Question

Find a unit vector perpendicular to the plane ABC, where the coordinates of A, B and Care A (3, −1, 2), B (1, −1, −3) and C (4, −3, 1).

Answer 1

\[\text{ The vector }  \vec{AB} \times \vec{AC} \text{ is perpendicular to the vectors } \vec{AB} \text{ and }  \vec{AC} . \]
\[ \therefore \text{ Required unit vector }  = \frac{\vec{AB} \times \vec{AC}}{\left| \vec{AB} \times \vec{AC} \right|}\]
\[\text{ Now, } \]
\[ \vec{AB} = \text{ Position vector of B - Position vector of A } \]
\[ = \left( \hat{ i }  - \hat{ j }  - 3 \hat{ k }  \right) - \left( 3 \hat{ i  } - \hat{ j }  + 2 \hat{ k }  \right)\]
\[ = - 2 \hat{ i }  + 0 \hat{ j }  - 5k\]
\[ \vec{AC} = \text{ Position vector of C - Position vector of A} \]
\[ = \left( 4 \hat{ i }  - 3 \hat{ j } + \hat{ k }  \right) - \left( 3 \hat{ i }  - \hat{ j }  + 2 \hat{ k }  \right)\]
\[ = \hat{ i }  - 2 \hat{ j }  - \hat{ k }  \]
\[ \therefore \vec{AB} \times \vec{AC} = \begin{vmatrix}\hat{ i }  & \hat{ j }  & \hat{ k } \\ - 2 & 0 & - 5 \\ 1 & - 2 & - 1\end{vmatrix}\]
\[ = \left( 0 - 10 \right) i - \left( 2 + 5 \right) j + \left( 4 - 0 \right) \hat{ k } \]
\[ = - 10 \hat{ i }  - 7 \hat{ j } + 4 \hat{ k }  \]
\[\left| \vec{AB} \times \vec{BC} \right| = \sqrt{\left( - 10 \right)^2 + \left( - 7 \right)^2 + 4^2}\]
\[ = \sqrt{165}\]
\[\text{ Unit vector perpendicular to the plane ABC } =\frac{\vec{AB} \times \vec{AC}}{\left| \vec{AB} \times \vec{AC} \right|} = \frac{- 10 \hat{ i }  - 7 \hat{ j }  + 4 \hat{ k } }{\sqrt{165}}\]