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can anyone help? Find the distance the point P(1, -3, -5), is to the plane through the three points Q(4, -4, 0), R(8, 1, 3), and S(3, 0, 2).

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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
that line will pierce the plane at the shortest distance ... since it creates a perp line; and the distance between the given point and the piercing point is the solution

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so i find |PR| and |PS|?
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
normal I find is: <2,11,-21> we can either make that a unit vector or just use it as is to find the piercing point P(1, -3, -5) x = 1 + 2t y = -3+11t z = -5 - 21t lets use any of the 3 given points in the plane to define the plane equation with; ill use S(3,0,2) i spose 1(x-3)+11(y-0)-21(z-2)=0 and using the xyz values in terms of t 1(1 + 2t-3)+11(-3+11t-0)-21(-5 - 21t-2)=0 and the calculations give me t = -28/121 use that to define the piercing point if you want and distance it from the given P
or we can unit the normal vector and just solve for t :) divide normal by sqrt(566) x = 1 + 2t/sqrt(566) y = -3+11t/sqrt(566) z = -5 - 21t/sqrt(566) 1(1 + 2t/sqrt(566)-3)+11(-3+11t/sqrt(566)-0)-21(-5 - 21t/sqrt(566)-2)=0 do it your way and lets see if you get something close to 5 :)
why sqrt9566)
the normal is <2,11,-21> which has a length of sqrt(566) so to make a unit normal we have to divide all the parts by its length
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?
my idea, and granted its not like the textbook since I cant recall the textbook formula :) is to define the normal of the plane you want to measure from. Create a line equation using the P point so that you can determine the point on the plane that this line will pierce thru at. The distance between point P and the piercing point is the distance P is from the plane.
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.
Q(4, -4, 0) R(8, 1, 3) -S( 3, 0, 2) -S(3, 0, 2) ------------------------- <1,-4-2> <5,1,1> these are the 2 plane vectors I came up with they cross to get <2,11,-21> which has a length of sqrt(566)
ok but how do i get the distance of P then?
do i minus 92,11,-21) from the point i was given?
if we apply the normal as a direction vector to the stated point P we create a line equation L = (1,-3,-5) + t<2,11,-21> we can use this parametrically as: x = 1 +2t y = -3 +11t z = -5 -21t do you agree?
ya but i still dont understand where i get the distance of P?
do you agree that this line equation is perpendicular to the plane, and that it will pierce thru the plane at some point?
does it not pierce through at the point (1,-3,-5) ?
no, that is the given point that is floating someplace above the plane
lets take the normal vector and define the plane equation with it 2(x-xo) + 11(y-yo) - 21(z-zo) = 0 I will use the one of the points that are already on the plane, I chose point S (3,0,2) 2(x-3) + 11(y-0) - 21(z-2) = 0 do you agree that this equation will define all xyz point on the plane?
so do i sub in that point into the line to get the distance to the plane?
ya i do
now for any given point on the line we have the xyz equation for them in terms of "t" so lets put the line equation point values into the plane equation and solve for t x = 1 +2t y = -3 +11t z = -5 -21t 2(x-3) + 11(y-0) - 21(z-2) = 0 2((1 +2t)-3) + 11((-3 +11t)-0) - 21((-5 -21t)-2) = 0 this is a longer way to do it, if I could recall the formula it would be simpler prolly, but this works fine as well do you agree that the value of t will define us a point x,y,z that is common to both the line and the plane?
the formula seems to be: \[D=\frac{ax_o+by_o+cz_o+d}{\sqrt{a^2+b^2+c^2}}\]
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?
a,b,c is the normal vector components: <2,11,-21> xo,yo,zo is the stated point: (1,-3,-5) d is the constant on the plane equation, so i found d=36 from the plane equation giving me:\[D=\frac{2(1)+11(-3)-21(-5)+36}{\sqrt{2^2+11^2+21^2}}=5\sqrt{\frac{2}{283}}=4.6237~ or~ so\]
