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

$$Acubeofunitedgelength$$$$isheldbeforeaplane.Provethat$$$$thesumofthesquaresofthe$$$$projectedlengthsofedgesofthecube$$$$ontheplane(irrespectiveof$$$$theorientationofthecube)is8.$$

Answered by mr W last updated on 16/Jun/19

Commented by mr W last updated on 16/Jun/19

$$sayO1234567istheunitcube.$$$$sayABCistheplanewith$$$$A(a,0,0),B(0,b,0)andC(0,0,c)$$$$eqn.oftheplaneis$$$$\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$$$$\overrightarrow{n}=(\frac{1}{a},\frac{1}{b},\frac{1}{c})$$$$\mid n\mid =\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}$$$$fortwopointsP(x_1,y_1,z_1)andQ(x_2,y_2,z_2)$$$$\overrightarrow{PQ}=(\mathrm{\Delta }x,\mathrm{\Delta }y,\mathrm{\Delta }z)with$$$$\mathrm{\Delta }x=x_2-x_1,\mathrm{\Delta }y=y_2-y_1,\mathrm{\Delta }z=z_2-z_1$$$$\mid PQ\mid =\sqrt{{\left(\mathrm{\Delta }x\right)}^2+{\left(\mathrm{\Delta }y\right)}^2+{\left(\mathrm{\Delta }z\right)}^2}$$$$\cos \theta =\frac{\overrightarrow{PQ}.\overrightarrow{n}}{\mid PQ\mid \times \mid n\mid }=\frac{\frac{\mathrm{\Delta }x}{a}+\frac{\mathrm{\Delta }y}{b}+\frac{\mathrm{\Delta }z}{c}}{\sqrt{{\left(\mathrm{\Delta }x\right)}^2+{\left(\mathrm{\Delta }y\right)}^2+{\left(\mathrm{\Delta }z\right)}^2}\times \sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}}$$$$lengthofprojectionofPQonplaneis$$$$\mid P^{\prime }{Q}^{\prime }\mid =\mid PQ\mid \sin \theta$$$$\Rightarrow \mid P^{\prime }{Q}^{\prime }{\mid }^2=\mid PQ{\mid }^2{\sin }^2\theta =\mid PQ{\mid }^2(1-{\cos }^2\theta )$$$$\Rightarrow \mid P^{\prime }{Q}^{\prime }{\mid }^2={\left(\mathrm{\Delta }x\right)}^2+{\left(\mathrm{\Delta }y\right)}^2+{\left(\mathrm{\Delta }z\right)}^2-\frac{{(\frac{\mathrm{\Delta }x}{a}+\frac{\mathrm{\Delta }y}{b}+\frac{\mathrm{\Delta }z}{c})}^2}{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}$$$$nowweapplythisforall12edgesofthe$$$$unitcube.$$$$edgeO1=32=45=76:\left(4edges\right)$$$$\mathrm{\Delta }x=1,\mathrm{\Delta }y=0,\mathrm{\Delta }z=0$$$$\mid O^{\prime }{1}^{\prime }{\mid }^2=1-\frac{\frac{1}{a^2}}{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}$$$$$$$$edge12=O3=74=65:\left(4edges\right)$$$$\mathrm{\Delta }x=0,\mathrm{\Delta }y=1,\mathrm{\Delta }z=0$$$$\mid 1^{\prime }{2}^{\prime }{\mid }^2=1-\frac{\frac{1}{b^2}}{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}$$$$$$$$edge25=34=O7=15:\left(4edges\right)$$$$\mathrm{\Delta }x=0,\mathrm{\Delta }y=0,\mathrm{\Delta }z=1$$$$\mid 2^{\prime }{5}^{\prime }{\mid }^2=1-\frac{\frac{1}{c^2}}{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}$$$$$$$$sumforalledges:$$$$4(1-\frac{\frac{1}{a^2}}{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}+1-\frac{\frac{1}{b^2}}{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}+1-\frac{\frac{1}{c^2}}{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}})$$$$=4(3-\frac{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}})$$$$=4(3-1)$$$$=8$$$$theresultisindependentfromthe$$$$orientationoftheplane.$$

