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Geometria Anal´ıtica e A´lgebra Linear - GA21NB Professor: Geovani Raulino Gabarito Lista de Exerc´ıcios 05 - Vetores 1. (a) V (b) F (c) V (d) V (e) V (f) V (g) F (h) V (i) V (j) V (k) F (l) V (m) V (n) V (o) F 2. (a) −→ OC + −→ CH = −→ OH = −→ AE (b) −→ EH + −→ FG = −→ AO + −→ OC = −→ AC (c) 2 −→ AE + 2 −→ AF = −→ AD + −→ DC = −→ AC (d) −→ EH + −→ EF = −→ AO + −→ OB = −→ AB (e) −→ AO + 1 2 −→ BC = −→ AO + −→ OH = −→ AH (f) −→ EO + −→ BG = −→ AF + −→ FO = −→ AO (g) 2 −→ OE + 2 −→ OC = −→ CD + −→ AC = −→ AC + −→ CD = −→ AD (h) 1 2 −→ BC + −→ EH = −→ AE + −→ EH = −→ AH (i) −→ FE + −→ FG = −→ FG + −→ FE = −→ AO + −→ OD = −→ AD (j) −→ OG − −→ HO = −→ OG + −→ OH = −→ AF + −→ FO = −→ AO (k) −→ AF + −→ FO + −→ AO = −→ AO + −→ AO = 2 −→ AO = −→ AC (l) −→ OG − −→ HC − −→ CG = −→ OG + −→ CH + −→ GC = −→ AF + −→ FA + −→ AE = −→ AA + −→ AE = −→ AE (m) −→ DH + −→ HG + −→ GC = −→ DG + −→ GC = −→ DC = −→ AB (n) −→ FG + −→ FE + −→ HG = −→ AO + −→ OD + −→ DO = −→ AD + −→ DO = −→ AO (o) −→ DO + −→ OH + −→ FB = −→ DH + −→ FB = −→ AF + −→ FB = −→ AB 3. (a) 120◦ (b) 120◦ (c) 60◦ (d) 60◦ 4. Considere o triaˆngulo ABC como na figura onde M e N sa˜o os pontos me´dios de AB e AC, respectivamente. Queremos provar que −→ MN // −→ BC, ou seja, −→ MN = α −→ BC, com α ∈ R. Observe que, pela figura, temos −→ MN = −→ MA + −→ AN = 1 2 −→ BA + 1 2 −→ AC = 1 2 ( −→ BA + −→ AC) = 1 2 −→ BC . Portanto, −→ MN // −→ BC e ‖ −→ MN ‖ = 1 2 ‖ −→ BC ‖. 5. (a) (3,-5) (b) (-5,4) (c) (1,−1 2 ) (d) (13 2 ,−9) 6. (a) −→x = (−15 2 , 15 2 ) (b) −→x = (23 5 ,−11 5 ) 7. (a) (-4,1) (b) (2,5) (c) (-5,-30) 8. (a) √ 2 (b) 5 (c) 10 (d) √ 13 (e) 2 √ 13 (f) 1 9. m = 3 ou m = −13 5 10. −→v = (−6, 3,−9) 11. (a) 2 (b) 38 (c) 4 (d) -181 12. −→ AB · −−→BC = 0 13. α = -5 14. (a) 120◦ (b) 150◦ 15. m = 3 ou m = -6 16. (a) ( 8 9 ,−4 9 ,−8 9 ) (b) ( −6 5 , 0,−2 5 ) 17. ~x = (1, 0, 1) 18. −→ AB ×−→AC = (12,−3, 10) 19. ~u.~v = 5 ou −5. 20. Ap = 3 √ 10.
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