Question about equation of plane, parametric equations and stuff like that: http://f.imgtmp.com/HIhr6.jpg
Help would be GREATLY appreciated!

- s3a

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- amistre64

personally id create 2 vectors and cross them for the normal vector

- amistre64

can you type up the points for me, its hard to keep track of them

- s3a

that's to get the first part?

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## More answers

- amistre64

yeah, plane equation

- s3a

P(-4,-6,2) and Q(-8,-7,0), R(7,3,5)

- amistre64

good, now what im gonna do is define 2 vectors by subtracting one point from the other 2

- s3a

wait, first let me finish part 1

- 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

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

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

2 secs, i was searching for something. now i can read properly.

- 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

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

in effect yes.

- s3a

ya i get that part now thank you. :)

- s3a

what about the line?

- amistre64

gotta dbl chk my plane equation first, im prone to error at my age :)

- amistre64

wolfram likes it :)

- amistre64

so line PQ ... let me see

- amistre64

its a vector addition of P and Q right?

- amistre64

|dw:1327363860012:dw|

- amistre64

vector subtraction that is and the points themselvs act as defining the vectors

- s3a

i just pictured like a toddler helping me on the other side :P

- s3a

and sorry technical issues make it hard to respond fast here

- 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

i'm guessing your older rather than younger

- amistre64

the best thing about today is that im one step closer to death :)

- s3a

there is a comma between P and Q if that matters.

- s3a

lol ... well eat healthy, etc

- 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

if you plug in P for the x y z parts you should come out to t being the same for all parametrics

- amistre64

P=-4,-6,2
-4 = -8 +4t
4 = 4t
t = 1
-7+1 = -6
0+2 = 2... that works for me :)

- s3a

ok so you used Q to anchor and P works too. just confirmed.

- s3a

sorry for being delayed.

- amistre64

'sok, this computers getting slow as well

- amistre64

now the next one is perp to L and hits R

- s3a

wait before that.

- s3a

i didn't get what you did when you found a value for t and why you did it.

- 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

just saying.

- 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

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

if we anchored to P, then we would have used Q to test it out with
:)

- s3a

i see it now :D

- s3a

so what about the last part?

- 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

|dw:1327364753446:dw|

- s3a

passing through means parallel but more precise?

- s3a

i mean covering it wholly

- 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

we want a plane perp to L, not parallel

- 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

o R is a point.

- s3a

makes sense since a point R is given :P

- amistre64

yeah, it was what (7,3,5) right?

- 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

R(7,3,5) yes

- s3a

o too late lol

- amistre64

im quick for an old man lol

- s3a

lol eye of the tiger

- amistre64

im sure well be going over these in calc3 this term.

- amistre64

trying to get me a masters in math before i die :)

- s3a

go for it lol and eat healthy!

- amistre64

:) ill try. good luck

- amistre64

we got the part 3 right?

- s3a

i think that;'s wrong. brb

- 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

I'm at part three if you're going to help.

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