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  • How to solve this problem- - Vector Algebra - JEE Main-4

If the volume of parallelopiped formed by the vectors \hat{i}+\lambda \hat{j}+\hat{k},\: \: \: \hat{j}+\lambda \hat{k}  and \lambda \hat{i}+\hat{k} is minimum, then \lambda is equal to : 


 

  • Option 1)

    \sqrt{3} 

  • Option 2)

     \frac{1}{\sqrt{3}} 

  • Option 3)

     -\frac{1}{\sqrt{3}}  

  • Option 4)

    -\sqrt{3}

 

Answers (1)

best_answer

vector \hat{i}+\lambda \hat{j}+\hat{k}

           \hat{j}+\lambda \hat{k}

           \lambda \hat{i}+\hat{k}

 

Volume of parellelopiped = \begin{vmatrix} 1 &\lambda &1 \\ 0 &1 &\lambda \\ \lambda &0 &1 \end{vmatrix}

f(x)=\left | 1\left ( 1 \right ) -\lambda \left ( -\lambda ^{2} \right )+1\left ( -\lambda \right )\right |

           =\left | \lambda ^{3}-\lambda +1 \right |

\because Question is asking the minimum value of parallelopied and the corresponding value of \lambda : the minimum value is zero

\because Cubic always has atleast one real root.

Hence the answer of this question will be the root of this equation. But none of the options satisfies the cubic. 

So the problem setter to the question in local minimum value of volume. 

so corrcet option (2)


Option 1)

\sqrt{3} 

Option 2)

 \frac{1}{\sqrt{3}} 

Option 3)

 -\frac{1}{\sqrt{3}}  

Option 4)

-\sqrt{3}

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