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

Given: direction cosines of a variable line in two adjacent positions are l, m, n and $l+\delta l, m+\delta m,n+\delta n,$

We have to prove 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 know, the relationships between direction cosines is given as

$l^{2}+ m^{2}+ n^{2}=1 ....(1)$

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

$\Rightarrow l^{2}+(\delta l)^{2}+2(l)(\delta l)+m^{2}+(\delta m)^{2}+2(m)(\delta m)+n^{2}+(\delta n)^{2}+2(n)(\delta n)=1\\ \Rightarrow l^{2}+m^{2}+n^{2}+\left (\delta l \right )^{2}+\left (\delta m \right )^{2}+\left (\delta n \right )^{2}+2l\delta l+2m\delta m+2n\delta n=1\\ \Rightarrow 1+\delta l^{2}+\delta m^{2}+ \delta n^{2}+2l\delta l+2m\delta m+2n\delta n=1\: \: \: \left [ from (i) \right ] \\ \Rightarrow 2l\delta l+2m\delta m+2n\delta n+\delta l^{2}+\delta m^{2}+ \delta n^{2}=1-1 \\ \Rightarrow 2 \left (l\delta l+m\delta m+n\delta n \right )=-\left (\delta l^{2}+\delta m^{2}+ \delta n^{2} \right )\\ \Rightarrow l\delta l+m\delta m+n\delta n =-\frac{1}{2}\left (\delta l^{2}+\delta m^{2}+ \delta n^{2} \right ).......(iii)$

Let

$\vec{a}=l\hat{i}+m\hat{j}+n\hat{k}\\ \Rightarrow \vec{b}=\left ( l+\delta l \right )\hat{i}+\left ( m+\delta m \right )\hat{j}+\left ( n+\delta n \right )\hat{k}$

We know, angle between two lines = $\cos \theta=\vec{a}. \vec{b}$

Here, the angle is very small because the line is variable in different although adjacent positions. According to the question, this small angle is $\delta \theta$

Therefore,

$\cos \delta \theta=\vec{a}. \vec{b}$

Substituting the values of the two vectors, we get

$\Rightarrow \cos \delta \theta = \left (l\hat{i}+m\hat{j}+n\hat{k} \right ).\left (\left ( l+\delta l \right )\hat{i}+\left ( m+\delta m \right )\hat{j}+\left ( n+\delta n \right )\hat{k} \right )$

The dot product of 2 vectors is calculated by obtaining the sum of the product of the coefficients of $\hat{i},\hat{j}\; and \; \hat{k}$

$\Rightarrow \cos \delta \theta =l \left ( l+\delta l \right )+m \left ( m+\delta m \right )+n \left ( n+\delta n \right )\\ \Rightarrow \cos \delta \theta = l^{2}+l \delta l+m^{2}+m \delta m+n^{2}+n \delta n\\ \Rightarrow \cos \delta \theta =l^{2}+m^{2}+n^{2}+l \delta l+m \delta m+n \delta n\\ \Rightarrow \cos \delta \theta =1+l \delta l+m \delta m+n \delta n\: \: \left [ from(i) \right ]\\ \Rightarrow \cos \delta \theta=1-\frac{1}{2}\left ( \delta l^{2}+\delta m^{2}+\delta n^{2} \right )\: \: \: \: \left [ \because from(ii) \right ]\\ \Rightarrow \frac{1}{2}\left ( \delta l^{2}+\delta m^{2}+\delta n^{2} \right )=1-\cos \delta \theta$

Or,

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

We know, $1 -\cos 2 \theta = 2\sin^{2} \theta$

On the left-hand side, the angle is $2 \theta$. On the right hand side, it becomes half, that is, $\frac{ 2 \theta}{2} =\theta$.

Similarly replacing $2 \theta$ by $\delta \theta$ in LHS, then making the angle on the RHS half,

We get:

$1 -\cos \delta \theta = 2\sin^{2} \frac{\delta \theta}{2}$

$\Rightarrow 2\sin^{2} \frac{\delta \theta}{2}=\frac{1}{2}\left ( \delta l^{2}+\delta m^{2}+\delta n^{2} \right )\\ \Rightarrow 2 \times 2\sin^{2} \frac{\delta \theta}{2}= \delta l^{2}+\delta m^{2}+\delta n^{2} \\ \Rightarrow 4\sin^{2} \frac{\delta \theta}{2}= \delta l^{2}+\delta m^{2}+\delta n^{2} \\ \Rightarrow 4\left ( \sin \frac{\delta \theta}{2} \right )^{2}= \delta l^{2}+\delta m^{2}+\delta n^{2} \\$

Since $\delta \theta$ is a very small angle, $\frac{\delta \theta}{2}$ will be much smaller. Hence $\sin \frac{\delta \theta}{2}$ will also be very small in value.

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

Hence, proved.