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16 March, 11:29

Find II 2e-3f II^2 assuming that e & f are unit vectors such that II e + f II=sqrt (3/2).

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  1. 16 March, 11:52
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    We are given that ||e|| = 1, ||f|| = 1.

    Since ||e + f|| = sqrt (3/2), we have

    3/2 = (e + f) dot (e + f)

    = (e dot e) + 2 (e dot f) + (f dot f)

    = ||e||^2 + 2 (e dot f) + ||f||^2

    = 1^2 + 2 (e dot f) + 1^2

    = 2 + 2 (e dot f).

    So e dot f = - 1/4.

    Therefore,

    ||2e - 3f||^2 = (2e - 3f) dot (2e - 3f)

    = 4 (e dot e) - 12 (e dot f) + 9 (f dot f)

    = 4||e||^2 - 12 (e dot f) + 9||f||^2

    = 4 (1) ^2 - 12 (-1/4) + 9 (1) ^2

    = 4 + 3 + 9

    = 16.
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