anonymous
  • anonymous
help please. coplanarity.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
All collinear points are also coplanar.
anonymous
  • anonymous
k?
anonymous
  • anonymous
I was referring to your previous question which you deleted for no reason @isuckatmath9999

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anonymous
  • anonymous
given [1,0,0] and [0,1,1], how would I check if a third point were coplanar? ie. 2,-3,2 or 5,-1,-1
dan815
  • dan815
hahaa do u like my statement
anonymous
  • anonymous
Take it away @dan815
dan815
  • dan815
hi okay since u arent given a plane equation or 2 vectors u can only see if these points are in the sme line
anonymous
  • anonymous
and yet i still have to
dan815
  • dan815
oh okay i see
dan815
  • dan815
write an equation for the line
anonymous
  • anonymous
@isuckatmath9999 two points in space determine a line, so find the equation of the line passing through those points.
dan815
  • dan815
|dw:1371021340450:dw|
dan815
  • dan815
2,-3,2 or 5,-1,-1 so lets see if we can find a t that satisfies either of these points
dan815
  • dan815
nope cant find one
anonymous
  • anonymous
In fact, you could also just find the vector between those two points, normalize it, and see that that the vector from one of your points to your 3rd is just a scalar multiple... let your points be P(1,0,0), Q(0,1,1), and R(2,-3,2):$$\vec{PQ}=(-1,1,1)\\\vec{PR}=(1,-3,2)$$Does \(\vec{PR}\) look like a scalar multiple of \(\vec{PQ}\)? Find if their components ratios are equal:$$\frac{-1}1=-1\\\frac1{-3}=-\frac13\\\frac12=\frac12$$... clearly not :-)
anonymous
  • anonymous
how about 1, 1, -4?
anonymous
  • anonymous
Ignore the normalizing part, it's not needed what so ever :-) hypothetically you could also normalize your vector \(\vec{PQ}\) and then compute its dot product with \(\vec{PR}\) and see if it corresponds to \(\|\vec{PR}\|\)
dan815
  • dan815
i thought u cant do this with normal because u can only form 2 vectors given the 3 points... so no matter what u will find a normal that will give u a plane containing those 2 pts, so u gotta do it that line method way
dan815
  • dan815
unless they are parallel
dan815
  • dan815
oh i see what u mean, thats a cool way
dan815
  • dan815
so if cross = 0, u can say they are linear
anonymous
  • anonymous
@isuckatmath9999 letting R(1,1,-4) we find:$$\vec{PR}=(0,1,-4)\\\|\vec{PQ}\|=\sqrt{(-1)^2+1^2+1^2}=\sqrt3\\\hat{PQ}=\frac1{\sqrt3}\left(-1,1,1\right)\\\hat{PQ}\cdot\vec{PR}=\frac1{\sqrt3}\left((-1)(0)+1(1)+(-4)(1)\right)=\frac{-3}{\sqrt3}=-\frac1{\sqrt3}\\\|\vec{PR}\|=\sqrt{0^2+1^2+(-4)^2}=\sqrt{17}$$... so clearly they're not collinear :-)
anonymous
  • anonymous
|dw:1371022164330:dw|
anonymous
  • anonymous
3,6,0
anonymous
  • anonymous
If all are collinear, we'd find that \(\vec{PR}\cdot\hat{PQ}=\|\vec{PR}\|\) :-)
anonymous
  • anonymous
$$\vec{PR}=(2,6,0)\\\vec{PR}\cdot\hat{PQ}=\frac1{\sqrt3}\left(-2+6\right)=\frac4{\sqrt3}\\\|\vec{PR}\|=\sqrt{2^2+6^2}=\sqrt{40}=2\sqrt{10}$$... nope!
anonymous
  • anonymous
3 6 0 lol
anonymous
  • anonymous
I just did that... I'm not interesting in doing your homework for you any further, especially if you can't understand my work after I explained it in detail.
anonymous
  • anonymous
\(\vec{PR}=\vec{R}-\vec{P}=(3,6,0)-(1,0,0)=(2,6,0)\) btw
anonymous
  • anonymous
OOH sorry I totally misread that. thank you

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