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Question Number 75445 by vishalbhardwaj last updated on 11/Dec/19

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   b/w the two positions is given by  δθ^2  = δl^2  + δm^2  + δn^2   ??

$$\mathrm{If}\:\mathrm{a}\:\mathrm{variable}\:\mathrm{line}\:\mathrm{in}\:\mathrm{two}\:\mathrm{adjacent} \\ $$$$\mathrm{positions}\:\mathrm{has}\:\mathrm{direction}\:\mathrm{cosines} \\ $$$${l},{m},\:{n}\:\mathrm{and}\:{l}+\delta{l},\:{m}+\delta{m},\:{n}+\delta{n} \\ $$$$,\:\mathrm{show}\:\mathrm{that}\:\mathrm{the}\:\mathrm{small}\:\mathrm{angle}\:\delta\theta\: \\ $$$$\mathrm{b}/\mathrm{w}\:\mathrm{the}\:\mathrm{two}\:\mathrm{positions}\:\mathrm{is}\:\mathrm{given}\:\mathrm{by} \\ $$$$\delta\theta^{\mathrm{2}} \:=\:\delta{l}^{\mathrm{2}} \:+\:\delta{m}^{\mathrm{2}} \:+\:\delta{n}^{\mathrm{2}} \:\:?? \\ $$

Answered by mr W last updated on 11/Dec/19

x_1 =l  y_1 =m  z_1 =n  x_2 =l+δl  y_2 =m+δm  z_2 =n+δn  (Δl)^2 =(Δx)^2 +(Δy)^2 +(Δz)^2   =(x_2 −x_1 )^2 +(y_2 −y_1 )^2 +(z_2 −z_2 )^2   =(δl)^2 +(δm)^2 +(δn)^2    ...(i)    (Δl)^2 =1^2 +1^2 −2×1×1×cos (δθ)  =2−2 [1−2 sin^2  (((δθ)/2))]  =4 [sin ((δθ)/2)]^2   ≈4 [((δθ)/2)]^2   =(δθ)^2    ...(ii)    ⇒(δθ)^2 =(δl)^2 +(δm)^2 +(δn)^2

$${x}_{\mathrm{1}} ={l} \\ $$$${y}_{\mathrm{1}} ={m} \\ $$$${z}_{\mathrm{1}} ={n} \\ $$$${x}_{\mathrm{2}} ={l}+\delta{l} \\ $$$${y}_{\mathrm{2}} ={m}+\delta{m} \\ $$$${z}_{\mathrm{2}} ={n}+\delta{n} \\ $$$$\left(\Delta{l}\right)^{\mathrm{2}} =\left(\Delta{x}\right)^{\mathrm{2}} +\left(\Delta{y}\right)^{\mathrm{2}} +\left(\Delta{z}\right)^{\mathrm{2}} \\ $$$$=\left({x}_{\mathrm{2}} −{x}_{\mathrm{1}} \right)^{\mathrm{2}} +\left({y}_{\mathrm{2}} −{y}_{\mathrm{1}} \right)^{\mathrm{2}} +\left({z}_{\mathrm{2}} −{z}_{\mathrm{2}} \right)^{\mathrm{2}} \\ $$$$=\left(\delta{l}\right)^{\mathrm{2}} +\left(\delta{m}\right)^{\mathrm{2}} +\left(\delta{n}\right)^{\mathrm{2}} \:\:\:...\left({i}\right) \\ $$$$ \\ $$$$\left(\Delta{l}\right)^{\mathrm{2}} =\mathrm{1}^{\mathrm{2}} +\mathrm{1}^{\mathrm{2}} −\mathrm{2}×\mathrm{1}×\mathrm{1}×\mathrm{cos}\:\left(\delta\theta\right) \\ $$$$=\mathrm{2}−\mathrm{2}\:\left[\mathrm{1}−\mathrm{2}\:\mathrm{sin}^{\mathrm{2}} \:\left(\frac{\delta\theta}{\mathrm{2}}\right)\right] \\ $$$$=\mathrm{4}\:\left[\mathrm{sin}\:\frac{\delta\theta}{\mathrm{2}}\right]^{\mathrm{2}} \\ $$$$\approx\mathrm{4}\:\left[\frac{\delta\theta}{\mathrm{2}}\right]^{\mathrm{2}} \\ $$$$=\left(\delta\theta\right)^{\mathrm{2}} \:\:\:...\left({ii}\right) \\ $$$$ \\ $$$$\Rightarrow\left(\delta\theta\right)^{\mathrm{2}} =\left(\delta{l}\right)^{\mathrm{2}} +\left(\delta{m}\right)^{\mathrm{2}} +\left(\delta{n}\right)^{\mathrm{2}} \\ $$

