anonymous
  • anonymous
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).
Mathematics
  • Stacey Warren - Expert brainly.com
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katieb
  • katieb
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amistre64
  • amistre64
find 2 vectors from the given points to cross and find a normal to the plane
amistre64
  • amistre64
then you can define a line using the given point and the unit normal vector
amistre64
  • amistre64
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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anonymous
  • anonymous
so i find |PR| and |PS|?
amistre64
  • amistre64
|dw:1359740566102:dw|
amistre64
  • amistre64
im sure theres a formula for this, but as i said prior, I got no idea what it would be :)
anonymous
  • anonymous
ok thanks :) so i find the plane and then use the normal vector to find the distance of the point?
amistre64
  • amistre64
Q(4, -4, 0) R(8, 1, 3) -S( 3, 0, 2) -S(3, 0, 2) ------------------------- <1,-4-2> <5,1,1> yes :)
anonymous
  • anonymous
howd you get -S?
anonymous
  • anonymous
i get where ya got it now :) but how would i find is the point is on it?
amistre64
  • amistre64
-S is just subtracting the point S from the other 2 points to define the 2 vectors to cross
amistre64
  • amistre64
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
amistre64
  • amistre64
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 :)
anonymous
  • anonymous
why sqrt9566)
amistre64
  • amistre64
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
amistre64
  • amistre64
then its just a matter of finding out how many t units there are
anonymous
  • anonymous
i still dont understand how to find the distance of P to the plane?
amistre64
  • amistre64
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.
amistre64
  • amistre64
|dw:1359741666062:dw|
anonymous
  • anonymous
the normal of the plane is sqrt(566), how do i create a line for P?
amistre64
  • amistre64
thats the length of the vector that is found when crossing the plane vectors.
amistre64
  • amistre64
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)
anonymous
  • anonymous
ok but how do i get the distance of P then?
anonymous
  • anonymous
do i minus 92,11,-21) from the point i was given?
amistre64
  • amistre64
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?
anonymous
  • anonymous
ya but i still dont understand where i get the distance of P?
amistre64
  • amistre64
do you agree that this line equation is perpendicular to the plane, and that it will pierce thru the plane at some point?
anonymous
  • anonymous
does it not pierce through at the point (1,-3,-5) ?
amistre64
  • amistre64
no, that is the given point that is floating someplace above the plane
amistre64
  • amistre64
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?
anonymous
  • anonymous
so do i sub in that point into the line to get the distance to the plane?
anonymous
  • anonymous
ya i do
amistre64
  • amistre64
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?
amistre64
  • amistre64
the formula seems to be: http://mathworld.wolfram.com/Point-PlaneDistance.html \[D=\frac{ax_o+by_o+cz_o+d}{\sqrt{a^2+b^2+c^2}}\]
anonymous
  • anonymous
i worked that out and got this? 2(-3-6t) +11-21(10+42t)=0 is this right?
anonymous
  • anonymous
i dont understand how to use that formula?
amistre64
  • amistre64
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:\[D=\frac{2(1)+11(-3)-21(-5)+36}{\sqrt{2^2+11^2+21^2}}=5\sqrt{\frac{2}{283}}=4.6237~ or~ so\]
amistre64
  • amistre64
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
anonymous
  • anonymous
ya your ans right :) i tried working it out myself and got it wrong though :/
amistre64
  • amistre64
i got help from the wolf so i didnt have to kill an eraser ;)
anonymous
  • anonymous
wolfram alpha?
amistre64
  • amistre64
yep
amistre64
  • amistre64
maybe, my smarts tend to fade exponentially as the day progresses ;)
anonymous
  • anonymous
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).
anonymous
  • anonymous
do i just get the normal vector and use the same formula then?
amistre64
  • amistre64
you do something similar; but we dont have a plane to play with now.
anonymous
  • anonymous
its the one i tried earlier but couldnt get it!
amistre64
  • amistre64
can you define the vector from Q to R for me?
anonymous
  • anonymous
how do i define a vector? is that when i multiply them out?
amistre64
  • amistre64
a vector is defined by subtracting one point from the other.
anonymous
  • anonymous
so its (3,2,-3) i think :)
amistre64
  • amistre64
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>
amistre64
  • amistre64
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
amistre64
  • amistre64
now how do you find the angle between 2 vectors?
anonymous
  • anonymous
is it xi,yj,zk ?
