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

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• Ttwo lines with direction $a_1,b_1$ and $c_1$ and $a_2,b_2,c_2$ are perpendicular if $\theta=90^{\circ}$
• $(ie)\;a_1a_2+b_1b_2+c_1c_2=0$.
Let the direction cosines of the line $AB$ be
$\bigg( \large\frac{12}{13}, \frac{-3}{13},\frac{-4}{13} \bigg)$
Let the direction cosines of the line $BC$ be
$\bigg( \large\frac{4}{13}, \frac{12}{13},\frac{3}{13} \bigg)$
On substituting for $a_1,a_2,a_3\:and\:b_1,b_2,b_3,$ we get
Therefore $a_1.a_2+b_1b_2+c_1c_2=\bigg(\large\frac{12}{13} \times \frac{4}{13} +\frac{-3}{13} \times \frac{2}{13}+\frac{-4}{13} \times \frac{3}{12}\bigg)$.
On simplifying we get,
$=\bigg(\large \frac{48}{169}-\frac{36}{169}-\frac{12}{169}\bigg)$
$=0$
Hence $AB \perp BC$-----(1)
Let the direction cosines of $CA$ be
$\large\bigg( \frac{3}{13}, \frac{-4}{13},\frac{12}{13} \bigg)$
The direction cosines of the line $AB$ be
$\bigg( \large\frac{12}{13}, \frac{-3}{13},\frac{-4}{13} \bigg)$
On substituting the value
Therefore $a_1.a_2+b_1b_2+c_1c_2=\bigg(\large\frac{3}{13} \times \frac{12}{13} +\frac{-4}{13} \times \frac{3}{13}+\frac{12}{13} \times \frac{-4}{13}\bigg)$.
$= \large\bigg(\frac{36}{169}-\frac{2}{169}-\frac{48}{169}\bigg)$
$=0$
Hence $CA \perp AB$-----(2)
Therefore from equ(1) and equ(2) we obtain that $AB,BC,$ and $CA$ are mutually perpendicular.
Hence proved.

answered Jun 4, 2013 by