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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
 one year ago
 one year 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
 one year ago
 one year ago

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colorfulBest ResponseYou've already chosen the best response.1
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
 one year ago

colorfulBest ResponseYou've already chosen the best response.1
dw:1337148180943:dwso here are our three points if we can find\[\vec u=\overrightarrow{pq}=pq\]and\[\vec v=\overrightarrow{qr}=rq\]
 one year ago

colorfulBest ResponseYou've already chosen the best response.1
dw: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
 one year ago

colorfulBest ResponseYou've already chosen the best response.1
\[\vec n=\vec u\times\vec v\]
 one year ago

itzmashyBest ResponseYou've already chosen the best response.0
dw:1337148472550:dw
 one year ago

colorfulBest ResponseYou've already chosen the best response.1
yeah, that's a good graph :)
 one year ago

colorfulBest ResponseYou've already chosen the best response.1
all that stuff I said still works though
 one year ago

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

colorfulBest ResponseYou've already chosen the best response.1
typo, my bad I sort of invented my own notation there on the left
 one year ago

itzmashyBest ResponseYou've already chosen the best response.0
ohh okay so it's suppsoed to he pq = qp, right?
 one year ago

colorfulBest ResponseYou've already chosen the best response.1
you 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
 one year ago

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

colorfulBest ResponseYou've already chosen the best response.1
no that's not how you take a cross product...
 one year ago

itzmashyBest ResponseYou've already chosen the best response.0
oh wait.. so.. nope you lost me again lol i'm sorry..
 one year ago

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

itzmashyBest ResponseYou've already chosen the best response.0
and then crossing them?
 one year ago

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

colorfulBest ResponseYou've already chosen the best response.1
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
 one year ago

itzmashyBest ResponseYou've already chosen the best response.0
so, pr = < 4, 0, 2> pq = <4, 5, 0> pr x pq = <16, 0, 0> ?!
 one year ago

itzmashyBest ResponseYou've already chosen the best response.0
but 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
 one year ago

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

colorfulBest ResponseYou've already chosen the best response.1
let 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)
 one year ago

colorfulBest ResponseYou've already chosen the best response.1
...so if you want a vector that points from p to r you write rp
 one year ago

itzmashyBest ResponseYou've already chosen the best response.0
okay! perfect :D thank you!!! one more question?
 one year ago

itzmashyBest ResponseYou've already chosen the best response.0
find an equation of the plane containing points P Q and R
 one year ago

colorfulBest ResponseYou've already chosen the best response.1
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
 one year ago

colorfulBest ResponseYou've already chosen the best response.1
the normal vector\[\vec n=\langle P,Q,R\rangle\]implies the plane\[Px+Qy+Rz=d\]where \(d\) is some constant
 one year ago

itzmashyBest ResponseYou've already chosen the best response.0
and the normal vector was (16, 0 , 0)? so its 16x + 0 + 0 = d?
 one year ago

colorfulBest ResponseYou've already chosen the best response.1
oh I didn't check, but that seems wrong...
 one year ago

itzmashyBest ResponseYou've already chosen the best response.0
i crossed it in incorrectly lol hold on
 one year ago

colorfulBest ResponseYou've already chosen the best response.1
yeah I got <2,20> on the first try, but I'm tired so let me try again
 one year ago

colorfulBest ResponseYou've already chosen the best response.1
but wait that way still be wrong...
 one year ago

itzmashyBest ResponseYou've already chosen the best response.0
really? pellet.. i got <10 8 20> =\
 one year ago

itzmashyBest ResponseYou've already chosen the best response.0
whats your pq and pr? =\
 one year ago

colorfulBest ResponseYou've already chosen the best response.1
let me try again\[\vec u\times\vec v=\left\begin{matrix}\hat i&\hat j&\hat k\\4&5&0\\4&0&2\end{matrix}\right\]
 one year ago

itzmashyBest ResponseYou've already chosen the best response.0
this is what i got: (100)i  (80)j +(0+20)k is that wrong? :(
 one year ago

colorfulBest ResponseYou've already chosen the best response.1
\[\vec u\times\vec v=\left\begin{matrix}\hat i&\hat j&\hat k\\4&5&0\\4&0&2\end{matrix}\right=10\hat i+20\hat k8\hat j\]ok now we agree :)
 one year ago

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

colorfulBest ResponseYou've already chosen the best response.1
my bad told you I'm tired
 one year ago

itzmashyBest ResponseYou've already chosen the best response.0
no worries!!! i totally understand hahha
 one year ago

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

colorfulBest ResponseYou've already chosen the best response.1
yes, that's exactly what you do I know the method, I just can't execute it when I'm sleep deprived :P
 one year ago

colorfulBest ResponseYou've already chosen the best response.1
...of course you could have picked any point and you would have gotten the same d
 one year ago

colorfulBest ResponseYou've already chosen the best response.1
and you can simplify it obviously too 5x +4y +10z =20
 one year ago

itzmashyBest ResponseYou've already chosen the best response.0
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?
 one year ago

colorfulBest ResponseYou've already chosen the best response.1
isn't it positive? you said so yourself, and I agree
 one year ago

itzmashyBest ResponseYou've already chosen the best response.0
oops yeah hehe :D thats what i wrote down!
 one year ago

itzmashyBest ResponseYou've already chosen the best response.0
yayyyy!!!! i think i get it!!! thank you soooo much!!!!!!!!
 one year ago

colorfulBest ResponseYou've already chosen the best response.1
welcome, and g'night :)
 one year ago
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