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Vector Methods To Find An Area

  1. Mar 2, 2014 #1
    Hello,

    after searching around on the internet about this problem, it looks like it is Cramer's rule that I want to use, though it wasn't shown to us under that name.

    My textbook doesn't cover the material required for this problem, so i'd really like to run what I have done past you guys :)

    1. The problem statement, all variables and given/known data

    The triangle ABC has vertices A(-1,3,0), B(-3,0,7), C(-1,2,3). Find AB, AC, CB.

    Find also the area of the triangle, using a vector method.

    2. Relevant equations



    3. The attempt at a solution

    I found the vectors easily enough.

    AB = -2i - 3j + 7k

    AC = -j + 3k

    CB = -2i - 2j + 4k

    Area of a triangle = 1/2 base x height

    [itex]A[/itex] = [itex]( \frac{1}{2})[/itex] (AB x AC)

    I don't know how to make matrices in latex, but I use what looks to me like cramer's rule (we were just told how to do it, as we haven't covered determinants yet) and come to an answer that;

    A = -i + 3j + K

    I hope it's clear to you, but if not, if you can show me how to display matrices i'll gladly show the working.

    Thanks!
     
  2. jcsd
  3. Mar 2, 2014 #2

    micromass

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    Area is supposed to be a number no? You seem to get that area is a vector.
     
  4. Mar 2, 2014 #3
    Hm, yes.

    Area = |AB x AC| <--- Which I didn't actually say, so thanks for pulling that up.

    So I should give my answer as '3'?
     
  5. Mar 2, 2014 #4

    micromass

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    That's the area of the parallellogram formed by AB and AC, no? You want the triangle, so the factor 1/2 is necessary.
     
  6. Mar 2, 2014 #5
    Ack, yes.

    The factor of a half was included in my calculations, but I forgot to show it in my reply to your comment.

    A = 1/2 |AB x AC|

    = 1/2( -2 + 6 + 2)
    = 3
     
  7. Mar 2, 2014 #6

    micromass

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    That's a weird way of calculating the length. Isn't the length of a vector given by

    [tex]|(x,y,z)| = \sqrt{x^2 + y^2 + z^2}[/tex]
     
  8. Mar 2, 2014 #7
    Yes...

    Is the method I tried to use wrong? (excusing the mistakes we have already discussed)

    When using the method described above, my answers don't agree.
     
  9. Mar 2, 2014 #8

    micromass

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    You want to calculate the length of the vector ##(-1,3,1)##. You seem to do this by calculating

    [tex]-1 + 3 + 1[/tex]

    I don't see where that comes from. Isn't the length rather equal to

    [tex]\sqrt{(-1)^2 + 3^2 + 1^2}[/tex]
     
  10. Mar 2, 2014 #9
    I agree with you.

    Area of a triangle = 1/2 base x height

    |AB| = [itex]\sqrt{(-2)^{2} + (-3)^{2} + (7)^{2}}[/itex] = [itex]\sqrt{62}[/itex]

    |AC| = [itex]\sqrt{(0)^{2} + (-1)^{2} + (3)^{2}}[/itex] = [itex]\sqrt{10}[/itex]

    ∴ A = 1/2([itex]\sqrt{10}[/itex] [itex]\sqrt{62}[/itex]) = [itex]\sqrt{155}[/itex] ≈ 12.45

    Can this problem be solved using the cross product, or am I barking up the wrong tree?

    Thank you for the help by the way.
     
  11. Mar 2, 2014 #10

    micromass

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    Right now you're going wrong. You can't do this because you don't know that AC is actually the height. You just know it's a side from the triangle.

    You were actually doing very well before. We established that the area of the triangle is given by

    [tex]\frac{1}{2}|\mathbf{AC}\times\mathbf{AB}| = \frac{1}{2}|(-1,3,1)|[/tex]

    The only thing I disagreed with is how you practically calculated the length.
     
  12. Mar 2, 2014 #11
    Ok, i've got myself in a bit of a muddle here.

    Any side can be my base, so i'll say AB is. AC is therefore, the diagonal to the 'top'.

    The height is |AC|sinθ, where θ is the angle between AB and AC.

    AB.AC = 24

    AB.AC = |AB||AC| cosθ

    |AB||AC| cosθ = 24

    θ = arccos ([itex]\frac{24}{\sqrt{10}\sqrt{62}}[/itex])

    = 15.45°

    Height, |AC|sinθ = [itex]\sqrt{10}[/itex] sin(15.45)
    ≈ 0.84

    Now I should be able to say that the area = 1/2 base x height

    = 1/2 * [itex]\sqrt{62}[/itex] * 0.84
    = 3.32
     
  13. Mar 2, 2014 #12

    micromass

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    You're really making it difficult for yourself. You almost had the right answer in post 5, the only thing was that you calculated the distance wrongly.
     
  14. Mar 2, 2014 #13
    A parallelogram can be divided into 2 congruent triangles, so the area of the triangle should be half the area of the parallelogram as worked out by doing the cross product.

    Area = [itex]\frac{1}{2}[/itex] |AB|x|AC| = [itex]\frac{1}{2}[/itex] |-2, -3, 7|x|0, -1, 3|

    = [itex]\frac{1}{2}[/itex] (-2, -6, 2)

    = [itex]\frac{1}{2}[/itex] [itex]\sqrt{(-2)^{2} + (-6)^{2} + (2)^{2}}[/itex]

    = [itex]\sqrt{11}[/itex]

    That was a lot simpler, but i'll take some solace from the fact that it agrees with my longwinded route. (fingers crossed that this is correct!)
     
  15. Mar 2, 2014 #14

    micromass

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    Yep, you've got it!! Congratulations!
     
  16. Mar 2, 2014 #15
    Thanks so much for your patience.

    The format here, where you allow the question askers to arrive at the solution themselves is very beneficial and definitely more conducive to understanding than simply providing a solution.
     
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