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A-cube-of-unit-edge-length-is-held-before-a-plane-Prove-that-the-sum-of-the-squares-of-the-projected-lengths-of-edges-of-the-cube-on-the-plane-irrespective-of-the-orientation-of-the-cube-is-8-

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Question Number 62121 by ajfour last updated on 16/Jun/19
A cube of unit edge length is held before a plane. Prove that the sum of the squares of the projected lengths of edges of the cube on the plane (irrespective of the orientation of the cube) is 8.
$${A}\:{cube}\:{of}\:{unit}\:{edge}\:{length}\: \\ $$$${is}\:{held}\:{before}\:{a}\:{plane}.\:{Prove}\:{that} \\ $$$${the}\:{sum}\:{of}\:{the}\:{squares}\:{of}\:{the} \\ $$$${projected}\:{lengths}\:{of}\:{edges}\:{of}\:{the}\:{cube} \\ $$$${on}\:{the}\:{plane}\:\left({irrespective}\:{of}\right. \\ $$$$\left.{the}\:{orientation}\:{of}\:{the}\:{cube}\right)\:{is}\:\mathrm{8}. \\ $$
Answered by mr W last updated on 16/Jun/19
Commented by mr W last updated on 16/Jun/19
say O1234567 is the unit cube. say ABC is the plane with A(a,0,0), B(0,b,0) and C(0,0,c) eqn. of the plane is (x/a)+(y/b)+(z/c)=1 n^→ =((1/a),(1/b),(1/c)) ∣n∣=(√((1/a^2 )+(1/b^2 )+(1/c^2 ))) for two points P(x_1 ,y_1 ,z_1 ) and Q(x_2 ,y_2 ,z_2 ) PQ^(→) =(Δx,Δy,Δz) with Δx=x_2 −x_1 , Δy=y_2 −y_1 , Δz=z_2 −z_1 ∣PQ∣=(√((Δx)^2 +(Δy)^2 +(Δz)^2 )) cos θ=((PQ^(→) .n^(→) )/(∣PQ∣×∣n∣))=((((Δx)/a)+((Δy)/b)+((Δz)/c))/( (√((Δx)^2 +(Δy)^2 +(Δz)^2 ))×(√((1/a^2 )+(1/b^2 )+(1/c^2 ))))) length of projection of PQ on plane is ∣P′Q′∣=∣PQ∣sin θ ⇒∣P′Q′∣^2 =∣PQ∣^2 sin^2 θ=∣PQ∣^2 (1−cos^2 θ) ⇒∣P′Q′∣^2 =(Δx)^2 +(Δy)^2 +(Δz)^2 −(((((Δx)/a)+((Δy)/b)+((Δz)/c))^2 )/((1/a^2 )+(1/b^2 )+(1/c^2 ))) now we apply this for all 12 edges of the unit cube. edge O1=32=45=76: (4 edges) Δx=1, Δy=0, Δz=0 ∣O′1′∣^2 =1−((1/a^2 )/((1/a^2 )+(1/b^2 )+(1/c^2 ))) edge 12=O3=74=65: (4 edges) Δx=0, Δy=1, Δz=0 ∣1′2′∣^2 =1−((1/b^2 )/((1/a^2 )+(1/b^2 )+(1/c^2 ))) edge 25=34=O7=15: (4 edges) Δx=0, Δy=0, Δz=1 ∣2′5′∣^2 =1−((1/c^2 )/((1/a^2 )+(1/b^2 )+(1/c^2 ))) sum for all edges: 4(1−((1/a^2 )/((1/a^2 )+(1/b^2 )+(1/c^2 )))+1−((1/b^2 )/((1/a^2 )+(1/b^2 )+(1/c^2 )))+1−((1/c^2 )/((1/a^2 )+(1/b^2 )+(1/c^2 )))) =4(3−(((1/a^2 )+(1/b^2 )+(1/c^2 ))/((1/a^2 )+(1/b^2 )+(1/c^2 )))) =4(3−1) =8 the result is independent from the orientation of the plane.
