Find the Cartesian equation of plane Π containing the points A(6 , 2 , 1) and B(3, -1, 1) and perpendicular to the plane Π2 (x + 2y - z - 6 = 0).
The information we are given to solve for the equation of the plane are the points A and B, as well as the equation of a plane which lies perpendicular to the plane Π we are solving for.Method 1: Since A and B both lie on the plane Π, the vector AB (-3 , -3 , 0) that connects them will be parallel to the plane Π. The plane Π2 which lies perpendicular to the plane Π has the normal vector (1 , 2 , -1). This vector must also be parallel to the plane Π, as it is the normal vector of a perpendicular plane. To find the Cartesian form of plane Π calculate the cross product between the vector AB and the normal vector of Π2. (-3 , -3 , 0) x (1 , 2 , -1) = (3 , -3 , -3). We now simplify this by dividing by the common factor 3 to find the equation x - y - z = d. To find d plug the co-ordinates of either point A or B into the equation to find d = 3 , hence x - y -z = 3.Method 2:Let plane Π be equal to ax + by + cz = dAs the normal vector to the perpendicular plane is (1 , 2 , -1) we know that:a + 2b - c = 0By plugging points A and B into the equation for Π we know that:6a + 2b + c = d3a - b + c = dWe now have three simultaneous equations with three unknowns, which can be solved for through row reduction to find a = d/3 b = -d/3 c = -d/3a = -b = -cBy assigning an appropriate value to any of the unknowns findx - y - z = 3 or an equivalent expression.
