Выражение векторов по трём некомпланарным векторам Дано abcda1b1c1d1 - прараллелепипед A1A =a A1D1=b AB1=c BF:FC = 1:3 Выразите через векторы a, b, c следующие векторы FE GF FC1 B1E
Вектор FE = FB + BE
FB = BF = (1/4)FC = (1/4)3b = (3/4)b
BE = AB + AE
AB = a
AE = 2AD + DE = 2b + c
BE = a + 2b + c
FE = (3/4)b + a + 2b + c = (7/4)b + a + c
Вектор GF = 1/4 FE = (7/16)b + 1/4a + 1/4*c
Вектор FC1 = -2 FC = -2 3b = -6b
Вектор B1E = BE - BB1
BE = a + 2b + c
BB1 = CB1 = (3/4)BC = (3/4)c
B1E = a + 2b + c - (3/4)c = a + 2b + (1/4)c
Вектор FE = FB + BE
FB = BF = (1/4)FC = (1/4)3b = (3/4)b
BE = AB + AE
AB = a
AE = 2AD + DE = 2b + c
BE = a + 2b + c
FE = (3/4)b + a + 2b + c = (7/4)b + a + c
Вектор GF = 1/4 FE = (7/16)b + 1/4a + 1/4*c
Вектор FC1 = -2 FC = -2 3b = -6b
Вектор B1E = BE - BB1
BE = a + 2b + c
BB1 = CB1 = (3/4)BC = (3/4)c
B1E = a + 2b + c - (3/4)c = a + 2b + (1/4)c