1. 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.

Answers (1)

GIven direction cosines of the three lines;

L_{1}\ \left ( \frac{12}{13}, \frac{-3}{13},\frac{-4}{13} \right )       L_{2}\ \left ( \frac{4}{13}, \frac{12}{13},\frac{3}{13} \right )      L_{3}\ \left ( \frac{3}{13}, \frac{-4}{13},\frac{12}{13} \right ) 

And we know that two lines with direction cosines l_{1},m_{1},n_{1}   and  l_{2},m_{2},n_{2} are perpendicular to each other, if l_{1}l_{2}+m_{1}m_{2}+n_{1}n_{2}=0

Hence we will check each pair of lines:

Lines L_{1}\ and\ L_{2};

l_{1}l_{2}+m_{1}m_{2}+n_{1}n_{2}= \left [ \frac{12}{13}\times\frac{4}{13} \right ]+\left [ \frac{-3}{13}\times\frac{12}{13} \right ]+\left [ \frac{-4}{13}\times \frac{3}{13} \right ]

= \left [ \frac{48}{169} \right ]-\left [ \frac{36}{169} \right ]-\left [ \frac{12}{169} \right ]= 0

\therefore the lines L_{1}\ and\ L_{2} are perpendicular.

Lines L_{2}\ and\ L_{3};

l_{1}l_{2}+m_{1}m_{2}+n_{1}n_{2}= \left [ \frac{4}{13}\times\frac{3}{13} \right ]+\left [ \frac{12}{13}\times\frac{-4}{13} \right ]+\left [ \frac{3}{13}\times \frac{12}{13} \right ]

= \left [ \frac{12}{169} \right ]-\left [ \frac{48}{169} \right ]+\left [ \frac{36}{169} \right ]= 0

\therefore the lines L_{2}\ and\ L_{3} are perpendicular.

Lines L_{3}\ and\ L_{1};

l_{1}l_{2}+m_{1}m_{2}+n_{1}n_{2}= \left [ \frac{3}{13}\times\frac{12}{13} \right ]+\left [ \frac{-4}{13}\times\frac{-3}{13} \right ]+\left [ \frac{12}{13}\times \frac{-4}{13} \right ]

= \left [ \frac{36}{169} \right ]+\left [ \frac{12}{169} \right ]-\left [ \frac{48}{169} \right ]= 0

\therefore the lines L_{3}\ and\ L_{1} are perpendicular.

Thus, we have all lines are mutually perpendicular to each other.

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