Manipal Engineering Manipal Engineering Solved Paper-2015

  • question_answer
    Let a.b and c be non-zero vectors such that no two are collinear and \[{{l}^{-}}\]a. If \[{{K}^{+}};{{l}^{-}}\] is the acute angle between the vectors b and c, then sin \[{{l}_{2}}\] equals

    A) \[l_{3}^{-}\]

    B) \[{{(1.0002)}^{3000}}\]

    C) \[(a.\hat{i})(a\times \hat{i})+(a.\hat{j})(a\times \hat{j})+(a.\hat{k})(a\times \hat{k})\]                                

    D) \[A=\{(x,y):{{x}^{2}}+{{y}^{2}}=25\}\]

    Correct Answer: A

    Solution :

    We have, \[(a\times b)\times c=\frac{1}{3}|b|c|a\] \[\Rightarrow \]\[(a.c)b-(b.c)a=\frac{1}{3}|b|c|a\] \[\Rightarrow \]\[(a.c)b-\left\{ (b.c)+\frac{1}{3}|b|\,c| \right\}a=0\] \[\Rightarrow \]\[a.c=0\]and\[b.c+\frac{1}{3}|b|\,c|=0\] [\[\because a,b\]are non-collinear] \[\Rightarrow \]\[|b|\,c|\cos \theta +\frac{1}{3}|b|\,c|=0\] \[\Rightarrow \]\[\cos \theta =-\frac{1}{3}\] \[\Rightarrow \]\[\sin \theta =\sqrt{1-{{\cos }^{2}}\theta }=\sqrt{1-{{\left( -\frac{1}{3} \right)}^{2}}}\] \[=\sqrt{1-\frac{1}{9}}=\sqrt{\frac{8}{9}}=\frac{2\sqrt{2}}{3}\]                                


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