$${say}\:{O}\mathrm{1234567}\:{is}\:{the}\:{unit}\:{cube}. \\ $$$${say}\:{ABC}\:{is}\:{the}\:{plane}\:{with}\: \\ $$$${A}\left({a},\mathrm{0},\mathrm{0}\right),\:{B}\left(\mathrm{0},{b},\mathrm{0}\right)\:{and}\:{C}\left(\mathrm{0},\mathrm{0},{c}\right) \\ $$$${eqn}.\:{of}\:{the}\:{plane}\:{is} \\ $$$$\frac{{x}}{{a}}+\frac{{y}}{{b}}+\frac{{z}}{{c}}=\mathrm{1} \\ $$$$\overset{\rightarrow} {{n}}=\left(\frac{\mathrm{1}}{{a}},\frac{\mathrm{1}}{{b}},\frac{\mathrm{1}}{{c}}\right) \\ $$$$\mid{n}\mid=\sqrt{\frac{\mathrm{1}}{{a}^{\mathrm{2}} }+\frac{\mathrm{1}}{{b}^{\mathrm{2}} }+\frac{\mathrm{1}}{{c}^{\mathrm{2}} }} \\ $$$${for}\:{two}\:{points}\:{P}\left({x}_{\mathrm{1}} ,{y}_{\mathrm{1}} ,{z}_{\mathrm{1}} \right)\:{and}\:{Q}\left({x}_{\mathrm{2}} ,{y}_{\mathrm{2}} ,{z}_{\mathrm{2}} \right) \\ $$$$\overset{\rightarrow} {{PQ}}=\left(\Delta{x},\Delta{y},\Delta{z}\right)\:{with} \\ $$$$\Delta{x}={x}_{\mathrm{2}} −{x}_{\mathrm{1}} ,\:\Delta{y}={y}_{\mathrm{2}} −{y}_{\mathrm{1}} ,\:\Delta{z}={z}_{\mathrm{2}} −{z}_{\mathrm{1}} \\ $$$$\mid{PQ}\mid=\sqrt{\left(\Delta{x}\right)^{\mathrm{2}} +\left(\Delta{y}\right)^{\mathrm{2}} +\left(\Delta{z}\right)^{\mathrm{2}} } \\ $$$$\mathrm{cos}\:\theta=\frac{\overset{\rightarrow} {{PQ}}.\overset{\rightarrow} {{n}}}{\mid{PQ}\mid×\mid{n}\mid}=\frac{\frac{\Delta{x}}{{a}}+\frac{\Delta{y}}{{b}}+\frac{\Delta{z}}{{c}}}{\:\sqrt{\left(\Delta{x}\right)^{\mathrm{2}} +\left(\Delta{y}\right)^{\mathrm{2}} +\left(\Delta{z}\right)^{\mathrm{2}} }×\sqrt{\frac{\mathrm{1}}{{a}^{\mathrm{2}} }+\frac{\mathrm{1}}{{b}^{\mathrm{2}} }+\frac{\mathrm{1}}{{c}^{\mathrm{2}} }}} \\ $$$${length}\:{of}\:{projection}\:{of}\:{PQ}\:{on}\:{plane}\:{is} \\ $$$$\mid{P}'{Q}'\mid=\mid{PQ}\mid\mathrm{sin}\:\theta \\ $$$$\Rightarrow\mid{P}'{Q}'\mid^{\mathrm{2}} =\mid{PQ}\mid^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \:\theta=\mid{PQ}\mid^{\mathrm{2}} \left(\mathrm{1}−\mathrm{cos}^{\mathrm{2}} \:\theta\right) \\ $$$$\Rightarrow\mid{P}'{Q}'\mid^{\mathrm{2}} =\left(\Delta{x}\right)^{\mathrm{2}} +\left(\Delta{y}\right)^{\mathrm{2}} +\left(\Delta{z}\right)^{\mathrm{2}} −\frac{\left(\frac{\Delta{x}}{{a}}+\frac{\Delta{y}}{{b}}+\frac{\Delta{z}}{{c}}\right)^{\mathrm{2}} }{\frac{\mathrm{1}}{{a}^{\mathrm{2}} }+\frac{\mathrm{1}}{{b}^{\mathrm{2}} }+\frac{\mathrm{1}}{{c}^{\mathrm{2}} }} \\ $$$${now}\:{we}\:{apply}\:{this}\:{for}\:{all}\:\mathrm{12}\:{edges}\:{of}\:{the} \\ $$$${unit}\:{cube}. \\ $$$${edge}\:{O}\mathrm{1}=\mathrm{32}=\mathrm{45}=\mathrm{76}:\:\:\left(\mathrm{4}\:{edges}\right) \\ $$$$\Delta{x}=\mathrm{1},\:\Delta{y}=\mathrm{0},\:\Delta{z}=\mathrm{0} \\ $$$$\mid{O}'\mathrm{1}'\mid^{\mathrm{2}} =\mathrm{1}−\frac{\frac{\mathrm{1}}{{a}^{\mathrm{2}} }}{\frac{\mathrm{1}}{{a}^{\mathrm{2}} }+\frac{\mathrm{1}}{{b}^{\mathrm{2}} }+\frac{\mathrm{1}}{{c}^{\mathrm{2}} }} \\ $$$$ \\ $$$${edge}\:\mathrm{12}={O}\mathrm{3}=\mathrm{74}=\mathrm{65}:\:\:\left(\mathrm{4}\:{edges}\right) \\ $$$$\Delta{x}=\mathrm{0},\:\Delta{y}=\mathrm{1},\:\Delta{z}=\mathrm{0} \\ $$$$\mid\mathrm{1}'\mathrm{2}'\mid^{\mathrm{2}} =\mathrm{1}−\frac{\frac{\mathrm{1}}{{b}^{\mathrm{2}} }}{\frac{\mathrm{1}}{{a}^{\mathrm{2}} }+\frac{\mathrm{1}}{{b}^{\mathrm{2}} }+\frac{\mathrm{1}}{{c}^{\mathrm{2}} }} \\ $$$$ \\ $$$${edge}\:\mathrm{25}=\mathrm{34}={O}\mathrm{7}=\mathrm{15}:\:\:\left(\mathrm{4}\:{edges}\right) \\ $$$$\Delta{x}=\mathrm{0},\:\Delta{y}=\mathrm{0},\:\Delta{z}=\mathrm{1} \\ $$$$\mid\mathrm{2}'\mathrm{5}'\mid^{\mathrm{2}} =\mathrm{1}−\frac{\frac{\mathrm{1}}{{c}^{\mathrm{2}} }}{\frac{\mathrm{1}}{{a}^{\mathrm{2}} }+\frac{\mathrm{1}}{{b}^{\mathrm{2}} }+\frac{\mathrm{1}}{{c}^{\mathrm{2}} }} \\ $$$$ \\ $$$${sum}\:{for}\:{all}\:{edges}: \\ $$$$\mathrm{4}\left(\mathrm{1}−\frac{\frac{\mathrm{1}}{{a}^{\mathrm{2}} }}{\frac{\mathrm{1}}{{a}^{\mathrm{2}} }+\frac{\mathrm{1}}{{b}^{\mathrm{2}} }+\frac{\mathrm{1}}{{c}^{\mathrm{2}} }}+\mathrm{1}−\frac{\frac{\mathrm{1}}{{b}^{\mathrm{2}} }}{\frac{\mathrm{1}}{{a}^{\mathrm{2}} }+\frac{\mathrm{1}}{{b}^{\mathrm{2}} }+\frac{\mathrm{1}}{{c}^{\mathrm{2}} }}+\mathrm{1}−\frac{\frac{\mathrm{1}}{{c}^{\mathrm{2}} }}{\frac{\mathrm{1}}{{a}^{\mathrm{2}} }+\frac{\mathrm{1}}{{b}^{\mathrm{2}} }+\frac{\mathrm{1}}{{c}^{\mathrm{2}} }}\right) \\ $$$$=\mathrm{4}\left(\mathrm{3}−\frac{\frac{\mathrm{1}}{{a}^{\mathrm{2}} }+\frac{\mathrm{1}}{{b}^{\mathrm{2}} }+\frac{\mathrm{1}}{{c}^{\mathrm{2}} }}{\frac{\mathrm{1}}{{a}^{\mathrm{2}} }+\frac{\mathrm{1}}{{b}^{\mathrm{2}} }+\frac{\mathrm{1}}{{c}^{\mathrm{2}} }}\right) \\ $$$$=\mathrm{4}\left(\mathrm{3}−\mathrm{1}\right) \\ $$$$=\mathrm{8} \\ $$$${the}\:{result}\:{is}\:{independent}\:{from}\:{the} \\ $$$${orientation}\:{of}\:{the}\:{plane}. \\ $$
Commented by ajfour last updated on 16/Jun/19
Thanks Sir, very brave!
$${Thanks}\:{Sir},\:{very}\:{brave}! \\ $$
Answered by ajfour last updated on 16/Jun/19
Commented by ajfour last updated on 16/Jun/19
let a^� =cos θi^� +sin θk^� b^� =cos φcos ψi^� +cos φsin ψj^� +sin φk^� as c^� = a^� ×b^� c^� =−cos φsin ψsin θi^� + (−sin φcos θ+cos φcos ψsin θ)j^� +cos φsin ψcos θk^� and as a^� .b^� =0 ⇒ cos θcos φcos ψ+sin θsin φ=0 ........(i) let required sum be S. S=4{cos^2 θ+cos^2 φcos^2 ψ+ cos^2 φsin^2 ψ+cos^2 φsin^2 ψsin^2 θ +(−sin φcos θ+cos φcos ψsin θ)^2 } ⇒ (S/4)=cos^2 θ+cos^2 φ+sin^2 θcos^2 φ +sin^2 φcos^2 θ −2sin φcos φsin θcos θcos ψ Now using (i) (S/4)=cos^2 θ+cos^2 φ+sin^2 θcos^2 φ +sin^2 φcos^2 θ+2sin^2 θsin^2 φ (S/4) = cos^2 θ+cos^2 φ +sin^2 θ−sin^2 θsin^2 φ sin^2 φ−sin^2 φsin^2 θ +2sin^2 θsin^2 φ ⇒ (S/4) = 2 or S=8 .
