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Can someone explain - Vector Algebra - JEE Main-2

Let \vec{a}=3\hat{i}+2\hat{j}+2\hat{k} and  \vec{b}=\hat{i}+2\hat{j}-2\hat{k} be two vectors. If a vector perpendicular to both the vectors \vec{a}+\vec{b}  and  \vec{a}-\vec{b} has magnitude 12 then one such vector is : 



 

  • Option 1)

    4\left ( -2\hat{i}-2\hat{j}+\hat{k} \right )

  • Option 2)

    4\left ( 2\hat{i}-2\hat{j}-\hat{k} \right )

  • Option 3)

    4\left ( 2\hat{i}+2\hat{j}-\hat{k} \right )

     

  • Option 4)

    4\left ( 2\hat{i}+2\hat{j}+\hat{k} \right )

 
Answers (1)
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V Vakul

\vec{a}=3\hat{i}+2\hat{j}+2\hat{k} ,                      \vec{b}=\hat{i}+2\hat{j}-2\hat{k}

\left ( \vec{a} +\vec{b}\right )\times \left ( \vec{a}-\vec{b} \right )=2\left ( \vec{b} \times \vec{a}\right )

                                         =2\begin{vmatrix} \hat{i} &\hat{j} &\hat{k} \\ 1 &2 &-2 \\ 3 &2 &2 \end{vmatrix}

                                        =2\left ( 8\hat{i}-8\hat{j}+4\hat{k} \right )

                                        =8\left ( 2\hat{i}-2\hat{j}-\hat{k} \right )

\hat{h}=\frac{\vec{\alpha }\times \vec{\beta }}{\left | \vec{\alpha }\times \vec{\beta } \right |}=\frac{2\hat{i}-2\hat{j}-\hat{k}}{3}

required vector = \pm 12\cdot \frac{\left ( 2\hat{i}-2\hat{j}-\hat{j} \right )}{3}

                        =\pm 4\left ( 2\hat{i}-2\hat{j}-\hat{k} \right )         


Option 1)

4\left ( -2\hat{i}-2\hat{j}+\hat{k} \right )

Option 2)

4\left ( 2\hat{i}-2\hat{j}-\hat{k} \right )

Option 3)

4\left ( 2\hat{i}+2\hat{j}-\hat{k} \right )

 

Option 4)

4\left ( 2\hat{i}+2\hat{j}+\hat{k} \right )

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