We are given that,

l, m, n and l + δl, m + δm, n + δn are direction cosines of a variable line in two adjacent positions.

We need to show that, the small angle δθ between the two positions is given by

δθ² = δl² + δm² + δn²

We know the relationship between direction cosines, that is,

l² + m² + n² = 1 …(i)

Also,

(l + δl)² + (m + δm)² + (n + δn)² = 1

⇒ l² + (δl)² + 2(l)(δl) + m² + (δm)² + 2(m)(δm) + n² + (δn)² + 2(n)(δn) = 1

⇒ l² + m² + n² + (δl)² + (δm)² + (δn)² + 2lδl + 2mδm + 2nδn = 1

⇒ 1 + δl² + δm² + δn² + 2lδl + 2mδm + 2nδn = 1 [from (i)]

⇒ 2lδl + 2mδm + 2nδn + δl² + δm² + δn² = 1 – 1

⇒ 2(lδl + mδm + nδn) = -( δl² + δm² + δn²)

…(ii)

Let and are unit vectors along the line with direction cosines l, m, n and (l + δl), (m + δm), (n + δn) respectively.

That is,

We know that,

Angle between two lines is given by

Where, .

Here, the angle is very small as the line is variable in different but adjacent positions. According to the question, the small angle is δθ.

So,

Angle between two lines is given by δθ,

Substituting the values of and , we get

The dot multiplication of two vectors is calculated by summing up the multiplication of coefficients of , and .

⇒ cos δθ = l(l + δl) + m(m + δm) + n(n + δn)

⇒ cos δθ = l² + lδl + m² + mδm + n² + nδn

⇒ cos δθ = l² + m² + n² + lδl + mδm + nδn

⇒ cos δθ = 1 + lδl + mδm + nδn [∵, from (i)]

[∵, from (ii)]

Or,

Since, we know that

1 – cos 2θ = 2sin² θ

In L.H.S., the angle is 2θ. Then, in R.H.S, the angle becomes half, that is,

Similarly, first replace 2θ by δθ in L.H.S.

Then, make the angle half in R.H.S., that is, .

We get,

Note that, δθ is very small angle, so will be very smaller.

This also means, has a very small value.

⇒ δθ² = δl² + δm² + δn²

Thus, we have showed the required.