Here's the question you clicked on:
itzmashy
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
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
|dw:1337148180943:dw|so here are our three points if we can find\[\vec u=\overrightarrow{pq}=p-q\]and\[\vec v=\overrightarrow{qr}=r-q\]
|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
\[\vec n=\vec u\times\vec v\]
yeah, that's a good graph :)
all that stuff I said still works though
wait.. \[pq = p-q?\] but then you have \[qr = r -q\] i'm confused here:(
typo, my bad I sort of invented my own notation there on the left
ohh okay so it's suppsoed to he pq = q-p, right?
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
so then \[u x v = (-4 x 0, -5 x 5, 2 x 0) = < 0, -10, 0> ?\]
no that's not how you take a cross product...
oh wait.. so.. nope you lost me again lol i'm sorry..
so you're picking one point as the middle point, and then creating two vectors from it?
and then crossing them?
and if i'm finding a vector PR, then its r -p, right? second point, minus first point? o.O
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
so, pr = < -4, 0, 2> pq = <-4, 5, 0> pr x pq = <-16, 0, 0> ?!
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
so getting the cross product of two vectors on a plane, gets me a vector perpendicular to the plane?
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)
...so if you want a vector that points from p to r you write r-p
okay! perfect :D thank you!!! one more question?
find an equation of the plane containing points P Q and R
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
the normal vector\[\vec n=\langle P,Q,R\rangle\]implies the plane\[Px+Qy+Rz=d\]where \(d\) is some constant
and the normal vector was (16, 0 , 0)? so its 16x + 0 + 0 = d?
oh I didn't check, but that seems wrong...
i crossed it in incorrectly lol hold on
yeah I got <2,-20> on the first try, but I'm tired so let me try again
but wait that way still be wrong...
really? pellet.. i got <10 8 20> =\
whats your pq and pr? =\
let me try again\[\vec u\times\vec v=\left|\begin{matrix}\hat i&\hat j&\hat k\\-4&5&0\\-4&0&2\end{matrix}\right|\]
this is what i got: (10-0)i - (-8-0)j +(0+20)k is that wrong? :(
\[\vec u\times\vec v=\left|\begin{matrix}\hat i&\hat j&\hat k\\-4&5&0\\-4&0&2\end{matrix}\right|=10\hat i+20\hat k-8\hat j\]ok now we agree :)
but.. but.. i got 10i + 8j + 20k, doesn't the - distribute to the 8? :(
my bad told you I'm tired
no worries!!! i totally understand hahha
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?
yes, that's exactly what you do I know the method, I just can't execute it when I'm sleep deprived :P
...of course you could have picked any point and you would have gotten the same d
and you can simplify it obviously too 5x +4y +10z =20
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?
isn't it positive? you said so yourself, and I agree
oops yeah hehe :D thats what i wrote down!
yayyyy!!!! i think i get it!!! thank you soooo much!!!!!!!!
welcome, and g'night :)