Find the Equation of the plane

Question

Find the equation of the plane which contains the point ( 2/3, −1, 3) and has normal
vector 3i − 2j + 9k. Also, find another two points on this plane which are different
from ( 2/3, −1, 3).

Answer 1

Let v be the vector < 3, -2, 9 >.  v is perpendicular to the plane.

 

Let (x,y,z) be a point in the plane.

 

Then w = < x-2/3, y+1, z-3 > is a vector parallel to the plane.

 

So, an equation of the plane is v • w = 0

 

      3(x-2/3) - 2(y+1) + 9(z-3) = 0

 

Simplifying, we get 3x - 2y + 9z = 31

 

To find points in the plane, plug in values for x and y and then solve for z.  For example, if x = 1 and y = -1, then 5 + 9z = 31.  So, z = 26/9.

 

So, the point (1, -1, 26/9) is a point in the plane.