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 1

The direction cosines of the three lines are

$$\begin{gathered} l_1=\frac{12}{13},m_1=-\frac{3}{13},n_1=-\frac{4}{13} \\ {l}_2=\frac{4}{13},m_2=\frac{12}{13},n_1=\frac{3}{13} \\ {l}_3=\frac{3}{13},m_3=-\frac{4}{13},n_3=\frac{12}{13} \end{gathered}$$

$$\begin{gathered} \therefore l_1{l}_2+m_1{m}_2+n_1{n}_2=\frac{48-36-12}{169}=0 \\ \mathrm{Also}, \\ {l}_2{l}_3+m_2{m}_3+n_2{n}_3=\frac{12-48+36}{169}=0 \\ {l}_1{l}_3+m_1{m}_3+n_1{n}_3=\frac{36+12-48}{169}=0 \end{gathered}$$

Hence, the given lines are perpendicular to each other.