Trabajo_colaborativo_1_algebra_lineal.docx

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ESCUELA DE CIENCIAS BASICAS TECNOLOGIA E INGENIERIA 100408- ALGEBRA LINEAL ACT. No. 6 - TRABAJO COLABORATIVO No. 1 1 Reconocimiento de la Unidad 1: Estimado estudiante, se espera que a travΓ©s de esta actividad se realice el proceso de transferencia de los temas de la primera unidad. Esta actividad es de carΓ‘cter grupal. 1. Dados los siguientes vectores dados en forma polar: 𝒂. |𝒖| = πŸ“; 𝜽 = πŸπŸ‘πŸ“Β° 𝑒 = (5 π‘π‘œπ‘ 135Β°)𝑖 + ( 5 sen 135Β°)𝑗 𝑒 = ( βˆ’3.535)𝑖 + ( 3.535)𝑗 𝑒 = ( βˆ’3.535 , 3.535 ) 𝒃. |𝒗| = πŸ‘; 𝜽 = πŸ”πŸŽΒ° 𝑣 = (3 cos 60 Β°)𝑖 + ( 3 sen 60Β°)𝑗 𝑣 = (1.5)𝑖 + ( 2.598 )𝑗 𝑣 = (1.5 , 2.598 )

Realice analΓ­ticamente, las operaciones siguientes: βƒ— +𝒗 βƒ— 1.1. πŸπ’– βƒ— +𝒗 βƒ— = 2( βˆ’3.535 , 3.535 ) + (1.5 , 2.598 ) πŸπ’– βƒ— + ⃗𝒗 = (βˆ’7.07 , 7.07 ) + (1.5 , 2.59 8) πŸπ’– βƒ— +𝒗 βƒ— = (βˆ’7.07 + 1.5 , 7.07 + 2.598) πŸπ’– ο€ 

βƒ— +𝒗 βƒ— = (βˆ’5.57 , 9.668) πŸπ’–

βƒ— βˆ’π’– βƒ— 1.2. 𝒗 βƒ— βˆ’π’– βƒ— = (1.5 , 2.259 ) + ( 3.535 , βˆ’3.535 ) 𝒗 βƒ— βˆ’π’– βƒ— = ( 1.5 + 3.535 , 2.598 βˆ’ 3.535 ) 𝒗 βƒ— βˆ’π’– βƒ— = ( 5.035 , βˆ’0.9371 ) 𝒗 βƒ— βˆ’ πŸ’π’– βƒ— 1.3 3𝒗

βƒ— βˆ’ πŸ’π’– βƒ— = 3( (1.5 , 2.598 ) + 4( 3.535 , βˆ’3.535 ) 3𝒗 βƒ— βˆ’ πŸ’π’– βƒ— = (4.5 , 6.777 ) + ( 14.14 , βˆ’14.14 ) 3𝒗 βƒ— βˆ’ πŸ’π’– βƒ— = (4.5 + 14.14 , 6.777 βˆ’ 14.14) 3𝒗 βƒ— βˆ’ πŸ’π’– βƒ— = (18.64 , βˆ’6.347) 3𝒗 2. Encuentre el Γ‘ngulo entre los siguientes vectores: 2.1. 𝑒 βƒ— = 2𝑖 + 9𝑗 𝑦 𝑣 = βˆ’10𝑖 βˆ’ 4𝑗 𝑒 βƒ— = 𝑐1 𝑖 + 𝑐2𝑗 𝑐3 π‘˜ π‘ π‘œπ‘› π‘π‘œπ‘›π‘ π‘‘π‘Žπ‘›π‘‘π‘’π‘  𝑒𝑛 π‘‘π‘’π‘Ÿπ‘šπ‘–π‘›π‘œπ‘  𝑑𝑒 π‘™π‘œπ‘  π‘£π‘’π‘π‘‘π‘œπ‘Ÿπ‘’π‘  π‘π‘Žπ‘ π‘’ 𝑖, 𝑗 𝑦 π‘˜ (2, 9) = 2( 1, 0) + 9(0, 1) (2, 9) = 2𝑖 + 9𝑗 𝑒 βƒ— = (2, 9) 𝑣 = ( βˆ’10, βˆ’4) 𝑒 βƒ— . 𝑣 = (2, 9) . ( βˆ’10, βˆ’4) = (βˆ’20, ) + ( βˆ’36) = (βˆ’56) |𝑒| = βˆšπ‘Ž2 + 𝑏 2 |𝑒| = √(2)2 + (9)2 = √85 |𝑣| = √(βˆ’10)2 + (βˆ’4)2 = √116 Por lo tanto cos πœƒ = cos πœƒ = cos πœƒ =

𝑒 βƒ— .𝑣 |𝑒| |𝑣| (βˆ’56) √85 . √116 (βˆ’56)

=

(βˆ’56) √9860

√9860 (βˆ’56) πœƒ = cosβˆ’1 ( ) √9860 πœƒ = 124.33Β° 2.2. 𝑀 βƒ—βƒ— = βˆ’2𝑖 βˆ’ 3𝑗 𝑦 𝑒 βƒ— = βˆ’7𝑖 βˆ’ 5𝑗 𝑀 βƒ—βƒ— = (βˆ’2, βˆ’3) 𝑒 βƒ— = ( βˆ’7, βˆ’5) 𝑀 βƒ—βƒ— . 𝑒 βƒ— = (βˆ’2, βˆ’3) . ( βˆ’7, βˆ’5) = (14) + ( 15) = 29 |𝑀| = βˆšπ‘Ž2 + 𝑏 2 |𝑀| = √(βˆ’2)2 + (βˆ’3)2 = √13 |𝑒| = √(βˆ’7)2 + (βˆ’5)2 = √74 Por lo tanto cos πœƒ =

𝑒 βƒ— .𝑣 |𝑀| |𝑒|

cos πœƒ = cos πœƒ =

29 √13 . √74 29

=

√962 29 πœƒ = cosβˆ’1 ( ) √962 πœƒ = 20.77Β°

29 √962

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