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

Both positions are unit vectors of direction cosines, so $l^2+m^2+n^2=1$ and $(l+\delta l)^2+(m+\delta m)^2+(n+\delta n)^2=1$.

Subtract the first from the expanded second:

Dot product of the two directions equals $\cos\delta\theta$:

$\cos\delta\theta = l(l+\delta l)+m(m+\delta m)+n(n+\delta n) = 1 + (l\,\delta l+m\,\delta m+n\,\delta n).$ …

The picture

Direction cosines $(l,m,n)$ are the components of a unit vector along the line, so the point $(l,m,n)$ lives on the unit sphere. As the line turns slightly, its direction cosines shift to $(l+\delta l,\,m+\delta m,\,n+\delta n)$, another point on the same sphere. The tiny angle $\delta\theta$ between the two positions is the angle between these two unit vectors.

Step 1 - the unit-length identity, for both positions

Expand the second and use the first:

Step 2 - the angle from the dot product

The two direction vectors are unit, so their dot product is $\cos\delta\theta$:

Since $l^2+m^2+n^2=1$, …

Method: Relating a small turn of a line to the change in its direction cosines

This is a proof pattern, not a numeric problem: connect a tiny angle $\delta\theta$ between two nearby positions of a line to the changes $\delta l,\delta m,\delta n$ in its direction cosines. Two facts do all the work.

Steps

Step 1: Use the unit-length identity for both positions.

Direction cosines satisfy $l^2+m^2+n^2=1$. Write it for the original and for the shifted triple, expand the second, and subtract. Keep the second-order term $\delta l^2+\delta m^2+\delta n^2$ — it does not vanish.

Step 2: Get the angle from the dot product of the two unit directions.

Since both are unit vectors, their dot product is $\cos\delta\theta$, giving $\cos\delta\theta = 1 + (l\,\delta l+m\,\delta m+n\,\delta n)$.

Step 3: Bring in the small-angle expansion. …

Common Mistakes

Mistake 1: Discarding $l\,\delta l+m\,\delta m+n\,\delta n$ as zero.

Why it's wrong: it is zero only to first order, but this is a second-order result; that very term equals $-\tfrac12\delta\theta^2$ and carries the whole answer. Correct approach: keep it, and evaluate it from the dot-product/small-angle expansion.

Mistake 2: Dropping the $\delta l^2+\delta m^2+\delta n^2$ term when subtracting the two unit identities. …