Commented by mr W last updated on 11/Dec/19

Commented by mr W last updated on 11/Dec/19

P_1 (x_1 ,y_1 ,z_1 )  P_2 (x_2 ,y_2 ,z_2 )  let OP_1 =OP_2 =1  x_1 =1 cos α=l  y_1 =1 cos β=m  z_1 =1 cos γ=n  similarly  x_2 =l+δl  y_2 =m+δm  z_2 =n+δn  let Δl=P_1 P_2   Δl=(√((x_2 −x_1 )^2 +(y_2 −y_1 )^2 +(z_2 −z_1 )^2 ))  =(√((δl)^2 +(δm)^2 +(δn)^2 ))  ⇒(Δl)^2 =(δl)^2 +(δm)^2 +(δn)^2     on the other side,  (P_1 P_2 )^2 =(OP_1 )^2 +(OP_2 )^2 −2(OP_1 )(OP_2 )cos ∠P_1 OP_2   i.e.  (Δl)^2 =1^2 +1^2 −2×1×1 cos (δθ)  =2−2 cos (δθ)  =2−2(1−2 sin^2  ((δθ)/2))  =4 (sin ((δθ)/2))^2   ≈4 (((δθ)/2))^2   since δθ is small, sin ((δθ)/2)≈((δθ)/2)  =(δθ)^2     ⇒(δθ)^2 =(δl)^2 +(δm)^2 +(δn)^2

$${P}_{\mathrm{1}} \left({x}_{\mathrm{1}} ,{y}_{\mathrm{1}} ,{z}_{\mathrm{1}} \right) \\ $$$${P}_{\mathrm{2}} \left({x}_{\mathrm{2}} ,{y}_{\mathrm{2}} ,{z}_{\mathrm{2}} \right) \\ $$$${let}\:{OP}_{\mathrm{1}} ={OP}_{\mathrm{2}} =\mathrm{1} \\ $$$${x}_{\mathrm{1}} =\mathrm{1}\:\mathrm{cos}\:\alpha={l} \\ $$$${y}_{\mathrm{1}} =\mathrm{1}\:\mathrm{cos}\:\beta={m} \\ $$$${z}_{\mathrm{1}} =\mathrm{1}\:\mathrm{cos}\:\gamma={n} \\ $$$${similarly} \\ $$$${x}_{\mathrm{2}} ={l}+\delta{l} \\ $$$${y}_{\mathrm{2}} ={m}+\delta{m} \\ $$$${z}_{\mathrm{2}} ={n}+\delta{n} \\ $$$${let}\:\Delta{l}={P}_{\mathrm{1}} {P}_{\mathrm{2}} \\ $$$$\Delta{l}=\sqrt{\left({x}_{\mathrm{2}} −{x}_{\mathrm{1}} \right)^{\mathrm{2}} +\left({y}_{\mathrm{2}} −{y}_{\mathrm{1}} \right)^{\mathrm{2}} +\left({z}_{\mathrm{2}} −{z}_{\mathrm{1}} \right)^{\mathrm{2}} } \\ $$$$=\sqrt{\left(\delta{l}\right)^{\mathrm{2}} +\left(\delta{m}\right)^{\mathrm{2}} +\left(\delta{n}\right)^{\mathrm{2}} } \\ $$$$\Rightarrow\left(\Delta{l}\right)^{\mathrm{2}} =\left(\delta{l}\right)^{\mathrm{2}} +\left(\delta{m}\right)^{\mathrm{2}} +\left(\delta{n}\right)^{\mathrm{2}} \\ $$$$ \\ $$$${on}\:{the}\:{other}\:{side}, \\ $$$$\left({P}_{\mathrm{1}} {P}_{\mathrm{2}} \right)^{\mathrm{2}} =\left({OP}_{\mathrm{1}} \right)^{\mathrm{2}} +\left({OP}_{\mathrm{2}} \right)^{\mathrm{2}} −\mathrm{2}\left({OP}_{\mathrm{1}} \right)\left({OP}_{\mathrm{2}} \right)\mathrm{cos}\:\angle{P}_{\mathrm{1}} {OP}_{\mathrm{2}} \\ $$$${i}.{e}. \\ $$$$\left(\Delta{l}\right)^{\mathrm{2}} =\mathrm{1}^{\mathrm{2}} +\mathrm{1}^{\mathrm{2}} −\mathrm{2}×\mathrm{1}×\mathrm{1}\:\mathrm{cos}\:\left(\delta\theta\right) \\ $$$$=\mathrm{2}−\mathrm{2}\:\mathrm{cos}\:\left(\delta\theta\right) \\ $$$$=\mathrm{2}−\mathrm{2}\left(\mathrm{1}−\mathrm{2}\:\mathrm{sin}^{\mathrm{2}} \:\frac{\delta\theta}{\mathrm{2}}\right) \\ $$$$=\mathrm{4}\:\left(\mathrm{sin}\:\frac{\delta\theta}{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$$\approx\mathrm{4}\:\left(\frac{\delta\theta}{\mathrm{2}}\right)^{\mathrm{2}} \:\:{since}\:\delta\theta\:{is}\:{small},\:\mathrm{sin}\:\frac{\delta\theta}{\mathrm{2}}\approx\frac{\delta\theta}{\mathrm{2}} \\ $$$$=\left(\delta\theta\right)^{\mathrm{2}} \\ $$$$ \\ $$$$\Rightarrow\left(\delta\theta\right)^{\mathrm{2}} =\left(\delta{l}\right)^{\mathrm{2}} +\left(\delta{m}\right)^{\mathrm{2}} +\left(\delta{n}\right)^{\mathrm{2}} \\ $$

