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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
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.
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