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Home  >>  CBSE XII  >>  Math  >>  Three Dimensional Geometry
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If a line has the direction ratios $-18, 12, -4,$ then what are its direction cosines?

$\begin{array}{1 1} \bigg( \large\frac{9}{11},\large \frac{6}{11}, \large\frac{-2}{11} \bigg) \\ \bigg( \large\frac{-9}{11},\large \frac{6}{11}, \large\frac{-2}{11} \bigg) \\ \bigg( \large\frac{-9}{11},\large \frac{6}{11}, \large\frac{2}{11} \bigg) \\ \bigg( \large\frac{-9}{11},\large \frac{-6}{11}, \large\frac{-2}{11} \bigg) \end{array} $

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  • If $l,m,n$ are the direction cosines and $a,b,c $ are direction ration of a line, their $ \large\frac{a}{\lambda l},\frac{b}{\lambda m}$ and $ \large\frac{c}{\lambda n}$ for any non zero $ \lambda \in R$
Given dircetion ration are $(-18,12,-4)$
If $l,m,n$ are the direction ration, then its direction cosines are
$\large\frac{l}{\sqrt {l^2+m^2+n^2}}$$,\large\frac{m}{\sqrt {l^2+m^2+n^2}}$$\large\frac{n}{\sqrt {l^2+m^2+n^2}}$
Here $l=-18,m=12\; and\; n=-4$
Therfore $\sqrt {l^2+m^2+n^2}=\sqrt {(-18)^2+(12)^2+(-4)^2}$
$\qquad=\sqrt {(324)+(144)+(16)}$
$\qquad=\sqrt {484}$
Therefore The direction cosines are $ \bigg( \large\frac{-18}{22},\large \frac{12}{22}, \large\frac{-4}{22} \bigg) $
$= \bigg( \large\frac{-9}{11},\large \frac{6}{11}, \large\frac{-2}{11} \bigg) $
answered Jun 3, 2013 by meena.p

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