Question

If a variable line in two adjacent positions has direction cosines l, m, n and l + δl, m + δm, n + δn, show that the small angle δθ between the two positions is given by δθ² = δl² + δm² + δn²

Answer 1

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)

We know that,

Angle between two lines is given by

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,

⇒ 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)]

We get,

Thus, we have showed the required.