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mags093
 3 years 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).
mags093
 3 years 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).

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amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1find 2 vectors from the given points to cross and find a normal to the plane

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1then you can define a line using the given point and the unit normal vector

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1that 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

mags093
 3 years ago
Best ResponseYou've already chosen the best response.0so i find PR and PS?

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1dw:1359740566102:dw

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1im sure theres a formula for this, but as i said prior, I got no idea what it would be :)

mags093
 3 years ago
Best ResponseYou've already chosen the best response.0ok thanks :) so i find the plane and then use the normal vector to find the distance of the point?

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1Q(4, 4, 0) R(8, 1, 3) S( 3, 0, 2) S(3, 0, 2)  <1,42> <5,1,1> yes :)

mags093
 3 years ago
Best ResponseYou've already chosen the best response.0i get where ya got it now :) but how would i find is the point is on it?

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1S is just subtracting the point S from the other 2 points to define the 2 vectors to cross

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1normal 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

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1or 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 :)

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1the 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
 3 years ago
Best ResponseYou've already chosen the best response.1then its just a matter of finding out how many t units there are

mags093
 3 years ago
Best ResponseYou've already chosen the best response.0i still dont understand how to find the distance of P to the plane?

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1my 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
 3 years ago
Best ResponseYou've already chosen the best response.1dw:1359741666062:dw

mags093
 3 years ago
Best ResponseYou've already chosen the best response.0the normal of the plane is sqrt(566), how do i create a line for P?

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1thats the length of the vector that is found when crossing the plane vectors.

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1Q(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)

mags093
 3 years ago
Best ResponseYou've already chosen the best response.0ok but how do i get the distance of P then?

mags093
 3 years ago
Best ResponseYou've already chosen the best response.0do i minus 92,11,21) from the point i was given?

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1if 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?

mags093
 3 years ago
Best ResponseYou've already chosen the best response.0ya but i still dont understand where i get the distance of P?

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1do you agree that this line equation is perpendicular to the plane, and that it will pierce thru the plane at some point?

mags093
 3 years ago
Best ResponseYou've already chosen the best response.0does it not pierce through at the point (1,3,5) ?

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1no, that is the given point that is floating someplace above the plane

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1lets 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?

mags093
 3 years ago
Best ResponseYou've already chosen the best response.0so do i sub in that point into the line to get the distance to the plane?

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1now 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?

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1the formula seems to be: http://mathworld.wolfram.com/PointPlaneDistance.html \[D=\frac{ax_o+by_o+cz_o+d}{\sqrt{a^2+b^2+c^2}}\]

mags093
 3 years ago
Best ResponseYou've already chosen the best response.0i worked that out and got this? 2(36t) +1121(10+42t)=0 is this right?

mags093
 3 years ago
Best ResponseYou've already chosen the best response.0i dont understand how to use that formula?

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1a,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
 3 years ago
Best ResponseYou've already chosen the best response.1if 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

mags093
 3 years ago
Best ResponseYou've already chosen the best response.0ya your ans right :) i tried working it out myself and got it wrong though :/

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1i got help from the wolf so i didnt have to kill an eraser ;)

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1maybe, my smarts tend to fade exponentially as the day progresses ;)

mags093
 3 years ago
Best ResponseYou've already chosen the best response.0ok 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).

mags093
 3 years ago
Best ResponseYou've already chosen the best response.0do i just get the normal vector and use the same formula then?

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1you do something similar; but we dont have a plane to play with now.

mags093
 3 years ago
Best ResponseYou've already chosen the best response.0its the one i tried earlier but couldnt get it!

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1can you define the vector from Q to R for me?

mags093
 3 years ago
Best ResponseYou've already chosen the best response.0how do i define a vector? is that when i multiply them out?

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1a vector is defined by subtracting one point from the other.

mags093
 3 years ago
Best ResponseYou've already chosen the best response.0so its (3,2,3) i think :)

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1if 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
 3 years ago
Best ResponseYou've already chosen the best response.1lets 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
 3 years ago
Best ResponseYou've already chosen the best response.1now how do you find the angle between 2 vectors?

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1dw:1359744085632:dw

mags093
 3 years ago
Best ResponseYou've already chosen the best response.0is it the length of the vectors * coz theta?

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1\[cos\alpha=\frac{\vec u \cdot \vec v}{\vec u~ \vec v}\] \[\alpha=cos^{1}\frac{\vec u \cdot \vec v}{\vec u~ \vec v}\]

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1u = <0,3,0>; u = 3 v = <3,2,3>; v = sqrt(20)  u.v= 0+6+0 = 6

mags093
 3 years ago
Best ResponseYou've already chosen the best response.0so its cos^1(6) and then we find the angle ya?

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1this 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
 3 years ago
Best ResponseYou've already chosen the best response.1http://www.wolframalpha.com/input/?i=3sin%28arccos%282%2Fsqrt%2820%29%29%29

mags093
 3 years ago
Best ResponseYou've already chosen the best response.0so the angle should be 6/sqrt(5) ? it says thats wrong?

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1you are not asked to find an angle are you, just the distance between a point and a line

mags093
 3 years ago
Best ResponseYou've already chosen the best response.0oh ya just the distance not the angle

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1http://mathworld.wolfram.com/PointLineDistance3Dimensional.html this is the distance formula for this then if they need an exact answer :)

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1\[d=\frac{uxv}{u}\]

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1or i think that shows v for my setup

amistre64
 3 years ago
Best 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

mags093
 3 years ago
Best ResponseYou've already chosen the best response.0ok so x1= Q and x2=R what do i need the t for?

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1the t is just demonstrating the process of how they derived the formula.

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1the formula is the key, and knowing which vector parts to put in which place is important as well :)

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1x0 = p x1 = q x2 = r so you need the vectors from: x0x1; x0x2; and x2x1

mags093
 3 years ago
Best ResponseYou've already chosen the best response.0the line = (1,2,5) + t(3,2,3) right?

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1almost 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

mags093
 3 years ago
Best ResponseYou've already chosen the best response.0is z not equal to 5 3(t)?

mags093
 3 years ago
Best ResponseYou've already chosen the best response.0oh ya you changed that :)

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1it 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

mags093
 3 years ago
Best ResponseYou've already chosen the best response.0where did you getbthe new x,y,z from? :)

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1the new xyz is the point where the line QR pierces thru the plane

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1by putting the line parts for xyz into the plane equation, you define a point that is both on the line and in the plane

mags093
 3 years ago
Best ResponseYou've already chosen the best response.0oh ok :) it says its wrong though? :/

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1yeah, i prolly tripped up on the math along the way :/

mags093
 3 years ago
Best ResponseYou've already chosen the best response.0oh ok :) ill try it now :) i did it the d formula and got 8+d/sqrt(4) ?

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1cos a = u.v/uv u.v = 6; uv = 3 sqrt20 cos a = 2/sqrt(20) is the correct answer by chance, 6/sqrt(5) ??

mags093
 3 years ago
Best ResponseYou've already chosen the best response.0is it suppose to be 103(3/2) for z?

mags093
 3 years ago
Best ResponseYou've already chosen the best response.0is z= 10 3(3/2) rather than 105(3/2) ??

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1and with any luck, we get something like 9/sqrt(11) ... crosses fingers

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1distance 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
 3 years ago
Best ResponseYou've already chosen the best response.1Q(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)

mags093
 3 years ago
Best ResponseYou've already chosen the best response.0thats right :) howd you figure it out?

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1i 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

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1im going to bed after that one :) good luck
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