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Thread: Angle in a 3-D Parallelogram

  1. #1
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    Angle in a 3-D Parallelogram

    Hello! I am in Calc III, and I have a question. The answer sheet gives me a different answer than I have, but I feel that I am correct. Could you all point me one way or another?

    Parallelogram with points: A(2, -1, 4); B(1,0,-1); C(1,2,3); D(2,1,8)

    The question is.... what is the interior angle at B?

    The answer sheet says use vectors AB and BC and comes up with 140.8 degrees.

    I used vectors BA and BC and got 39.23 degrees.

    Which one is correct? Thanks!
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  2. #2
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    Quote Originally Posted by SSVicious View Post
    Hello! I am in Calc III, and I have a question. The answer sheet gives me a different answer than I have, but I feel that I am correct. Could you all point me one way or another?


    Parallelogram with points: A(2, -1, 4); B(1,0,-1); C(1,2,3); D(2,1,8)

    The question is.... what is the interior angle at B?

    The answer sheet says use vectors AB and BC and comes up with 140.8 degrees.

    I used vectors BA and BC and got 39.23 degrees.

    Which one is correct? Thanks!
    Here is the problem the angles AB BA are no equal (why?) but
    $\displaystyle 180-39.23=...$
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  3. #3
    Member mfetch22's Avatar
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    Quote Originally Posted by SSVicious View Post
    Hello! I am in Calc III, and I have a question. The answer sheet gives me a different answer than I have, but I feel that I am correct. Could you all point me one way or another?
    Quote Originally Posted by SSVicious View Post

    Parallelogram with points: A(2, -1, 4); B(1,0,-1); C(1,2,3); D(2,1,8)

    The question is.... what is the interior angle at B?

    The answer sheet says use vectors AB and BC and comes up with 140.8 degrees.

    I used vectors BA and BC and got 39.23 degrees.

    Which one is correct? Thanks!
    I did the calculation and got 140.768 or 140.8. Heres what I did, I found the vector from $\displaystyle B$ to $\displaystyle A$ and called it vector $\displaystyle U$. I found $\displaystyle U$ by subtracting each corresponding point in $\displaystyle B$ from each point in $\displaystyle A$ to get vector $\displaystyle U = (-1, 1, -5)$.

    I then did the same thing with from points $\displaystyle B$ to $\displaystyle C$ and called this vector $\displaystyle V$. The subtraction resulted in: $\displaystyle V = (0, 2, 4)$.

    Then I computed the dot product, divided it by the product of the magnitudes of $\displaystyle U$ and $\displaystyle V$ and took the inverse cosine of this value, to find the angle between the vectors, like this:

    $\displaystyle cos(\theta) = \frac{VU}{|{V}||{U}|}$

    I found the angle to be 140.8

    I could be wrong, but check my work to make sure, dont assume I am right. Hope this helps
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