# Vectors in 3 space

• Dec 4th 2006, 05:45 PM
pakman
Vectors in 3 space
Find the equation of the plane through point (-1, -2, 3) and perpendicular to both the planes x - 3y + 2z = 7 and 2x - 2y - z = -3. Please help, final tomorrow and I'm confused on this. I know how to find the equation given 3 points but not with 2 planes. Thanks!
• Dec 4th 2006, 05:54 PM
ThePerfectHacker
Quote:

Originally Posted by pakman
Find the equation of the plane through point (-1, -2, 3) and perpendicular to both the planes x - 3y + 2z = 7 and 2x - 2y - z = -3. Please help, final tomorrow and I'm confused on this. I know how to find the equation given 3 points but not with 2 planes. Thanks!

If two planes are parrallel then their normals are parrallel.
These two plane normals are not parallel. *)
Thus, the planes are not parallel.
Assume you can draw a third plane (this problem) parallel to both.
Then the first one is parallel with third one.
And the second one is parallel with third one.
Thus, the first one is parallel with second one.

*)The vectors are not scalar multiples of each other.
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Do you mean that I should find two seperate planes, one parallel with first one and another parallel with second one. So they pass through the point (-1,-2,3).
• Dec 4th 2006, 06:00 PM
pakman
I copied the question directly from the book, it was a suggested problem to prepare for the final. The answer to the problem is 7x + 5y + 4z = -5. It says perpendicular though?
• Dec 4th 2006, 06:08 PM
pakman
Actually I just found the problem in my notes.

So we have the point is (-1, -2, 3) we need the normal vector

a = <1, -3, 2> and b = <2,-2,-1> derived from the plane equations provided

then use cross products...
v = a x b

I don't know how to show it using ASCII but it'll come out to (3-(-4))i - (-1-4)j + (-2-(-6))k = <7,5,4> = v = normal vector

plane equation is 7(x+1) + 5(y+2) + 4(z-3) = 0