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anonymous
 4 years ago
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
anonymous
 4 years ago
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

This Question is Closed

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I 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
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1337148180943:dwso here are our three points if we can find\[\vec u=\overrightarrow{pq}=pq\]and\[\vec v=\overrightarrow{qr}=rq\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1337148370560:dw\[\vec u=pq=\langle4,5,0\rangle\]\[\vec v=rq=\langle0,5,2\rangle\]the cross product of these vectors will be normal to the plane

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\vec n=\vec u\times\vec v\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1337148472550:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yeah, that's a good graph :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0all that stuff I said still works though

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0wait.. \[pq = pq?\] but then you have \[qr = r q\] i'm confused here:(

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0typo, my bad I sort of invented my own notation there on the left

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ohh okay so it's suppsoed to he pq = qp, right?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0you can do it many ways, but they need a common vertex the vector pq points toward p from q the vector rq points toward r from q so this should just be\[pq=pq\]\[rq=rq\]as in the distance from p and r to q, respectively

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so then \[u x v = (4 x 0, 5 x 5, 2 x 0) = < 0, 10, 0> ?\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0no that's not how you take a cross product...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0oh wait.. so.. nope you lost me again lol i'm sorry..

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so you're picking one point as the middle point, and then creating two vectors from it?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0and then crossing them?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0and if i'm finding a vector PR, then its r p, right? second point, minus first point? o.O

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I 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
 4 years ago
Best ResponseYou've already chosen the best response.0so, pr = < 4, 0, 2> pq = <4, 5, 0> pr x pq = <16, 0, 0> ?!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0but when i'm finding any vector, like for example PR, then how would i find it? r  p, or pr? just in general, not related to this problem

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so getting the cross product of two vectors on a plane, gets me a vector perpendicular to the plane?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0let me check you on that cross product... as far as what minus what is concerned, in ab the vector points toward a so it is vector=(to point)(from point)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0...so if you want a vector that points from p to r you write rp

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0okay! perfect :D thank you!!! one more question?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0find an equation of the plane containing points P Q and R

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0thankfully 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
 4 years ago
Best ResponseYou've already chosen the best response.0the normal vector\[\vec n=\langle P,Q,R\rangle\]implies the plane\[Px+Qy+Rz=d\]where \(d\) is some constant

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0and the normal vector was (16, 0 , 0)? so its 16x + 0 + 0 = d?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0oh I didn't check, but that seems wrong...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i crossed it in incorrectly lol hold on

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yeah I got <2,20> on the first try, but I'm tired so let me try again

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0but wait that way still be wrong...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0really? pellet.. i got <10 8 20> =\

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0whats your pq and pr? =\

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0let 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
 4 years ago
Best ResponseYou've already chosen the best response.0this is what i got: (100)i  (80)j +(0+20)k is that wrong? :(

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\vec u\times\vec v=\left\begin{matrix}\hat i&\hat j&\hat k\\4&5&0\\4&0&2\end{matrix}\right=10\hat i+20\hat k8\hat j\]ok now we agree :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0but.. but.. i got 10i + 8j + 20k, doesn't the  distribute to the 8? :(

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0my bad told you I'm tired

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0no worries!!! i totally understand hahha

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so then i think i'm supposed to plug in the variables and a point so.. 10(x4)+8(y0)+20(z0)=0?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yes, that's exactly what you do I know the method, I just can't execute it when I'm sleep deprived :P

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0...of course you could have picked any point and you would have gotten the same d

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0and you can simplify it obviously too 5x +4y +10z =20

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so 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
 4 years ago
Best ResponseYou've already chosen the best response.0isn't it positive? you said so yourself, and I agree

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0oops yeah hehe :D thats what i wrote down!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yayyyy!!!! i think i get it!!! thank you soooo much!!!!!!!!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0welcome, and g'night :)
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