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Home » Show that the three lines with direction cosines Are mutually perpendicular.
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.
CBSE | Class 12 | Exercise 11.2 | Maths | NCERT | Three Dimensional Geometry
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.

Solution:

Consider the direction cosines of L_{1}, L_{2} and L_{3} be l_{1}, m_{1}, n_{1} ; l_{2}, m_{2}, n_{2} and l_{3}, m_{3}, n_{3}.
It is known that
If \mathrm{f}_{1}, \mathrm{~m}_{1}, \mathrm{n}_{1} and \mathrm{l}_{2}, \mathrm{~m}_{2}, \mathrm{n}_{2} are the direction cosines of two lines;
And \theta is the acute angle between the two lines;
Therefore \cos \theta=\|_{1} l_{2}+m_{1} m_{2}+n_{1} n_{2} \mid
If two lines are perpendicular, then the angle between the two is \theta=90^{\circ}
For perpendicular lines, \left|I_{1}\right|_{2}+m_{1} m_{2}+n_{1} n_{2} \mid=\cos 90^{\circ}=0, i.e. \left|I_{1} l_{2}+m_{1} m_{2}+n_{1} n_{2}\right|=0
Therefore, in order to check if the three lines are mutually perpendicular, we compute \left|I_{1} l_{2}+m_{1} m_{2}+n_{1} n_{2}\right| for all the pairs of the three lines.
Firstly let’s compute, \left|I_{1} l_{2}+m_{1} m_{2}+n_{1} n_{2}\right|
\begin{array}{l} \left|1_{1} 1_{2}+m_{1} m_{2}+n_{1} n_{2}\right|=\left|\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)\right|=\frac{48}{13}+\left(\frac{-36}{13}\right)+\left(\frac{-12}{13}\right) \\ =\frac{48+(-48)}{13}=0 \end{array}
Therefore, L_{1} \perp L_{2} \ldots \ldots (1)
In the similar way,
Let’s compute, \mid \mathrm{l}_{2} 1_{3}+\mathrm{m}_{2} \mathrm{~m}_{3}+\mathrm{n}_{2} \mathrm{n}_{3} \mathrm{l}
\begin{array}{l} \left|1_{2} 1_{3}+m_{2} m_{3}+n_{2} n_{3}\right|=\left|\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)\right|=\frac{12}{13}+\left(\frac{-48}{13}\right)+\frac{36}{13} \\ =\frac{(-48)+48}{13}=0 \end{array}
So, L_{2} \perp L_{3} \ldots . .(2)
In the similar way,
Let’s compute, \left|\mathrm{I}_{3}\right|_{1}+\mathrm{m}_{3} \mathrm{~m}_{1}+\mathrm{n}_{3} \mathrm{n}_{1} \mathrm{l}
\begin{array}{l} \left|1_{3} 1_{1}+\mathrm{m}_{3} \mathrm{~m}_{1}+\mathrm{n}_{3} \mathrm{n}_{1}\right|=\left|\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)\right|=\frac{36}{13}+\frac{12}{13}+\left(\frac{-48}{13}\right) \\ =\frac{48+(-48)}{13}=0 \end{array}
So, L_{1} \perp L_{3} \ldots . .(3)
As a result, by (1),(2) and (3), the lines are perpendicular.
\mathrm{L}_{1}, \mathrm{~L}_{2} and \mathrm{L}_{3} are mutually perpendicular.

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