JEE Main & Advanced JEE Main Paper (Held On 12 April 2014)

  • question_answer
    If \[\hat{x},\hat{y}\]and \[\hat{z}\] are three unit vectors in three-dimensional space, then the minimum value of\[|\hat{x}+\hat{y}{{|}^{2}}+|\hat{y}+\hat{z}{{|}^{2}}+|\hat{z}+\hat{x}{{|}^{2}}\]   [JEE Main Online Paper ( Held On 12 Apirl  2014 )

    A) \[\frac{3}{2}\]                                   

    B) 3

    C) \[3\sqrt{3}\]                                      

    D) 6

    Correct Answer: B

    Solution :

                    \[{{(\hat{x}+\hat{y}+\hat{z})}^{2}}\ge 0\] \[\Rightarrow \]\[3+2\Sigma \hat{x},\hat{y}\ge 0\]\[\Rightarrow \]\[2\Sigma \hat{x}.\hat{y}\ge -3\] Now, \[|\hat{x}+\hat{y}{{|}^{2}}+|\hat{y}+\hat{z}{{|}^{2}}+|\hat{z}+\hat{x}{{|}^{2}}\] \[=6+2\Sigma \hat{x}.\hat{y}\ge 6+(-3)\] \[\Rightarrow \]\[|\hat{x}+\hat{y}{{|}^{2}}|\hat{y}+\hat{z}{{|}^{2}}+|\hat{z}+\hat{x}{{|}^{2}}\ge 3\]

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