Posted by Soroban on September 20, 2002 at 14:59:46:
In Reply to: Lines and Planes in Space posted by Kearney on September 19, 2002 at 13:07:36:
: Find an equation of the plane.
: The plane pass through the points (2,2,1) and (-2,1,-1) and is perpendicular to the plane 2x - 3y + z = 3.
You know the Equation of a Plane: a(x-x1) + b(y-y1) + c(y-y1) = 0.
We need a point (x1,y1,z1), and the normal to the plane [a,b,c].
We have TWO points, but we need the normal direction.
The vector through the two points is: V = [4,1,2].
The normal of the given plane is: n = [2,-3,1].
I drew a rough sketch and saw that the normal we want is
perpendicular to vector V and normal n.
The cross-product Vxn is perpendicular to both V and n: N = [1,0,-2].
(I checked this: the dot products v*N and n*N equal 0.)
There ~~ just drop the values into the formula!
I got: x - 2z = 0.
Interesting - parallel to the y-axis. I didn't see that coming.
Is there an early-warning system available? Anyone?
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