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Question
Show that the three lines with direction cosines \[\frac{12}{13}, \frac{- 3}{13}, \frac{- 4}{13}; \frac{4}{13}, \frac{12}{13}, \frac{3}{13}; \frac{3}{13}, \frac{- 4}{13}, \frac{12}{13}\] are mutually perpendicular.
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Solution
The direction cosines of the three lines are
\[l_1 = \frac{12}{13}, m_1 = - \frac{3}{13}, n_1 = - \frac{4}{13}\]
\[ l_2 = \frac{4}{13}, m_2 = \frac{12}{13}, n_1 = \frac{3}{13}\]
\[ l_3 = \frac{3}{13}, m_3 = - \frac{4}{13}, n_3 = \frac{12}{13}\]
\[\therefore l_1 l_2 + m_1 m_2 + n_1 n_2 = \frac{48 - 36 - 12}{169} = 0\]
Also,
\[ l_2 l_3 + m_2 m_3 + n_2 n_3 = \frac{12 - 48 + 36}{169} = 0\]
\[ l_1 l_3 + m_1 m_3 + n_1 n_3 = \frac{36 + 12 - 48}{169} = 0\]
Hence, the given lines are perpendicular to each other.
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