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Let `vecu = 2hati - hatj + hatk, vecv = -3hatj + 2hatk` be vectors in `R^(3)` and `vecw` be a unit vector in the xy-plane. Then the maximum possible value of `|(vecu xx vecv).vecw|` is-A. `sqrt(5)`B. `sqrt(12)`C. `sqrt(13)`D. `sqrt(17)` |
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Answer» Correct Answer - D `vecuxxvecv = (2hati - hatj + hatk)xx(-3hatj + 2hatk)` `-6hatk - 4hatj - 2hati + 3hati = hati - 4hatj - 6hatk` Let `vecw = a hati + bhatj , a^(2) + b^(2) = 1 , a cos theta , b = sintheta` `vecu xx vecv.vecw = a - 4b = costheta - 4 sin theta` max. value `= sqrt(1^(2) +(-4)^(2)) = sqrt(17)` |
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