Posted by Gabe on Sunday, May 30, 2010 at 5:43pm.
If the plane is perpendicular to
L: (3,-1,-4)+t(-2,3,1)
in which the vector (-2,3,1) is parallel to the line, and therefore perpendicular to the plane P, then
the required plane is:
-2x+3y+z+k=0
where k is a constant to be determined.
Since the required plan passes through P(1,2,2), we substitute P in the plane to get:
-2(1)+3(2)+(2)+k=0
or k=-6
Therefore the plane is given by:
-2x+3y+z-6=0
Check by joining any two distinct points on the plane to form a vector and prove that the vector is perpendicular to line L.
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