Unidad 3. Espacios Vectoriales Básicos 3.4 Producto escalar, vectorial , Triple producto escalr

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Este producto sólo está denido para vectores en R 3. Denición 1. El producto vectorial (o producto cruz) de dos vectores asocia a dos vectores u, v ∈ R 3 un nuevo vector u × v de la forma siguiente u × v = (a, b, c) × (r, s, t) = (bt − sc, cr − at,

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  ❯♥✐❞❛❞✸✳❊♣❛❝✐♦❱❡❝♦✐❛❧❡❇✐❝♦  ✸✳✹♦❞✉❝♦❡❝❛❧❛✱✈❡❝♦✐❛❧✱❚✐♣❧❡♣♦❞✉❝♦❡❝❛❧ ♦❞✉❝♦❱❡❝♦✐❛❧ ❊❡♣♦❞✉❝♦❧♦❡❞❡✜♥✐❞♦♣❛❛✈❡❝♦❡❡♥  R 3 ✳ ❉❡✜♥✐❝✐♥✶✳  ❊❧♣♦❞✉❝♦✈❡❝♦✐❛❧✭♦♣♦❞✉❝♦❝✉③✮❞❡❞♦✈❡❝♦❡❛♦❝✐❛❛❞♦✈❡❝♦❡  u,v  ∈ R 3 ✉♥ ♥✉❡✈♦✈❡❝♦  u × v ❞❡❧❛❢♦♠❛✐❣✉✐❡♥❡  u × v  = ( a,b,c ) × ( r,s,t ) = ( bt − sc,cr − at,as − br ) ▲❛♠❛♥❡❛♣❝✐❝❛❞❡❝❛❧❝✉❧❛❡❧♣♦❞✉❝♦✈❡❝♦✐❛❧❡✉❧❛❞❡❧❛✐❣✉❛❧❞❛❞❡✐❣✉✐❡♥❡✱❡♥❧❛❧✐♠❛❞❡ ❧❛❝✉❛❧❡❡❢❛❝✐❧✐❛✐❡❧❞❡❡♠✐♥❛♥❡❡❞❡❛♦❧❧❛❡♣❡❝♦❛❧♣✐♠❡❡♥❣❧♥✳❙✐  a  = ( a 1 ,a 2 ,a 3 ) ②   b  = ( b 1 ,b 2 ,b 3 ) ❡♥♦♥❝❡  a × b  = ( a 2 b 3  − b 2 a 3 ,b 1 a 3  − a 1 b 3 ,a 1 b 2  − a 2 b 1 )= ( a 2 b 3  − b 2 a 3 )(1 , 0 , 0) + ( b 1 a 3  − a 1 b 3 )(0 , 1 , 0) + ( a 1 b 2  − a 2 b 1 )(0 , 0 , 1)=  a 2  a 3 b 2  b 3  ˆ i −  a 1  a 3 b 1  b 3  ˆ  j  +  a 1  a 2 b 1  b 2  ˆ k =  ˆ i  ˆ  j  ˆ ka 1  a 2  a 3 b 1  b 2  b 3  ❞♦♥❞❡  ˆ i  = (1 , 0 , 0) ,  ˆ  j  = (0 , 1 , 0) ,  ˆ k  = (0 , 0 , 1) ❊❥❡♠♣❧♦  ❊❧♣♦❞✉❝♦✈❡❝♦✐❛❧❞❡  (1 , 0 , 0) ❝♦♥   (0 , 1 , 0) ♥♦❞❛  (1 , 0 , 0) × (0 , 1 , 0) =  ˆ i  ˆ  j  ˆ k 1 0 00 1 0  = ˆ i  0 01 0  − ˆ  j  1 00 0  + ˆ k  1 00 1  = ˆ i (0) − ˆ  j (0) + ˆ k (1) = ˆ k  = (0 , 0 , 1) ❊❥❡♠♣❧♦  ❙✐❡♥❡♠♦❧♦✈❡❝♦❡  a  = ( a 1 ,a 2 ,a 3 ) ②   b  = ( b 1 ,b 2 , 3 3 ) ❡♥♦♥❝❡  ( a 1 ,a 2 ,a 3 ) × ( b 1 ,b 2 , 3 3 ) = ( a 2 b 3  − a 3 b 2 ,a 3 b 1  − a 1 b 3 ,a 1 b 2  − a 2 b 1 ) ②❤❛❝❡♠♦  [( a 1 ,a 2 ,a 3 ) × ( b 1 ,b 2 , 3 3 )] · ( a 1 ,a 2 ,a 3 ) = ( a 2 b 3  − a 3 b 2 ,a 3 b 1  − a 1 b 3 ,a 1 b 2  − a 2 b 1 ) · ( a 1 ,a 2 ,a 3 )= ( a 2 b 3  − a 3 b 2 ) a 1  + ( a 3 b 1  − a 1 b 3 ) a 2  + ( a 1 b 2  − a 2 b 1 ) a 3 =  a 2 b 3 a 1  − a 3 b 2 a 1  +  a 3 b 1 a 2  − a 1 b 3 a 2  +  a 1 b 2 a 3  − a 2 b 1 a 3 = 0 ❍❛❝❡♠♦❛♠❜✐♥  [( a 1 ,a 2 ,a 3 ) × ( b 1 ,b 2 , 3 3 )] · ( b 1 ,b 2 ,b 3 ) = ( a 2 b 3  − a 3 b 2 ,a 3 b 1  − a 1 b 3 ,a 1 b 2  − a 2 b 1 ) · ( b 1 ,b 2 ,b 3 )= ( a 2 b 3  − a 3 b 2 ) b 1  + ( a 3 b 1  − a 1 b 3 ) b 2  + ( a 1 b 2  − a 2 b 1 ) b 3 =  a 2 b 3 b 1  − a 3 b 2 b 1  +  a 3 b 1 b 2  − a 1 b 3 b 2  +  a 1 b 2 b 3  − a 2 b 1 b 3 = 0 ❋❛❝✉❧❛❞❞❡❈✐❡♥❝✐❛❯◆❆▼ ●❡♦♠❡❛❆♥❛❧✐❝❛■ ♦❢✳❊❡❜❛♥❘✉❜♥❍✉❛❞♦❈✉③  ✶   ❯♥✐❞❛❞✸✳❊♣❛❝✐♦❱❡❝♦✐❛❧❡❇✐❝♦  ✸✳✹♦❞✉❝♦❡❝❛❧❛✱✈❡❝♦✐❛❧✱❚✐♣❧❡♣♦❞✉❝♦❡❝❛❧ ❈♦♥❝❧✉✐♠♦✉❡  [ a × b ] · a  = 0[ a × b ] · b  = 0 ♦❧♦✉❡❡❧♣♦❞✉❝♦✈❡❝♦✐❛❧❞❡❞♦✈❡❝♦❡❡✉♥❡❝❡✈❡❝♦✱❡❧❝✉❛❧❡♣❡♣❡♥❞✐❝✉❧❛❛❧♦ ✉❡❧❡❞✐❡♦♥♦✐❣❡♥✳ ❊❥❡♠♣❧♦  ❙✐❡♥❡♠♦❧♦✈❡❝♦❡  a  = ( a 1 ,a 2 ,a 3 ) ②   b  =  λa  = ( λa 1 ,λa 2 ,λa 3 ) ❡♥♦♥❝❡  [( a 1 ,a 2 ,a 3 ) × ( λa 1 ,λa 2 ,λa 3 ) = ( a 2 λa 3  − a 3 λa 2 ,a 3 λa 1  − a 1 λa 3 ,a 1 λa 2  − a 2 λa 1 )=  λ ( a 2 a 3  − a 3 a 2 ,a 3 a 1  − a 1 a 3 ,a 1 a 2  − a 2 a 1 )=  λ (0 , 0 , 0)= (0 , 0 , 0) ❈♦♥❝❧✉✐♠♦✉❡❡❧♣♦❞✉❝♦✈❡❝♦✐❛❧❞❡❞♦✈❡❝♦❡♣❛❛❧❡❧♦❡❡❧✈❡❝♦❝❡♦ ♦♣✐❡❞❛❞❡❞❡❧♦❞✉❝♦❱❡❝♦✐❛❧ ✶✳❊❛♥✐❝♦♥♠✉❛✐✈♦✱❡❞❡❝✐✱ a × b  =  − b × a ✷✳❙❡❞✐✐❜✉②❡♦❜❡❧❛✉♠❛✱❡♦❡✱ a × ( b  +  c ) =  a × b  +  a × c ❞♦♥❞❡✱❝♦♠♦❞❡❝♦✉♠❜❡✱♣✐♠❡♦♦❜❡♥❡♠♦❧♦♣♦❞✉❝♦②❞❡♣✉✉♠❛♠♦✳✸✳❙❛❝❛❡❝❛❧❛❡✱❡❞❡❝✐✱ ( λa ) × b  =  λ ( a × b ) ✹✳❊❧♣♦❞✉❝♦✈❡❝♦✐❛❧❞❡❞♦✈❡❝♦❡❡✉♥✈❡❝♦♣❡♣❡♥❞✐❝✉❧❛❛❝❛❞❛✉♥♦❞❡✉❢❛❝♦❡✱❡♦❡  ( a × b ) · a  = 0  y  ( a × b ) · b  = 0 ✺✳▲❛♥♦♠❛❞❡❧♣♦❞✉❝♦✈❡❝♦✐❛❧❞❡❞♦✈❡❝♦❡❡✐❣✉❛❧❛❧♣♦❞✉❝♦❞❡❧❛♥♦♠❛❞❡❧♦❢❛❝♦❡ ♣♦❡❧✈❛❧♦❛❜♦❧✉♦❞❡❧❡♥♦❞❡❧♥❣✉❧♦❡♥❡❡❧❧♦✱  a × b   =   a   b   sen  θ ab ❋❛❝✉❧❛❞❞❡❈✐❡♥❝✐❛❯◆❆▼ ●❡♦♠❡❛❆♥❛❧✐❝❛■ ♦❢✳❊❡❜❛♥❘✉❜♥❍✉❛❞♦❈✉③  ✷   ❯♥✐❞❛❞✸✳❊♣❛❝✐♦❱❡❝♦✐❛❧❡❇✐❝♦  ✸✳✹♦❞✉❝♦❡❝❛❧❛✱✈❡❝♦✐❛❧✱❚✐♣❧❡♣♦❞✉❝♦❡❝❛❧  ♦♣✐❡❞❛❞❡  ❚❡♥❡♠♦❧❛✐❣✉✐❡♥❡ ✶✳  a × b  2 =   a  2  a  2 − ( a · b ) 2 ❉❡♠♦❛❝✐♥✳   a × b  2 = ( a × b ) · ( a × b )= [( a 1 ,a 2 ,a 3 ) × ( b 1 ,b 2 ,b 3 )] · [( a 1 ,a 2 ,a 3 ) × ( b 1 ,b 2 ,b 3 )]= ( a 2 b 3  − b 2 a 3 ,b 1 a 3  − a 1 b 3 ,a 1 b 2  − a 2 b 1 ) · ( a 2 b 3  − b 2 a 3 ,b 1 a 3  − a 1 b 3 ,a 1 b 2  − a 2 b 1 )= ( a 2 b 3  − b 2 a 3 ) 2 + ( b 1 a 3  − a 1 b 3 ) 2 + ( a 1 b 2  − a 2 b 1 ) 2 =  a 22 b 23  − 2 a 2 b 3 a 3 b 2  +  a 23 b 22  +  b 21 a 23  − 2 a 1 b 3 a 3 b 1  +  a 21 b 22  +  a 21 b 22  − 2 a 1 b 2 a 2 b 1  +  a 22 b 21 ♦♦♦❧❛❞♦   a  2  b  2 = ( a · a )( b · b )= ( a 21  +  a 22  +  a 23 )( b 21  +  b 22  +  b 23 )=  a 21 b 21  +  a 21 b 22  +  a 21 b 23  +  a 22 b 21  +  a 22 b 22  +  a 22 b 23  +  a 23 b 21  +  a 23 b 22  +  a 23 b 23 ❨❛♠❜✐♥  ( a · b ) 2 = ( a 1 b 1  +  a 2 b 2  +  a 3 b 3 ) 2 =  a 21 b 21  + 2 a 1 b 1 a 2 b 1  + 2 a 1 b 1 a 3 b 3  +  a 22 b 22  + 2 a 2 b 2 a 3 b 3  +  a 23 b 23 ❍❛❝✐❡♥❞♦♦♣❡❛❝✐♦♥❡   a × b  2 =   a  2  a  2 − ( a · b ) 2 ✷✳  a × b   =   a   b   sen  θ ab ❉❡♠♦❛❝✐♥✳   a × b  2 =   a  2  b  2 − ( a · b ) 2 =   a  2  b  2 − (  a   b   cos  θ ab ) 2 =   a  2  b  2 − a  2  b  2 cos 2 θ ab )=   a  2  b  2 (1 − cos 2 θ ab )=   a  2  b  2 sen 2 θ ab ♦❧♦❛♥♦   a × b   =   a   b   sen  θ ab ✸✳②♣♦❛♥♦✱❡❡❧❡❛❞❡❧♣❛❛❧❡❧♦❣❛♠♦❞❡❧❛❋✐❣✉❛ ❋❛❝✉❧❛❞❞❡❈✐❡♥❝✐❛❯◆❆▼ ●❡♦♠❡❛❆♥❛❧✐❝❛■ ♦❢✳❊❡❜❛♥❘✉❜♥❍✉❛❞♦❈✉③  ✸   ❯♥✐❞❛❞✸✳❊♣❛❝✐♦❱❡❝♦✐❛❧❡❇✐❝♦  ✸✳✹♦❞✉❝♦❡❝❛❧❛✱✈❡❝♦✐❛❧✱❚✐♣❧❡♣♦❞✉❝♦❡❝❛❧ ➪❡❛❂✭❜❛❡✮✭❤✮ ❜❛❡❂   a  h  =   b   sen θ ab ➪❡❛❂   a   b   sen θ ab ➪❡❛❂   a × b  ❡♥♠❜♦❧♦✱  a × b   =   a   b   sen  θ ab ❋❛❝✉❧❛❞❞❡❈✐❡♥❝✐❛❯◆❆▼ ●❡♦♠❡❛❆♥❛❧✐❝❛■ ♦❢✳❊❡❜❛♥❘✉❜♥❍✉❛❞♦❈✉③  ✹ 
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