Question

If \[\vec{a} = \hat{ i }  + \hat{ j }  - \hat{ k }  , \vec{b} = - \hat{ i }  + 2\hat{ j }  + 2 \hat{ k }  \text{ and }  \vec{c} = - \hat{ i } + 2 \hat{ j }  - \hat{ k }  ,\]  then a unit vector normal to the vectors \[\vec{a} + \vec{b} \text{ and }  \vec{b} - \vec{c}\]  is

 

विकल्प

• \[\hat{ i } \]

• \[\hat{ j } \]

• \[\hat{ k } \]

• none of these 

Answer 1

\[\vec{a} + \vec{b} = 0 \hat{ i }  + 3 \hat{ j }  + \hat{ k }  \]

\[ \vec{b} - \vec{c} = 0 \hat{ i  }- 0 \hat { j }  + 3 \hat{ k }  \]

\[\left( \vec{a} + \vec{b} \right) \times \left( \vec{b} - \vec{c} \right) = \begin{vmatrix}\hat{ i  } & \hat{ j }  & \hat{ k }  \\ 0 & 3 & 1 \\ 0 & 0 & 3\end{vmatrix}\]

\[ = 9 \hat{ i }  \]

\[\left| \left( \vec{a} + \vec{b} \right) \times \left( \vec{b} - \vec{c} \right) \right| = 9 \left| \hat{ i }  \right|\]

\[ = 9\left( 1 \right)\]

\[ = 9\]

\[\text{ Unit vector perpendicular to both } \vec{a} + \vec{b} \text{ and  } \vec{b} - \vec{c} = \frac{\left( \vec{a} + \vec{b} \right) \times \left( \vec{b} - \vec{c} \right)}{\left| \left( \vec{a} + \vec{b} \right) \times \left( \vec{b} - \vec{c} \right) \right|}\]

\[ = \frac{9 \hat{ i } }{9}\]

\[ = \hat{ i } \]