Find conditions on the variable “k” under which the intersection of the given set of planes is:
a) a point
b) a line
c) does not exist

Plane 1: kx – y +z = -3
Plane 2: x + y – kz= 1
Plane 3 : 3x + 2y – 2z = 0

Question

Find conditions on the variable “k” under which the intersection of the given set of planes is:
a) a point
b) a line
c) does not exist

Plane 1: kx – y +z = -3
Plane 2: x + y – kz= 1
Plane 3 : 3x + 2y – 2z = 0

anonymous · Answer

b) determinant $$\left[\begin{matrix}k &-1&1 \ 1 & 1&-k \ 3&2&-2\end{matrix}\right]=0$$
It happens for k=1 and k=-3/2

poofypenguin · Answer

ummm... i'm not quite sure what your solution means?

anonymous · Answer

if the planes intersect in a line, it means that their perpendicular vectors all belong to one plane. So it means this vectors are dependent. That's why determinant formed with them have to be equal 0

poofypenguin · Answer

i'm unframiliar with the term determinant... my school does not include matix in the curiculum...

anonymous · Answer

then look here, maybe helps... http://answers.yahoo.com/question/index?qid=20080825150238AAe2UdK

poofypenguin · Answer

i've got those basics already... i'm just not sure how to solve for k