Question

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

Answer 1

Hint: We will be using the basic concept of vectors and 3-D geometry to solve the problem. We will be using the method of finding angles between two lines in 3-D to further simplify the problem.

Complete step by step answer:
We know that the angle between two lines in 3-D is,
$\cos \theta =\dfrac{{{a}_{1}}{{a}_{2}}+{{b}_{1}}{{b}_{2}}+{{c}_{1}}{{c}_{2}}}{\sqrt{{{a}_{1}}^{2}+{{b}_{1}}^{2}+{{c}_{1}}^{2}}\sqrt{{{a}_{2}}^{2}+{{b}_{2}}^{2}+{{c}_{2}}^{2}}}.............\left( 1 \right)$
Where ${{a}_{1}},{{b}_{1}},{{c}_{1}}\ and\ {{a}_{2}},{{b}_{2}},{{c}_{2}}$ are the direction ratios of two lines.
Now, we have to show three lines to be mutually perpendicular.
Let, first line be ${{L}_{1}}$ having direction cosines as,
${{l}_{1}}=\dfrac{12}{13},{{m}_{1}}=\dfrac{-3}{13},{{n}_{1}}=\dfrac{-4}{13}$
Second line be ${{L}_{2}}$ having direction cosines as,
${{l}_{2}}=\dfrac{4}{13},{{m}_{2}}=\dfrac{12}{13},{{n}_{2}}=\dfrac{3}{13}$
Third line be ${{L}_{3}}$ having direction cosines as,
${{l}_{2}}=\dfrac{4}{13},{{m}_{2}}=\dfrac{12}{13},{{n}_{2}}=\dfrac{3}{13}$
Now, from equation (1) we can see that for two lines to be perpendicular, $\theta $ show be $90{}^\circ $therefore,
$\begin{align}
  & \cos 90=\dfrac{{{a}_{1}}{{a}_{2}}+{{b}_{1}}{{b}_{2}}+{{c}_{1}}{{c}_{2}}}{\sqrt{{{a}_{1}}^{2}+{{b}_{1}}^{2}+{{c}_{1}}^{2}}\sqrt{{{a}_{2}}^{2}+{{b}_{2}}^{2}+{{c}_{2}}^{2}}}=0 \\
 & {{a}_{1}}{{a}_{2}}+{{b}_{1}}{{b}_{2}}+{{c}_{1}}{{c}_{2}}=0 \\
\end{align}$
Now, we take ${{L}_{1}}\ \And \ {{L}_{2}}$ and will find the value of ${{l}_{1}}{{l}_{2}}+{{m}_{1}}{{m}_{2}}+{{n}_{1}}{{n}_{2}}$. So,
$\begin{align}
  & {{l}_{1}}{{l}_{2}}+{{m}_{1}}{{m}_{2}}+{{n}_{1}}{{n}_{2}}=\dfrac{12}{13}\left( \dfrac{4}{13} \right)+\left( \dfrac{-3}{13} \right)\dfrac{12}{13}+\dfrac{-4}{13}\left( \dfrac{3}{13} \right) \\
 & =\dfrac{48}{169}-\dfrac{36}{169}-\dfrac{12}{169} \\
 & =\dfrac{48-48}{169} \\
 & =0 \\
\end{align}$
Hence, ${{L}_{1}}\ \And \ {{L}_{2}}$ are perpendicular. Similarly we have to do for ${{L}_{2}},{{L}_{3}}\ \And \ {{L}_{3}},{{L}_{1}}$.
Now, we will repeat the same process for ${{L}_{2}},{{L}_{3}}$.
$\begin{align}
  & {{l}_{2}}{{l}_{3}}+{{m}_{2}}{{m}_{3}}+{{n}_{2}}{{n}_{3}}=\dfrac{4}{13}\left( \dfrac{3}{13} \right)+\dfrac{12}{13}\left( \dfrac{-4}{13} \right)+\dfrac{3}{13}\left( \dfrac{12}{13} \right) \\
 & =\dfrac{12}{169}-\dfrac{48}{169}+\dfrac{36}{169} \\
 & =\dfrac{48-48}{169} \\
 & =0 \\
\end{align}$
Hence, ${{L}_{2}},{{L}_{3}}$are perpendicular. Now, for ${{L}_{3}},{{L}_{1}}$.
$\begin{align}
  & {{l}_{1}}{{l}_{3}}+{{m}_{1}}{{m}_{3}}+{{n}_{1}}{{n}_{3}}=\dfrac{12}{13}\left( \dfrac{3}{13} \right)+\left( \dfrac{-3}{13} \right)\left( \dfrac{-4}{13} \right)+\left( \dfrac{-4}{13} \right)\dfrac{12}{13} \\
 & =\dfrac{36}{169}+\dfrac{12}{169}-\dfrac{48}{169} \\
 & =\dfrac{48-48}{169} \\
 & =0 \\
\end{align}$
Hence, ${{L}_{3}},{{L}_{1}}$ are also perpendicular. Since, ${{L}_{1}},{{L}_{2}};{{L}_{2}},{{L}_{3}};{{L}_{3}},{{L}_{1}}$ are perpendicular this shows that ${{L}_{1}},{{L}_{2}},{{L}_{3}}$ are mutually perpendicular or the three lines with given direction cosines are mutually perpendicular.

Note: These type of question can be easily solved if the formula to find angle between true lines is remembered that is,
$\cos \theta =\dfrac{{{a}_{1}}{{a}_{2}}+{{b}_{1}}{{b}_{2}}+{{c}_{1}}{{c}_{2}}}{\sqrt{{{a}_{1}}^{2}+{{b}_{1}}^{2}+{{c}_{1}}^{2}}\sqrt{{{a}_{2}}^{2}+{{b}_{2}}^{2}+{{c}_{2}}^{2}}}$
Where ${{a}_{1}},{{b}_{1}},{{c}_{1}}\ and\ {{a}_{2}},{{b}_{2}},{{c}_{2}}$ are the direction cosines.