Commented by vishalbhardwaj last updated on 11/Dec/19

sir please explain in detail

$$\mathrm{sir}\:\mathrm{please}\:\mathrm{explain}\:\mathrm{in}\:\mathrm{detail} \\ $$

Commented by peter frank last updated on 11/Dec/19

thank you

$${thank}\:{you} \\ $$

Commented by vishalbhardwaj last updated on 11/Dec/19

but sir it is not given that the   length of line is unit

$$\mathrm{but}\:\mathrm{sir}\:\mathrm{it}\:\mathrm{is}\:\mathrm{not}\:\mathrm{given}\:\mathrm{that}\:\mathrm{the}\: \\ $$$$\mathrm{length}\:\mathrm{of}\:\mathrm{line}\:\mathrm{is}\:\mathrm{unit} \\ $$

Commented by vishalbhardwaj last updated on 12/Dec/19

And sir this is a particular case  because it is not given that   lines are started from origin,   Infact they can be started from  any point in space except origin  , Please give that general case  to understand.

$$\mathrm{And}\:\mathrm{sir}\:\mathrm{this}\:\mathrm{is}\:\mathrm{a}\:\mathrm{particular}\:\mathrm{case} \\ $$$$\mathrm{because}\:\mathrm{it}\:\mathrm{is}\:\mathrm{not}\:\mathrm{given}\:\mathrm{that}\: \\ $$$$\mathrm{lines}\:\mathrm{are}\:\mathrm{started}\:\mathrm{from}\:\mathrm{origin},\: \\ $$$$\mathrm{Infact}\:\mathrm{they}\:\mathrm{can}\:\mathrm{be}\:\mathrm{started}\:\mathrm{from} \\ $$$$\mathrm{any}\:\mathrm{point}\:\mathrm{in}\:\mathrm{space}\:\mathrm{except}\:\mathrm{origin} \\ $$$$,\:\mathrm{Please}\:\mathrm{give}\:\mathrm{that}\:\mathrm{general}\:\mathrm{case} \\ $$$$\mathrm{to}\:\mathrm{understand}. \\ $$

Commented by mr W last updated on 12/Dec/19

since we only need to find the (small)  change of the direction of the line,  we can consider the change of a  unit length segment of it and we  can translate the segment such  that one end of the segment is at  the same point. the translation  doesn′t affect the direction of the  line.

$${since}\:{we}\:{only}\:{need}\:{to}\:{find}\:{the}\:\left({small}\right) \\ $$$${change}\:{of}\:{the}\:{direction}\:{of}\:{the}\:{line}, \\ $$$${we}\:{can}\:{consider}\:{the}\:{change}\:{of}\:{a} \\ $$$${unit}\:{length}\:{segment}\:{of}\:{it}\:{and}\:{we} \\ $$$${can}\:{translate}\:{the}\:{segment}\:{such} \\ $$$${that}\:{one}\:{end}\:{of}\:{the}\:{segment}\:{is}\:{at} \\ $$$${the}\:{same}\:{point}.\:{the}\:{translation} \\ $$$${doesn}'{t}\:{affect}\:{the}\:{direction}\:{of}\:{the} \\ $$$${line}. \\ $$

Commented by mr W last updated on 12/Dec/19

Commented by vishalbhardwaj last updated on 12/Dec/19

please sir if it is possible to you  to use general case please post  it

$$\mathrm{please}\:\mathrm{sir}\:\mathrm{if}\:\mathrm{it}\:\mathrm{is}\:\mathrm{possible}\:\mathrm{to}\:\mathrm{you} \\ $$$$\mathrm{to}\:\mathrm{use}\:\mathrm{general}\:\mathrm{case}\:\mathrm{please}\:\mathrm{post} \\ $$$$\mathrm{it} \\ $$

Commented by mr W last updated on 12/Dec/19

i have posted what i think is the  correct answer to the question sir.

$${i}\:{have}\:{posted}\:{what}\:{i}\:{think}\:{is}\:{the} \\ $$$${correct}\:{answer}\:{to}\:{the}\:{question}\:{sir}. \\ $$

Answered by vishalbhardwaj last updated on 12/Dec/19

Answered by vishalbhardwaj last updated on 12/Dec/19

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