Vector
posted by Gabe on .
Determine the caresian equaiton of a plane that pass through point P(1,2,2) and is perpendicular to the line (x,y,z)= 3,1,4 + t(2,3,1)

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+z6=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.