Find an equation of the plane through the point and perpendicular to the given line?

Find an equation of the plane through the point and perpendicular to the given vector.?

• Find an equation of the plane through the point and perpendicular to the given vector. (6, 1, 3) ‹-5, 1, 4›

• Answer:

  All planes normal to <-5,1,4> have the form -5 x + y + 4 z = constant (this you can see by considering the plane through the origin (constant = 0). Then any point in the plane has position vector <x, y, z>. The inner product with the normal vector then indeed equals zero: <x, y, z> dot <-5, 1, 4> = -5x + y + 4z = 0 To accomodate the point (6 , 1, 3) we need the constant to be -30 + 1 + 12 = -17 So 5 x - y - 4 z = 17