anonymous
  • anonymous
a plane contains points p(4,0,0) q (0,5,0) and r(0,0,2) a - sketch the plane b - find a vector perpendicular to the plane
Mathematics
  • Stacey Warren - Expert brainly.com
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
I don't know about sketching it, but if you want a vector perpendicular to it you need to find two vectors, \(\vec u\) and \(\vec v\) that lie in the plane you can do that by finding the vectors between the various points with a common vertex
anonymous
  • anonymous
|dw:1337148180943:dw|so here are our three points if we can find\[\vec u=\overrightarrow{pq}=p-q\]and\[\vec v=\overrightarrow{qr}=r-q\]
anonymous
  • anonymous
|dw:1337148370560:dw|\[\vec u=p-q=\langle4,-5,0\rangle\]\[\vec v=r-q=\langle0,-5,2\rangle\]the cross product of these vectors will be normal to the plane

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anonymous
  • anonymous
\[\vec n=\vec u\times\vec v\]
anonymous
  • anonymous
|dw:1337148472550:dw|
anonymous
  • anonymous
yeah, that's a good graph :)
anonymous
  • anonymous
all that stuff I said still works though
anonymous
  • anonymous
wait.. \[pq = p-q?\] but then you have \[qr = r -q\] i'm confused here:(
anonymous
  • anonymous
typo, my bad I sort of invented my own notation there on the left
anonymous
  • anonymous
ohh okay so it's suppsoed to he pq = q-p, right?
anonymous
  • anonymous
you can do it many ways, but they need a common vertex the vector p-q points toward p from q the vector r-q points toward r from q so this should just be\[pq=p-q\]\[rq=r-q\]as in the distance from p and r to q, respectively
anonymous
  • anonymous
so then \[u x v = (-4 x 0, -5 x 5, 2 x 0) = < 0, -10, 0> ?\]
anonymous
  • anonymous
no that's not how you take a cross product...
anonymous
  • anonymous
oh wait.. so.. nope you lost me again lol i'm sorry..
anonymous
  • anonymous
so you're picking one point as the middle point, and then creating two vectors from it?
anonymous
  • anonymous
and then crossing them?
anonymous
  • anonymous
exactly :)
anonymous
  • anonymous
and if i'm finding a vector PR, then its r -p, right? second point, minus first point? o.O
anonymous
  • anonymous
I sort of am making up the notation... whether it is supposed to be written PR or RP I really don't know all that matters is that you're consistent in your choice of vertex
anonymous
  • anonymous
so, pr = < -4, 0, 2> pq = <-4, 5, 0> pr x pq = <-16, 0, 0> ?!
anonymous
  • anonymous
but when i'm finding any vector, like for example PR, then how would i find it? r - p, or p-r? just in general, not related to this problem
anonymous
  • anonymous
so getting the cross product of two vectors on a plane, gets me a vector perpendicular to the plane?
anonymous
  • anonymous
let me check you on that cross product... as far as what minus what is concerned, in a-b the vector points toward a so it is vector=(to point)-(from point)
anonymous
  • anonymous
...so if you want a vector that points from p to r you write r-p
anonymous
  • anonymous
okay! perfect :D thank you!!! one more question?
anonymous
  • anonymous
sure :)
anonymous
  • anonymous
find an equation of the plane containing points P Q and R
anonymous
  • anonymous
thankfully once you have the normal vector to the plane this is a piece of cake just put the numbers in the components of the normal vector as the coefficients
anonymous
  • anonymous
the normal vector\[\vec n=\langle P,Q,R\rangle\]implies the plane\[Px+Qy+Rz=d\]where \(d\) is some constant
anonymous
  • anonymous
and the normal vector was (16, 0 , 0)? so its 16x + 0 + 0 = d?
anonymous
  • anonymous
oh I didn't check, but that seems wrong...
anonymous
  • anonymous
hold on...
anonymous
  • anonymous
i crossed it in incorrectly lol hold on
anonymous
  • anonymous
yeah I got <2,-20> on the first try, but I'm tired so let me try again
anonymous
  • anonymous
sorry <2,0,-20>
anonymous
  • anonymous
but wait that way still be wrong...
anonymous
  • anonymous
really? pellet.. i got <10 8 20> =\
anonymous
  • anonymous
whats your pq and pr? =\
anonymous
  • anonymous
let me try again\[\vec u\times\vec v=\left|\begin{matrix}\hat i&\hat j&\hat k\\-4&5&0\\-4&0&2\end{matrix}\right|\]
anonymous
  • anonymous
this is what i got: (10-0)i - (-8-0)j +(0+20)k is that wrong? :(
anonymous
  • anonymous
\[\vec u\times\vec v=\left|\begin{matrix}\hat i&\hat j&\hat k\\-4&5&0\\-4&0&2\end{matrix}\right|=10\hat i+20\hat k-8\hat j\]ok now we agree :)
anonymous
  • anonymous
so that is\[\vec n\]
anonymous
  • anonymous
but.. but.. i got 10i + 8j + 20k, doesn't the - distribute to the 8? :(
anonymous
  • anonymous
or.. it doesnt??
anonymous
  • anonymous
my bad told you I'm tired
anonymous
  • anonymous
no worries!!! i totally understand hahha
anonymous
  • anonymous
so then i think i'm supposed to plug in the variables and a point so.. 10(x-4)+8(y-0)+20(z-0)=0?
anonymous
  • anonymous
10x +8y +20z =40?
anonymous
  • anonymous
yes, that's exactly what you do I know the method, I just can't execute it when I'm sleep deprived :P
anonymous
  • anonymous
looks right though
anonymous
  • anonymous
...of course you could have picked any point and you would have gotten the same d
anonymous
  • anonymous
and you can simplify it obviously too 5x +4y +10z =20
anonymous
  • anonymous
so the vector perpendicular to the plane is < 10, 8, 20>, and the equation of the plane containing points p q and r is 10x+8y+20z = -40?
anonymous
  • anonymous
isn't it positive? you said so yourself, and I agree
anonymous
  • anonymous
the 40 I mean
anonymous
  • anonymous
oops yeah hehe :D thats what i wrote down!
anonymous
  • anonymous
yayyyy!!!! i think i get it!!! thank you soooo much!!!!!!!!
anonymous
  • anonymous
welcome, and g'night :)
anonymous
  • anonymous
sleep well!~!!!!

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