Question

[Calculus:Systems of equations:Solutions to square systems]

I can't understand the 3 minutes (43:00 to 46:00) in the folowing video.
https://youtu.be/YBajUR3EFSM?t=2586

In Homogeneous equation Ax=0, if the det(A) ≠ zero, then N1,N2,N3 -of the three planes- are coplanar, and the solution is a normal vector to the plane contains N1,N2,N3. The question is why that vector has to be contained in the three planes?! The professor says it's because that vector goes through the origin. Ax=0 tell us that x has to be normal to the plain contains N1,N2,N3, but why passing through the origin?!

Answer 1

@mathmale

Answer 2

the equation of a plane can be written as
N1 * X = c
where X represents the point <x,y,z> and * means dot product (only here!) and "c" is some constant

for example if N1= <2,3,1>
we have the equation
2x+3y+z = c

notice the point (0,0,0) is not a solution (i.e. is not contained in the plane)
unless c=0

but in your case
Ax= 0
all 3 planes contain the origin

Answer 3

also, at the end of Lecture 2, you are told the fact that
\[ det(A,B,C) = \vec{A} \cdot ( \vec{B} \times \vec{C})\]

B cross C generates a vector "D" perpendicular to both B and C i.e. there is a plane that contains both B and C and "D" will be normal to that plane
thus for get(A,B,C)=0
we must have
\[ A\cdot D = 0\]
and A is perpendicular to D, and must also lie in the same plane that contains B and C

Here, we assume all the vectors have their tail at 0,0,0

Answer 4

Thank you, I haven't thought about that way, now I can see it.