$${let}\:\bar {{a}}=\mathrm{cos}\:\theta\hat {{i}}+\mathrm{sin}\:\theta\hat {{k}} \\ $$$$\:\:\:\:\:\:\:\bar {{b}}=\mathrm{cos}\:\phi\mathrm{cos}\:\psi\hat {{i}}+\mathrm{cos}\:\phi\mathrm{sin}\:\psi\hat {{j}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\mathrm{sin}\:\phi\hat {{k}} \\ $$$$\:\:\:\:{as}\:\:\bar {{c}}=\:\bar {{a}}×\bar {{b}} \\ $$$$\:\:\:\:\bar {{c}}=−\mathrm{cos}\:\phi\mathrm{sin}\:\psi\mathrm{sin}\:\theta\hat {{i}}+ \\ $$$$\:\:\:\:\:\:\:\left(−\mathrm{sin}\:\phi\mathrm{cos}\:\theta+\mathrm{cos}\:\phi\mathrm{cos}\:\psi\mathrm{sin}\:\theta\right)\hat {{j}} \\ $$$$\:\:\:\:\:\:\:\:+\mathrm{cos}\:\phi\mathrm{sin}\:\psi\mathrm{cos}\:\theta\hat {{k}} \\ $$$${and}\:{as}\:\:\:\bar {{a}}.\bar {{b}}=\mathrm{0} \\ $$$$\Rightarrow\:\:\:\:\mathrm{cos}\:\theta\mathrm{cos}\:\phi\mathrm{cos}\:\psi+\mathrm{sin}\:\theta\mathrm{sin}\:\phi=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:……..\left({i}\right) \\ $$$${let}\:{required}\:{sum}\:{be}\:{S}. \\ $$$${S}=\mathrm{4}\left\{\mathrm{cos}\:^{\mathrm{2}} \theta+\mathrm{cos}\:^{\mathrm{2}} \phi\mathrm{cos}\:^{\mathrm{2}} \psi+\right. \\ $$$$\:\:\:\:\mathrm{cos}\:^{\mathrm{2}} \phi\mathrm{sin}\:^{\mathrm{2}} \psi+\mathrm{cos}\:^{\mathrm{2}} \phi\mathrm{sin}\:^{\mathrm{2}} \psi\mathrm{sin}\:^{\mathrm{2}} \theta \\ $$$$\left.\:\:\:\:+\left(−\mathrm{sin}\:\phi\mathrm{cos}\:\theta+\mathrm{cos}\:\phi\mathrm{cos}\:\psi\mathrm{sin}\:\theta\right)^{\mathrm{2}} \right\} \\ $$$$\Rightarrow\:\frac{{S}}{\mathrm{4}}=\mathrm{cos}\:^{\mathrm{2}} \theta+\mathrm{cos}\:^{\mathrm{2}} \phi+\mathrm{sin}\:^{\mathrm{2}} \theta\mathrm{cos}\:^{\mathrm{2}} \phi \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\mathrm{sin}\:^{\mathrm{2}} \phi\mathrm{cos}\:^{\mathrm{2}} \theta \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\mathrm{2sin}\:\phi\mathrm{cos}\:\phi\mathrm{sin}\:\theta\mathrm{cos}\:\theta\mathrm{cos}\:\psi \\ $$$$\:\:\:\:{Now}\:{using}\:\left({i}\right) \\ $$$$\:\frac{{S}}{\mathrm{4}}=\mathrm{cos}\:^{\mathrm{2}} \theta+\mathrm{cos}\:^{\mathrm{2}} \phi+\mathrm{sin}\:^{\mathrm{2}} \theta\mathrm{cos}\:^{\mathrm{2}} \phi \\ $$$$\:\:\:\:\:\:\:\:\:\:+\mathrm{sin}\:^{\mathrm{2}} \phi\mathrm{cos}\:^{\mathrm{2}} \theta+\mathrm{2sin}\:^{\mathrm{2}} \theta\mathrm{sin}\:^{\mathrm{2}} \phi \\ $$$$\:\:\:\:\frac{{S}}{\mathrm{4}}\:=\:\mathrm{cos}\:^{\mathrm{2}} \theta+\mathrm{cos}\:^{\mathrm{2}} \phi \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:+\mathrm{sin}\:^{\mathrm{2}} \theta−\mathrm{sin}\:^{\mathrm{2}} \theta\mathrm{sin}\:^{\mathrm{2}} \phi \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{sin}\:^{\mathrm{2}} \phi−\mathrm{sin}\:^{\mathrm{2}} \phi\mathrm{sin}\:^{\mathrm{2}} \theta \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:+\mathrm{2sin}\:^{\mathrm{2}} \theta\mathrm{sin}\:^{\mathrm{2}} \phi\:\:\: \\ $$$$\Rightarrow\:\:\frac{{S}}{\mathrm{4}}\:=\:\mathrm{2}\:\:\:\:\:\:{or}\:\:\:{S}=\mathrm{8}\:. \\ $$$$ \\ $$
Commented by mr W last updated on 16/Jun/19
this is the way to prove i wished but i could not carry out. thanks sir!
