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Math Help - 3D Vector Geometry Problem

  1. #1
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    3D Vector Geometry Problem

    Greeting MathHelpForum,

    I'm currently working through grade 12 Calculus and Vectors in an independent-learning course. I've just recently finished the calculus portion of the course, finding it relatively painless on the whole, and so I assumed the vector portion would be a walk in the park, especially since I've done some work with vectors in grade 12 physics. That does not seem to be case however; I've had a few questions now that have stumped me for quite a while, and I really can't seem to wrap my head around this one in particular. I don't have any teachers or instructors for this course than I can go to for help, so if anyone here could point out where I'm going wrong it would be greatly appreciated. The question is as follows:

    The vertices of a figure are given by A(-8, 4, 2), B(-6, 3, 5) and C(-10, 5, -9). What type of figure is ABC? Justify your answer. (Hint: Calculate the magnitudes of the sides of the figure.)

    and here is my attempt at an answer:

    AB = (-8, 4, -2) (-6, 3, 5) = (-2, 1, -7)
    |AB| = √[(-2)2 + (1)2 + (-7)2] = √(54) ≈ 7.35
    BC = (-6, 3, 5) (-10, 5, -9) = (4, -2, 14)
    |BC| = √[(4)2 + (-2)2 + (14)2] = √(216) ≈ 14.7
    CA = (-10, 5, -9) (-8, 4, -2) = (-2, 1, -7)
    |CA| = √[(-2)2 + (1)2 + (-7)2] = √(54) ≈ 7.35

    Can I simply leave my answer here, and state that ABC is an isosceles triangle? I know, at this point, that ABC has three vertices and three sides, two of which are the same, is this enough information? To me, it seemed that I should go a step further, and prove that ABC is a triangle by showing that all internal angles add up to 180. Since I have all three sides, I should be able to use Cosine Law to determine each angle, but when I apply Cosine Law I find the following:

    Let the internal angles of this figure be a, b and c
    cos a = (|BC|2 + |CA|2 |AB|2) / 2|BC||CA|
    = [(14.7)2 + (7.35)2 (7.35)2] / 2(14.7)(7.35)
    = 216.09 / 216.09 = 1
    a = cos-1 (1) = 0

    cos b = (|AB|2 + |CA|2 |BC|2) / 2|AB||CA|
    = [(7.35)2 + (7.35)2 (14.7)2] / 2(7.35)(7.35)
    = -108.045 / 108.045 = -1
    b = cos-1 (-1) = 180

    cos c = (|AB|2 + |BC|2 |CA|2) / 2|AB||BC|
    = [(7.35)2 + (14.7)2 (7.35)2] / 2(7.35)(14.7)
    = 216.09 / 216.09 = 1
    a = cos-1 (1) = 0

    So, by my calculations, the three angles DO add up to 180, but only because two of the angles are zero... I'm not sure what to make of this result, can anyone enlighten me?
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  2. #2
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    Re: 3D Vector Geometry Problem

    Quote Originally Posted by Lethargic View Post
    Greeting MathHelpForum,

    I'm currently working through grade 12 Calculus and Vectors in an independent-learning course. I've just recently finished the calculus portion of the course, finding it relatively painless on the whole, and so I assumed the vector portion would be a walk in the park, especially since I've done some work with vectors in grade 12 physics. That does not seem to be case however; I've had a few questions now that have stumped me for quite a while, and I really can't seem to wrap my head around this one in particular. I don't have any teachers or instructors for this course than I can go to for help, so if anyone here could point out where I'm going wrong it would be greatly appreciated. The question is as follows:

    The vertices of a figure are given by A(-8, 4, 2), B(-6, 3, 5) and C(-10, 5, -9). What type of figure is ABC? Justify your answer. (Hint: Calculate the magnitudes of the sides of the figure.)

    and here is my attempt at an answer:

    AB = (-8, 4, -2) (-6, 3, 5) = (-2, 1, -7)
    |AB| = √[(-2)2 + (1)2 + (-7)2] = √(54) ≈ 7.35
    A = (-8,4,2) not (-8,4,-2) as you've written when calculating |AB| and |AC|

    fix this and redo it.
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  3. #3
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    Re: 3D Vector Geometry Problem

    Quote Originally Posted by Lethargic View Post
    The vertices of a figure are given by A(-8, 4, 2), B(-6, 3, 5) and C(-10, 5, -9). What type of figure is ABC? Justify your answer. (Hint: Calculate the magnitudes of the sides of the figure.)
    and here is my attempt at an answer:
    AB = (-8, 4, -2) (-6, 3, 5) = (-2, 1, -7)
    |AB| = √[(-2)2 + (1)2 + (-7)2] = √(54) ≈ 7.35
    BC = (-6, 3, 5) (-10, 5, -9) = (4, -2, 14)
    |BC| = √[(4)2 + (-2)2 + (14)2] = √(216) ≈ 14.7
    CA = (-10, 5, -9) (-8, 4, -2) = (-2, 1, -7)
    |CA| = √[(-2)2 + (1)2 + (-7)2] = √(54) ≈ 7.35
    Notice that -2<-2,1,-7>=<4,-1,14> that is -2\overrightarrow{AB} =\overrightarrow {BC}  .
    When two vectors are multiples of each other then they are parallel.

    In this case  {A,~B,~\&~C} are co-linear. So there is no triangle.

    You should have seen that 2\sqrt {54}  = \sqrt {216}
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  4. #4
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    Re: 3D Vector Geometry Problem

    In answer to the question 'can I simply leave my answer here'?, the answer is no you can't. Why can't the three points lie on straight line ? In fact they do ! But only because of a typo. Check your co-ordinates for the point A.
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  5. #5
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    Re: 3D Vector Geometry Problem

    Thank you all so much! I can't believe I didn't catch that typo! Everything works out much better with the correct values, it is a triangle after all
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