1.设P,Q两点在标架O;可1,,]下的坐标分别是(2,2,1),(-1,-1,3)试画出PQ点的位置 解:见附图. 第1题图 2.对于平行四边形ABCD,求A,D,AD,DB在标架C;AC,BD下的坐标 解:A=-AC=(-1)AC+0BD,点A坐标为(-1,0) Cd=2CA+ 2BD BD, 点D坐标为(-2 万D坐标为(号 B=-BD=0AC+(-1BD DB坐标为(0,-1) 3.设a,b,乙的坐标分别是(1,5,2),(0,-3,4,(-2,3,-1).求向量2a+c,-3a+2b+4 的坐标 解:2a+=2(1,5,2)+(-2,3,-1)=(2,10,4)+(-2,3,-1)=(0,13,3) 3a+2b+4=-3(1,5,2)+2(0,-3,4)+4(-2,3,-1) (-3-15,-6)+(0,-6,8)+(_8,12,-4 4.证明三角形的三条中线交于一点(重心) 证明设D,E,F分别是边BC,CA,AB上的中点AD与BE交于G,AD与CF交于G.则 AG=kAD=kl-AB+-Ac AB+-AC 若建立仿射标架A(小则点G坐标为(,)又 BG=mBE=m(BA+BC)=m-AB+3(AC-AB mAC-mAB 所以BG坐标为(-m2)但xG=万B+BG所以 (会)=00+(-m)
1–3 1.  P, Q 7kUV [O; −→e1 , −→e2 , −→e3 ] WU (2, 2, 1), (−1, −1, 3). X% P, Q /Y. : _Z . . . / . . . u@ . l x x x x x   & C  55 ;: EF r r ~ ~  . . . / . . uur . l x x x x x G  G  G  G    P P P . l . l . l . l . l . l . l . l . l . l . l O −→e1 −→e2 −→e3 P(2, 2, 1) Q(−1, −1, 3) ￾ 1
0100 100 10 10 010 010010010100100100 7                 Puuuuuu               7 uuuuuuP  '^\ ] \ \ \ \ \ ] \ \ \ \ \ ] \ \ \ A C B D O ￾ 2
2. < l86ABCD, sA, D, −→AD, −→DBkUV[C; −→AC, −→BD] WU. : −→CA = − −→AC = (−1)−→AC + 0 −→BD,  A WU" (−1, 0); −→CD = 1 2 −→CA + 1 2 −→BD = − 1 2 −→AC + 1 2 −→BD,  D WU" ³ − 1 2 , 1 2 ´ ; −→AD = 1 2 −→AC + 1 2 −→BD, −→AD WU" ³ 1 2 , 1 2 ´ ; −→DB = − −→BD = 0 −→AC + (−1)−→BD, −→DB WU" (0, −1). 3.  −→a , −→b , −→c WU (1, 5, 2), (0, −3, 4), (−2, 3, −1). s 2 −→a + −→c , −3 −→a + 2 −→b + 4−→c WU. : 2−→a + −→c = 2(1, 5, 2) + (−2, 3, −1) = (2, 10, 4) + (−2, 3, −1) = (0, 13, 3). −3 −→a + 2 −→b + 4−→c = −3(1, 5, 2) + 2(0, −3, 4) + 4(−2, 3, −1) = (−3, −15, −6) + (0, −6, 8) + (−8, 12, −4) = (−11, −9, −2). 4. ST45641t<H ($). :  D, E, F 8 BC, CA, AB y. AD B BE < G, AD B CF < G0 . J −→AG = k −→AD = k µ 1 2 −→AB + 1 2 −→AC¶ = k 2 −→AB + k 2 −→AC. [+\]UV [A; −→AB, −→AC], J G WU" ³ k 2 , k 2 ´ . Q −→BG = m −→BE = m µ 1 2 −→BA + 1 2 −→BC¶ = m · − 1 2 −→AB + 1 2 ( −→AC − −→AB) ¸ = m 2 −→AC − m −→AB, #$ −→BG WU" ³ −m, m 2 ´ . q −→AG = −→AB + −→BG, #$, µ k 2 , k 2 ¶ = (1, 0) + ³ −m, m 2 ´ , · 9 ·