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Math Help - scalar products

  1. #1
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    scalar products

    The vectors AB and AC are (-1, 0, 3) and ( 2, 4, 3) respectively.
    The vector AD is the sum of vector AB and AC. Determine the acute angle between the diagonals of the parallelogram defined by the points A,B,C and D.

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  2. #2
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    Hello, Oasis1993!

    Given vectors: . \overrightarrow{AB} = \langle -1,0,3\rangle,\;\;\overrightarrow{AC} = \langle 2, 4, 3\rangle

    . . . . . . . . . . . \overrightarrow{AD} \:=\:\overrightarrow{AB} + \overrightarrow{AC}

    Determine the acute angle between the diagonals
    of the parallelogram defined by the points A,B,C, D.
    Code:
              (-1,0,3)           (1,4,6)
                B o - - - - - - - - o D
                 /  *           *  /
                /     *     *     /
               /        * θ      /
              /     *     *     /
             /  *           *  /
          A o - - - - - - - - o C
         (0,0,0)           (2,4,3)

    We have: . \begin{array}{c}\vec u \:=\: \overrightarrow{AD} \:=\:\langle 1,4,6\rangle \\ \\[-3mm]<br />
\vec v \:=\:\overrightarrow{BC} \:=\:\langle 3,4,0\rangle \end{array}


    \cos\theta \:=\:\frac{u \cdot v}{|u||v|} \;=\;\frac{\langle 1,4,6\rangle\cdot\langle3,4,0\rangle}{\sqrt{1+16+3  6}\,\sqrt{9+16+0}} \:=\:\frac{19}{5\sqrt{53}} \:=\:0.521970143


    Therefore: . \theta \;\approx\;58.5^o

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  3. #3
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    Hello! Thank you very much for your answre but im confused on how you got BC?
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  4. #4
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    AB+BC=AC, therefore AB-AC=-BC.
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