point of intersection by solving both equation, it is clear both pass through $(0,0,0)$
  3. The equation is $r(t)=(0,0,0)+t(-31,29,34)$
Semsem
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