Question

Show that the three lines with direction cosines
$\frac{12}{13},\frac{-3}{13},\frac{-4}{13}:\frac{4}{13},\frac{12}{13},\frac{3}{13};\frac{-4}{13},\frac{12}{13}$ are mutually perpendicular

Answer 1

We have,

$l_1=\frac{12}{13},m_1=-\frac{3}{13},n_1=-\frac{4}{13}$

$l_2=\frac{4}{13},m_2=\frac{12}{13},n_2=\frac{3}{13}$

Then, we know that,

Two lines with direction cosines $l_1,m_1,n_1\text{and}{l}_2,m_2,n_2$ are perpendicular to each other.

Then,

$l_1{l}_2+m_1{m}_2+n_1{n}_2=0$

$(\frac{12}{13}\times \frac{4}{13})+(\frac{-3}{13}\times \frac{12}{13})+(\frac{-4}{13}\times \frac{3}{13})$

$=\frac{48}{169}+\left(\frac{-36}{169}\right)+\left(\frac{-12}{169}\right)$

$=\frac{48-36-12}{169}$

$=\frac{48-48}{169}$

$=\frac{0}{169}$

$=0$

Hence they are perpendicular lines.

Now,

$l_3=\frac{3}{13},m_3=\frac{-4}{13},n_3=\frac{12}{13}$

Then,

Two lines with direction cosines $l_1,m_1,n_1\text{and}{l}_2,m_2,n_2$ are perpendicular to each other.

$l_2{l}_3+m_2{m}_3+n_2{n}_3=0$

$(\frac{4}{13}\times \frac{3}{13})+(\frac{12}{13}\times \frac{-4}{13})+(\frac{3}{13}\times \frac{12}{13})$

$=\frac{12}{169}+\left(\frac{-48}{169}\right)+\left(\frac{36}{169}\right)$

$=\frac{12-48+36}{169}$

$=\frac{48-48}{169}$

$=\frac{0}{169}$

$=0$

Hence, Hence they are perpendicular lines.

Hence, all 3 lines are perpendicular to each other.