www.770333.com:2.4.1平面向量数量积的物理背景及其意义19_图文

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2.4 第 二 章 平 面 向 平 面 量 的 数 向量 量积 2.4.1 平面 向量 数量 积的 物理 背景 及其 含义 课前预习·巧设计 名师课堂·一点通 创新演练·大冲关 读教材·填要点 小问题·大思维 考点一 考点二 考点三 解题高手 NO.1课堂强化 NO.2课下检测 [读教材·填要点] 1.平面向量数量积的定义 已知两非零向量a与b,它们的夹角为θ,则把数量 |a||b|·cos θ 叫做a与b的数量积 (或内积 ),记作 a·b ,即 a·b=|a||b|cos θ . 规定零向量与任一向量的数量积为 0 . 2.向量的数量积的几何意义 (1)投影:|a|cos θ(|b|cos θ)叫做向量 a在方b 向上( b在a 方向上)的投影. (2)几何意义:数量积a·b等于a的长度|a|与b在a的方向 上的投影|b|cos θ 的乘积. 3.向量的数量积的性质 设 a 与 b 都是非零向量, θ 为 a 与 b 的夹角. (1)a⊥b? a·b=0 . (2)当 a 与 b 同向时,a·b= |a||b| ; 当 a 与 b 反向时,a·b=- |a||b| . (3)a·a= |a|2 或|a|= a·a= a2 . a·b (4)cos θ= |a||b| . (5)|a·b| ≤ |a||b|. 4.向量数量积的运算律 (1)a·b= b·a (交换律). (2)(λa)·b= λ(a·b) = a·(λb)(结合律). (3)(a+b)·c= a·c+b·c (分配律). [小问题·大思维] 1.向量的数量积与数乘向量的运算结果有何区别? 提示:向量的数量积a·b是一个实数;数乘向量λa是一个 向量. 2.投影是向量还是数量? 提示:投影是数量而不是向量,它可正、可负、可为零. 3.对于向量a,b,c,等式(a·b)·c=a·(b·c)一定成立吗? 提示:不一定成立,∵若(a·b)·c≠0,则它与c共线,而 a·(b·c)≠0时与a共线,而a与c不一定共线,故该等式不一定成 立. 4.若a,b是非零向量,则|a·b|=|a||b|一定成立吗? 提示:不一定.因为a·b=|a||b|cos θ,所以只有|cos θ| =1,即a,b共线时才成立. 5.若a,b,c是非零向量,且a·c=b·c,则a=b一定成 立吗? 提示:不一定.由a·c=b·c可得c·(a-b)=0?a-b=0 或c⊥(a-b). [研一题] [例1] 已知a,b的夹角为θ,|a|=2,|b|=3,分别在下 列条件下求a·b. (1)θ=135°; (2)a∥b; (3)a⊥b. [自主解答] (1)a·b=|a||b|cos θ=2×3×cos 135°=-3 2. (2)当 a∥b 时,θ=0°或 180°. 若 θ=0°,则 a·b=|a||b|cos 0°=|a||b|=6; 若 θ=180°,则 a·b=|a||b|cos 180°=-|a||b|=-6. (3)当 a⊥b 时,a·b=0. 若本例条件变为“θ=120°”,试求(2a-b)·(3a+2b). 解:∵(2a-b)·(3a+2b) =6a2+4a·b-3a·b-2b2 =6a2+a·b-2b2 =6×4+2×3×cos 120°-2×9 =24+6×(-12)-18 =3. [悟一法] 求平面向量数量积的步骤是:①求a与b的夹角θ, θ∈[0,π];②分别求|a|和|b|;③求数量积,即a·b= |a||b|·cos θ,要特别注意书写时a与b之间用实心圆点“·”连接, 而不能用“×”连接,也不能省去. [通一类] 1.在等边△ABC 中,边长为 1,求 AB·AC , AB·BC . 解:根据两向量夹角的定义, AB与 AC 的夹角是 60°, AB与 BC 的夹角是 180°-60°=120°, 则 AB·AC =| AB|| AC |cos 60°=12, AB·BC =| AB|| BC |cos 120° =-12. [研一题] [例2] 已知单位向量e1,e2的夹角为60°,求向量a=e1 +e2,b=e2-2e1的夹角. [自主解答] 由单位向量 e1,e2 的夹角为 60°, 得 e1·e2=cos 60°=12, 所以 a·b=(e1+e2)·(e2-2e1) =-2e1·e1-e1·e2+e2·e2 =-2-12+1=-32. ① 又|a|2=|e1+e2|2=|e1|2+2e1·e2+|e2|2=3, |b|2=|e2-2e1|2=4|e1|2-4e1·e2+|e2|2=3, 所以|a|=|b|= 3. ② 设 a 与 b 的夹角为 θ. 由①②可得 cos θ=|aa|·|bb|= -32 3× 3=-12. 又 0°≤θ≤180°,所以 θ 为 120°. [悟一法] 1.求向量夹角的方法: (1)求出 a·b,|a|,|b|,代入公式 cos θ=|aa|·|bb|求解. (2)用同一个量表示 a·b,|a|,|b|,代入公式求解. (3)借助向量运算的几何意义,数形结合求夹角. 2.要注意夹角 θ 的范围 θ∈[0,π],当 cos θ>0 时,θ∈[0,π2); 当 cos θ<0 时,θ∈(π2,π],当 cos θ=0 时,θ=π2. [通一类] 2.已知a、b是两个非零向量,且|a|=|b|=|a+b|,求a与a-b 的夹角. 解:法一:设 a 与 a-b 的夹角为 θ. ∵|a|=|b|,∴|a|2=|b|2. ① 又∵|b|=|a+b|, ∴|b|2=|a+b|2=(a+b)2=|a|2+|b|2+2a·b. ② 由①②,得 a·b=-12|b|2, ∴|a-b|2=(a-b)2=|a|2+|b|2-2a·b=3|b|2, ∴|a-b|= 3|b|. ∴cos θ=a|a·?|

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