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June 26, 2016
Posted by **Nicole** on Wednesday, July 9, 2008 at 5:32pm.

x − y + 3z = 4

x + y + 2z = 2

3x + y + 7z = 9

- Calculus -
**Reiny**, Wednesday, July 9, 2008 at 10:31pmadding #1 and #2 I got 2x + 5z = 6

adding #1 and #3 I got 4x + 10z = 13

You should realize that we cannot solve this any further, since both x and z would drop out

so lets find the line of intersection of plane #1 and #2

from 2x + 5z = 6

let z=2, then x = -2 and from #2, y = 0

let z=4, then x = -7 and from #2 y = 1

so we have two points on the line of intersection of the first two planes, namely

(-2,0,2) and (-7,1,4) from which we can get a direction vector (5,-1,-2)

and parametric equations of that line could be

x = -2+5t

y = -t

z = 2-2t

subbing that into the third equation 3x + y + 7z = 0 we get

-6 + 15t - t + 14 - 14t = 9

8=9 which is a contradiction.

Had this been a TRUE statement then all 3 planes would have intersected in the same line, like the spine of an open book

but the statement was FALSE, so the 3 planes must intersect in 3 different but parallel lines,

to find the other two line equations follow my procedure above for the other two possible pairs of planes.

As a check they should all have the same direction numbers.

BTW, this question is the hardest of the possible types of intersection of 3 planes.