Consider a variable rotating line passing through a fixed point. The angle between two successive/adjacent positions of the line is a small angle $\delta \theta$. If the change in the direction cosines of the line in these two positions is $\delta l$, $\delta m$, $\delta n$ respectively, prove that:

$$(\delta \theta)^2=(\delta l)^2+(\delta m)^2+(\delta n)^2$$

This sort of resembles the formula for the length of a vector expressed in terms of the projections on 3 orthogonal axes. Does this above expression somehow result from the fact that we can treat angular vector variables in the same manner as all other kinds of vectors (by all other kinds of vectors, I mean displacement, velocity, acceleration, etc, and by angular vector variables, I mean angular displacement, angular velocity ($\omega$), etc)? How can I interpret this expression? I am curious because I found this to be a surprisingly concise and elegant formula for describing small angles of a variable line in 3D.

EDIT: I found an answer for this but I still have some doubts; let the initial direction cosines be $(l,m,n)$ and final direction cosines be $(l+\delta l, m+\delta m, n+ \delta n)$.

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

$\implies 2(l\delta l + m \delta m + n \delta n) = - ((\delta l)^2+(\delta m)^2+(\delta n)^2)...(i)$

Taking the dot product, we get $\cos(\delta \theta)=l(l+\delta l)+m(m+\delta m)+n(n+\delta n)$

$\implies \cos(\delta \theta)=1+l\delta l + m \delta m + n \delta n$

Since $\delta \theta$ is small, we may approximate $cos(\delta \theta)$ as $\left(1-\dfrac{(\delta \theta)^2}{2}\right)$. Now, using (i), it is easy to see that, hence $(\delta \theta)^2=(\delta l)^2+(\delta m)^2+(\delta n)^2$...(ii).

I decided to take things a step further and write $\delta l = \cos(\alpha+\delta \alpha)-\cos(\alpha)$, where $\delta \alpha$ is the small angle change of the line with the x-axis. Similar relations may be written for $\delta m$ and $\delta n$. In the end, I was hoping to find a relation of the Pythagorean form containing $(\delta \alpha)^2+(\delta \beta)^2+(\delta \gamma)^2$ which I could perhaps interpret better, but I got a complicated mess of terms which eventually boils down to the following expression upon neglecting all terms above the second order;

$$(\delta \theta)^2=(\delta \alpha)^2 \sin^2 \alpha + (\delta \beta)^2 \sin^2 \beta + (\delta \gamma)^2 \sin^2 \gamma ...(iii)$$

which is a simple expression but it doesn't give any insight as to why $\delta \theta$ obeys such a simple law. Does anyone have any insight into this problem or perhaps a more clever alternative way of arriving at equation (iii)?

  • 1
    $\begingroup$ Thanks for the bounty. Now I also understand the second question you have about your equation (iii). So $l=\cos \alpha$, where $\alpha$ is indeed the angle of the line with the $x$-axis. Then $\delta l = -\sin \alpha\, \delta \alpha$, and your equations (ii) and (iii) are equivalent. $\endgroup$ Commented Feb 18 at 16:46

1 Answer 1


Without loss of generality, we take the fixed point to be the origin. Let $L$ and $L'$ be two close lines in ${\mathbb R}^3$ passing through the origin with angle $\delta\theta$ between them.

Let $P=L\cap S^2$ and $P'=L'\cap S^2$ be the intersection points of the two lines with the unit sphere $S^2$. (I think these lines are oriented, so there is one intersection point as you travel from the origin in the positive direction.)

Write $P=(x, y, z)$ and $P'=(x',y',z')$. By definition, the direction cosines of $L$ are $l=x,m=y,n=z$ since $x=\vec{OP}\cdot e_1=\cos \angle(L,x-\text{axis})=l$, where $e_1=(1, 0, 0)$ is the standard basis in the $x$-direction. Similarly for the other two. So we indeed have $ P=(l, m, n),\ P'=(l',m',n'), $ and $$ \vec{PP'}=(l'-l,m'-m,n'-n)=(\delta l,\delta m, \delta n). $$ Note that $|\vec{PP'}|^2=\delta l^2 + \delta m^2 + \delta n^2$ by the Euclidean distance formula.

Note that $\delta\theta$ is the arc-length of the great circle arc $\overset{\large\frown}{PP'}$ on the sphere connecting $P$ to $P'$ by the definition of radian angle, since $S^2$ is the unit sphere. Then your desired equality is nothing but infinitely we have $$ |\overset{\large\frown}{PP'}|\approx |\vec{PP'}|, $$ where $|\cdot|$ denote arc-lengths. Upon squaring, we have $$ \delta\theta^2 = \delta l^2 + \delta m^2 + \delta n^2, $$ where the equal sign is in the sense of equality in the first order.


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