Let \vec{\alpha }=3\hat{i}+\hat{j} and \vec{\beta }=2\hat{i}-\hat{j}+3\hat{k}. If 

\vec{\beta }=\vec{\beta _{1}}-\vec{\beta _{2}}, where \vec{\beta _{1}} is parallel to \vec{\alpha } and \vec{\beta _{2}} is perpendicular to \vec{\alpha }, then \vec{\beta _{1}}\times \vec{\beta _{2}} is equal to :

  • Option 1)

      -3\hat{i}+9\hat{j}+5\hat{k}           

  • Option 2)

     -3\hat{i}-9\hat{j}-5\hat{k}

  • Option 3)

     \frac{1}{2}\left ( -3\hat{i}+9\hat{j}+5\hat{k} \right )

  • Option 4)

    \frac{1}{2}\left ( 3\hat{i}-9\hat{j}+5\hat{k} \right )

 

Answers (1)

\vec{\beta _{1}}=\frac{\vec{\alpha \cdot \vec{\beta }}}{\left | \vec{\alpha } \right |^{2}}\cdot \vec{\alpha }=\frac{5}{10}\vec{\alpha }=\frac{3}{2}\hat{i}+\frac{1}{2}\hat{j}

\vec{\beta _{2}}=\vec{\beta _{1}}-\vec{\beta }=-\frac{1}{2}\hat{i}+\frac{3}{2}\hat{j}-3\hat{k}

\therefore \vec{\beta _{1}}\times \vec{\beta _{2}}=\begin{vmatrix} \hat{i}& \hat{j}& \hat{k}\\ \frac{3}{2}& \frac{1}{2}& 0\\ \frac{-1}{2}& \frac{3}{2}& -3 \end{vmatrix}

                     =-\frac{3}{2}\hat{i}+\frac{9}{2}\hat{j}+\frac{5}{2}k

                     =\frac{1}{2}\left ( -3\hat{i}+9\hat{j}+5\hat{k} \right )

 


Option 1)

  -3\hat{i}+9\hat{j}+5\hat{k}           

Option 2)

 -3\hat{i}-9\hat{j}-5\hat{k}

Option 3)

 \frac{1}{2}\left ( -3\hat{i}+9\hat{j}+5\hat{k} \right )

Option 4)

\frac{1}{2}\left ( 3\hat{i}-9\hat{j}+5\hat{k} \right )

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