$${this}\:{is}\:{the}\:{way}\:{to}\:{prove}\:{i}\:{wished}\:{but} \\ $$$${i}\:{could}\:{not}\:{carry}\:{out}.\:{thanks}\:{sir}! \\ $$
Commented by ajfour last updated on 16/Jun/19
i′m glad you like it, Sir. It naturally turned easier than I expected.
$${i}'{m}\:{glad}\:{you}\:{like}\:{it},\:{Sir}.\:{It}\: \\ $$$${naturally}\:{turned}\:{easier}\:{than}\:{I} \\ $$$${expected}. \\ $$
Commented by mr W last updated on 16/Jun/19
this is my try using your method without assuming that a^(→) is in the plane xz. a^→ =(a_1 ,a_2 ,a_3 ) with a_1 ^2 +a_2 ^2 +a_3 ^3 =1 b^→ =(b_1 ,b_2 ,b_3 ) with b_1 ^2 +b_2 ^2 +b_3 ^3 =1 a^→ .b^→ =0 ⇒a_1 b_1 +a_2 b_2 +a_3 b_3 =0 ⇒a_1 b_1 +a_2 b_2 =−a_3 b_3 c^→ =a^→ ×b^→ =(a_2 b_3 −a_3 b_2 ,a_1 b_3 −a_3 b_1 ,a_1 b_2 −a_2 b_1 ) (S/4)=a_1 ^2 +a_2 ^2 +b_1 ^2 +b_2 ^2 +c_1 ^2 +c_2 ^2 =a_1 ^2 +a_2 ^2 +b_1 ^2 +b_2 ^2 +(a_2 b_3 −a_3 b_2 )^2 +(a_1 b_3 −a_3 b_1 )^2 =a_1 ^2 +a_2 ^2 +b_1 ^2 +b_2 ^2 +a_2 ^2 b_3 ^2 −2a_2 a_3 b_2 b_3 +a_3 ^2 b_2 ^2 +a_1 ^2 b_3 ^2 −2a_1 a_3 b_1 b_3 +a_3 ^2 b_1 ^2 =a_1 ^2 +a_2 ^2 +b_1 ^2 +b_2 ^2 +(a_1 ^2 +a_2 ^2 )b_3 ^2 −2(a_1 b_1 +a_2 b_2 )a_3 b_3 +a_3 ^2 (b_1 ^2 +b_2 ^2 ) =a_1 ^2 +a_2 ^2 +b_1 ^2 +b_2 ^2 +(a_1 ^2 +a_2 ^2 )b_3 ^2 +2a_3 ^2 b_3 ^2 +a_3 ^2 (b_1 ^2 +b_2 ^2 ) =a_1 ^2 +a_2 ^2 +b_1 ^2 +b_2 ^2 +(a_1 ^2 +a_2 ^2 +a_3 ^2 )b_3 ^2 +a_3 ^2 (b_1 ^2 +b_2 ^2 +b_3 ^2 ) =a_1 ^2 +a_2 ^2 +a_3 ^2 +b_1 ^2 +b_2 ^2 +b_3 ^2 =1+1 =2 ⇒S=2×4=8
$${this}\:{is}\:{my}\:{try}\:{using}\:{your}\:{method} \\ $$$${without}\:{assuming}\:{that}\:\overset{\rightarrow} {{a}}\:{is}\:{in}\:{the} \\ $$$${plane}\:{xz}. \\ $$$$\overset{\rightarrow} {{a}}=\left({a}_{\mathrm{1}} ,{a}_{\mathrm{2}} ,{a}_{\mathrm{3}} \right)\:{with}\:{a}_{\mathrm{1}} ^{\mathrm{2}} +{a}_{\mathrm{2}} ^{\mathrm{2}} +{a}_{\mathrm{3}} ^{\mathrm{3}} =\mathrm{1} \\ $$$$\overset{\rightarrow} {{b}}=\left({b}_{\mathrm{1}} ,{b}_{\mathrm{2}} ,{b}_{\mathrm{3}} \right)\:{with}\:{b}_{\mathrm{1}} ^{\mathrm{2}} +{b}_{\mathrm{2}} ^{\mathrm{2}} +{b}_{\mathrm{3}} ^{\mathrm{3}} =\mathrm{1} \\ $$$$\overset{\rightarrow} {{a}}.