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.

Answer

let $\text{l}_1=\frac{12}{13},\text{m}_1=-\frac{3}{13},\text{n}_1=-\frac{4}{13}$
$\text{l}_2=\frac{4}{13},\text{m}_2=\frac{12}{13},\text{n}_2=\frac{3}{13}$
$\text{l}_3=\frac{3}{13},\text{m}_3=-\frac{4}{13},\text{n}_3=\frac{12}{13}$
$\text{l}_1\text{l}_2+\text{m}_1\text{m}_2+\text{n}_1\text{n}_2$
$=\frac{12}{13}\times\frac{4}{13}+\big(-\frac{3}{13}\big)\times\frac{12}{13}+\big(-\frac{4}{13}\big)\times\frac{3}{13}$
$=\frac{48-36-13}{169}=0$
$\text{l}_2\text{l}_3+\text{m}_2\text{m}_3+\text{n}_2\text{n}_3$
$=\frac{4}{13}\times\frac{3}{13}+\frac{12}{13}\times\Big(-\frac{4}{13}\Big)+\frac{3}{13}\times\frac{12}{13}$
$=\frac{12-48+36}{169}=0$
$\text{l}_1\text{l}_3+\text{m}_1\text{m}_3+\text{n}_1\text{n}_3$
$=\frac{12}{13}\times\frac{3}{13}+\Big(-\frac{3}{13}\Big)\times\Big(-\frac{4}{13}\Big)+\Big(-\frac{4}{13}\Big)\times\frac{12}{13}$
$=\frac{36+12-48}{169}=0$
$\therefore$ The lines are mutually perpendicular.

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