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

$A=(\frac{12}{13},\frac{-13}{13},\frac{-4}{13})$

$B=(\frac{4}{13},\frac{12}{13},\frac{3}{13})$

$C=(\frac{3}{13},\frac{-4}{13},\frac{12}{13})$

For two lines to be perpendicular,

their direction cosines should be of the form

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

when $\left(l_1{m}_1{n}_1\right)$ and $\left(l_2{m}_2{n}_2\right)$ are the

direction cosines of each line

Now, consider line A and line B(${\theta }_{AB}=$ angle between A and B)

$\cos \left({\theta }_{AB}\right)=\left(\frac{12}{13}\right)\left(\frac{4}{13}\right)+\left(\frac{-3}{13}\right)\left(\frac{12}{13}\right)+\left(\frac{-4}{13}\right)\left(\frac{3}{13}\right)$

$=\frac{48-36-12}{169}=0$

$\cos \left({\theta }_{AB}\right)=0\Rightarrow {\theta }_{AB}={90}^o$

$\cos \left({\theta }_{BC}\right)=\left(\frac{4}{13}\right)\left(\frac{3}{13}\right)+\left(\frac{12}{13}\right)\left(\frac{-4}{13}\right)+\left(\frac{3}{13}\right)\left(\frac{12}{13}\right)$

$=\frac{12-48+36}{169}=0$

$\Rightarrow {\theta }_{BC}={90}^o$

$\cos \left({\theta }_{AC}\right)=\frac{36+12-48}{169}=0$

${\theta }_{AC}={90}^o$

$\therefore$ A, B and C we mutually perpendicular.