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Question-51535

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Question Number 51535 by ajfour last updated on 28/Dec/18
Commented by ajfour last updated on 28/Dec/18
If BC be parallel to x-axis, find coordinates of A.
$${If}\:{BC}\:{be}\:{parallel}\:{to}\:{x}-{axis},\:{find} \\ $$$${coordinates}\:{of}\:{A}. \\ $$
Answered by mr W last updated on 28/Dec/18
B((a/2), −(√(R^2 −(a^2 /4))), 0) C(−(a/2), −(√(R^2 −(a^2 /4))), 0) x_A ^2 +y_A ^2 +z_A ^2 =R^2 ...(i) (x_A −(a/2))^2 +(y_A +(√(R^2 −(a^2 /4))))^2 +z_A ^2 =c^2 ...(ii) (x_A +(a/2))^2 +(y_A +(√(R^2 −(a^2 /4))))^2 +z_A ^2 =b^2 ...(iii) (iii)−(ii): 2ax_A =b^2 −c^2 ⇒x_A =((b^2 −c^2 )/(2a)) (iii)−(i): (√(R^2 −(a^2 /4)))(2y_A +(√(R^2 −(a^2 /4))))+(a/2)(2x_A +(a/2))=b^2 −R^2 (√(4R^2 −a^2 ))(4y_A +(√(4R^2 −a^2 )))=2b^2 +2c^2 −a^2 −4R^2 2y_A (√(4R^2 −a^2 ))=b^2 +c^2 −4R^2 ⇒y_A =((b^2 +c^2 −4R^2 )/(2(√(4R^2 −a^2 )))) ⇒z_A =(√(R^2 −x_A ^2 −y_A ^2 ))
$${B}\left(\frac{{a}}{\mathrm{2}},\:−\sqrt{{R}^{\mathrm{2}} −\frac{{a}^{\mathrm{2}} }{\mathrm{4}}},\:\mathrm{0}\right) \\ $$$${C}\left(−\frac{{a}}{\mathrm{2}},\:−\sqrt{{R}^{\mathrm{2}} −\frac{{a}^{\mathrm{2}} }{\mathrm{4}}},\:\mathrm{0}\right) \\ $$$${x}_{{A}} ^{\mathrm{2}} +{y}_{{A}} ^{\mathrm{2}} +{z}_{{A}} ^{\mathrm{2}} ={R}^{\mathrm{2}} \:\:\:…\left({i}\right) \\ $$$$\left({x}_{{A}} −\frac{{a}}{\mathrm{2}}\right)^{\mathrm{2}} +\left({y}_{{A}} +\sqrt{{R}^{\mathrm{2}} −\frac{{a}^{\mathrm{2}} }{\mathrm{4}}}\right)^{\mathrm{2}} +{z}_{{A}} ^{\mathrm{2}} ={c}^{\mathrm{2}} \:\:\:…\left({ii}\right) \\ $$$$\left({x}_{{A}} +\frac{{a}}{\mathrm{2}}\right)^{\mathrm{2}} +\left({y}_{{A}} +\sqrt{{R}^{\mathrm{2}} −\frac{{a}^{\mathrm{2}} }{\mathrm{4}}}\right)^{\mathrm{2}} +{z}_{{A}} ^{\mathrm{2}} ={b}^{\mathrm{2}} \:\:\:…\left({iii}\right) \\ $$$$ \\ $$$$\left({iii}\right)−\left({ii}\right): \\ $$$$\mathrm{2}{ax}_{{A}} ={b}^{\mathrm{2}} −{c}^{\mathrm{2}} \\ $$$$\Rightarrow{x}_{{A}} =\frac{{b}^{\mathrm{2}} −{c}^{\mathrm{2}} }{\mathrm{2}{a}} \\ $$$$\left({iii}\right)−\left({i}\right): \\ $$$$\sqrt{{R}^{\mathrm{2}} −\frac{{a}^{\mathrm{2}} }{\mathrm{4}}}\left(\mathrm{2}{y}_{{A}} +\sqrt{{R}^{\mathrm{2}} −\frac{{a}^{\mathrm{2}} }{\mathrm{4}}}\right)+\frac{{a}}{\mathrm{2}}\left(\mathrm{2}{x}_{{A}} +\frac{{a}}{\mathrm{2}}\right)={b}^{\mathrm{2}} −{R}^{\mathrm{2}} \\ $$$$\sqrt{\mathrm{4}{R}^{\mathrm{2}} −{a}^{\mathrm{2}} }\left(\mathrm{4}{y}_{{A}} +\sqrt{\mathrm{4}{R}^{\mathrm{2}} −{a}^{\mathrm{2}} }\right)=\mathrm{2}{b}^{\mathrm{2}} +\mathrm{2}{c}^{\mathrm{2}} −{a}^{\mathrm{2}} −\mathrm{4}{R}^{\mathrm{2}} \\ $$$$\mathrm{2}{y}_{{A}} \sqrt{\mathrm{4}{R}^{\mathrm{2}} −{a}^{\mathrm{2}} }={b}^{\mathrm{2}} +{c}^{\mathrm{2}} −\mathrm{4}{R}^{\mathrm{2}} \\ $$$$\Rightarrow{y}_{{A}} =\frac{{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −\mathrm{4}{R}^{\mathrm{2}} }{\mathrm{2}\sqrt{\mathrm{4}{R}^{\mathrm{2}} −{a}^{\mathrm{2}} }} \\ $$$$\Rightarrow{z}_{{A}} =\sqrt{{R}^{\mathrm{2}} −{x}_{{A}} ^{\mathrm{2}} −{y}_{{A}} ^{\mathrm{2}} } \\ $$
Commented by ajfour last updated on 28/Dec/18
Beautiful expression! whats the logic sir ?
$${Beautiful}\:{expression}!\:{whats}\:{the} \\ $$$${logic}\:{sir}\:? \\ $$
Commented by ajfour last updated on 28/Dec/18
Wonderfully presented Sir. Thanks. Easy one for you, i think.
$${Wonderfully}\:\:{presented}\:{Sir}.\:{Thanks}. \\ $$$${Easy}\:{one}\:{for}\:{you},\:{i}\:{think}. \\ $$
Commented by mr W last updated on 28/Dec/18
thank you sir! a related question sir: which relation must the sides a,b,c have such that the triangle can be placed into a sphere with radius R?
$${thank}\:{you}\:{sir}! \\ $$$${a}\:{related}\:{question}\:{sir}: \\ $$$${which}\:{relation}\:{must}\:{the}\:{sides}\:{a},{b},{c} \\ $$$${have}\:{such}\:{that} \\ $$$${the}\:{triangle}\:{can}\:{be}\:{placed}\:{into}\:{a} \\ $$$${sphere}\:{with}\:{radius}\:{R}? \\ $$
Commented by mr W last updated on 28/Dec/18
i think the condition that the triangle can be placed into the sphere is ((abc)/( (√((a+b+c)(−a+b+c)(a−b+c)(a+b−c)))))≤R
$${i}\:{think}\:{the}\:{condition}\:{that}\:{the}\:{triangle} \\ $$$${can}\:{be}\:{placed}\:{into}\:{the}\:{sphere}\:{is} \\ $$$$\frac{{abc}}{\:\sqrt{\left({a}+{b}+{c}\right)\left(−{a}+{b}+{c}\right)\left({a}−{b}+{c}\right)\left({a}+{b}−{c}\right)}}\leqslant{R} \\ $$
Commented by mr W last updated on 28/Dec/18
the curcumcircle of the triangle must fit into the sphere, i.e. the circumradius of the triangle must be less than or equal to the radius of the sphere.
$${the}\:{curcumcircle}\:{of}\:{the}\:{triangle}\:{must} \\ $$$${fit}\:{into}\:{the}\:{sphere},\:{i}.{e}.\:{the}\:{circumradius} \\ $$$${of}\:{the}\:{triangle}\:{must}\:{be}\:{less}\:{than}\:{or} \\ $$$${equal}\:{to}\:{the}\:{radius}\:{of}\:{the}\:{sphere}. \\ $$

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