Vectors, Volume, and plane equations.

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Homework Statement


Four planes intersect in such a way that there is no common point between all four of them but that each set of three has a single common point. The position vectors of the four common points,A,B,C,D are, <-4,-3,-3> ,<7,5,-2>,<1,-2,5>,<-4,6,-1> respectively.
(a) Find the volume enclosed by the four planes. (b) Find the cartesian equations of the four planes.


Homework Equations



When I drew this out I got something that reminded me of a tetrahedron. But it looked like there were two of them. So, in general I for part a i just did V = 1/6[abc] + 1/6[abd]
I would write it but I don't have my papers with me.

The Attempt at a Solution



For the Cartesian equations I don't know what to use. Draw the picture and you will see that there are more then 4 faces so it is confusing me. How do I decifer which ones to use?
Thanks
 

Answers and Replies

  • #2
verty
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How did you get that formula for volume?
 
  • #3
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When I drew the picture it looked to me like two tetrahedrons. So I just did 1/6[abc]+1/6[abd]. Because the tetrahedron is 1/6th the parrelliped formed when you do a tripple scalar product.
 
  • #4
LCKurtz
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If you connect the four points all the ways possible you get one tetrahedron with four faces. But you don't use the position vectors to the points in your calculations. You have to use vectors along the sides.
 
  • #5
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Ah. So it isn't two then. How did you know that then? I connected them but obviously wrong. That is my problem.
 
  • #6
haruspex
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Ah. So it isn't two then. How did you know that then? I connected them but obviously wrong. That is my problem.
If O is the origin, the given vectors are OA, OB, OC, OD. How would you obtain the vector AB from OA and OB?
 
  • #7
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OK, I did

c-a = < 1,-2,5> - < 4,-3,-3> = < -3,1,8>
b-a = < 7 , 5 , -2 > - < 4,-3,-3> = < 3, 8 ,1>
d-a = <-4,6,-1> - < 4,-3,-3> = < 8,9,2>


Then I did the box product and divided by 6.

< -3,1,8> dot < 8,9,2> cross < 3, 8 ,1> = < -3(9-16) -1(8-6) + 8(64-27)> = 105/2 units^3

V = [ (c-a)(d-a) (b-a)] = 315/6 = 105/2 units^3


?
 
  • #8
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Hey this is just a suggestion but check the distance between the points, try to find some type of symmetry which can produce the vol
 
  • #10
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<-8,9,2>

Thanks. Is the volume correct. The other parts
 
  • #11
LCKurtz
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Homework Statement


Four planes intersect in such a way that there is no common point between all four of them but that each set of three has a single common point. The position vectors of the four common points,A,B,C,D are, <-4,-3,-3> ,<7,5,-2>,<1,-2,5>,<-4,6,-1> respectively.

OK, I did

c-a = < 1,-2,5> - < 4,-3,-3> = < -3,1,8>
b-a = < 7 , 5 , -2 > - < 4,-3,-3> = < 3, 8 ,1>
d-a = <-4,6,-1> - < 4,-3,-3> = < 8,9,2>

Don't you have ##a## wrong in all three?
 
  • #12
haruspex
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Don't you have ##a## wrong in all three?

Yes, I didn't check that. I believe there are also errors in the triple product calculation.
 
  • #13
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I wrote it wrong origonally it should be a =<4,-3,-3> not a = <-4,-3,-3> Sorry, all the others are correct I double checked though. And I redid the calculation and I got 693/6 units cubed = 231/2 units^3.
 
  • #14
haruspex
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I wrote it wrong origonally it should be a =<4,-3,-3> not a = <-4,-3,-3> Sorry, all the others are correct I double checked though. And I redid the calculation and I got 693/6 units cubed = 231/2 units^3.

That's what I got.
 
  • #15
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Very cool. I will work on part b now. Thank you for the help haruspex and LCKurtz
 
  • #16
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OK, I think I worked out part b
I will show how I found each of the 4 planes. But in general I need a vector perpendicular to the plane, general points X = < x,y,z> and a point on the plane for each.


(b-a) = <7,5,-2> - < 4,-3,-3> = <8,8,1>

(d-a) = <-8,9,2>

(b-a) cross (d-a) = <7,-14,91>
A point on the plane is <-4,6,-1> = d

Then,
<x,y,z> dot <7,-14,91> = <-4,6,-1> dot <7,-14,91>

7x -14y+91z = -203
implies -x +2y -13z = 29

The next one

(c-a) = <1,-2,5> - < 4,-3,-3> = <-3,1-8>

(b-a) = <3,8,1>

(c-a) cross (b-a) = <-63,27,-27>
A point on the plane is <1,-2,5> = c

Then,
<x,y,z> dot <-63,27,-27> = <1,-2,5> dot <-63,27,-27>
-63x +27y-27z = -252
implies 7x -3y+ 3z = 28


The next one

(c-a) = <1,-2,5> - < 4,-3,-3> = <-3,1-8>

(d-a) = <-8,9,2>

(c-a) cross (d-a) = <-70,-58,-19>
A point on the plane is <1,-2,5> = d

Then,
<x,y,z> dot <-70,-58,-19> = <1,-2,5> dot <-70,-58,-19>

implies 70x +58y +19z = 49


The last one

(c-a) = <1,-2,5> -<-4,6,-1> = <5,-8,6>
(b-a) = <7,5,-2> - <-4,6,-1> = <11,-1,-1>


(c-a) cross (b-a) = <14,71,83>

14x +71y +83z = 287


Think that is right.
 
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  • #17
haruspex
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(b-a) = <7,5,-2> - < 4,-3,-3> = <8,8,1>
<3,8,1>, which is what you seem to have used later
7x -14y+91z = -203
implies -x +7y -13z = 29
Another typo?
You can check the answers easily enough by seeing whether the given points satisfy the equations for the planes they're in.
 
  • #18
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<3,8,1>, which is what you seem to have used laterAnother typo?
You can check the answers easily enough by seeing whether the given points satisfy the equations for the planes they're in.

Yeah it was a typo. How do you mean check? Process seem right?
 
  • #19
haruspex
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Yeah it was a typo. How do you mean check? Process seem right?
Yes, the method's fine.
If you have an equation for a plane that is supposed to pass through A, B, C, you can substitute the coordinates of A for x, y, z in the equation and see whether the point satisfies the equation. Just a check that you've made no arithmetic errors.
 
  • #20
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I found one so far I edited it. I will double check them before again before I have to submit it.
Thanks
 

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