Linear Combinations of Equations of Planes Innitely many planes pass through a given line in space. In problems involving a plane passing through the line of intersection of two planes, the following approach is very effective. Consider the planes .727] and .'z with these equations; .7712 3xy+z2=0 :72: x+2y4z+1=0 We can tell by inspection that the normal vectors of It] and JZ'2 are . Hence, these two planes have a . Suppose we combine their equations as follows. Multiply ill by a scalar, s: 3(3): y + z 2) = 0 Multiply J12 by a scalar, t: t(x + 2 y - 42 + 1) = 0 Example # 2: Find the equation of the plane passing through the line of intersection of the planes 3x y + z 2 = 0 and x + 2y 42 +1 = 0, and satises the given conditions a) The plane passes through he point A (3,1,3) b) The plane is also parallel to the plane 5x + 3 y 7z 6 = 0 Solution: Three planes can interact in . Today we will be looking at Type I III and tomorrow we will look at Type IV V. Type I: Three Parallel Planes When 3 planes are parallel, the 3 normals are \".3 1123 v n2 n1 1!] n. V2 _ n3 The equations below represent this situation. We can tell this because all three normal vectors [3,-2,1] , [9,-6,3] > > > > and [6,-4,2] are collinear. That is, 3 m = m and 2 m = n 3. The planes are because their equations do not satisfy these relationships. The planes are distinct, separated because their constant term does not follow the same scalar multiples as the normals