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

Given that l, m, n and l + δl, m + δm, n + δn, are the direction cosines of a variable line in two positions

∴ l² + m² + n² = 1  ......(i)

And (l + δl)² + (m + δm)² + (n + δn)² = 1  ......(ii)

⇒ l² + δl² + 2l.δl + m² + δm² + 2m.δm + n² + δn² + 2n.δn = 1

⇒ (l² + m² + n²) + (δl² + δm² + δn²) + 2(l.δl + m.δm + n.δn) = 1

⇒ 1 + (δl² + δm² + δn²) + 2(l.δl + m.δm + n.δn) = 1

⇒ l.δl + m.δm + n.δn =`-1/2(δl^2 + δm^2 + δn^2)`

Let `vec"a"` and `vec"b"` be the unit vectors along a line with d’cosines l, m, n and d (l + δl), (m + δm), (n + δn).

∴ `vec"a" = lhat"i" + mhat"j" + nhat"k"` and `vec"b" = (l + δl)hat"i" + (m + δm)hat"j" + (n + δn)hat"k"`

`cosδtheta = (vec"a"*vec"b")/(|vec"a"||vec"b"|)`

`cosδtheta = ((lhat"i" + mhat"j" + nhat"k").[(l + δl)hat"i" + (m + δm)hat"j" + (n + δn)hat"k"])/(1.1)`  .....`[because |vec"a"| = |vec"b"| = 1]`

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

⇒ cos δθ = l² + l.δl + m² + m.δm + n² + n.δn

⇒ cos δθ = (l² + m² + n²) + (l.δl + m.δm + n.δn)

⇒ cos δθ = `1 - 1/2(δl^2 + δm^2 + δn^2)`

⇒ `1 - cosδtheta = 1/2 (δl^2 + δm^2 + δn^2)`

⇒ `2sin^2  (δtheta)/2 = 1/2 (δ1^2 + δm^2 + δn^2)`

⇒ `4sin^2  (δtheta)/2 = δl^2 + δm^2 + δn^2`

⇒ `4((δtheta)/2)^2 = δl^2 + δm^2 + δn^2`  ......`[(because (δtheta)/2  "is very small so"","),(sin  (δtheta)/2 = (δtheta)/2)]`

⇒ `(δtheta)^2 = δl^2 + δm^2 + δn^2`

Hence proved.