For the following two (02) items: A plane P is parallel to the line having direction ratios 1, 3, 2 and contains the line of intersection of the planes 6x + 4y - 5z = 2 and x - 2y + 3z = 0 . What is the equation of the plane P ?

Step-by-step solution

The equation of the plane passing through the line of intersection of the given planes is (6x + 4y - 5z - 2) + (x - 2y + 3z) = 0 (6 + )x + (4 - 2 )y + (-5 + 3 )z - 2 = 0 Since this plane is parallel to the line with direction ratios 1, 3, 2 , the normal to the plane is perpendicular to the line. Thus, 1(6 + ) + 3(4 - 2 ) + 2(-5 + 3 ) = 0 6 + + 12 - 6 - 10 + 6 = 0 + 8 = 0 = -8 Substituting = -8 in the equation of the plane, we get (6 - 8)x + (4 - 2(-8))y + (-5 + 3(-8))z - 2 = 0 -2x + 20y - 29z - 2 = 0 2x - 20y + 29z