CET Karnataka Engineering CET - Karnataka Engineering Solved Paper-2006

  • question_answer
    If a and b are the two vectors such that \[|\vec{a}|=3\sqrt{3},\,|\vec{b}|=4\] and \[|\vec{a}+\vec{b}|=\sqrt{7}\], then the angle between a and b is:

    A)  \[{{120}^{o}}\]                

    B)  \[{{60}^{o}}\]

    C)  \[{{30}^{o}}\]                                   

    D)  \[{{150}^{o}}\]  

    Correct Answer: D

    Solution :

    Given that, \[|\vec{a}|=3\sqrt{3},\] \[|\vec{b}|=4\]and \[|\vec{a}+\vec{b}|=\sqrt{7}\] Now, \[|\vec{a}+\vec{b}{{|}^{2}}=|\vec{a}{{|}^{2}}+|\vec{b}{{|}^{2}}+2|\vec{a}||\vec{b}|\cos \theta \] \[\Rightarrow \]               \[{{(\sqrt{7})}^{2}}={{(3\sqrt{3})}^{2}}+{{4}^{2}}+2(3\sqrt{3})\,(4)\,\cos \theta \] \[\Rightarrow \]               \[7=27+16+24\sqrt{3}\,\cos \theta \] \[\Rightarrow \]               \[24\sqrt{3}\,\cos \theta =-36\] \[\Rightarrow \]               \[\cos \theta =-\frac{3}{2\sqrt{3}}=-\frac{\sqrt{3}}{2}\] \[\Rightarrow \]               \[\theta ={{150}^{o}}\]


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