Question Number 80146 by Khyati last updated on 31/Jan/20
$${Two}\:{system}\:{of}\:{rectangular}\:{axes}\:{have} \\ $$$${the}\:{same}\:{origin}.\:{If}\:{a}\:{plane}\:{cuts}\:{them} \\ $$$${at}\:{distance}\:{a},\:{b},\:{c}\:{and}\:{p},\:{q},\:{r} \\ $$$${respectively},\:{then}\:{prove}\:{with}\:{the}\:{help} \\ $$$${of}\:{an}\:{appropriate}\:{diagram}\:{that}\:: \\ $$$$\frac{\mathrm{1}}{{a}^{\mathrm{2}} }\:+\:\frac{\mathrm{1}}{{b}^{\mathrm{2}} }\:+\:\frac{\mathrm{1}}{{c}^{\mathrm{2}} }\:=\:\frac{\mathrm{1}}{{p}^{\mathrm{2}} }\:+\:\frac{\mathrm{1}}{{q}^{\mathrm{2}} }\:+\:\frac{\mathrm{1}}{{r}^{\mathrm{2}} } \\ $$
Answered by mr W last updated on 31/Jan/20
Commented by mr W last updated on 31/Jan/20
![say the origin is point O and the plane is P. the distance from O to the plane is h, which is constant. let′s look at an arbitrary rectangular axis system with origin at O and the axes intersect the plane at A,B,C with distances a,b,c to the origin. O, A, B, C build a tetrahedron. AB=(√(a^2 +b^2 )) BC=(√(b^2 +c^2 )) CA=(√(c^2 +a^2 )) the volume of the tetrahedron is V. V=(1/3)×((ab)/2)×c=((abc)/6) or V=(1/3)×Δ_(ABC) ×h with Δ_(ABC) =area ABC Δ_(ABC) =(1/2)×AB×AC×sin ∠BAC Δ_(ABC) =((√((a^2 +b^2 )(c^2 +a^2 )))/2)×sin ∠BAC BC^2 =AB^2 +AC^2 −2×AB×AC×cos ∠BAC b^2 +c^2 =a^2 +b^2 +c^2 +a^2 −2(√((a^2 +b^2 )(c^2 +a^2 )))×cos ∠BAC a^2 =(√((a^2 +b^2 )(c^2 +a^2 )))×cos ∠BAC ⇒cos ∠BAC=(a^2 /( (√((a^2 +b^2 )(c^2 +a^2 ))))) ⇒sin ∠BAC=((√((a^2 +b^2 )(c^2 +a^2 )−a^4 ))/( (√((a^2 +b^2 )(c^2 +a^2 ))))) ⇒sin ∠BAC=((√(b^2 c^2 +a^2 (b^2 +c^2 )))/( (√((a^2 +b^2 )(c^2 +a^2 ))))) Δ_(ABC) =((√((a^2 +b^2 )(c^2 +a^2 )))/2)×((√(b^2 c^2 +a^2 (b^2 +c^2 )))/( (√((a^2 +b^2 )(c^2 +a^2 ))))) ⇒Δ_(ABC) =((√(b^2 c^2 +a^2 (b^2 +c^2 )))/2) V=(1/3)×((√(b^2 c^2 +a^2 (b^2 +c^2 )))/2)×h=((h(√(b^2 c^2 +a^2 (b^2 +c^2 ))))/6) ⇒((h(√(b^2 c^2 +a^2 (b^2 +c^2 ))))/6)=((abc)/6) ⇒h^2 [b^2 c^2 +a^2 (b^2 +c^2 )]=a^2 b^2 c^2 ⇒(1/a^2 )+(1/b^2 )+(1/c^2 )=(1/h^2 ) since h is constant, we get also for an other system: ⇒(1/p^2 )+(1/q^2 )+(1/r^2 )=(1/h^2 ) therefore: ⇒(1/a^2 )+(1/b^2 )+(1/c^2 )=(1/p^2 )+(1/q^2 )+(1/r^2 )](https://www.tinkutara.com/question/Q80153.png)
$${say}\:{the}\:{origin}\:{is}\:{point}\:{O}\:{and}\:{the}\:{plane} \\ $$$${is}\:{P}.\: \\ $$$${the}\:{distance}\:{from}\:{O}\:{to}\:{the}\:{plane}\:{is}\:{h}, \\ $$$${which}\:{is}\:{constant}. \\ $$$$ \\ $$$${let}'{s}\:{look}\:{at}\:{an}\:{arbitrary}\:{rectangular} \\ $$$${axis}\:{system}\:{with}\:{origin}\:{at}\:\:{O}\:{and}\:{the} \\ $$$${axes}\:{intersect}\:{the}\:{plane}\:{at}\:{A},{B},{C} \\ $$$${with}\:{distances}\:{a},{b},{c}\:{to}\:{the}\:{origin}. \\ $$$$ \\ $$$${O},\:{A},\:{B},\:{C}\:{build}\:{a}\:{tetrahedron}. \\ $$$${AB}=\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} } \\ $$$${BC}=\sqrt{{b}^{\mathrm{2}} +{c}^{\mathrm{2}} } \\ $$$${CA}=\sqrt{{c}^{\mathrm{2}} +{a}^{\mathrm{2}} } \\ $$$${the}\:{volume}\:{of}\:{the}\:{tetrahedron}\:{is}\:{V}. \\ $$$${V}=\frac{\mathrm{1}}{\mathrm{3}}×\frac{{ab}}{\mathrm{2}}×{c}=\frac{{abc}}{\mathrm{6}} \\ $$$${or} \\ $$$${V}=\frac{\mathrm{1}}{\mathrm{3}}×\Delta_{{ABC}} ×{h}\:{with}\:\Delta_{{ABC}} ={area}\:{ABC} \\ $$$$ \\ $$$$\Delta_{{ABC}} =\frac{\mathrm{1}}{\mathrm{2}}×{AB}×{AC}×\mathrm{sin}\:\angle{BAC} \\ $$$$\Delta_{{ABC}} =\frac{\sqrt{\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)\left({c}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)}}{\mathrm{2}}×\mathrm{sin}\:\angle{BAC} \\ $$$${BC}^{\mathrm{2}} ={AB}^{\mathrm{2}} +{AC}^{\mathrm{2}} −\mathrm{2}×{AB}×{AC}×\mathrm{cos}\:\angle{BAC} \\ $$$${b}^{\mathrm{2}} +{c}^{\mathrm{2}} ={a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +{a}^{\mathrm{2}} −\mathrm{2}\sqrt{\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)\left({c}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)}×\mathrm{cos}\:\angle{BAC} \\ $$$${a}^{\mathrm{2}} =\sqrt{\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)\left({c}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)}×\mathrm{cos}\:\angle{BAC} \\ $$$$\Rightarrow\mathrm{cos}\:\angle{BAC}=\frac{{a}^{\mathrm{2}} }{\:\sqrt{\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)\left({c}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)}} \\ $$$$\Rightarrow\mathrm{sin}\:\angle{BAC}=\frac{\sqrt{\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)\left({c}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)−{a}^{\mathrm{4}} }}{\:\sqrt{\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)\left({c}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)}} \\ $$$$\Rightarrow\mathrm{sin}\:\angle{BAC}=\frac{\sqrt{{b}^{\mathrm{2}} {c}^{\mathrm{2}} +{a}^{\mathrm{2}} \left({b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)}}{\:\sqrt{\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)\left({c}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)}} \\ $$$$\Delta_{{ABC}} =\frac{\sqrt{\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)\left({c}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)}}{\mathrm{2}}×\frac{\sqrt{{b}^{\mathrm{2}} {c}^{\mathrm{2}} +{a}^{\mathrm{2}} \left({b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)}}{\:\sqrt{\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)\left({c}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)}} \\ $$$$\Rightarrow\Delta_{{ABC}} =\frac{\sqrt{{b}^{\mathrm{2}} {c}^{\mathrm{2}} +{a}^{\mathrm{2}} \left({b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)}}{\mathrm{2}} \\ $$$${V}=\frac{\mathrm{1}}{\mathrm{3}}×\frac{\sqrt{{b}^{\mathrm{2}} {c}^{\mathrm{2}} +{a}^{\mathrm{2}} \left({b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)}}{\mathrm{2}}×{h}=\frac{{h}\sqrt{{b}^{\mathrm{2}} {c}^{\mathrm{2}} +{a}^{\mathrm{2}} \left({b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)}}{\mathrm{6}} \\ $$$$ \\ $$$$\Rightarrow\frac{{h}\sqrt{{b}^{\mathrm{2}} {c}^{\mathrm{2}} +{a}^{\mathrm{2}} \left({b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)}}{\mathrm{6}}=\frac{{abc}}{\mathrm{6}} \\ $$$$\Rightarrow{h}^{\mathrm{2}} \left[{b}^{\mathrm{2}} {c}^{\mathrm{2}} +{a}^{\mathrm{2}} \left({b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)\right]={a}^{\mathrm{2}} {b}^{\mathrm{2}} {c}^{\mathrm{2}} \\ $$$$\Rightarrow\frac{\mathrm{1}}{{a}^{\mathrm{2}} }+\frac{\mathrm{1}}{{b}^{\mathrm{2}} }+\frac{\mathrm{1}}{{c}^{\mathrm{2}} }=\frac{\mathrm{1}}{{h}^{\mathrm{2}} } \\ $$$$ \\ $$$${since}\:{h}\:{is}\:{constant},\:{we}\:{get}\:{also}\:{for} \\ $$$${an}\:{other}\:{system}: \\ $$$$\Rightarrow\frac{\mathrm{1}}{{p}^{\mathrm{2}} }+\frac{\mathrm{1}}{{q}^{\mathrm{2}} }+\frac{\mathrm{1}}{{r}^{\mathrm{2}} }=\frac{\mathrm{1}}{{h}^{\mathrm{2}} } \\ $$$$ \\ $$$${therefore}: \\ $$$$\Rightarrow\frac{\mathrm{1}}{{a}^{\mathrm{2}} }+\frac{\mathrm{1}}{{b}^{\mathrm{2}} }+\frac{\mathrm{1}}{{c}^{\mathrm{2}} }=\frac{\mathrm{1}}{{p}^{\mathrm{2}} }+\frac{\mathrm{1}}{{q}^{\mathrm{2}} }+\frac{\mathrm{1}}{{r}^{\mathrm{2}} } \\ $$
Commented by mr W last updated on 31/Jan/20
$${An}\:{other}\:{way}: \\ $$$${eqn}.\:{of}\:{plane}\:{in}\:{system}\:\mathrm{1}: \\ $$$$\frac{{x}}{{a}}+\frac{{y}}{{b}}+\frac{{z}}{{c}}−\mathrm{1}=\mathrm{0} \\ $$$${distance}\:{from}\:{origin}\:{to}\:{plane}\:{is} \\ $$$${h}=\frac{\mid\frac{\mathrm{0}}{{a}}+\frac{\mathrm{0}}{{b}}+\frac{\mathrm{0}}{{c}}−\mathrm{1}\mid}{\:\sqrt{\frac{\mathrm{1}}{{a}^{\mathrm{2}} }+\frac{\mathrm{1}}{{b}^{\mathrm{2}} }+\frac{\mathrm{1}}{{c}^{\mathrm{2}} }}}=\frac{\mathrm{1}}{\:\sqrt{\frac{\mathrm{1}}{{a}^{\mathrm{2}} }+\frac{\mathrm{1}}{{b}^{\mathrm{2}} }+\frac{\mathrm{1}}{{c}^{\mathrm{2}} }}} \\ $$$$ \\ $$$${similarly}\:{in}\:{system}\:\mathrm{2}\:{the} \\ $$$${distance}\:{from}\:{origin}\:{to}\:{plane}\:{is} \\ $$$${h}=\frac{\mathrm{1}}{\:\sqrt{\frac{\mathrm{1}}{{p}^{\mathrm{2}} }+\frac{\mathrm{1}}{{p}^{\mathrm{2}} }+\frac{\mathrm{1}}{{r}^{\mathrm{2}} }}} \\ $$$${since}\:{h}\:{is}\:{in}\:{both}\:{systems}\:{the}\:{same}, \\ $$$$\frac{\mathrm{1}}{\:\sqrt{\frac{\mathrm{1}}{{a}^{\mathrm{2}} }+\frac{\mathrm{1}}{{b}^{\mathrm{2}} }+\frac{\mathrm{1}}{{c}^{\mathrm{2}} }}}=\frac{\mathrm{1}}{\:\sqrt{\frac{\mathrm{1}}{{p}^{\mathrm{2}} }+\frac{\mathrm{1}}{{p}^{\mathrm{2}} }+\frac{\mathrm{1}}{{r}^{\mathrm{2}} }}} \\ $$$$\Rightarrow\frac{\mathrm{1}}{{a}^{\mathrm{2}} }+\frac{\mathrm{1}}{{b}^{\mathrm{2}} }+\frac{\mathrm{1}}{{c}^{\mathrm{2}} }=\frac{\mathrm{1}}{{p}^{\mathrm{2}} }+\frac{\mathrm{1}}{{p}^{\mathrm{2}} }+\frac{\mathrm{1}}{{r}^{\mathrm{2}} } \\ $$
Commented by TawaTawa last updated on 31/Jan/20
$$\mathrm{Weldon}\:\mathrm{sir},\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you} \\ $$