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Evaluate : [(cosα cosβ , cosα sinβ , -sinα ),(-sin β , cosβ , 0),(sinα cosβ , sinα sinβ , cosα )] |
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Answer» Expanding along the first row, we get Δ = cosα cosβ(cosβ cosα – 0) – cosα sinβ(–sinβ cosα) – sinα(–sinα sin2β – sinα cos2β) = cos2α cos2β + cos2α sin2β + sin2α sin2β+ sin2α cos2β = cos2α(cos2β + sin2β) + sin2α(sin2β + cos2β) = cos2α· 1 + sin2α· 1 = 1 |
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