Type IV: Three Planes Intersect in a Line One normal vector is a of the other 2 normals and the also satisfy this same linear combination. All 3 normals are still coplanar. H4 \"1; 2xy+322=0 m=[2,1,3] ie: :14: x3y+22+10=0 X4=[1,3,2] its: 5x5y+82+6=0 n6=[5,5,8] The normal vector of .7176 , [5,-5,8], is a linear combination of the other two normal vectors: . The planes intersect in a single line because the equations satisfy this same linear combination. That is Example # 1: Show that the following three planes intersect at a line and nd the parametric equations of that line of intersection. -> m: 3x2y+z1=O n1=[3,2,1] n2: x+3y22+7=0 22=[1,3,2] _. 113210x3y+z+4=0 n3=[10,3,l] Type V: Three Planes Intersect at a Single Point This is the only situation out of the 5 types where the normal vectors are . It is not possible to express one of the normal vectors as a linear combination of the other two normals. To determine if there is just one point of intersection, use the check for coplanar vectors. If then the normals are not coplanar and there is a single point of intersection. For Type IV & V only, the set of 3 equations for planes because all 3 planes have a common intersection. These systems of equations are called Example # 2: Show that the following 3 planes intersect at a single point and then determine the coordinates of the point of intersection. .71'12 x+y+z+5=0 ;1=[l,1,l] J72: x+2y+32+4=0 ;2=[l,2,3] 1:3: x+2y+z+7=0 n3=[1,2,l] Solution