```Question 206224
Find an equation of the line intersection of the two planes:
x+2y-z = -1 and 3x-2y+5z = 5
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The vector v1 = (1,2,-1) is normal to x+2y-z = -1.
The vector v2 = (3,-2,5) is normal to 3x-2y+5z = 5.
These vectors are not parallel, so the planes do intersect.
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The cross product gives a vector that's perpendicular to both above.
v3 = v1Xv2 = i - j - k (Cross product of v1 and v2), or (1,-1,-1)
v3 gives the direction of the line of intersection.
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Next, we need a point that is on both planes.
+x+2y-z = -1
3x-2y+5z = 5
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Set one dimension to zero, and get 2 linear eqns:
x=0
2y - z = -1
2y -5z = -5
----------- Subtract 2 from 1
4z = 4
z = 1
y = 0
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So the point with the position vector (0,0,1) lies on the line of intersection.
The eqn of the line of intersection is (0,0,1) + t(1,-1,-1)
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To check a few points:
+x+2y-z = -1
3x-2y+5z = 5
t = 1 --> (1,-1,0) is on both planes.
t = 4 --> (4,-4,-3) is also on both planes.
2 points is sufficient to fix a line, so it's correct.

```