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

(A) 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.
$\therefore l^{2}+m^{2}+n^{2}=1 \dots (i)$
And $(l+\delta l)^{2}+(m+\delta m)^{2}+(n+\delta n)^{2}=1 \dots (ii)$
$\Rightarrow l^{2}+m^{2}+n^{2}+\delta l^{2}+\delta m^{2}+\delta n^{2}+2(l \delta l+m \delta m+n \delta n)=1$
$\Rightarrow \delta l^{2}+\delta m^{2}+\delta n^{2}=-2(l \delta l+m \delta m+n \delta n) \left[\because l^{2}+m^{2}+n^{2}=1\right]$
$\Rightarrow l \delta l+m \delta m+n \delta n=-\frac{1}{2}(\delta l^{2}+\delta m^{2}+\delta n^{2}) \dots (iii)$
Now,$\vec{a}$ and $\vec{b}$ are unit vectors along a line with direction cosines $l, m, n$ and $(l+\delta l), (m+\delta m), (n+\delta n)$ respectively.
$\therefore \vec{a}=l \hat{i}+m \hat{j}+n \hat{k}$ and $\vec{b}=(l+\delta l) \hat{i}+(m+\delta m) \hat{j}+(n+\delta n) \hat{k}$
$\Rightarrow \cos \delta \theta = \vec{a} \cdot \vec{b} = l(l+\delta l)+m(m+\delta m)+n(n+\delta n)$
$= (l^{2}+m^{2}+n^{2})+(l \delta l+m \delta m+n \delta n)$
$= 1-\frac{1}{2}(\delta l^{2}+\delta m^{2}+\delta n^{2}) \text{ [using Eq. } (iii)]$
$\Rightarrow 2(1-\cos \delta \theta) = \delta l^{2}+\delta m^{2}+\delta n^{2}$
$\Rightarrow 2(2 \sin^{2} \frac{\delta \theta}{2}) = \delta l^{2}+\delta m^{2}+\delta n^{2} \left[\because 1-\cos \theta = 2 \sin^{2} \frac{\theta}{2}\right]$
$\Rightarrow 4(\frac{\delta \theta}{2})^{2} = \delta l^{2}+\delta m^{2}+\delta n^{2} \left[\text{since } \delta \theta \text{ is small, } \sin \frac{\delta \theta}{2} \approx \frac{\delta \theta}{2}\right]$
$\therefore \delta \theta^{2} = \delta l^{2}+\delta m^{2}+\delta n^{2}$