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Area of A Triangle in 3D

  1. Aug 30, 2013 #1
    1. The problem statement, all variables and given/known data
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    2. Relevant equations
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    3. The attempt at a solution
    I actually need someone to check my work for 1.1 and 1.2. Is what I have done in 1.3 correct I mean it does not seem right?
     

    Attached Files:

  2. jcsd
  3. Aug 30, 2013 #2

    gneill

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    For 1.1 (b) it looks like they want you to take the magnitude of B in the denominator... otherwise the expression wouldn't make sense.

    For 1.1 (c), note that the items ux, uy, and uz are the unit vectors of the vector space. So when you take the cross product, don't introduce i, j, or k. The top row of your determinant should be |ux uy uz|. The same applies to 1.1 (d)

    For 1.2, consider what vectors being parallel or perpendicular implies for their dot and cross products.

    For 1.3, form two vectors that define the sides of the triangle and carry out the cross product. Then ##A = \frac{1}{2}|V_1 \times V_2|##.
     
  4. Aug 30, 2013 #3
    Ok so the denominator for (b) should be 23.065. For (c) 8UxUyUz. For (d) -26UxUz. on 1.2 that is precisley what had me lost. I know that if they are parallel, then one vector must be a multiple of another. If perpendicular the dot product must result in a 0 and with alpha and beta only being tied to two terms. I struggle with how I can make this happen on the Z value for the parallel part. For part 1.3 A=48.489
     

    Attached Files:

  5. Aug 30, 2013 #4
    AB dot AC = AB dot AC cos(theta) is how I find the angle right?
     
  6. Aug 30, 2013 #5

    gneill

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    Sure. So what then are the components of the resulting vector?
    I don't understand that. The result of a cross product should be a vector.
    Part d contains a cross product followed by a dot product. The result should be a single scalar value.
    For the parallel vectors you can write x*A = B, for some scalar value x. Expand that and you'll see three equations. Solve for x (simple!). Use x to find ##\alpha## and ##\beta##.
    Show your work; that's not the value I'm seeing.

    EDIT: Sorry, I didn't see your attachments. Looking at your work for 1.3, I see that you've done something bizarre to find the vectors defining the triangle: you're multiplying the point components! A vector from point A to point B is given by VAB = B - A.

    For your cross products, the Ux, Uy, Uz stuff should ONLY appear in the top line of the determinant! Ux, Uy, and Uz designate the unit vectors of the vector space (or coordinate system axes). A vector written as: 3Ux + 5Uy -6Uz has Ux component 3, Uy component 5, and Uz component -6. It might also be written as: (3, 5, -6), with the understanding that those are the Ux, Uy, Uz components.
     
    Last edited: Aug 30, 2013
  7. Aug 30, 2013 #6

    gneill

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    For two vectors A and B, ##A \cdot B = |A||B| cos(\theta)##

    then ##cos(\theta) = \frac{A \cdot B}{|A||B|}##

    Pay attention to the vector directions: you want to take the angle between pairs such that the vectors are diverging from the triangle vertex. It may help to sketch the triangle "flat" using the vector lengths as the sides. Identify the "points" with the vertices and the vector directions along the triangle sides. You'll know something went wrong if the angle sum is not 180° :smile:
     
  8. Aug 30, 2013 #7
    For (b) I get 2.948Ux+3.728Uy-1.213Uz. Did I do the cross product incorrect or something I am not understanding on part (c). Once again on part (d) I did the cross product and then used the dot product I am confused on what to do now?
     
  9. Aug 30, 2013 #8

    gneill

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    Your part (b) result looks fine.

    For the cross products, take a look at the EDITs I added to post #5.
     
  10. Aug 30, 2013 #9
    ok so X(Ux+3Uy-2Uz)=4Ux+Uy+8Uz then XUx+3XUy-2XUz = 4Ux+Uy+8Uz then Uz(-2x-8)=Ux(4-x)+(1-3x)Uy
     
  11. Aug 30, 2013 #10
    How do I view the edits it still looks the same on my comp and BTW I cant read the code in #6
     
  12. Aug 30, 2013 #11

    gneill

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    What happened to the parameters ##\alpha## and ##\beta##? You should have:

    ##X(\alpha, 3, -2) = (4, \beta, 8)##

    from which you have three equations if you equate the like components.
     
  13. Aug 30, 2013 #12

    gneill

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    If you refresh the page, the edits at the bottom of the post should appear.

    By "code" I presume you mean the mathematical expressions rendered in LaTex? Perhaps your browser is not compatible with the Physics Forums' usage of LaTex? What type of machine/operating system are you using?
     
  14. Aug 30, 2013 #13
    ok for c I get 6Uy+2Uz and for d I get -8
     
  15. Aug 30, 2013 #14
    Ok after doing as you suggested if I did it right alpha=-1 and beta=-12 also x=-4
     
  16. Aug 30, 2013 #15
    yah thanks the refresh worked
     
  17. Aug 30, 2013 #16

    gneill

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    Check the signs on your expansion of the cross product for c. -8 for part d looks fine.
     
  18. Aug 30, 2013 #17

    gneill

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    Right :smile:
     
  19. Aug 30, 2013 #18
    ok my signs were off on (c) 6uy-2Uz
     
  20. Aug 30, 2013 #19

    gneill

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    I still see a sign problem with the first term....!
     
  21. Aug 30, 2013 #20
    doh! so you are saying it is -6Uy-2Uz. Are you sure, I have checked it 3 times and feel confident in my initial answer?
     

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