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A. accent B.accompany C.accept D. access

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设 a → b → c → 是任意的非零向量则下列结论正确的是
0 ⋅ a → = 0    a → ⋅ b → c → = a → b → ⋅ c →    a → ⋅ b → = 0 ⇒ a → = 0 → 或 b → = 0 →    a → + b → ⋅ a → - b → = | a → | 2 - | b → | 2   
设 a → b → 都是非零向量下列四个条件中使 a → | a → |
a → = - b →    a → / / b →    a → = 2 b →    a → / / b → 且 | a → | = | b → |   
若 { a → b → c → } 为空间的一组基底则下列各项中能构成基底
a → , a → + b → , a → - b →    b → , a → + b → , a → - b →    c → ， a → + b → ， a → - b →    a → + b → ， a → - b → ， a → + 2 b →   
如图在正五边形 A B C D E 中若 A B ⃗ = a → B C ⃗
给出下列结论①若 a → ≠ 0 → a → ⋅ b → = 0 则 b
设 a → = A B ⃗ + C D ⃗ + B C ⃗
①②  ①③  ①③⑤  ③④⑤  
设 a → b → 是两个非零向量.1若 | a → + b → | = | a
若 a → ≠ 0 → b → ≠ 0 → | a → | = | b →
已知三个向量 a → b → c → 不共面并且 p → = a → +
对于向量 a → b → c → 和实数 λ 下列说法中正确的是.
若 a → ⋅ b → = 0 ，则 a → = 0 → 或 b → = 0 →   若 λ a → = 0 → ，则 λ = 0 或 a → = 0 →   若 a → 2 = b → 2 ，则 a → = b → 或 a → = - b →   若 a → ⋅ b → = a → ⋅ c → ，则 b → = c →   
已知三个向量 a ⃗ b ⃗ c ⃗ 不共面并且 p ⃗ = a ⃗ +
下列各式正确的是
| a → ⋅ b → | = | a → | | b → |    a → ⋅ b → 2 = a → 2 ⋅ b → 2   若 a → ⊥ b → - c → 则 a → ⋅ b → = a → ⋅ c →   若 a → ⋅ b → = a → ⋅ c → 则 b → = c →   
下列等式中错误的是
a → ⋅ b → = b → ⋅ a →    λ a → ⋅ b → = λ a → ⋅ b → λ ∈ R    a → + b → ⋅ c → = a → ⋅ c → + b → ⋅ c →    a → ⋅ b → ⋅ c → = a → ⋅ b → ⋅ c →   
设 a → b → c → 是任意的非零平面向量且它们相互不共线下列命题① a
①②  ②③  ③④  ②④  
设 a → = A B ⃗ + C D ⃗ + B C ⃗
①②⑥  ①③⑥  ①③⑤  ③④⑤⑥  
设 a → = A B ⃗ + C D ⃗ + B C ⃗
①②  ①③  ①③⑤  ③④⑤  
下列命题①若 a → ⋅ b → = 0 → 则 a → = 0 → 或 b
对任意向量 a → b → 下列关系式中不恒成立的是
| a → ⋅ b → | ≤ | a → | | b → |    | a → - b → | ≤ | | a → | - | b → | |    a → + b → 2 = | a → + b → | 2    a → + b → a → - b → = a → 2 - b → 2   
判断下列向量 a → 与 b → 不共线的是
a → = - 2 e → , b → = 2 e →    a → = e → 1 - e → 2 , b → = 2 e → 1 + 2 e → 2    a → = e → 1 + e → 2 , b → = 3 e → 1 + 3 e → 2    a → = 5 e → , b → = 11 e →   
如图已知 O A ⃗ = a → O B ⃗ = b → O C
a → + b → + c → + d → = 0    a → - b → + c → - d → = 0    a → + b → - c → + d → = 0    a → - b → - c → + d → = 0   

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