## mags093 2 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).

1. amistre64

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

2. amistre64

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

3. 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

4. mags093

so i find |PR| and |PS|?

5. amistre64

|dw:1359740566102:dw|

6. amistre64

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

7. mags093

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

8. amistre64

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

howd you get -S?

10. mags093

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

11. amistre64

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

12. 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

13. 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 :)

14. mags093

why sqrt9566)

15. 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

16. amistre64

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

17. mags093

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

18. 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.

19. amistre64

|dw:1359741666062:dw|

20. mags093

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

21. amistre64

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

22. 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)

23. mags093

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

24. mags093

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

25. 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?

26. mags093

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

27. amistre64

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

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

29. amistre64

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

30. 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?

31. mags093

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

32. mags093

ya i do

33. 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?

34. amistre64

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

35. mags093

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

36. mags093

i dont understand how to use that formula?

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

38. 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

39. mags093

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

40. amistre64

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

41. mags093

wolfram alpha?

42. amistre64

yep

43. amistre64

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

44. mags093

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

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

46. amistre64

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

47. mags093

its the one i tried earlier but couldnt get it!

48. amistre64

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

49. mags093

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

50. amistre64

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

51. mags093

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

52. 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>

53. 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

54. amistre64

now how do you find the angle between 2 vectors?

55. mags093

is it xi,yj,zk ?

56. amistre64

|dw:1359744085632:dw|

57. mags093

is it the length of the vectors * coz theta?

58. amistre64

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

59. amistre64

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

60. mags093

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

61. amistre64

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

62. amistre64
63. mags093

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

64. amistre64

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

65. mags093

oh ya just the distance not the angle

66. amistre64

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

67. amistre64

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

68. amistre64

or i think that shows |v| for my setup

69. 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

70. mags093

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

71. amistre64

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

72. amistre64

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

73. amistre64

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

74. mags093

|dw:1359745138247:dw|

75. mags093

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

76. 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

77. mags093

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

78. mags093

oh ya you changed that :)

79. 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

80. mags093

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

81. amistre64

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

82. 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

83. mags093

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

84. amistre64

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

85. mags093

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

86. 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) ??

87. mags093

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

88. mags093

no thats wrong :/

89. amistre64

....buggers :/

90. mags093

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

91. amistre64

yes

92. amistre64

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

93. mags093
94. mags093

could that be right?

95. mags093

still wrong :(

96. 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|

97. 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)

98. mags093

thats right :) howd you figure it out?

99. 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

100. mags093

oh ok thanks :)

101. amistre64

im going to bed after that one :) good luck

102. mags093

ok thanks :) bye