Algebra
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Resposta da 1ª lista de exercícios1) A=(2,1), B=(-3,-1), C=(4,4) e D=(5,-2), calcule os vetores:
a) AB + 2CD AB = B – A = (-3, -1) – (2,1) = (-5, -2)
CD = D – C = (5, -2) – (4, 4) = (1, -6)
AB + 2CD = (-5, -2) + 2 (1, -6) = (-5, -2) + (2, -12) = (3, -14)
b) 3AC – 2DB
AC = C – A = (4,4) – (2,1) = (2,3)
DB = B – D = (-3, -1) – (5, -2) = (-8, 1)
3AB – 2DB = 3(2,3) – 2(-8, 1) = (6,9) – (-16, -2) = (22, 11)
c) AB + 2AC – 3AD
AB = B – A = (-3, -1) – (2,1) = (-5, -2)
AC = C – A = (4,4) – (2,1) = (2,3)
AD = D – A = (5, -2) – (2,1) = (3, -3)
AB + 2AC – 3AD = (-5, -2) + 2(2,3) – 3(3, -3) = (-5, -2) + (4,6) – (-9,9) = (-1, 4) – (-9,9) = (8, 5)
d) AB + BC + CD + DA
AB = B – A = (-3, -1) – (2,1) = (-5, -2) BC = C – B = (4,4) – (-3, -1) = (7,5)
CD = D – C = (5, -2) – (4, 4) = (1, -6)
DA = A – D = (2,1) – (5, -2) = (3,3) OU DA = A – D = (2,1) – (5, -2) = (-3,3)
AB + BC + CD + DA = (-5, -2) + (7,5) + (1, -6) + (-3,3) = (0,0)
OU
AB + BC + CD + DA = (-5, -2) + (7,5) + (1, -6) + (3,3) = (6,0)
2) Se v = AB, sendo A=(3,2) e v=(5,8) calcule o ponto de B?
v = AB
(5,8) = (x,y) – (3,2)
(5,8) = (x-3,y-2)
x-3=5 y-2=8 x=5+3 y=2+8 x=8 y=10
B=(8,10)
3) De o ponto médio dos segmentos abaixo:
a) A= (2,1) e B= (6,9)
m = (2,1)+(6,9) = (8,10) = (4,5) 2 2
b) A= (-1, -4) e B= (7, -1)
m = (-1, -4)+(7, -1) = (6,-5) = 3, -5 2 2 2
c) A= (½, -1) e B= (5, -⅔)
m = (½, -1)+( 5,