Commented by ajfour last updated on 16/Jun/19

$$ThanksSir,verybrave!$$

Answered by ajfour last updated on 16/Jun/19

Commented by ajfour last updated on 16/Jun/19

$$let\overline{a}=\cos \theta \hat{i}+\sin \theta \hat{k}$$$$\overline{b}=\cos \varphi \cos \psi \hat{i}+\cos \varphi \sin \psi \hat{j}$$$$+\sin \varphi \hat{k}$$$$as\overline{c}=\overline{a}\times \overline{b}$$$$\overline{c}=-\cos \varphi \sin \psi \sin \theta \hat{i}+$$$$(-\sin \varphi \cos \theta +\cos \varphi \cos \psi \sin \theta )\hat{j}$$$$+\cos \varphi \sin \psi \cos \theta \hat{k}$$$$andas\overline{a}.\overline{b}=0$$$$\Rightarrow \cos \theta \cos \varphi \cos \psi +\sin \theta \sin \varphi =0$$$$\dots \dots ..\left(i\right)$$$$letrequiredsumbeS.$$$$S=4\{\cos {}^2\theta +\cos {}^2\varphi \cos {}^2\psi +$$$$\cos {}^2\varphi \sin {}^2\psi +\cos {}^2\varphi \sin {}^2\psi \sin {}^2\theta$$$$+{(-\sin \varphi \cos \theta +\cos \varphi \cos \psi \sin \theta )}^2\}$$$$\Rightarrow \frac{S}{4}=\cos {}^2\theta +\cos {}^2\varphi +\sin {}^2\theta \cos {}^2\varphi$$$$+\sin {}^2\varphi \cos {}^2\theta$$$$-2\sin \varphi \cos \varphi \sin \theta \cos \theta \cos \psi$$$$Nowusing\left(i\right)$$$$\frac{S}{4}=\cos {}^2\theta +\cos {}^2\varphi +\sin {}^2\theta \cos {}^2\varphi$$$$+\sin {}^2\varphi \cos {}^2\theta +2\sin {}^2\theta \sin {}^2\varphi$$$$\frac{S}{4}=\cos {}^2\theta +\cos {}^2\varphi$$$$+\sin {}^2\theta -\sin {}^2\theta \sin {}^2\varphi$$$$\sin {}^2\varphi -\sin {}^2\varphi \sin {}^2\theta$$$$+2\sin {}^2\theta \sin {}^2\varphi$$$$\Rightarrow \frac{S}{4}=2orS=8.$$$$$$

Commented by mr W last updated on 16/Jun/19

$$thisisthewaytoproveiwishedbut$$$$icouldnotcarryout.thankssir!$$

Commented by ajfour last updated on 16/Jun/19

$$i^{\prime }mgladyoulikeit,Sir.It$$$$naturallyturnedeasierthanI$$$$expected.$$

Commented by mr W last updated on 16/Jun/19

$$thisismytryusingyourmethod$$$$withoutassumingthat\overrightarrow{a}isinthe$$$$planexz.$$$$\overrightarrow{a}=(a_1,a_2,a_3)with{a}_1^2+a_2^2+a_3^3=1$$$$\overrightarrow{b}=(b_1,b_2,b_3)with{b}_1^2+b_2^2+b_3^3=1$$$$\overrightarrow{a}.\overrightarrow{b}=0$$$$\Rightarrow a_1{b}_1+a_2{b}_2+a_3{b}_3=0$$$$\Rightarrow a_1{b}_1+a_2{b}_2=-a_3{b}_3$$$$$$$$\overrightarrow{c}=\overrightarrow{a}\times \overrightarrow{b}=(a_2{b}_3-a_3{b}_2,a_1{b}_3-a_3{b}_1,a_1{b}_2-a_2{b}_1)$$$$\frac{S}{4}=a_1^2+a_2^2+b_1^2+b_2^2+c_1^2+c_2^2$$$$=a_1^2+a_2^2+b_1^2+b_2^2+{(a_2{b}_3-a_3{b}_2)}^2+{(a_1{b}_3-a_3{b}_1)}^2$$$$=a_1^2+a_2^2+b_1^2+b_2^2+a_2^2{b}_3^2-2a_2{a}_3{b}_2{b}_3+a_3^2{b}_2^2+a_1^2{b}_3^2-2a_1{a}_3{b}_1{b}_3+a_3^2{b}_1^2$$$$=a_1^2+a_2^2+b_1^2+b_2^2+(a_1^2+a_2^2)b_3^2-2(a_1{b}_1+a_2{b}_2)a_3{b}_3+a_3^2(b_1^2+b_2^2)$$$$=a_1^2+a_2^2+b_1^2+b_2^2+(a_1^2+a_2^2)b_3^2+2a_3^2{b}_3^2+a_3^2(b_1^2+b_2^2)$$$$=a_1^2+a_2^2+b_1^2+b_2^2+(a_1^2+a_2^2+a_3^2)b_3^2+a_3^2(b_1^2+b_2^2+b_3^2)$$$$=a_1^2+a_2^2+a_3^2+b_1^2+b_2^2+b_3^2$$$$=1+1$$$$=2$$$$\Rightarrow S=2\times 4=8$$

Commented by ajfour last updated on 16/Jun/19

$$Itmusthavebeenpossible,this$$$$generalway,\mathrm{\&}itsnotthatlengthy$$$$too,\mathcal{N}iceSir.$$

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