Line of intersection between two planes?

1. 3x - 5y + 5z = -41

2. 5x + 3y + 2z = -9

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Multiply first equation by -2, and second equation by 5. Then add:

-6x + 10y - 10z = 82

25x + 15y + 10z = -45

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19x + 25y = 37

25y = 37 - 19x

y = 37/25 - 19/25 x

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Multiply first equation by 3, and second equation by 5. Then add:

9x -15y + 15z = -123

25x + 15y + 10z = -45

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34x + 25z = -168

25z = -168 - 34x

z = -168/25 - 34/25 x

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Now, if we let x = t, then we get

y = 37/25 - 19/25 t

z = -168/25 - 34/25 t

Vector parametric equation of line: < t, 37/25 - 19/25 t, -168/25 - 34/25 t >

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Note 1: We can find a point on the line that has integer coordinate if we let t = -2

x = 0 - 2 = -2

y = 37/25 -2(-19/25) = 75/25 = 3

z = -168/25 -2(-34/25) = -100/25 = -4

Note 2: If v = direction of line, then so is k*v, where k is a constant.

So 25 * < 1, -19/25, -34/25 > = < 25, -19, -34 > is also a direction vector of line

Note 3: We can also find direction vector of line by taking cross product of normals of the two planes:

<5, 3, 2> x <3, -5, 5> = <25, -19, -34>

So an alternate equation to line is: < -2 + 25t, 3 - 19t, -4 - 34t >

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Check: when t = 0, and t = 1, we get points (-2, 3, -4) and (23, -16, -38)

Plugging both these values for x,y,z into the two equations of the planes, we find that both these points are on both planes. Therefore they are both on the line that is the intersection of the planes.