Question

Given \[\vec{a} = \frac{1}{7}\left( 2 \hat{ i } + 3 \hat{ j } + 6 \hat{ k }  \right), \vec{b} = \frac{1}{7}\left( 3 \hat{ i } - 6 \hat{ j }  + 2 \hat{ k }  \right), \vec{c} = \frac{1}{7}\left( 6 \hat{ i } + 2 \hat{ j }  - 3 \hat{ k }\right), \hat{ i } , \hat{ j }  , \hat{ k } \] being a right handed orthogonal system of unit vectors in space, show that \[\vec{a} , \vec{b} , \vec{c}\] is also another system.

 

 

Answer 1

\[\text{ Given } : \]
\[ \vec{a} = \frac{1}{7} \left( 2 \hat{ i }  + 3 \hat{ j }+ 6 \hat{ k }  \right)\]
\[ \vec{b} = \frac{1}{7} \left( 3 \hat{ i }  - 6 \hat{ j }  + 2 \hat { k  } \right)\]
\[ \vec{c} = \frac{1}{7}\left( 6 \hat{ i } + 2 \hat{ j }  - 3 \hat{ k}  \right)\]
\[ \vec{a} \times \vec{b} = \left( \frac{1}{7} \right) \left( \frac{1}{7} \right)\begin{vmatrix}\hat{ i }  & \hat{ j } & \hat{ k }  \\ 2 & 3 & 6 \\ 3 & - 6 & 2\end{vmatrix}\]
\[ = \frac{1}{49}\left( 42 \hat{ i }  + 14 \hat{ j }  - 21 \hat{ k } \right)\]
\[ = \frac{1}{49}\left[ 7 \left( 6 \hat{ i } + 2 \hat{ j } - 3 \hat{ k }  \right) \right]\]
\[ = \frac{1}{7}\left( 6 \hat{ i }  + 2 \hat{ j } - 3 \hat{ k } \right)\]
\[ = \vec{c} \]
\[ \vec{b} \times \vec{c} = \left( \frac{1}{7} \right) \left( \frac{1}{7} \right)\begin{vmatrix}\hat{ i } & \hat{ j }  & \hat{ k } \\ 3 & - 6 & 2 \\ 6 & 2 & - 3\end{vmatrix}\]
\[ = \frac{1}{49}\left( 14 \hat{ i }  + 21 \hat{ j }  + 42 \hat{ k } \right)\]
\[ = \frac{1}{49}\left[ 7 \left( 2 \hat{ i }  + 3 \hat{ j} + 6 \hat{ k }  \right) \right]\]
\[ = \frac{1}{7} \left( 2 \hat{ i }  + 3 \hat{ j }  + 6 \hat{ k  } \right)\]
\[ = \vec{a} \]
\[ \vec{c} \times \vec{a} = \left( \frac{1}{7} \right) \left( \frac{1}{7} \right)\begin{vmatrix}\hat{ i } & \hat{ j }  & k \\ 6 & 2 & - 3 \\ 2 & 3 & 6\end{vmatrix}\]
\[ = \frac{1}{49}\left( 21 \hat{ i } - 42 \hat{ j }  + 14 \hat{ k }  \right)\]
\[ = \frac{1}{49}\left[ 7 \left( 3 \hat{ i }  - 6 \hat{ j }  + 2 \hat{ k }  \right) \right]\]
\[ = \frac{1}{7} \left( 3 \hat{ i }  - 6 \hat{ j }  + 2 \hat { k } \right)\]
\[ = \vec{b} \]
\[\left| \vec{a} \right| = \frac{1}{7}\sqrt{4 + 9 + 36}\]
\[ = \frac{7}{7}\]
\[ = 1\]
\[\left| \vec{b} \right| = \frac{1}{7}\sqrt{9 + 36 + 4}\]
\[ = \frac{7}{7}\]
\[ = 1\]
\[\left| \vec{c} \right| = \frac{1}{7}\sqrt{36 + 4 + 9}\]
\[ = \frac{7}{7}\]
\[ = 1\]
\[\text{ Thus } , \vec{a} , \vec{b} \text{ and }  \vec{c} \text{ form a right handed orthogonal system of unit vectors. } \]