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Question Number 11309 by FilupS last updated on 20/Mar/17
ax^2 +by^2 +cz^2 =r^2 Point P=(a, b, c) Point Q=(l, m, n) Both points lie on the curve what is the shortest path from point P to Q, along the outside of the curve?
$${ax}^{\mathrm{2}} +{by}^{\mathrm{2}} +{cz}^{\mathrm{2}} ={r}^{\mathrm{2}} \\ $$$$\: \\ $$$$\mathrm{Point}\:{P}=\left({a},\:{b},\:{c}\right) \\ $$$$\mathrm{Point}\:{Q}=\left({l},\:{m},\:{n}\right) \\ $$$$\mathrm{Both}\:\mathrm{points}\:\mathrm{lie}\:\mathrm{on}\:\mathrm{the}\:\mathrm{curve} \\ $$$$\: \\ $$$$\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{shortest}\:\mathrm{path}\:\mathrm{from}\:\mathrm{point} \\ $$$${P}\:\mathrm{to}\:{Q},\:\mathrm{along}\:\mathrm{the}\:\mathrm{outside}\:\mathrm{of}\:\mathrm{the}\:\mathrm{curve}? \\ $$
Commented by mrW1 last updated on 20/Mar/17
This is a very difficult question. If a,b,c>0, ax^2 +by^2 +cz^2 =r^2 is a triaxial ellipsoid. Your question is then about geodesics on a triaxial ellipsoid.
$${This}\:{is}\:{a}\:{very}\:{difficult}\:{question}.\:{If} \\ $$$${a},{b},{c}>\mathrm{0},\:{ax}^{\mathrm{2}} +{by}^{\mathrm{2}} +{cz}^{\mathrm{2}} ={r}^{\mathrm{2}} \:{is}\:{a}\:{triaxial} \\ $$$${ellipsoid}.\:{Your}\:{question}\:{is}\:{then}\:{about} \\ $$$${geodesics}\:{on}\:{a}\:{triaxial}\:{ellipsoid}. \\ $$
Commented by FilupS last updated on 21/Mar/17
what if a=b=c=1?
$$\mathrm{what}\:\mathrm{if}\:{a}={b}={c}=\mathrm{1}? \\ $$
Commented by mrW1 last updated on 21/Mar/17
Then it is a sphere with radius r. The shortest path on a sphere is the great−circle distance. Point P(x_1 ,y_1 ,z_1 ) Point Q(x_2 ,y_2 ,z_2 ) Vector p=OP^(→) , q=OQ^(→) Angle between p and q =Δα ∣p∣=(√(x_1 ^2 +y_1 ^2 +z_1 ^2 ))=r ∣q∣=(√(x_2 ^2 +y_2 ^2 +z_2 ^2 ))=r cos Δα=((p∙q)/(∣p∣×∣q∣))=((x_1 x_2 +y_1 y_2 +z_1 z_2 )/r^2 ) Δα=cos^(−1) (((x_1 x_2 +y_1 y_2 +z_1 z_2 )/r^2 )) The shortest path from P to Q is d=Δα×r
$${Then}\:{it}\:{is}\:{a}\:{sphere}\:{with}\:{radius}\:{r}.\:{The}\:{shortest}\:{path} \\ $$$${on}\:{a}\:{sphere}\:{is}\:{the}\:{great}−{circle}\:{distance}. \\ $$$$ \\ $$$${Point}\:{P}\left({x}_{\mathrm{1}} ,{y}_{\mathrm{1}} ,{z}_{\mathrm{1}} \right) \\ $$$${Point}\:{Q}\left({x}_{\mathrm{2}} ,{y}_{\mathrm{2}} ,{z}_{\mathrm{2}} \right) \\ $$$${Vector}\:\boldsymbol{{p}}=\overset{\rightarrow} {{OP}},\:\boldsymbol{{q}}=\overset{\rightarrow} {{OQ}} \\ $$$${Angle}\:{between}\:\boldsymbol{{p}}\:{and}\:\boldsymbol{{q}}\:=\Delta\alpha \\ $$$$\mid\boldsymbol{{p}}\mid=\sqrt{{x}_{\mathrm{1}} ^{\mathrm{2}} +{y}_{\mathrm{1}} ^{\mathrm{2}} +{z}_{\mathrm{1}} ^{\mathrm{2}} }={r} \\ $$$$\mid\boldsymbol{{q}}\mid=\sqrt{{x}_{\mathrm{2}} ^{\mathrm{2}} +{y}_{\mathrm{2}} ^{\mathrm{2}} +{z}_{\mathrm{2}} ^{\mathrm{2}} }={r} \\ $$$$\mathrm{cos}\:\Delta\alpha=\frac{\boldsymbol{{p}}\centerdot\boldsymbol{{q}}}{\mid\boldsymbol{{p}}\mid×\mid\boldsymbol{{q}}\mid}=\frac{{x}_{\mathrm{1}} {x}_{\mathrm{2}} +{y}_{\mathrm{1}} {y}_{\mathrm{2}} +{z}_{\mathrm{1}} {z}_{\mathrm{2}} }{{r}^{\mathrm{2}} } \\ $$$$\Delta\alpha=\mathrm{cos}^{−\mathrm{1}} \left(\frac{{x}_{\mathrm{1}} {x}_{\mathrm{2}} +{y}_{\mathrm{1}} {y}_{\mathrm{2}} +{z}_{\mathrm{1}} {z}_{\mathrm{2}} }{{r}^{\mathrm{2}} }\right) \\ $$$$ \\ $$$${The}\:{shortest}\:{path}\:{from}\:{P}\:{to}\:{Q}\:{is} \\ $$$${d}=\Delta\alpha×{r} \\ $$
Commented by FilupS last updated on 21/Mar/17
fascinating!
$$\mathrm{fascinating}! \\ $$

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