Question

If a variable line in two adjacent position has direction cosines $l+\delta l,m+\delta m,n+\delta n$, then the small angle $\delta \theta$ between the two positions is given by

• A. $\delta {\theta }^2=4(\delta l^2+\delta m^2+\delta n^2)$
• B. $\delta {\theta }^2=2(\delta l^2+\delta m^2+\delta n^2)$
• C. $\delta {\theta }^2=(\delta l^2+\delta m^2+\delta n^2)$
• D. None of these

Answer 1

The correct option is C $\delta {\theta }^2=(\delta l^2+\delta m^2+\delta n^2)$

$[l,m,n]$ and $[l+\delta l,m+\delta m,n+\delta n]$ are d.c.'s

$\therefore l^2+m^2+n^2=1$ ...(1)

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

i.e., $(l^2+m^2+n^2)+({\delta l}^2+{\delta m}^2+{\delta n}^2)+2l\delta l+2m\delta m+2n\delta n=1$

${\delta l}^2+{\delta m}^2+{\delta n}^2=-2(l\delta l+m\delta m+n\delta n)$ from (1)...(2)

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

$=l^2+m^2+n^2+l\delta l+m\delta m+n\delta n$

$=1-\frac{1}{2}[{\delta l}^2+{\delta m}^2+{\delta n}^2]$ from (1) and (2)

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

$=2.2{\sin }^2\frac{1}{2}\delta \theta$

$=4{\left(\frac{1}{2}\delta \theta \right)}^2\sin \frac{1}{2}\delta \theta \approx \frac{1}{2}\delta \theta .$

$={\delta \theta }^2$