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 $\delta {\theta ^2} = \delta {l^2} + \delta {m^2} + \delta {n^2}$.

We have $l,m,n$ and $l + \delta l,m + \delta m,n + \delta n$ as

direction cosines of a variable line in two different positions.
$\therefore$ ${l^2} + {m^2} + {n^2} = 1$
……..(i)
and ${(l + \delta l)^2} + {(m + \delta m)^2} + {(n + \delta n)^2} = 1$

…..(ii)
$\Rightarrow$ ${l^2} + {m^2} + {n^2} + \delta {l^2} + \delta {m^2} + \delta {n^2} + 2(l\delta l + m\delta m + n\delta n) = 1$

$\Rightarrow$ $\delta {l^2} + \delta {m^2} + \delta {n^2} = - 2(l\delta l + m\delta m + n\delta n)$

$\Rightarrow$ $l\delta l + m\delta m + n\delta n = \frac{{ - 1}}{2}\left( {\delta {l^2} + \delta {m^2} + \delta {n^2}} \right)$

…..(iii)
Now, $\overrightarrow {\rm{a}}$ and $\overrightarrow {\rm{b}}$ are unit vectors along a line with direction

cosines $l,m,n$ and $(l + \delta l),(m + \delta m),(n + \delta n)$, respectively.

$\therefore$ $\overrightarrow {\rm{a}} = l\widehat {\rm{i}} + m\widehat {\rm{j}} + n\widehat {\rm{k}}$

and $\overrightarrow {\rm{b}} = (l + \delta l)\widehat {\rm{i}} + (m + \delta m)\widehat {\rm{j}} + (n + \delta n)\widehat {\rm{k}}$

$\Rightarrow$ $\cos \delta \theta = \frac{{\overrightarrow {\rm{a}} \cdot \overrightarrow {\rm{b}} }}{{|\overrightarrow {\rm{a}} ||\overrightarrow {\rm{b}} |}} = \overrightarrow {\rm{a}} \cdot \overrightarrow {\rm{b}}$

$\Rightarrow$ $\cos \delta \theta = l(l + \delta l) + m(m + \delta m) + n(n + \delta n)$

$= \left( {{l^2} + {m^2} + {n^2}} \right) + (l\delta l + m\delta m + n\delta n)$

$= 1 - \frac{1}{2}\left( {\delta {l^2} + \delta {m^2} + \delta {n^2}} \right)$

[using Eq. (iii)]
$\Rightarrow$ $2(1 - \cos \delta \theta ) = \left( {\delta {l^2} + \delta {m^2} + \delta {n^2}} \right)$

$\Rightarrow$ $2 \cdot 2{\sin ^2}\frac{{\delta \theta }}{2} = \delta {l^2} + \delta {m^2} + \delta {n^2}$

$\Rightarrow$ $4{\left( {\frac{{\delta \theta }}{2}} \right)^2} = \delta {l^2} + \delta {m^2} + \delta {n^2}$.

[since, $\frac{{\delta \theta }}{2}$ is

small, then $\sin \frac{{\delta \theta }}{2} = \frac{{\delta \theta }}{2}$]
$\therefore$ $\delta {\theta ^2} = \delta {l^2} + \delta {m^2} + \delta {n^2}$