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Question
If li, mi, ni, i = 1, 2, 3 denote the direction cosines of three mutually perpendicular vectors in space, prove that AAT = I, where \[A = \begin{bmatrix}l_1 & m_1 & n_1 \\ l_2 & m_2 & n_2 \\ l_3 & m_3 & n_3\end{bmatrix}\]
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Solution
Given : \[\left( l_1 , m_1 , n_1 \right), \left( l_2 , m_2 , n_2 \right), \left( l_3 , m_3 , n_3 \right)\]are the direction cosines of three mutually perpendicular vectors in space.
\[\left. \begin{array}l_1^2 + m_1^2 + n_1^2 = 1 \\ l_2^2 + m_2^2 + n_2^2 = 1 \\ l_3^2 + m_3^2 + n_3^2 = 1\end{array} \right\} . . . . . \left( i \right)\]
\[\left. \begin{array}l_1 l_2 + m_1 m_2 + n_1 n_2 = 0 \\ l_2 l_3 + m_2 m_3 + n_2 n_3 = 0 \\ l_3 l_1 + m_3 m_1 + n_3 n_1 = 0\end{array} \right\} . . . . . \left( ii \right)\]
Let \[A = \begin{bmatrix}l_1 & m_1 & n_1 \\ l_2 & m_2 & n_2 \\ l_3 & m_3 & n_3\end{bmatrix}\]
\[A = \begin{bmatrix}l_1 & m_1 & n_1 \\ l_2 & m_2 & n_2 \\ l_3 & m_3 & n_3\end{bmatrix}\]
\[A A^T = \begin{bmatrix}l_1 & m_1 & n_1 \\ l_2 & m_2 & n_2 \\ l_3 & m_3 & n_3\end{bmatrix}\begin{bmatrix}l_1 & l_2 & l_3 \\ m_1 & m_2 & m_3 \\ n_1 & n_2 & n_3\end{bmatrix}\]
\[ \Rightarrow A A^T = \begin{bmatrix}{l_1}^2 + {m_1}^2 + {n_1}^2 & l_1 l_2 + m_1 m_2 + n_1 n_2 & l_3 l_1 + m_3 m_1 + n_3 n_1 \\ l_1 l_2 + m_1 m_2 + n_1 n_2 & {l_2}^2 + {m_2}^2 + {n_2}^2 & l_2 l_3 + m_2 m_3 + n_2 n_3 \\ l_3 l_1 + m_3 m_1 + n_3 n_1 & l_2 l_3 + m_2 m_3 + n_2 n_3 & {l_3}^2 + {m_3}^2 + {n_3}^2\end{bmatrix}\]
\[\]
From (i) and (ii), we get
\[A A^T = \begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix} = I\]
Hence proved.
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