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personally id create 2 vectors and cross them for the normal vector
can you type up the points for me, its hard to keep track of them
that's to get the first part?
yeah, plane equation
P(-4,-6,2) and Q(-8,-7,0), R(7,3,5)
good, now what im gonna do is define 2 vectors by subtracting one point from the other 2
wait, first let me finish part 1
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
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
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
2 secs, i was searching for something. now i can read properly.
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
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?
in effect yes.
ya i get that part now thank you. :)
what about the line?
gotta dbl chk my plane equation first, im prone to error at my age :)
wolfram likes it :)
so line PQ ... let me see
its a vector addition of P and Q right?
vector subtraction that is and the points themselvs act as defining the vectors
i just pictured like a toddler helping me on the other side :P
and sorry technical issues make it hard to respond fast here
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
i'm guessing your older rather than younger
the best thing about today is that im one step closer to death :)
there is a comma between P and Q if that matters.
lol ... well eat healthy, etc
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
if you plug in P for the x y z parts you should come out to t being the same for all parametrics
P=-4,-6,2 -4 = -8 +4t 4 = 4t t = 1 -7+1 = -6 0+2 = 2... that works for me :)
ok so you used Q to anchor and P works too. just confirmed.
sorry for being delayed.
'sok, this computers getting slow as well
now the next one is perp to L and hits R
wait before that.
i didn't get what you did when you found a value for t and why you did it.
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.
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
when t=1 we get Px so i just subbed in t=1 for the others to see if they matched Py and Pz
if we anchored to P, then we would have used Q to test it out with :)
i see it now :D
so what about the last part?
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
passing through means parallel but more precise?
i mean covering it wholly
sooo, n<4,1,2>, calibrate with R(x,y,z) 4(x-Rx)+1(y-Ry)+2(z-Rz)=0 expanded of course
we want a plane perp to L, not parallel
which means L is our norm for this perp plane we need a point on that plane to anchor to, they say use R
o R is a point.
makes sense since a point R is given :P
yeah, it was what (7,3,5) right?
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?
o too late lol
im quick for an old man lol
lol eye of the tiger
im sure well be going over these in calc3 this term.
trying to get me a masters in math before i die :)
go for it lol and eat healthy!
:) ill try. good luck
we got the part 3 right?
i think that;'s wrong. brb
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
I'm at part three if you're going to help.