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The area of the parallelogram formed by the vectors p = (a, 1, -1) and q = (1, 1, 2) is root 35. Determine the values of 'a' for which this is true.

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Anonymous

  1. the area of the parallelogram is the magnitude of the cross product

    so [a,1,-1] x [1,1,2 ]
    = [ 3 , -2a-1 , a-1]
    then | [ 3 , -2a-1 , a-1]| = √35

    √(9 + 4a^2 + 4a + 1 + a^2 - 2a + 1) = √35
    5a^2 +2a + 11 = 35
    5a^2 + 2a - 24 = 0
    (a - 2)(5a + 12) = 0
    a = 2 or a = -12/5

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    Reiny

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