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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
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Solution
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
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