Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing
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).
 one year ago
 one year ago
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).
 one year ago
 one year ago

This Question is Closed

amistre64Best ResponseYou've already chosen the best response.1
find 2 vectors from the given points to cross and find a normal to the plane
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
then you can define a line using the given point and the unit normal vector
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
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
 one year ago

mags093Best ResponseYou've already chosen the best response.0
so i find PR and PS?
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
dw:1359740566102:dw
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
im sure theres a formula for this, but as i said prior, I got no idea what it would be :)
 one year ago

mags093Best ResponseYou've already chosen the best response.0
ok thanks :) so i find the plane and then use the normal vector to find the distance of the point?
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
Q(4, 4, 0) R(8, 1, 3) S( 3, 0, 2) S(3, 0, 2)  <1,42> <5,1,1> yes :)
 one year ago

mags093Best ResponseYou've already chosen the best response.0
i get where ya got it now :) but how would i find is the point is on it?
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
S is just subtracting the point S from the other 2 points to define the 2 vectors to cross
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
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(x3)+11(y0)21(z2)=0 and using the xyz values in terms of t 1(1 + 2t3)+11(3+11t0)21(5  21t2)=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
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
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 :)
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
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
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
then its just a matter of finding out how many t units there are
 one year ago

mags093Best ResponseYou've already chosen the best response.0
i still dont understand how to find the distance of P to the plane?
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
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.
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
dw:1359741666062:dw
 one year ago

mags093Best ResponseYou've already chosen the best response.0
the normal of the plane is sqrt(566), how do i create a line for P?
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
thats the length of the vector that is found when crossing the plane vectors.
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
Q(4, 4, 0) R(8, 1, 3) S( 3, 0, 2) S(3, 0, 2)  <1,42> <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)
 one year ago

mags093Best ResponseYou've already chosen the best response.0
ok but how do i get the distance of P then?
 one year ago

mags093Best ResponseYou've already chosen the best response.0
do i minus 92,11,21) from the point i was given?
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
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?
 one year ago

mags093Best ResponseYou've already chosen the best response.0
ya but i still dont understand where i get the distance of P?
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
do you agree that this line equation is perpendicular to the plane, and that it will pierce thru the plane at some point?
 one year ago

mags093Best ResponseYou've already chosen the best response.0
does it not pierce through at the point (1,3,5) ?
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
no, that is the given point that is floating someplace above the plane
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
lets take the normal vector and define the plane equation with it 2(xxo) + 11(yyo)  21(zzo) = 0 I will use the one of the points that are already on the plane, I chose point S (3,0,2) 2(x3) + 11(y0)  21(z2) = 0 do you agree that this equation will define all xyz point on the plane?
 one year ago

mags093Best ResponseYou've already chosen the best response.0
so do i sub in that point into the line to get the distance to the plane?
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
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(x3) + 11(y0)  21(z2) = 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?
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
the formula seems to be: http://mathworld.wolfram.com/PointPlaneDistance.html \[D=\frac{ax_o+by_o+cz_o+d}{\sqrt{a^2+b^2+c^2}}\]
 one year ago

mags093Best ResponseYou've already chosen the best response.0
i worked that out and got this? 2(36t) +1121(10+42t)=0 is this right?
 one year ago

mags093Best ResponseYou've already chosen the best response.0
i dont understand how to use that formula?
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
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:\[D=\frac{2(1)+11(3)21(5)+36}{\sqrt{2^2+11^2+21^2}}=5\sqrt{\frac{2}{283}}=4.6237~ or~ so\]
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
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
 one year ago

mags093Best ResponseYou've already chosen the best response.0
ya your ans right :) i tried working it out myself and got it wrong though :/
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
i got help from the wolf so i didnt have to kill an eraser ;)
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
maybe, my smarts tend to fade exponentially as the day progresses ;)
 one year ago

mags093Best ResponseYou've already chosen the best response.0
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).
 one year ago

mags093Best ResponseYou've already chosen the best response.0
do i just get the normal vector and use the same formula then?
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
you do something similar; but we dont have a plane to play with now.
 one year ago

mags093Best ResponseYou've already chosen the best response.0
its the one i tried earlier but couldnt get it!
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
can you define the vector from Q to R for me?
 one year ago

mags093Best ResponseYou've already chosen the best response.0
how do i define a vector? is that when i multiply them out?
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
a vector is defined by subtracting one point from the other.
 one year ago

