Example # 1: Find the equation of a new plane that passes through the point A (3,-1 ,2) and is perpendicular to the line of intersection ofthe planes .7171: x+y+3z12=0 and .75: 7xy+3Z2=0 Solution: Consider the planes J11 and 3:2 with these equations; 111: 3xy+z2=O IZZZ x+2y4z+l=0 We can tell by inspection that the normal vectors of :31 and yrz are . Hence, these two planes have a . Suppose we combine their equations as follows. Multiply ft] by a scalar, s: S(3x y + z 2) = 0 Multiply .772 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 + 2 y 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 5): + 3y 7z 6 = 0 Solution