Question

If a variable line in two adjacent positions has direction cosines l,m,n and $l+\delta l,m+\delta m,n+\delta n$, then show that the small angle $\delta \theta$ between the two positions is given by $\delta {\theta }^2=\delta l^2+\delta m^2+\delta n^2.$

Answer 1

We have l,m,n, and $\delta l,m+\delta m,n+\delta n$ as direction cosines of a variable line in two different positions. $\therefore l^2+m^2+n^2=1$ ...(i)
and $(l+\delta l{)}^2+(m+\delta m{)}^2+(n+\delta n{)}^2$ ...(ii)
$\Rightarrow l^2+m^2+n^2+\delta l^2+\delta m^2+\delta n^2+2(l\delta l+m\delta m+n\delta n)=1$ $\Rightarrow l\delta l+m\delta m+n\delta n=\frac{-1}{2}(\delta l^2+\delta m^2+\delta n^2)$ ...(iii)
Now, $\overrightarrow{a}and\overrightarrow{b}$ are unit vectors along a line with direction cosines l,m,n and $(l+\delta l),(m+\delta m),(n+\delta n)$, respectively.
$\therefore \overrightarrow{a}=l\hat{i}+m\hat{j}+n\hat{k}and\overrightarrow{b}=(l+\delta l)\hat{i}+(m+\delta m)\hat{j}+(n+\delta n)\hat{k}$
$\Rightarrow cos\delta \theta =\frac{\overrightarrow{a}.\overrightarrow{b}}{\overrightarrow{|a|}\overrightarrow{|b|}}=\overrightarrow{a}.\overrightarrow{b}[\because |\hat{a}|=|\hat{b}|=1]$
$\Rightarrow cos\delta \theta =l(l+\delta l)+m(m+\delta m)+n(n+\delta n)$
$\Rightarrow cos\delta \theta ==(l^2+m^2+n^2)+(l\delta l+m\delta m+n\delta n)$
$\Rightarrow cos\delta \theta ==1-\frac{1}{2}(\delta l^2+\delta m^2+\delta n^2)$ [using Eq.(iii)]
$\Rightarrow 2(1-cos\delta \theta )=(\delta l^2+\delta m^2+\delta n^2)$
$\Rightarrow 2.2si{n}^2\frac{\delta \theta }{2}=\delta l^2+\delta m^2+\delta n^2$ $[\because 1-cos\theta =2si{n}^2\frac{\theta }{2}]$
$\Rightarrow 4{\left(\frac{\delta \theta }{2}\right)}^2=\delta l^2+\delta m^2+\delta n^2$$[since,\frac{\delta \theta }{2}issmall,thensin\frac{\delta \theta }{2}=\frac{\delta \theta }{2}]$
$\therefore \delta {\theta }^2=\delta l^2+\delta m^2+\delta n^2$