u Si m 4 3 4 n 1 0 3 n 4 0 n Si m 4 3 4 n 1 0 3 n 4 0 n Calcula m para que v 7

U si m 4 3 4 n 1 0 3 n 4 0 n si m 4 3 4 n 1 0 3 n 4 0

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u = Si m = 4: (3, 4) ( n , 1) = 0 3 n 4 = 0 n = Si m = − 4: (3, 4) ( n , 1) = 0 3 n + 4 = 0 n = − Calcula m para que v = (7, 2) y w = ( m , 6): a) Sean perpendiculares. b) Sean paralelos. c) Tengan el mismo módulo. a) (7, 2) ( m , 6) = 7 m + 12 = 0 m = c) v = w = 53 = m 2 + 36 m = Dados a = (6, 2) y b = (16, 12), calcula el valor de m para que los vectores u = ma + b y v = ma b sean perpendiculares. ¿Hay una solución única? u = m (6, 2) + (16, 12) = (16 + 6 m , 12 2 m ) v = m (6, 2) (16, 12) = ( 16 + 6 m , 12 2 m ) Para que los vectores sean perpendiculares, su producto escalar debe ser cero. (16 + 6 m , 12 2 m ) ( 16 + 6 m , 12 2 m ) = 36 m 2 256 + 4 m 2 144 = 0 m = La solución no es única. ± 10 049 17 m 2 2 6 + 7 2 53 2 2 + − = ( ) b) 7 2 6 21 m m = = − 12 7 048 4 3 4 3 3 5 9 25 4 2 2 2 + = + = = ± m m m 047 Si a b = − = 4 5 47 5 a a a a a a 2 2 2 1 2 7 3 89 10 42 40 0 4 5 5 + = = = − = ( ) 89 046
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214 ¿Podrías conseguir un vector a tal que a b = 5, siendo b = (2, 1), y que sea perpendicular a c = (2, 6)? a b = 5 ( a , b ) (2, 1) = 5 2 a + b = 5 a c = 0 ( a , b ) (2, 6) = 0 2 a + 6 b = 0 Resolvemos el sistema: a = 3, b = − 1 Considera que A , B , C y D son los vértices de un cuadrado de lado 1 cm. Calcula los siguientes productos escalares. a) AB BC b) AC DB c) AD CB d) AC CB Tomamos el punto A como origen, por lo que obtenemos las siguientes coordenadas: A (0, 0), B (1, 0), C (1, 1) y D (0, 1) a) AB = (1, 0), BC = (0, 1) (1, 0) (0, 1) = 0 b) AC = (1, 1), DB = (1, 1) (1, 1) (1, 1) = 1 1 = 0 c) AD = (0, 1), CB = (0, 1) (0, 1) (0, 1) = − 1 d) AC = (1, 1), CB = (0, 1) (1, 1) (0, 1) = − 1 Por una traslación, el punto (1, 3) se transforma en (3, 7). ¿Cuál es la imagen del punto ( 2, 5) por la misma traslación? ¿Y la imagen de (4, 6)? Calculamos el vector de la traslación: A (1, 3), B (3, 7) AB = (2, 10) Hallamos la imagen de ( 2, 5): ( 2, 5) + (2, 10) = (0, 15) Determinamos la imagen de (4, 6): (4, 6) + (2, 10) = (6, 4) Halla el ángulo que forman los vectores. a) a = ( 1, 5) y b = (3, 2) c) e = ( 6, 4) y f = ( 9, 6) b) c = y d = d) g = (2, 5) y h = (4, 6) a) cos α = = b) cos α = = 2 4 4 60 = = 0,5 ° α c d c d 7 26 13 67 37 = = 0,38 ° ' 11,51" α a b a b ( ) 1 3 , 1 3 , ( ) 053 052 051 050 Geometría analítica
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215 c) cos α = = d) cos α = = Calcula m para que los vectores a = (8, 6) y b = ( m , 3) formen un ángulo de 60°. cos 60° = 255 m 2 1.472 m + 2.107 = 0 m = Encuentra un vector a que forme un ángulo de 30° con b = (3, 4) y tal que a = b . cos 30° = Resolvemos el sistema: Calcula a y b , si los vectores u = ( a, 4) y v = ( b, 14) forman un ángulo de 45° y u = 5. cos 45° = Resolvemos el sistema: 5 2 2 392 2 112 16 5 2 2 3 1 + + = + + = = b ab a b a ⎯→ ( ) + = ( ) 3 2 56 2 5 17 9 629 560 2 289 3 2 56 2 5 17 9 2 . b .629 560 2 289 a 2 16 5 + = 2 2 56 5 196 5 2 2 392 2 112 2 2 = + + + + + = + ab b b ab u v u v 056 50 40 3 3 1 75 3 1 200 0 2 3 3 1 5 1 29 2 b b b + + ( ) + = = − + ( ) ± .
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