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find 2 vectors from the given points to cross and find a normal to the plane

then you can define a line using the given point and the unit normal vector

so i find |PR| and |PS|?

|dw:1359740566102:dw|

im sure theres a formula for this, but as i said prior, I got no idea what it would be :)

ok thanks :) so i find the plane and then use the normal vector to find the distance of the point?

Q(4, -4, 0) R(8, 1, 3)
-S( 3, 0, 2) -S(3, 0, 2)
-------------------------
<1,-4-2> <5,1,1>
yes :)

howd you get -S?

i get where ya got it now :) but how would i find is the point is on it?

-S is just subtracting the point S from the other 2 points to define the 2 vectors to cross

why sqrt9566)

then its just a matter of finding out how many t units there are

i still dont understand how to find the distance of P to the plane?

|dw:1359741666062:dw|

the normal of the plane is sqrt(566), how do i create a line for P?

thats the length of the vector that is found when crossing the plane vectors.

ok but how do i get the distance of P then?

do i minus 92,11,-21) from the point i was given?

ya but i still dont understand where i get the distance of P?

does it not pierce through at the point (1,-3,-5) ?

no, that is the given point that is floating someplace above the plane

so do i sub in that point into the line to get the distance to the plane?

ya i do

i worked that out and got this? 2(-3-6t) +11-21(10+42t)=0 is this right?

i dont understand how to use that formula?

ya your ans right :) i tried working it out myself and got it wrong though :/

i got help from the wolf so i didnt have to kill an eraser ;)

wolfram alpha?

yep

maybe, my smarts tend to fade exponentially as the day progresses ;)

do i just get the normal vector and use the same formula then?

you do something similar; but we dont have a plane to play with now.

its the one i tried earlier but couldnt get it!

can you define the vector from Q to R for me?

how do i define a vector? is that when i multiply them out?

a vector is defined by subtracting one point from the other.

so its (3,2,-3) i think :)

now how do you find the angle between 2 vectors?

is it xi,yj,zk ?

|dw:1359744085632:dw|

is it the length of the vectors * coz theta?

u = <0,3,0>; |u| = 3
v = <3,2,-3>; |v| = sqrt(20)
---------
u.v= 0+6+0 = 6

so its cos^-1(6) and then we find the angle ya?

http://www.wolframalpha.com/input/?i=3sin%28arccos%282%2Fsqrt%2820%29%29%29

so the angle should be 6/sqrt(5) ? it says thats wrong?

you are not asked to find an angle are you, just the distance between a point and a line

oh ya just the distance not the angle

\[d=\frac{|uxv|}{|u|}\]

or i think that shows |v| for my setup

ok so x1= Q and x2=R what do i need the t for?

the t is just demonstrating the process of how they derived the formula.

the formula is the key, and knowing which vector parts to put in which place is important as well :)

x0 = p
x1 = q
x2 = r
so you need the vectors from:
x0-x1; x0-x2; and x2-x1

|dw:1359745138247:dw|

the line = (1,2,5) + t(3,2,-3) right?

is z not equal to 5 -3(t)?

oh ya you changed that :)

where did you getbthe new x,y,z from? :)

the new xyz is the point where the line QR pierces thru the plane

oh ok :) it says its wrong though? :/

yeah, i prolly tripped up on the math along the way :/

oh ok :) ill try it now :) i did it the d formula and got -8+d/sqrt(4) ?

is it suppose to be 10-3(3/2) for z?

no thats wrong :/

....buggers :/

is z= 10 -3(3/2) rather than 10-5(3/2) ??

yes

and with any luck, we get something like 9/sqrt(11) ... crosses fingers

could that be right?

still wrong :(

thats right :) howd you figure it out?

oh ok thanks :)

im going to bed after that one :) good luck

ok thanks :) bye