COSa=<OM OM +(√2) cosB=OM_√2 k·OM cos y OM kloM 2 B 2.求同时垂直于向量a={368}和y轴的单位向量 解:记b=axj=368=(80-3} 故同时垂直于向量a与y轴的单位向量为±=±80-3} 3.求与a=i++k平行且满足a·x=1的向量x 解:因a∥x,故可设x=mn=2,,,再由ax=1得2+2+2=1,即=1 从而x=11 333 4.a={00},b={010},c=(0,1),求ab,a·c,b·c,及axa,axb,axC, b×c 解:依题意,a=i,b=j,c=k,故 a·b=i·j=0,a·c=ik=0,b·c=jk=0 a×a=ixi=0,axb=ixj=k,axc=i×k=-j,bxc=j×k=i 5.a={12}b={22,1},求a·b及axb 解:a·b=1×2+1×2+2×1=6 a×b=112|=330 6.证明向量a={101}与向量b=(-11}垂直 证明:∵a·b=1×(-1)+0×1+1×1=0,2 1 1 ( 2) 1 1 cos 2 2 = + + = = OM OM i i , 2 2 cos = = OM OM j j , 2 1 cos = = OM OM k k . 3 π = , 4 π = , 3 π = . 2. 求同时垂直于向量 a = − 3,6,8 和 y 轴的单位向量. 解:记 8,0, 3 0 1 0 = = − 3 6 8 = − − i j k b a j , 故同时垂直于向量 a 与 y 轴的单位向量为 8,0, 3 73 1 = − − b b . 3. 求与 a = i + j + k 平行且满足 a x =1 的向量 x . 解:因 a// x , 故可设 x = a = ,, ,再由 a x =1 得 + + =1 ,即 3 1 = , 从而 = 3 1 , 3 1 , 3 1 x . 4. a = 1,0,0,b = 0,1,0,c = (0,0,1) ,求 a b ,a c ,b c ,及 aa,a b,ac , b c . 解:依题意, a = i , b = j , c = k ,故 a b = i j = 0 , a c = i k = 0, b c = j k = 0 . aa = i i = 0, a b = i j = k ,a c = i k = − j ,b c = j k = i . 5. a = 1,1,2,b = 2,2,1 ,求 a b 及 a b . 解: a b =12 +12 + 21= 6, 3,3,0 2 2 1 = 1 1 2 = − i j k a b . 6. 证明向量 a = 1,0,1 与向量 b = −1,1,1 垂直. 证明: a b = 1 (−1) + 0 1+11 = 0