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12 April, 14:18

Find the area of the parallelogram that has adjacent sides u = i-3j+2k and v = 3j-k

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  1. 12 April, 16:47
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    The area is the given by the magnitude of the cross product of the side vectors. That value is

    Area = ||u*v|| = ||u||·||v||·sin (θ)

    where θ is the angle between the vectors.

    The cross product of [1, - 3, 2] and [0, 3, - 1] is [-3, 1, 3}, which has magnitude

    √ ((-3) ² + 1² + 3²) = √19 ≈ 4.3589
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