Vectors and the Geometry of Space

Question

Find an equation of the plane through the point (4, 0, 1) and parallel to the plane −1x−1y+3z=5.

x + y + z =

Answer 1

Find an equation of the plane through the point (4, 0, 1) and parallel to the plane −1x−1y+3z=5.

SOLUTION

Lets call the given plane ( Π₁ ) : −1x−1y+3z=5.

The normal vector of ( Π₁ ) is n = < -1, -1, 3 >

Since the unknown plane , call it ( Π₂) , is parallel to ( Π₁ ) they must

have the same normal vector.

Hence the equation of ( Π₂) is as follows (-1)( x -4 ) + (-1)( y -0 ) +3( z -1) = 0

in other words -x - y +3z = -4 +3 Finally -x - y +3z = - 1