## mags093 Group Title 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

1. amistre64 Group Title

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

2. amistre64 Group Title

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

3. amistre64 Group Title

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

4. mags093 Group Title

so i find |PR| and |PS|?

5. amistre64 Group Title

|dw:1359740566102:dw|

6. amistre64 Group Title

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

7. mags093 Group Title

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

8. amistre64 Group Title

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

9. mags093 Group Title

howd you get -S?

10. mags093 Group Title

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

11. amistre64 Group Title

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

12. amistre64 Group Title

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

13. amistre64 Group Title

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 :)

14. mags093 Group Title

why sqrt9566)

15. amistre64 Group Title

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

16. amistre64 Group Title

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

17. mags093 Group Title

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

18. amistre64 Group Title

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.

19. amistre64 Group Title

|dw:1359741666062:dw|

20. mags093 Group Title

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

21. amistre64 Group Title

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

22. amistre64 Group Title

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)

23. mags093 Group Title

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

24. mags093 Group Title

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

25. amistre64 Group Title

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?

26. mags093 Group Title

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

27. amistre64 Group Title

do you agree that this line equation is perpendicular to the plane, and that it will pierce thru the plane at some point?

28. mags093 Group Title

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

29. amistre64 Group Title

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

30. amistre64 Group Title

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?

31. mags093 Group Title

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

32. mags093 Group Title

ya i do

33. amistre64 Group Title

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?

34. amistre64 Group Title

the formula seems to be: http://mathworld.wolfram.com/Point-PlaneDistance.html $D=\frac{ax_o+by_o+cz_o+d}{\sqrt{a^2+b^2+c^2}}$

35. mags093 Group Title

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

36. mags093 Group Title

i dont understand how to use that formula?

37. amistre64 Group Title

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:$D=\frac{2(1)+11(-3)-21(-5)+36}{\sqrt{2^2+11^2+21^2}}=5\sqrt{\frac{2}{283}}=4.6237~ or~ so$

38. amistre64 Group Title

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

39. mags093 Group Title

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

40. amistre64 Group Title

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

41. mags093 Group Title

wolfram alpha?

42. amistre64 Group Title

yep

43. amistre64 Group Title

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

44. mags093 Group Title

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).

45. mags093 Group Title

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

46. amistre64 Group Title

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

47. mags093 Group Title

its the one i tried earlier but couldnt get it!

48. amistre64 Group Title

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

49. mags093 Group Title

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

50. amistre64 Group Title

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

51. mags093 Group Title

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

52. amistre64 Group Title

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>

53. amistre64 Group Title

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

54. amistre64 Group Title

now how do you find the angle between 2 vectors?

55. mags093 Group Title

is it xi,yj,zk ?

56. amistre64 Group Title

|dw:1359744085632:dw|

57. mags093 Group Title

is it the length of the vectors * coz theta?

58. amistre64 Group Title

$cos\alpha=\frac{\vec u \cdot \vec v}{|\vec u|~| \vec v|}$ $\alpha=cos^{-1}\frac{\vec u \cdot \vec v}{|\vec u|~| \vec v|}$

59. amistre64 Group Title

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

60. mags093 Group Title

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

61. amistre64 Group Title

this gives us: $\alpha=cos^{-1}\frac{6}{3\sqrt(20)}$ the distance is equal to |u| sin a $|u|~sin \alpha=|u|~sin\left(cos^{-1}\frac{6}{3\sqrt(20)}\right)$ $3~sin \alpha=3~sin\left(cos^{-1}\frac{2}{\sqrt(20)}\right)$

62. amistre64 Group Title
63. mags093 Group Title

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

64. amistre64 Group Title

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

65. mags093 Group Title

oh ya just the distance not the angle

66. amistre64 Group Title

http://mathworld.wolfram.com/Point-LineDistance3-Dimensional.html this is the distance formula for this then if they need an exact answer :)

67. amistre64 Group Title

$d=\frac{|uxv|}{|u|}$

68. amistre64 Group Title

or i think that shows |v| for my setup

69. amistre64 Group Title

| (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

70. mags093 Group Title

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

71. amistre64 Group Title

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

72. amistre64 Group Title

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

73. amistre64 Group Title

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

74. mags093 Group Title

|dw:1359745138247:dw|

75. mags093 Group Title

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

76. amistre64 Group Title

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

77. mags093 Group Title

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

78. mags093 Group Title

oh ya you changed that :)

79. amistre64 Group Title

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

80. mags093 Group Title

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

81. amistre64 Group Title

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

82. amistre64 Group Title

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

83. mags093 Group Title

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

84. amistre64 Group Title

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

85. mags093 Group Title

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

86. amistre64 Group Title

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) ??

87. mags093 Group Title

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

88. mags093 Group Title

no thats wrong :/

89. amistre64 Group Title

....buggers :/

90. mags093 Group Title

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

91. amistre64 Group Title

yes

92. amistre64 Group Title

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

93. mags093 Group Title
94. mags093 Group Title

could that be right?

95. mags093 Group Title

still wrong :(

96. amistre64 Group Title

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|

97. amistre64 Group Title

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)

98. mags093 Group Title

thats right :) howd you figure it out?

99. amistre64 Group Title

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

100. mags093 Group Title

oh ok thanks :)

101. amistre64 Group Title

im going to bed after that one :) good luck

102. mags093 Group Title

ok thanks :) bye