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 have,  l, m, n and l + δl, m + δm, n + δn, as direction cosines of a variable line of two different positions. 

∴ l² + m²  + n²  = 1   ......(i)

And (l + δ1)² + (m + δm)² + (n + δn)² = 1  ......(ii)

⇒ l² + m² + n² + δl² + δn² + 2(lδl + mδm + nδn) = 1

⇒ δl² + δm² + δn² = – 2(lδl + ,m + nδn) .....[∵ l² + m² + n² = 1]

⇒ lδl + mδm + nδn = `(-1)/2` (δl² + δm² + δn²) ......(iii)

Now `veca` and `vecb` are unit vectors along a line with direction cosines l, m, n and (l + δl), (m + δm), (n + δn), respectively.

∴ `veca = lhati + mhatj + nhatk` and `vecb = (l + deltal)hati + (m + mdelta)hatj + (n + deltan)hatk`

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

= (l² + m² + n²) + (lδl + mδm + nδn)

= `1 - 1/2 (deltal^2 + deltam^2 + deltan^2)`  .....[Using equation (iii)]

⇒ 2(1 – cos δθ) = (δ1² + δm² + δn²)

⇒ `2.2 sin^2  (δtheta)/2` = δ1² + δm² + δn²

⇒ `4((deltatheta)/2)^2 = deltal^2 + deltam^2 + deltan^2`  .....`["Since" (deltatheta)/2  "is small," sin  (deltatheta)/2 = (deltatheta)/2]` 

⇒ δθ² + δl² + δm² + δn²