习题解答 第一章向量代数 习题1-1 1.如图,已知平行六面体ABCD-A1B1C1D1,E、F分别是棱BC、C1D1的中点.设AB=a, AD=b,AA1=.试用,b,表示下列向量 解:(1)因为 BC=AD. CC= AA. AC=AB+ BC+CC 所以 AC=0+b+c (2)因为BD=B+DD1,而 BD=AD-AB=b-C DD 所以 BD,= 6 (3)=A+DD+D1F,而 DD,=AA. D,=-D,C 所以 AF=a+b (4)EF=AF-AE-AF-(AB+BE)=AF-(AB+BE)=AF-(AB+Bc) 第1题图 第3(1)题图 2.要使下列各式成立向量a,b应满足什么条件? a+b|=|+|b (2)+b|=|d-|b; (3)|a-b|=|-|b| (4)|d-b|=|+|b1 解:(1)利用“三角形两边之和大于第三边可知7∥b且要使+b=7+b必须: 与b同向或云,b中至少有一为0￾
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1–1 1.  ,   ABCD−A1B1C1D1, E￾ F  BC￾ C1D1 .  −→AB = −→a , −→AD = −→b , −−→AA1 = −→c .  −→a , −→b , −→c  : (1) −−→AC1; (2) −−→BD1; (3) −→AF; (4) −→EF. : (1) !" −→BC = −→AD, −−→CC1 = −−→AA1, −−→AC1 = −→AB + −→BC + −−→CC1, #$ −−→AC1 = −→a + −→b + −→c . (2) !" −−→BD1 = −→BD + −−→DD1, % −→BD = −→AD − −→AB = −→b − −→a , −−→DD1 = −−→AA1. #$ −−→BD1 = −→b − −→a + −→c . (3) −→AF = −→AD + −−→DD1 + −−→D1F, % −−→DD1 = −−→AA1, −−→D1F = 1 2 −−−→ D1C1 = 1 2 −→AB, #$ −→AF = 1 2 −→a + −→b + −→c . (4) −→EF = −→AF − −→AE = −→AF − ( −→AB + −→BE) = −→AF − ( −→AB + −→BE) = −→AF − ³−→AB + 1 2 −→BC´ = 1 2 −→a + −→b + −→c − −→a − 1 2 −→b = − 1 2 −→a + 1 2 −→b + −→c . uuu       uuu      
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