mags093Best ResponseYou've already chosen the best response.0
so its (3,2,3) i think :)
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
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>
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
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
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
now how do you find the angle between 2 vectors?
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
dw:1359744085632:dw
 one year ago

mags093Best ResponseYou've already chosen the best response.0
is it the length of the vectors * coz theta?
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
\[cos\alpha=\frac{\vec u \cdot \vec v}{\vec u~ \vec v}\] \[\alpha=cos^{1}\frac{\vec u \cdot \vec v}{\vec u~ \vec v}\]
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
u = <0,3,0>; u = 3 v = <3,2,3>; v = sqrt(20)  u.v= 0+6+0 = 6
 one year ago

mags093Best ResponseYou've already chosen the best response.0
so its cos^1(6) and then we find the angle ya?
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
this gives us: \[\alpha=cos^{1}\frac{6}{3\sqrt(20)}\] the distance is equal to u sin a \[u~sin \alpha=u~sin\left(cos^{1}\frac{6}{3\sqrt(20)}\right)\] \[3~sin \alpha=3~sin\left(cos^{1}\frac{2}{\sqrt(20)}\right)\]
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
http://www.wolframalpha.com/input/?i=3sin%28arccos%282%2Fsqrt%2820%29%29%29
 one year ago

mags093Best ResponseYou've already chosen the best response.0
so the angle should be 6/sqrt(5) ? it says thats wrong?
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
you are not asked to find an angle are you, just the distance between a point and a line
 one year ago

mags093Best ResponseYou've already chosen the best response.0
oh ya just the distance not the angle
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
http://mathworld.wolfram.com/PointLineDistance3Dimensional.html this is the distance formula for this then if they need an exact answer :)
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
\[d=\frac{uxv}{u}\]
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
or i think that shows v for my setup
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
 (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
 one year ago

mags093Best ResponseYou've already chosen the best response.0
ok so x1= Q and x2=R what do i need the t for?
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
the t is just demonstrating the process of how they derived the formula.
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
the formula is the key, and knowing which vector parts to put in which place is important as well :)
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
x0 = p x1 = q x2 = r so you need the vectors from: x0x1; x0x2; and x2x1
 one year ago

mags093Best ResponseYou've already chosen the best response.0
the line = (1,2,5) + t(3,2,3) right?
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
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(x1)+2(y5)+3(z5) = 0 the line containing QR is then defined as: x=1+3t y=2+2t z=53t < 3 not 5 :) 3((1+3t)1)+2((2+2t)5)+3((53t)5) = 0 t = 3/2 x=2+3(3/2) = 11/2 y=4+2(3/2) = 10/2 z=105(3/2) = 5/2
 one year ago

mags093Best ResponseYou've already chosen the best response.0
is z not equal to 5 3(t)?
 one year ago

mags093Best ResponseYou've already chosen the best response.0
oh ya you changed that :)
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
it is, just typoed it ;) sqrt((111/2)^2+(210/2)^2+(5+5/2)^2) http://www.wolframalpha.com/input/?i=sqrt%28%28111%2F2%29%5E2%2B%28210%2F2%29%5E2%2B%285%2B5%2F2%29%5E2%29
 one year ago

mags093Best ResponseYou've already chosen the best response.0
where did you getbthe new x,y,z from? :)
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
the new xyz is the point where the line QR pierces thru the plane
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
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
 one year ago

mags093Best ResponseYou've already chosen the best response.0
oh ok :) it says its wrong though? :/
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
yeah, i prolly tripped up on the math along the way :/
 one year ago

mags093Best ResponseYou've already chosen the best response.0
oh ok :) ill try it now :) i did it the d formula and got 8+d/sqrt(4) ?
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
cos a = u.v/uv u.v = 6; uv = 3 sqrt20 cos a = 2/sqrt(20) is the correct answer by chance, 6/sqrt(5) ??
 one year ago

mags093Best ResponseYou've already chosen the best response.0
is it suppose to be 103(3/2) for z?
 one year ago

mags093Best ResponseYou've already chosen the best response.0
is z= 10 3(3/2) rather than 105(3/2) ??
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
and with any luck, we get something like 9/sqrt(11) ... crosses fingers
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
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
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
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)
 one year ago

mags093Best ResponseYou've already chosen the best response.0
thats right :) howd you figure it out?
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
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
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
im going to bed after that one :) good luck
 one year ago
See more questions >>>
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.