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Question:
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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.
$$
\frac{12}{13}, \frac{-3}{13}, \frac{-4}{13} ; \frac{4}{13}, \frac{12}{13}, \frac{3}{13} ; \frac{3}{13}, \frac{-4}{13}, \frac{12}{13}
$$
are mutually perpendicular.
Solution:
2205 Upvotes
Verified Answer
Let the lines be $\mathrm{L}_1, \mathrm{~L}_2$ and $\mathrm{L}_3$.
$\therefore$ For lines $\mathrm{L}_1$ and $\mathrm{L}_2$
$$
\begin{aligned}
&l_1 l_2+\mathrm{m}_1 \mathrm{~m}_2+\mathrm{n}_1 \mathrm{n}_2 \\
&=\frac{12}{13} \times \frac{4}{13}+\left(\frac{-3}{13}\right) \times \frac{12}{13}+\left(\frac{-4}{13}\right) \times \frac{3}{13}=0
\end{aligned}
$$
$$
\therefore \quad \mathrm{L}_1 \perp \mathrm{L}_2
$$
Similarly, Again for lines $\mathrm{L}_2$ and $\mathrm{L}_3$
$l_1 l_2+\mathrm{m}_1 \mathrm{~m}_2+\mathrm{n}_1 \mathrm{n}_2=0 \therefore \mathrm{L}_2 \perp \mathrm{L}_3$
Similarly, Again for lines $\mathrm{L}_3$ and $\mathrm{L}_1$ $l_1 l_2+\mathrm{m}_1 \mathrm{~m}_2+\mathrm{n}_1 \mathrm{n}_2=0 \therefore \mathrm{L}_1 \perp \mathrm{L}_3$
Hence the three lines are mutually perpendicular.
$\therefore$ For lines $\mathrm{L}_1$ and $\mathrm{L}_2$
$$
\begin{aligned}
&l_1 l_2+\mathrm{m}_1 \mathrm{~m}_2+\mathrm{n}_1 \mathrm{n}_2 \\
&=\frac{12}{13} \times \frac{4}{13}+\left(\frac{-3}{13}\right) \times \frac{12}{13}+\left(\frac{-4}{13}\right) \times \frac{3}{13}=0
\end{aligned}
$$
$$
\therefore \quad \mathrm{L}_1 \perp \mathrm{L}_2
$$
Similarly, Again for lines $\mathrm{L}_2$ and $\mathrm{L}_3$
$l_1 l_2+\mathrm{m}_1 \mathrm{~m}_2+\mathrm{n}_1 \mathrm{n}_2=0 \therefore \mathrm{L}_2 \perp \mathrm{L}_3$
Similarly, Again for lines $\mathrm{L}_3$ and $\mathrm{L}_1$ $l_1 l_2+\mathrm{m}_1 \mathrm{~m}_2+\mathrm{n}_1 \mathrm{n}_2=0 \therefore \mathrm{L}_1 \perp \mathrm{L}_3$
Hence the three lines are mutually perpendicular.
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