if we work out my idea : 2((1 +2t)-3) + 11((-3 +11t)-0) - 21((-5 -21t)-2) = 0 t = -55/283 which would define the point x,y,z that is common to both line and plane. or, if we unitify the normal to begin with by dividing it by its length sqrt(2^2+11^2+21^2) we get: 2((1 +2t/sqrt(566))-3) + 11((-3 +11t/sqrt(566))-0) - 21((-5 -21t/sqrt(566))-2) = 0 and t equals the distance required of 55 sqrt(2/283) just like the 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?
maybe, my smarts tend to fade exponentially as the day progresses ;)
ok well this is it anyway :) Find the distance the point P(1,5,5) is to the line through the two points Q(1,2,5), and R(4,4,2).
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 :)
if we move the points so that one of them is at the origin (0,0,0) the components of the vector are defined by the nonorigined point Q(1,2,5) R(4,4,2) -Q(1,2,5) -Q(1,2,5) --------------------- (0,0,0) (3,2,-3) so yes the vector between them is <3,2,-3>
lets move the point P as well to define its vector as QP Q(1,2,5) R(4,4,2) P(1,5,5) -Q(1,2,5) -Q(1,2,5) -Q(1,2,5) -------------------------------- (0,0,0) (3,2,-3) (0,3,0) all we have done is move things on the graph, we havent change the nature of the problem itself
now how do you find the angle between 2 vectors?
is it xi,yj,zk ?
is it the length of the vectors * coz theta?
\[cos\alpha=\frac{\vec u \cdot \vec v}{|\vec u|~| \vec v|}\] \[\alpha=cos^{-1}\frac{\vec u \cdot \vec v}{|\vec u|~| \vec v|}\]
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?
this gives us: \[\alpha=cos^{-1}\frac{6}{3\sqrt(20)}\] the distance is equal to |u| sin a \[|u|~sin \alpha=|u|~sin\left(cos^{-1}\frac{6}{3\sqrt(20)}\right)\] \[3~sin \alpha=3~sin\left(cos^{-1}\frac{2}{\sqrt(20)}\right)\]
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 this is the distance formula for this then if they need an exact answer :)
or i think that shows |v| for my setup
| (3,2,-3) x (0,3,0)| = |(9,0,9)| = sqrt(162) and |v| = sqrt(20) sqrt(162/20) = 9/sqrt(10) if I seen it right
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
the line = (1,2,5) + t(3,2,-3) right?
almost had that lol, ... lets do this :) define the direction vector from Q to R as the normal vector of a plane containing P QR = <3,2,-3> and P = (1,5,5) plane: 3(x-1)+2(y-5)+3(z-5) = 0 the line containing QR is then defined as: x=1+3t y=2+2t z=5-3t <--- -3 not -5 :) 3((1+3t)-1)+2((2+2t)-5)+3((5-3t)-5) = 0 t = 3/2 x=2+3(3/2) = 11/2 y=4+2(3/2) = 10/2 z=10-5(3/2) = -5/2
is z not equal to 5 -3(t)?
oh ya you changed that :)
it is, just typoed it ;) sqrt((1-11/2)^2+(2-10/2)^2+(5+5/2)^2)
where did you getbthe new x,y,z from? :)
the new xyz is the point where the line QR pierces thru the plane
by putting the line parts for xyz into the plane equation, you define a point that is both on the line and in 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) ?
cos a = u.v/|u||v| u.v = 6; |u||v| = 3 sqrt20 cos a = 2/sqrt(20) is the correct answer by chance, 6/sqrt(5) ??
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) ??
and with any luck, we get something like 9/sqrt(11) ... crosses fingers
could that be right?
still wrong :(
distance Q from plane containg P is:\[\frac{3(1)+2(2)-3(5)+2}{\sqrt{3^2+2^2+3^2}}\] and distance Q is from P is 3 giving me: |dw:1359746813395:dw|
Q(1,2,5) R(4,4,2) P(1,5,5) -P(1,5,5) -P(1,5,5) -P(1,5,5) -------------------------------- (0,-3,0) (3,-1,-3) (0,0,0) (0,-3,0) x (3,-1,-3) = (9,0,9) ; 9 sqrt(2) 9sqrt(2)/|QR| |QR| = sqrt(3^2+2^2+3^2) = sqrt(22) 9sqrt(2/22) = 9/sqrt(11)
thats right :) howd you figure it out?
i had to see which points the formula was refering to and i worked it out from there. I worked it out earlier using the distance of Q to the plane and the distance from Q to P; and pythaging it
oh ok thanks :)
im going to bed after that one :) good luck
ok thanks :) bye

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