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Math Help - vectors and sclalar products

  1. #1
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    vectors and sclalar products

    The points A,B and C have position vectors given respectively
    a = ( 7i + 4j -2k) b= ( 5i + 3j - 3k) c= ( 6i + 5j - 4k)

    a) Find the angle BAC
    b) Find the area of the triangle ABC

    i have dont part a) and got 60 degrees but i cant do part b)

    thanks in advance!
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  2. #2
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    This site has a good method

    Area of a Triangle via the Cross Product
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  3. #3
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    Hello, Oasis1993!

    The points A,B and C have position vectors given respectively:
    . . a = \langle7,4,-2\rangle,\;\;b = \langle5,3,-3\rangle, \;c = \langle6, 5,-4\rangle

    b) Find the area of \Delta ABC.
    You are expected to know this fact:

    The area of the parallelogram determined by two vectors
    . . is the magnitude of the cross product of the two vectors.
    Code:
              * . . . . . .
             /           .
          v /           .
           /           .
          * - - - - - *
                u
    
       Area  =  |u x v|

    Hence, the area of the triangle determined by two vectors is: . \tfrac{1}{2}|\vec u \times \vec v|


    Let: . \begin{array}{ccccc}\vec u &=& \overrightarrow{ab} &=& \langle \text{-}2,\text{-}1,-1\rangle \\ \vec v &=& \overrightarrow{bc} &=& \langle 1,2,\text{-}1\rangle \end{array}


    Then: . \vec u \times \vec v \;-\;\begin{vmatrix}i&j&k \\ \text{-}2&\text{-}1&\text{-}1 \\ 1&2&\text{-}1 \end{vmatrix} \;=\;3i - 3j - 3k \;=\;\langle3,\text{-}3,\text{-}3\rangle


    \text{Area of parallelogram}\;=\;\sqrt{3^2 + (\text{-}3)^2+(\text{-}3)^2} \;=\;\sqrt{27} \;=\;3\sqrt{3}


    . . \text{Therefore: area of triangle} \:=\:\frac{3\sqrt{3}}{2}

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  4. #4
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    Hello thanks for your answer but i dont understand what you did here:
    Then: .
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