Find the general form of the equation of the plane with the given characteristics. The plane passes through the points (5,5,-2)and (4,2,1)and is perpendicular to the plane -3x - 2y + 4z = 1.
A)6x + 5y + 7z - 41 = 0
B)3x + 2y - 4z + 33 = 0
C)6x - 5y + 7z - 9 = 0
D)3x + 2y - 4z = 0
E)7x + 6y + 5z - 55 = 0

Question

Quizplus · Accepted Answer

The normal vector of the given plane is (-3, -2, 4). A plane perpendicular to this one will have a normal vector parallel to (-3, -2, 4). The equation of the plane can be found using the point-normal form, and since it passes through (5,5,-2), substituting these values into the general form of the plane equation $ax + by + cz = d$ gives us the equation 6x + 5y + 7z - 41 = 0 after solving for the constants that satisfy both the direction being perpendicular and passing through the given points.

Find the General Form of the Equation of the Plane