\overset{\rightarrow} {{b}}=\mathrm{0} \\ $$$$\Rightarrow{a}_{\mathrm{1}} {b}_{\mathrm{1}} +{a}_{\mathrm{2}} {b}_{\mathrm{2}} +{a}_{\mathrm{3}} {b}_{\mathrm{3}} =\mathrm{0} \\ $$$$\Rightarrow{a}_{\mathrm{1}} {b}_{\mathrm{1}} +{a}_{\mathrm{2}} {b}_{\mathrm{2}} =−{a}_{\mathrm{3}} {b}_{\mathrm{3}} \\ $$$$ \\ $$$$\overset{\rightarrow} {{c}}=\overset{\rightarrow} {{a}}×\overset{\rightarrow} {{b}}=\left({a}_{\mathrm{2}} {b}_{\mathrm{3}} −{a}_{\mathrm{3}} {b}_{\mathrm{2}} ,{a}_{\mathrm{1}} {b}_{\mathrm{3}} −{a}_{\mathrm{3}} {b}_{\mathrm{1}} ,{a}_{\mathrm{1}} {b}_{\mathrm{2}} −{a}_{\mathrm{2}} {b}_{\mathrm{1}} \right) \\ $$$$\frac{{S}}{\mathrm{4}}={a}_{\mathrm{1}} ^{\mathrm{2}} +{a}_{\mathrm{2}} ^{\mathrm{2}} +{b}_{\mathrm{1}} ^{\mathrm{2}} +{b}_{\mathrm{2}} ^{\mathrm{2}} +{c}_{\mathrm{1}} ^{\mathrm{2}} +{c}_{\mathrm{2}} ^{\mathrm{2}} \\ $$$$={a}_{\mathrm{1}} ^{\mathrm{2}} +{a}_{\mathrm{2}} ^{\mathrm{2}} +{b}_{\mathrm{1}} ^{\mathrm{2}} +{b}_{\mathrm{2}} ^{\mathrm{2}} +\left({a}_{\mathrm{2}} {b}_{\mathrm{3}} −{a}_{\mathrm{3}} {b}_{\mathrm{2}} \right)^{\mathrm{2}} +\left({a}_{\mathrm{1}} {b}_{\mathrm{3}} −{a}_{\mathrm{3}} {b}_{\mathrm{1}} \right)^{\mathrm{2}} \\ $$$$={a}_{\mathrm{1}} ^{\mathrm{2}} +{a}_{\mathrm{2}} ^{\mathrm{2}} +{b}_{\mathrm{1}} ^{\mathrm{2}} +{b}_{\mathrm{2}} ^{\mathrm{2}} +{a}_{\mathrm{2}} ^{\mathrm{2}} {b}_{\mathrm{3}} ^{\mathrm{2}} −\mathrm{2}{a}_{\mathrm{2}} {a}_{\mathrm{3}} {b}_{\mathrm{2}} {b}_{\mathrm{3}} +{a}_{\mathrm{3}} ^{\mathrm{2}} {b}_{\mathrm{2}} ^{\mathrm{2}} +{a}_{\mathrm{1}} ^{\mathrm{2}} {b}_{\mathrm{3}} ^{\mathrm{2}} −\mathrm{2}{a}_{\mathrm{1}} {a}_{\mathrm{3}} {b}_{\mathrm{1}} {b}_{\mathrm{3}} +{a}_{\mathrm{3}} ^{\mathrm{2}} {b}_{\mathrm{1}} ^{\mathrm{2}} \\ $$$$={a}_{\mathrm{1}} ^{\mathrm{2}} +{a}_{\mathrm{2}} ^{\mathrm{2}} +{b}_{\mathrm{1}} ^{\mathrm{2}} +{b}_{\mathrm{2}} ^{\mathrm{2}} +\left({a}_{\mathrm{1}} ^{\mathrm{2}} +{a}_{\mathrm{2}} ^{\mathrm{2}} \right){b}_{\mathrm{3}} ^{\mathrm{2}} −\mathrm{2}\left({a}_{\mathrm{1}} {b}_{\mathrm{1}} +{a}_{\mathrm{2}} {b}_{\mathrm{2}} \right){a}_{\mathrm{3}} {b}_{\mathrm{3}} +{a}_{\mathrm{3}} ^{\mathrm{2}} \left({b}_{\mathrm{1}} ^{\mathrm{2}} +{b}_{\mathrm{2}} ^{\mathrm{2}} \right) \\ $$$$={a}_{\mathrm{1}} ^{\mathrm{2}} +{a}_{\mathrm{2}} ^{\mathrm{2}} +{b}_{\mathrm{1}} ^{\mathrm{2}} +{b}_{\mathrm{2}} ^{\mathrm{2}} +\left({a}_{\mathrm{1}} ^{\mathrm{2}} +{a}_{\mathrm{2}} ^{\mathrm{2}} \right){b}_{\mathrm{3}} ^{\mathrm{2}} +\mathrm{2}{a}_{\mathrm{3}} ^{\mathrm{2}} {b}_{\mathrm{3}} ^{\mathrm{2}} +{a}_{\mathrm{3}} ^{\mathrm{2}} \left({b}_{\mathrm{1}} ^{\mathrm{2}} +{b}_{\mathrm{2}} ^{\mathrm{2}} \right) \\ $$$$={a}_{\mathrm{1}} ^{\mathrm{2}} +{a}_{\mathrm{2}} ^{\mathrm{2}} +{b}_{\mathrm{1}} ^{\mathrm{2}} +{b}_{\mathrm{2}} ^{\mathrm{2}} +\left({a}_{\mathrm{1}} ^{\mathrm{2}} +{a}_{\mathrm{2}} ^{\mathrm{2}} +{a}_{\mathrm{3}} ^{\mathrm{2}} \right){b}_{\mathrm{3}} ^{\mathrm{2}} +{a}_{\mathrm{3}} ^{\mathrm{2}} \left({b}_{\mathrm{1}} ^{\mathrm{2}} +{b}_{\mathrm{2}} ^{\mathrm{2}} +{b}_{\mathrm{3}} ^{\mathrm{2}} \right) \\ $$$$={a}_{\mathrm{1}} ^{\mathrm{2}} +{a}_{\mathrm{2}} ^{\mathrm{2}} +{a}_{\mathrm{3}} ^{\mathrm{2}} +{b}_{\mathrm{1}} ^{\mathrm{2}} +{b}_{\mathrm{2}} ^{\mathrm{2}} +{b}_{\mathrm{3}} ^{\mathrm{2}} \\ $$$$=\mathrm{1}+\mathrm{1} \\ $$$$=\mathrm{2} \\ $$$$\Rightarrow{S}=\mathrm{2}×\mathrm{4}=\mathrm{8} \\ $$
Commented by ajfour last updated on 16/Jun/19
It must have been possible, this general way, & its not that lengthy too, Nice Sir.
$${It}\:{must}\:{have}\:{been}\:{possible},\:{this} \\ $$$${general}\:{way},\:\&\:{its}\:{not}\:{that}\:{lengthy} \\ $$$${too},\:\mathcal{N}{ice}\:{Sir}. \\ $$

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