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Intersection Of Planes

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Wong | 13:29 Wed 20th Feb 2013 | Science
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Find an equation for the line of intersection of the following two planes.
x+2y-3z=2 and 2x+3y-5z=3

Many thanks.
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Step 1: Find the cross product of (1,2,-3) and (2,3,-5)
step 2: find a point that lies on both planes, substitute in and solve simultaneously
I have worked it out but it's very easy to slip up with all the positives and negatives
I got the equation as r = (0,1,0) +t(-1,-1,-1) but I'd need to check it when i have more time
Yes- if you choose a value for t and substitute it in you can then check that the resultant point lies on either plane.
eg if we use t=-2 then :
r = (0,1,0) - 2(-1,-1,-1) = (2,3,2)
Try this in x+2y-3z=2 and 2x+3y-5z=3
2+6- 6 does indeed equal 2 and 4+9 -10 does indeed equal 3.

It also works for all other values tried for t .
Yup that looks the right idea to me. The trick is to find a vector pointing in the direction of the line of intersection. This will lie inside both planes, and therefore has to be perpendicular to the normals of the two planes. Then you can find a vector normal to any two others by taking their cross product.

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