Question

Find an equation of the plane through the point (1, -5, 3) and perpendicular to the vector (2, -3, -3).

Answer 1

2x - 3y - 3z = 8
   We've been given the normal vector to the plane <2, -3, -3> and a point within the plane (1, -5, 3). In general if you've been given both the normal vector <a,b,c> and a point (e,f,g) within the plane, the expression for the plane will be: ax + by + cz = d
 and you can compute d by:
 d = ae + bf + cg
   So let's calculate d:
 d = ae + bf + cg
 d = 2*1 + -3*-5 + -3*3
 d = 2 + 15 + -9
 d = 8
   And the equation for the plane is 2x - 3y - 3z = 8