设a、b、c是任意的非零平面向量,且互相不共线,则

2024-11-15 15:05:26
推荐回答(2个)
回答1:

1 错误。是向量数量积的常见考点。
a·b和c·a均是没有方向的数值,因此题式即为两不共线向量之差为零向量,这是不可能的。由此可知向量的数量积不满足乘法结合律。
2 正确。考虑三角形三边的关系,两边之差小于第三边。
3 错误。
[(b·c)a-(c·a)b]·c
=(b·c)(a·c)-(c·a)(b·c)
=0,故两向量垂直。
4 正确。关键:a^2=|a|^2
(3a+2b)·(3a-2b)
=9a·a+6a·b-6a·b-4b·b
=9|a|^2-4|b|^2

回答2:

解析:因为b、c不是共线向量,所以①是假命题.
②中的命题为假命题.
∵[(b·c)a-(c·a)b]·c
=(b·c)(a·c)-(c·a)(b·c)=0,
∴(b·c)a-(c·a)b与c垂直,所以③中的命题是真命题.由(3a+2b)·(3a-2b)=9a2-4b2=9|a|2-4|b|2知④中的命题为真命题.∴选C.
答案:C

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