8-2
P(4,2,1)2. 已知两点与.(1)求向量的模、方向余弦和方向角;(2)求P(3,0,2)PP1212
与向量同向与反向的单位向量;(3)求点使得. PPP,,2PPPP1212
解:(1),向量的模为:PP,1,2,1,2, PP,{,1,,2,1}PP121212
112向量,的方向余弦为:,,. cos,,,cos,,PPcos,,12222
,,,23向量的方向角:,,. ,,,,,,PP12343
121,(2)与向量a,{cos,,cos,,cos,},{,,,,}同向的单位向量:, PP12222
121, 与向量,a,,{cos,,cos,,cos,},{,,,}反向的单位向量:. PP12222(3)设{x,4,y,2,z,1},,2{x,3,y,z,2}P(x,y,z),则,
10,x,,3x,4,,2x,6,,1025,2即,P(,,),故,所以点的坐标为. Py,2,,2y,y,,3333,,z,1,,2z,4,5,z,,3,
,,5.试求b,{,,,6,2},,a,{,2,3,,}、,使与平行.
,,,23,解:由b,,,1与平行得:,,,所以、. ,,4a,,62
8-3
,,b,,{1,1,3}a,,{2,3,1}c,,{1,2,0}a,b5.已知,,.求(1);
,,,,,,,,,,,,(2)(2a),(b,c)(a,b)c(a,c),b(a,b),c;(3);(4);(5).
ijk
解:(1); ab,,,,,,231{8,5,1}
113,
ijk
(2); (2)()2231{20,12,4}abc,,,,,,,
013
()(233){8,16,0}abcc,,,,,,;
ijk (3) (4); ()231{2,1,1}{1,1,3}2132acbb,,,,,,,,,,,,,,
120,
ijk
(5). ()344{8,4,2}abc,,,,,,
120,
11.已知||26b,||72ab,,,, ,求:. ||3a,ab,
12解:||72||||sin326sinabab,,,,,,,,,得, sin,,13
52 , cos1sin,,,,,,,13
abab,,,,||||cos30,.