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Manipal Engineering Manipal Engineering Solved Paper-2015

  • question_answer
    Let \[\int_{{}}^{{}}{\frac{1-{{x}^{2}}}{(1+{{x}^{2}})\sqrt{1+{{x}^{4}}}}dx}\]and \[\sqrt{2}{{\sin }^{-1}}\left\{ \frac{\sqrt{2}x}{{{x}^{2}}+1} \right\}+C\]vector such that \[\frac{1}{\sqrt{2}}{{\sin }^{-1}}\left\{ \frac{\sqrt{2}x}{{{x}^{2}}+1} \right\}+C\]and the angle between \[\frac{1}{2}{{\sin }^{-1}}\left\{ \frac{\sqrt{2}x}{{{x}^{2}}+1} \right\}+C\] and c is 30°. Then, \[\Delta ABC,\left| \begin{matrix}    1 & a & b  \\    1 & c & a  \\    1 & b & c  \\ \end{matrix} \right|=0,\]is equal to

    A) \[{{\sin }^{2}}A+{{\sin }^{2}}B\text{ }+{{\sin }^{2}}C\]                                    

    B) \[\frac{3\sqrt{3}}{2}\]

    C) 2                                             

    D) 3

    Correct Answer: B

    Solution :

    We have, \[a\times b=\left| \begin{matrix}    {\hat{i}} & {\hat{j}} & {\hat{k}}  \\    2 & 1 & -2  \\    1 & 1 & 0  \\ \end{matrix} \right|=2\hat{i}-2\hat{j}+\hat{k}\]and \[|a\times b|=\sqrt{{{2}^{2}}+{{2}^{2}}+{{1}^{2}}}=\sqrt{4+4+1}=3\] \[\therefore \]\[|(a\times b)\times c|=|a\times b|\,\,|c|\sin {{30}^{o}}\] \[=\frac{3}{2}|c|\] Given,   \[|c-a|=2\sqrt{2}\] \[\Rightarrow \]\[|c-a{{|}^{2}}=8\] \[\Rightarrow \]\[|c{{|}^{2}}+|a{{|}^{2}}-2(a.c)=8\] \[\Rightarrow \]\[|c{{|}^{2}}+9-2|c|=8\] \[\Rightarrow \]\[|c{{|}^{2}}-2|c|+1=0\]\[\Rightarrow \]\[|c|=1\] Hence, \[|(a\times b)\times c|=\frac{3}{2}\]


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