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Three Dimensional Geometry — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSThree Dimensional Geometry3 Marks
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

We know that
If $l_1, m_1, n_1$ and $l_2, m_2, n_2$ are the direction cosines of two lines; and $\theta$ is the acute angle between the two lines; then $\cos  \theta = |l_1l_2 + m_1m_2 + n_1n_2|$
If two lines are perpendicular, then the angle between the two is $\theta = 90^\circ$
$\Rightarrow$ For perpendicular lines, $| l_1l_2 + m_1m_2 + n_1n_2 | = \cos 90^\circ = 0,$ i.e.
$| l_1l_2 + m_1m_2 + n_1n_2 | = 0$
So, in order to check if the three lines are mutually perpendicular, we compute |$ l_1l_2 + m_1m_2 + n_1n_2 $| for all the pairs of the three lines.
Now let the direction cosines of $L_1, L_2$ and $L_3$ be $l_1, m_1, n_1; l_2, m_2, n_2$ and $l_3, m_3, n_3.$
First, consider
$\left|1_{1} 1_{2}+\mathrm{m}_{1} \mathrm{m}_{2}+\mathrm{n}_{1} \mathrm{n}_{2}\right| = |\left(\frac{12}{13} \times \frac{4}{13}\right)+\left(\frac{-3}{13} \times \frac{12}{13}\right)+\left(\frac{-4}{13} \times \frac{3}{13}\right)| = \frac{48}{13}+\left(\frac{-36}{13}\right)+\left(\frac{-12}{13}\right)$
$= \frac{48+(-48)}{13}=0$
$\Rightarrow L_{1 } \perp L_2 ……(i)$
Next, We take,
$\left|1_{2} 1_{3}+\mathrm{m}_{2} \mathrm{m}_{3}+\mathrm{n}_{2} \mathrm{n}_{3}\right| = |\left(\frac{4}{13} \times \frac{3}{13}\right)+\left(\frac{12}{13} \times \frac{-4}{13}\right)+\left(\frac{3}{13} \times \frac{12}{13}\right)| = \frac{12}{13}+\left(\frac{-48}{13}\right)+\frac{36}{13}$
$= \frac{48+(-48)}{13}=0$
$\Rightarrow L_{2 }\perp L_3 …(ii)$
Thereafter,
$\left|1_{3} 1_{1}+\mathrm{m}_{3} \mathrm{m}_{1}+\mathrm{n}_{3} \mathrm{n}_{1}\right| = |\left(\frac{3}{13} \times \frac{12}{13}\right)+\left(\frac{-4}{13} \times \frac{-3}{13}\right)+\left(\frac{12}{13} \times \frac{-4}{13}\right)| = \frac{36}{13}+\frac{12}{13}+\left(\frac{-48}{13}\right)$
= $\frac{48+(-48)}{13}=0$
$\Rightarrow L_{1 } \perp L_3 …(iii)$
$\therefore$ By $(i), (ii)$ and $(iii),$ we conclude ,
$L_1, L_2$ and $L_3$ are mutually perpendicular.

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