amistre64
  • amistre64
|dw:1359744085632:dw|
anonymous
  • anonymous
is it the length of the vectors * coz theta?
amistre64
  • amistre64
\[cos\alpha=\frac{\vec u \cdot \vec v}{|\vec u|~| \vec v|}\] \[\alpha=cos^{-1}\frac{\vec u \cdot \vec v}{|\vec u|~| \vec v|}\]
amistre64
  • amistre64
u = <0,3,0>; |u| = 3 v = <3,2,-3>; |v| = sqrt(20) --------- u.v= 0+6+0 = 6
anonymous
  • anonymous
so its cos^-1(6) and then we find the angle ya?
amistre64
  • amistre64
this gives us: \[\alpha=cos^{-1}\frac{6}{3\sqrt(20)}\] the distance is equal to |u| sin a \[|u|~sin \alpha=|u|~sin\left(cos^{-1}\frac{6}{3\sqrt(20)}\right)\] \[3~sin \alpha=3~sin\left(cos^{-1}\frac{2}{\sqrt(20)}\right)\]
amistre64
  • amistre64
http://www.wolframalpha.com/input/?i=3sin%28arccos%282%2Fsqrt%2820%29%29%29
anonymous
  • anonymous
so the angle should be 6/sqrt(5) ? it says thats wrong?
amistre64
  • amistre64
you are not asked to find an angle are you, just the distance between a point and a line
anonymous
  • anonymous
oh ya just the distance not the angle
amistre64
  • amistre64
http://mathworld.wolfram.com/Point-LineDistance3-Dimensional.html this is the distance formula for this then if they need an exact answer :)
amistre64
  • amistre64
\[d=\frac{|uxv|}{|u|}\]
amistre64
  • amistre64
or i think that shows |v| for my setup
amistre64
  • amistre64
| (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
anonymous
  • anonymous
ok so x1= Q and x2=R what do i need the t for?
amistre64
  • amistre64
the t is just demonstrating the process of how they derived the formula.
amistre64
  • amistre64
the formula is the key, and knowing which vector parts to put in which place is important as well :)
amistre64
  • amistre64
x0 = p x1 = q x2 = r so you need the vectors from: x0-x1; x0-x2; and x2-x1
anonymous
  • anonymous
|dw:1359745138247:dw|
anonymous
  • anonymous
the line = (1,2,5) + t(3,2,-3) right?
amistre64
  • amistre64
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
anonymous
  • anonymous
is z not equal to 5 -3(t)?
anonymous
  • anonymous
oh ya you changed that :)
amistre64
  • amistre64
it is, just typoed it ;) sqrt((1-11/2)^2+(2-10/2)^2+(5+5/2)^2) http://www.wolframalpha.com/input/?i=sqrt%28%281-11%2F2%29%5E2%2B%282-10%2F2%29%5E2%2B%285%2B5%2F2%29%5E2%29
anonymous
  • anonymous
where did you getbthe new x,y,z from? :)
amistre64
  • amistre64
the new xyz is the point where the line QR pierces thru the plane
amistre64
  • amistre64
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
anonymous
  • anonymous
oh ok :) it says its wrong though? :/
amistre64
  • amistre64
yeah, i prolly tripped up on the math along the way :/
anonymous
  • anonymous
oh ok :) ill try it now :) i did it the d formula and got -8+d/sqrt(4) ?
amistre64
  • amistre64
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) ??
anonymous
  • anonymous
is it suppose to be 10-3(3/2) for z?
anonymous
  • anonymous
no thats wrong :/
amistre64
  • amistre64
....buggers :/
anonymous
  • anonymous
is z= 10 -3(3/2) rather than 10-5(3/2) ??
amistre64
  • amistre64
yes
amistre64
  • amistre64
and with any luck, we get something like 9/sqrt(11) ... crosses fingers
anonymous
  • anonymous
http://www.wolframalpha.com/input/?i=sqrt%28%281-13%2F2%29%5E2%2B%282-7%29%5E2%2B%285-11%2F2%29%5E2%29
anonymous
  • anonymous
could that be right?
anonymous
  • anonymous
still wrong :(
amistre64
  • amistre64
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|
amistre64
  • amistre64
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)
anonymous
  • anonymous
thats right :) howd you figure it out?
amistre64
  • amistre64
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
anonymous
  • anonymous
oh ok thanks :)
amistre64
  • amistre64
im going to bed after that one :) good luck
anonymous
  • anonymous
ok thanks :) bye

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