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Question-126414




Question Number 126414 by sdfg last updated on 20/Dec/20
Answered by physicstutes last updated on 20/Dec/20
 Q_2 . normally we should be computing ∣A × B∣    the unit vector n could be any unit vector they are millions of them so  i guess we can not get a specific A× B vector but we can get ∣A×B∣ thus:   ∣A×B∣ = ∣A∣∣B∣ sin θ ∣ n^∧ ∣                  = ((1/( (√3))))(4) sin 60 ,  since ∣n∣ = 1                  = ((1/( (√3))))(4)(((√(3 ))/2)) =  2   Q_3 . 0 since A, B, C are perpendicular irrespectively.
$$\:\mathrm{Q}_{\mathrm{2}} .\:\mathrm{normally}\:\mathrm{we}\:\mathrm{should}\:\mathrm{be}\:\mathrm{computing}\:\mid\boldsymbol{\mathrm{A}}\:×\:\boldsymbol{\mathrm{B}}\mid\: \\ $$$$\:\mathrm{the}\:\mathrm{unit}\:\mathrm{vector}\:\boldsymbol{\mathrm{n}}\:\mathrm{could}\:\mathrm{be}\:\mathrm{any}\:\mathrm{unit}\:\mathrm{vector}\:\mathrm{they}\:\mathrm{are}\:\mathrm{millions}\:\mathrm{of}\:\mathrm{them}\:\mathrm{so} \\ $$$$\mathrm{i}\:\mathrm{guess}\:\mathrm{we}\:\mathrm{can}\:\mathrm{not}\:\mathrm{get}\:\mathrm{a}\:\mathrm{specific}\:\boldsymbol{\mathrm{A}}×\:\boldsymbol{\mathrm{B}}\:\mathrm{vector}\:\mathrm{but}\:\mathrm{we}\:\mathrm{can}\:\mathrm{get}\:\mid\boldsymbol{\mathrm{A}}×\boldsymbol{\mathrm{B}}\mid\:\mathrm{thus}: \\ $$$$\:\mid\boldsymbol{\mathrm{A}}×\boldsymbol{\mathrm{B}}\mid\:=\:\mid\boldsymbol{\mathrm{A}}\mid\mid\boldsymbol{\mathrm{B}}\mid\:\mathrm{sin}\:\theta\:\mid\:\overset{\wedge} {\boldsymbol{\mathrm{n}}}\mid \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)\left(\mathrm{4}\right)\:\mathrm{sin}\:\mathrm{60}\:,\:\:\mathrm{since}\:\mid\boldsymbol{\mathrm{n}}\mid\:=\:\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\left(\frac{\mathrm{1}}{\cancel{\:\sqrt{\mathrm{3}}}}\right)\left(\mathrm{4}\right)\left(\frac{\cancel{\sqrt{\mathrm{3}\:}}}{\mathrm{2}}\right)\:=\:\:\mathrm{2} \\ $$$$\:\mathrm{Q}_{\mathrm{3}} .\:\mathrm{0}\:\mathrm{since}\:\boldsymbol{\mathrm{A}},\:\boldsymbol{\mathrm{B}},\:\boldsymbol{\mathrm{C}}\:\mathrm{are}\:\mathrm{perpendicular}\:\mathrm{irrespectively}. \\ $$

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