JEE Main & Advanced Mathematics Vector Algebra Question Bank Scaler or Dot product of two vectors and its application

  • question_answer
    The angle between the vectors \[3\,\mathbf{i}+\mathbf{j}+2\,\mathbf{k}\] and \[2\,\mathbf{i}-2\,\mathbf{j}+4\,\mathbf{k}\] is                                [MP PET 1990]

    A)             \[{{\cos }^{-1}}\frac{2}{\sqrt{7}}\]

    B)             \[{{\sin }^{-1}}\frac{2}{\sqrt{7}}\]

    C)             \[{{\cos }^{-1}}\frac{2}{\sqrt{5}}\]

    D)             \[{{\sin }^{-1}}\frac{2}{\sqrt{5}}\]

    Correct Answer: B

    Solution :

               \[\cos \theta =\frac{3(2)+(1)(-2)+2(4)}{\sqrt{9+1+4}\sqrt{4+4+16}}\]\[=\frac{12}{\sqrt{14}\sqrt{24}}=\frac{6}{\sqrt{14}\sqrt{6}}\]                                 \[\Rightarrow \cos \theta =\frac{\sqrt{3}}{\sqrt{7}}\Rightarrow \sin \theta =\frac{2}{\sqrt{7}}\] Þ \[\theta ={{\sin }^{-1}}\left( \frac{2}{\sqrt{7}} \right)\].


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