s3a
  • s3a
Question about equation of plane, parametric equations and stuff like that: http://f.imgtmp.com/HIhr6.jpg Help would be GREATLY appreciated!
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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amistre64
  • amistre64
personally id create 2 vectors and cross them for the normal vector
amistre64
  • amistre64
can you type up the points for me, its hard to keep track of them
s3a
  • s3a
that's to get the first part?

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amistre64
  • amistre64
yeah, plane equation
s3a
  • s3a
P(-4,-6,2) and Q(-8,-7,0), R(7,3,5)
amistre64
  • amistre64
good, now what im gonna do is define 2 vectors by subtracting one point from the other 2
s3a
  • s3a
wait, first let me finish part 1
amistre64
  • amistre64
Q(-8,-7,0) R( 7, 3, 5) - P(-4,-6,2) - P(-4,-6,2) ------------------------- -<4, 1, 2> <11,9,3> these are 2 vectors that reside on your plane between your points
amistre64
  • amistre64
lets cross them for a normal vector <4, 1, 2> <11,9,3> ---------- x = (3-18) = -15 y = -(12-22) = 10 z = (36-11) = 25 <-15,10,25> is the normal vector which we can scale down to work with -------- -5 <3,-2,-5> looks good to me
amistre64
  • amistre64
now, given any point on the plane; they gave you 3 to play with; we can construct the equation as such: 3(x-Px)-2(y-Py)-5(z-Pz)=0
s3a
  • s3a
2 secs, i was searching for something. now i can read properly.
amistre64
  • amistre64
id use a simple point like (-8,-7,0) 3(x--8)-2(y--7)-5(z-0)=0 3x+24-2y-14-5z=0 3x-2y-5z+10=0 if i did it right that should be a match to your 3 points
s3a
  • s3a
ok so <3,-2,-5> is the normal vector and 3(x-Px)-2(y-Py)-5(z-Pz)=0 is from dot producting it?
amistre64
  • amistre64
in effect yes.
s3a
  • s3a
ya i get that part now thank you. :)
s3a
  • s3a
what about the line?
amistre64
  • amistre64
gotta dbl chk my plane equation first, im prone to error at my age :)
amistre64
  • amistre64
wolfram likes it :)
amistre64
  • amistre64
so line PQ ... let me see
amistre64
  • amistre64
its a vector addition of P and Q right?
amistre64
  • amistre64
|dw:1327363860012:dw|
amistre64
  • amistre64
vector subtraction that is and the points themselvs act as defining the vectors
s3a
  • s3a
i just pictured like a toddler helping me on the other side :P
s3a
  • s3a
and sorry technical issues make it hard to respond fast here
amistre64
  • amistre64
so, P-Q, i did above = <4,1,2> and this needs to be scaled to any length to create the line t<4,1,2> now all we have to do is attach it to either P or Q
s3a
  • s3a
i'm guessing your older rather than younger
amistre64
  • amistre64
the best thing about today is that im one step closer to death :)
s3a
  • s3a
there is a comma between P and Q if that matters.
s3a
  • s3a
lol ... well eat healthy, etc
amistre64
  • amistre64
using -8,-7,0 to anchor our line vector to we get: x = -8 + 4t y = -7 + 1t z = 0 + 2t should be the parametric stuff
amistre64
  • amistre64
if you plug in P for the x y z parts you should come out to t being the same for all parametrics
amistre64
  • amistre64
P=-4,-6,2 -4 = -8 +4t 4 = 4t t = 1 -7+1 = -6 0+2 = 2... that works for me :)
s3a
  • s3a
ok so you used Q to anchor and P works too. just confirmed.
s3a
  • s3a
sorry for being delayed.
amistre64
  • amistre64
'sok, this computers getting slow as well
amistre64
  • amistre64
now the next one is perp to L and hits R
s3a
  • s3a
wait before that.
s3a
  • s3a
i didn't get what you did when you found a value for t and why you did it.
s3a
  • s3a
also, i don't recommend you use CRT. rather use LCD screens because they don't harm your eyes or are negligible in comparison. plasmas make radiation that dissipates before harming you so it's acceptable but i'd go with lcd anyways.
s3a
  • s3a
just saying.
amistre64
  • amistre64
since im anchored to Q and it has to go thru P then some value of "t" has to remain the same for each parametric in order to come up with the P values. I solved for Px in place of x to see if we can get a t that worked
amistre64
  • amistre64
when t=1 we get Px so i just subbed in t=1 for the others to see if they matched Py and Pz
amistre64
  • amistre64
if we anchored to P, then we would have used Q to test it out with :)
s3a
  • s3a
i see it now :D
s3a
  • s3a
so what about the last part?
amistre64
  • amistre64
the last part I believe uses this line vector as a normal vector of this other plane; and this other plane has to hit R so we should just place Rx Ry Rz in the plane equation
amistre64
  • amistre64
|dw:1327364753446:dw|
s3a
  • s3a
passing through means parallel but more precise?
s3a
  • s3a
i mean covering it wholly
amistre64
  • amistre64
sooo, n<4,1,2>, calibrate with R(x,y,z) 4(x-Rx)+1(y-Ry)+2(z-Rz)=0 expanded of course
amistre64
  • amistre64
we want a plane perp to L, not parallel
amistre64
  • amistre64
which means L is our norm for this perp plane we need a point on that plane to anchor to, they say use R
s3a
  • s3a
o R is a point.
s3a
  • s3a
makes sense since a point R is given :P
amistre64
  • amistre64
yeah, it was what (7,3,5) right?
amistre64
  • amistre64
4(x-7)+1(y-3)+2(z-5)=0 4x-28+y-3+2z-10=0 4x+y+2z-10-28-3=0 4x+y+2z-41=0 maybe?
s3a
  • s3a
R(7,3,5) yes
s3a
  • s3a
o too late lol
amistre64
  • amistre64
im quick for an old man lol
s3a
  • s3a
lol eye of the tiger
amistre64
  • amistre64
im sure well be going over these in calc3 this term.
amistre64
  • amistre64
trying to get me a masters in math before i die :)
s3a
  • s3a
go for it lol and eat healthy!
amistre64
  • amistre64
:) ill try. good luck
amistre64
  • amistre64
we got the part 3 right?
s3a
  • s3a
i think that;'s wrong. brb
amistre64
  • amistre64
pretty confident its good. the line vector from part 2 is our normal, and we anchored it with pointR; unless i missed the multiplication and adding parts :) that happens
s3a
  • s3a
I'm at part three if you're going to help.

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