Question

If \[\vec{a}\] is any vector, then \[\left( \vec{a} \times \hat{ i }  \right)^2 + \left( \vec{a} \times \hat{ j } \right)^2 + \left( \vec{a} \times \hat{ k }  \right)^2 =\]

विकल्प

• \[\vec{a}^2\]

   

• \[2 \vec{a}^2\]

• \[3 \vec{a}^2\]

   

• \[4 \vec{a}^2\]

Answer 1

\[2 \vec{a}^2\] 

\[\text{ Let }  \vec{a} = a_1 \hat{ i }  + a_2 \hat{ j }  + a_3 \hat{ k }  \]

\[ \vec{a} \times \hat{ i }  = \begin{vmatrix}\hat{ i }  & \hat{ j }  & \hat{ k }  \\ a_1 & a_2 & a_3 \\ 1 & 0 & 0\end{vmatrix}\]

\[ = a_3 \hat{ j }  - a_2 \hat{ k }  \]

\[ \Rightarrow \left( \vec{a} \times \hat{ i } \right)^2 = \left( a_3 \hat{ j }  - a_2 \hat{ k }  \right)^2 \]

\[ = {a_3}^2 \left| \hat{ j  } \right|^2 + {a_2}^2 \left| \hat{ k } \right|^2 - 2 a_3 a_2 \left( \hat{ j } . \hat{ k }  \right)\]

\[ = {a_3}^2 + {a_2}^2 (\because \hat{ j }  . \hat{ k }  =0) . . . (1)\]

\[ \therefore \vec{a} \times \hat{ j } = \begin{vmatrix}\hat{ i } & \hat{ j } & \hat{ k } \\ a_1 & a_2 & a_3 \\ 0 & 1 & 0\end{vmatrix}\]

\[ = - a_3 \hat{ i } + a_1 \hat{ k }  \]

\[ \Rightarrow \left( \vec{a} \times \hat{ j }  \right)^2 = \left( - a_3 \hat{ i } + a_1 \hat{ k } \right)^2 \]

\[ = {a_3}^2 \left| \hat{ i } \right|^2 + {a_1}^2 \left| \hat{ k }  \right|^2 - 2 a_3 a_2 \left( \hat{ i} . \hat{ k } \right)\]

\[ = {a_3}^2 + {a_1}^2 (\because \hat{ i } .\hat{ k } =0) . . . (2)\]

\[ \therefore \vec{a} \times \hat{ k }  = \begin{vmatrix}\hat{ i }  & \hat{ j } & \hat{ k }  \\ a_1 & a_2 & a_3 \\ 0 & 0 & 1\end{vmatrix}\]

\[ = a_2 \hat{ i  } - a_1 \hat{ j }   \]

\[ \Rightarrow \left( \vec{a} \times k \right)^2 = \left( a_2 \hat{ i} - a_1 \hat{ j}  \right)^2 \]

\[ = {a_2}^2 \left| \hat{ i }  \right|^2 + {a_1}^2 \left| j \right|^2 + 2 a_1 a_2 \left( \hat{ i }  . \hat{ j },\right)\]

\[ = {a_2}^2 + {a_1}^2 (\because \hat{ i }  . \hat{ j }  =0) . . . (3)\]

\[\text{ Adding (1), (2) and (3), we get } \]

\[ \left( \vec{a} \times \hat{ i }  \right)^2 + \left( \vec{a} \times \hat{ j }  \right)^2 + \left( \vec{a} \times k \right)^2 = {a_3}^2 + {a_2}^2 + {a_3}^2 + {a_1}^2 + {a_2}^2 + {a_1}^2 \]

\[ = 2 \left( {a_1}^2 + {a_2}^2 + {a_3}^2 \right)\]

\[ = 2 \vec{a}^2 (\because\left| \vec{a} \right|=\sqrt{{a_1}^2 + {a_2}^2 + {a_3}^2})\]