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12th Class Mathematics Sample Paper Mathematics Olympiad - Sample Paper-1

  • question_answer
    If \[\left| \overrightarrow{a} \right|=3,\left| \overrightarrow{b} \right|=1,\left| \overrightarrow{c} \right|=4\] and\[\overrightarrow{\mathbf{a}}+\mathbf{\vec{b}}+\mathbf{\vec{c}}=\mathbf{0}\],then \[\mathbf{\vec{a}}.\mathbf{\vec{b}}+\mathbf{\vec{b}}.\mathbf{\vec{c}}+\mathbf{\vec{c}}.\mathbf{\vec{a}}\] =?

    A)   10                  

    B)  \[-10\]          

    C)  13                               

    D)  \[-13\]

    Correct Answer: D

    Solution :

    [d] \[\left| \overrightarrow{a} \right|=3,\left| \overrightarrow{b} \right|=1,\left| \overrightarrow{c} \right|=4\] \[\because \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=0.\] Squaring both sides, we have \[{{\left| \overrightarrow{a} \right|}^{2}}+{{\left| \overrightarrow{b} \right|}^{2}}+{{\left| \overrightarrow{c} \right|}^{2}}+2\left( \overrightarrow{a}.\overrightarrow{b}+\overrightarrow{b}.\overrightarrow{c}+\overrightarrow{c}.\overrightarrow{a} \right)=0\] \[\Rightarrow 9+1+16+2\left( \overrightarrow{a}.\overrightarrow{b}+\overrightarrow{b}.\overrightarrow{c}+\overrightarrow{c}.\overrightarrow{a} \right)=0\]          \[\Rightarrow \overrightarrow{a}.\overrightarrow{b}+\overrightarrow{b}.\overrightarrow{c}+\overrightarrow{c}.\overrightarrow{a}=\frac{-26}{2}=-13\] Hence, option [d] is correct.


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