through the point (4,-3.2) and is parallel to the line of intersection of the planes: i) 2x +3y + 2z = 6 and ii) 4x - 3y + z =11
pre.s. i pologise if this does n't belong to this forum and would appreciate your help directing my question to the right forum in this case
Points P(0,1,3/2), Q(2,0,1), Z(1,4/3,0) are on the first plane.
vector PQ is <2,-1,-1/2> and QZ is <-1,4/3,-1> by forming their cross product I find the normal to the first plane to be <1,3,2> .
Points P'(0,1,14), Q'(5/2,0,1), Z'(1,-7/3,0) are on the second plane.
vector P'Q' is <5/2,-1,-13> and Q'Z' is <-3/2,-7/3,-1> by forming their cross product I find the normal to the second plane to be <-4,3,-1>
By forming the cross product of the normals I find the direction vector of the line i am interested in to be <3,1,-9>
and hence the cartesian equation of the wanted line to be
(x-4)/3 = y + 3 = (z-2)/-9
However the book gives (x-4)/3 = (y + 3) = (z-2)/-6
I have gone through it a miriad times and can't find my mistake I would absolutely appreciate it if I can get some help