已知△ABC及一点O,求证:O为△ABC的重心的充要条件是overrightarrow (OA)+overrightarrow (OB)+overrightar
设 overrightarrow(a) = 3, 5, -2,overrightarrow(b) = 2, 1, 4,且已知 lambda overri
10证明:若 overrightarrow (a)times overrightarrow (b)+overrightarrow (b)times overri
16.设a,b是共线的单位向量,则 overrightarrow (a)cdot overrightarrow (b)=()-|||-
[题目]设 |overrightarrow (a)|=4 |overrightarrow (b)|=3 , langle overrightarrow (a),
设未知向量 overrightarrow(x)与 overrightarrow(a) = (2, -1, 2)共线,且满足 overrightarrow(a)
向量overrightarrow(a)、overrightarrow(b)、overrightarrow(c)两两垂直,且|overrightarrow(a)|
[题目]设 |overrightarrow (a)|=3,, |overrightarrow (b)|=4,, |overrightarrow (c)|=5,
设overrightarrow(a)=(-1,1,2),overrightarrow(b)=(2,-2,-4),则向量|overrightarrow(a)×ov
设overrightarrow(a)=(1,-1,2),overrightarrow(b)=(3,-2,-1),则overrightarrow(a)×overr