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-A1B1CD,E、F分别是棱BC、CD1的中点.设A店=a, 而=万AA=亡.试用云，万，亡表示下列向量 (1)AC:(2)BD:(3)A正：(4E. 解：()因为 BC=AD.CC=AA.AC=AB+BC+CCi. 所以 A可=a+万+元 (2因为可=面+D可，而 BD=AD-AB=-DD:=AA. 所以 BD=6-元+元 (3)A正=AD+DD+D正，而 D=4, F-号DG=5a孤， 所以 菲-云+可+己 (④EF-A正-A正=A正-(A店+B)=A-(4店+B)=A正-(A店+号C)=分云+ 万+元-可-号万=-石+号万+元 6 第1题图 第31)题图 2.要使下列各式成立向量云，应满足什么条件？ (1)1a+b1=1a1+161 (2)1a+1=|a1-6 (3)1a-1=a1- (4a-1=1a1+6 解(1)利用三角形两边之和大于第三边”可知下.且要使云+1-1+1必须：云 与同向，或a,中至少有一为0 1…￾
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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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