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Which pairs of planes are parallel and distinct and which are conincident? 2x+4y-7z-2=0 4x+6y-14z-8=0

Mathematics
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I tried looking through my notes and textbook for this one, but unless there is a direct connection, I don't understand how to apply it to this question.
I managed to answer some unrelated questions beforehand, but I don't know how I managed to solve them...
So I figured out the normals were [2,3,7] and [4,6,-14]

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Other answers:

I can see that I just have to multiply 2,3,7 by 2 to get the other plane. So would that be conidental or parallel and distinct?
that should be parallel and distinct.. if u can just multiply them by a scalar, but i suck at this, so idk.. @Mertsj @Zarkon
Wait. Maybe I'll try something out here.
how is [2,3,7] the normal vector?
Er, I just looked at my notes and inferred that...
if 2x+4y-7z-2=0 is your equation then <2,4,7> is the normal vector
<2,4,-7>
Zarkon to the rescue!!!!
Ok, got that. For the other then, it must be 4,6,-14?
yes
So from this, how would I determine the answer? Or do I need more?
if the planes are parrallel then the cross product of their normal vectors should be 0
two plane are parallel iff their normal vectors are parellel
I'll try that cross product thing. Thanks guys. I'lll cry out if I have a problem.
two non zero vectors \(\vec{v}\) and \(\vec{u}\) are parallel iff \(\vec{v}=c\vec{u}\) for some scalar \(c\)
Pardon?
is one of the vectors a constant multiple of the other?
The first one, multiplied by 2, equals the second one.
really?
IT seemed so...
if you have the equations 2x+4y-7z-2=0 4x+6y-14z-8=0 then you have the normal vectors <2,4,-7> and <4,6,-14> are these constant multiples of each other?
That 4y was suppose to be a 3y....
I did mistype that. No wonder my confusion. And probably yours...
then <4,6,-14>=2<2,3,-7> and so they are parallel...and therefore their planes are parallel
And distinct??
yes
Awesome. Thank you. This was relatively simple after all...

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