EXAMPLE 3: Find an equation of SOLUTION: Let M be the required plane and (a, b, c) be a normal vector of the plane perpendicular to each of M. Let M, be the plane having equation x- y + z = 0. By Theorem 18.4.3 a the planes x - y + z = 0 and 2x + y normal vector of M, is (1, -1, 1). Because M and M, are perpendicular, it - 4z-5 = 0 and containing the follows from Eq. (7) that point (4, 0, -2). (a, b, c) . (1, -1, 1) =0 or, equivalently, 4-b+c=0 (8) Let M, be the plane having the equation 2x + y -4z -5=0. A normal vector of M, is (2, 1, -4). Because M and My are perpendicular, we have (a, b, c) . (2, 1, -4) = 0 or, equivalently, 2a + b - 40 = 0 (9) 834 VECTORS IN THREE-DIMENSIONAL SPACE AND SOLID ANALYTIC GEOMETRY Solving Eqs. (8) and (9) simultaneously for b and c in terms of a, we get b = 2a and c = a. Therefore, a normal vector of M is (a, 2a, a). Because (4,0, -2) is a point in M, it follows from Theorem 18.4.2 that an equation of M is a(x - 4) + 2a(y -0) + a(z+ 2) =0 or, equivalently, x + 2y +z-2=0