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Math Help - Vector Problem. Finding the angles and sides of a triangle in 3d as well as the area.

  1. #1
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    Vector Problem. Finding the angles and sides of a triangle in 3d as well as the area.

    Problem:

    Find all angles and the lengths of all the sides in the triangle with the corners P(1,2,3) Q(0,2,1) R(-2,4-2). Hence compute its area.

    Solution so far:

    Vectors (u = u2 - u1)

    PQ = (-1,0,-2), PR = (-3,2,-5)
    QP = (1,0,2), QR =(-2,2-3)
    RQ = (2,-2,3), RP = (3,-2,5)

    Lengths of vectors |u| = sqrt(u1^2 + u2^2 + u3^2)
    hence,

    |PQ| = sqrt(5)
    |PR| = sqrt(38)
    |QP| = sqrt(5)
    |QR| = sqrt(17)
    |RQ| = sqrt(17)
    |RP| = sqrt(38)

    Therefore the sides of the triangle are sqrt(5), sqrt(38) and sqrt(17)

    Area of parallelogram = the magnitude of the cross product of PQ and PR
    PQ cross PR = (4, 1 ,-2)

    |PQ cross PR| = sqrt(21)

    The area of the triangle is therefore 1/2 sqrt(21)

    Angles between vectors
    PQ and PR => cosine(a) = (PQ.PR)/(|PQ||PR|)
    therefore a = cos^-1 (PQ.PR)/(|PQ||PR|)

    Now if I calculate the angle between all the vectors then the sum of those angles should be 180 degrees, right...

    Are there any problems with my workings?

    Basically, am I doing anything wrong?

    Feedback would be greatly appreciated.
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  2. #2
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    You've done about twice as much work as what's needed.

    Your calculation of the lengths look correct (though you didn't need to do it twice).

    I would simply apply Heron's formula to find the area of the triangle, and use the Cosine rule to find all the angles...
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  3. #3
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    But would I not require all those vector lengths in order to apply the law of cosines to calculate the angles?

    That is angles between PQ and PR, QP and QR, RP and RQ
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    What I am saying is, the length of PR is the same as RP, the length of PQ is the same as QP, etc... Like I said, your calculations of the lengths look correct, but you only needed to do it three times, not six.

    So you really only need to work out the three lengths of the triangle. That is enough to be able to apply Heron's formula to find the area, and the cosine rule three times to find the angles. But your logic is correct, as using dot products is another valid method (in fact, the dot product is proven using the cosine rule). And yes, the angles should add to 180^{\circ}.
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  5. #5
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    Cool, thankyou very much.
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