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December 19, 2014

December 19, 2014

Posted by **Gabe** on Sunday, May 30, 2010 at 5:43pm.

- Vector -
**MathMate**, Sunday, May 30, 2010 at 5:58pmIf 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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