9. 向量的内积运算
定义：有向量 α , β \pmb{\alpha},\pmb{\beta} α,β， α ⋅ β = ∣ α ∣ ∣ β ∣ ⋅ cos < α , β > \pmb{\alpha}\cdot\pmb{\beta}=|\pmb{\alpha}||\pmb{\beta}|\cdot\cos<\pmb{\alpha},\pmb{\beta}> α⋅β=∣α∣∣β∣⋅cos<α,β>（结果是一个数）
- α ⋅ β = 0 ⇔ α ⊥ β \pmb{\alpha}\cdot\pmb{\beta}=0\Leftrightarrow\pmb{\alpha}\bot\pmb{\beta} α⋅β=0⇔α⊥β
- α ⋅ α = ∣ α ∣ 2 ⇒ α ⋅ α \pmb{\alpha}\cdot\pmb{\alpha}=|\pmb{\alpha}|^{2}\Rightarrow\sqrt{\pmb{\alpha}\cdot\pmb{\alpha}} α⋅α=∣α∣2⇒α⋅α
- < α , β > = arccos α ⋅ β ∣ α ∣ ∣ β ∣ <\pmb{\alpha},\pmb{\beta}>=\arccos\frac{\pmb{\alpha}\cdot\pmb{\beta}}{|\pmb{\alpha}||\pmb{\beta}|} <α,β>=arccos∣α∣∣β∣α⋅β
- 【交换律】 α ⋅ β = β ⋅ α \pmb{\alpha}\cdot\pmb{\beta}=\pmb{\beta}\cdot\pmb{\alpha} α⋅β=β⋅α
- P β α = ∣ α ∣ cos < α , β > β 0 = ∣ α ∣ cos < α , β > β ∣ β ∣ ⇒ α ⋅ β = P β α β 0 ∣ β ∣ P_{\pmb{\beta}}\pmb{\alpha}=|\alpha|\cos<\pmb{\alpha},\pmb{\beta}>\pmb{\beta}_{0}=|\alpha|\cos<\pmb{\alpha},\pmb{\beta}>\frac{\pmb{\beta}}{|\pmb{\beta}|}\Rightarrow\pmb{\alpha}\cdot\pmb{\beta}=\frac{P_{\pmb{\beta}}\pmb{\alpha}}{\pmb{\beta}_{0}}|\pmb{\beta}| Pβα=∣α∣cos<α,β>β0=∣α∣cos<α,β>∣β∣β⇒α⋅β=β0Pβα∣β∣
- 【定理1.3】内积具有双线性性
（1） α ⋅ ( β + γ ) = α ⋅ β + α ⋅ γ \pmb{\alpha}\cdot(\pmb{\beta}+\pmb{\gamma})=\pmb{\alpha}\cdot\pmb{\beta}+\pmb{\alpha}\cdot\pmb{\gamma} α⋅(β+γ)=α⋅β+α⋅γ
( α + γ ) ⋅ β = α ⋅ β + γ ⋅ α (\pmb{\alpha}+\pmb{\gamma})\cdot\pmb{\beta}=\pmb{\alpha}\cdot\pmb{\beta}+\pmb{\gamma}\cdot\pmb{\alpha} (α+γ)⋅β=α⋅β+γ⋅α
（2） α ⋅ ( λ β ) = λ α β = ( λ α ) β \pmb{\alpha}\cdot(\lambda \pmb{\beta})=\lambda \pmb{\alpha}\pmb{\beta}=(\lambda\pmb{\alpha})\pmb{\beta} α⋅(λβ)=λαβ=(λα)β
【证】（1） ( α + γ ) ⋅ β = P β ( α + γ ) β 0 ∣ β ∣ = P β α + P β γ β 0 ∣ β ∣ ( β 0 与分母的向量都平行 ) = P β α β 0 + P β γ β 0 = α ⋅ β + α ⋅ γ (\pmb{\alpha}+\pmb{\gamma})\cdot\pmb{\beta}=\frac{P_{\pmb{\beta}}(\pmb{\alpha}+\pmb{\gamma})}{\pmb{\beta}_{0}}|\pmb{\beta}|=\frac{P_{\pmb{\beta}}\pmb{\alpha}+P_{\pmb{\beta}}\pmb{\gamma}}{\pmb{\beta}_{0}}|\pmb{\beta}|(\pmb{\beta}_{0}与分母的向量都平行)=\frac{P_{\pmb{\beta}}\pmb{\alpha}}{\pmb{\beta}_{0}}+\frac{P_{\pmb{\beta}}\pmb{\gamma}}{\pmb{\beta}_{0}}=\pmb{\alpha}\cdot\pmb{\beta}+\pmb{\alpha}\cdot\pmb{\gamma} (α+γ)⋅β=β0Pβ(α+γ)∣β∣=β0Pβα+Pβγ∣β∣(β0与分母的向量都平行)=β0Pβα+β0Pβγ=α⋅β+α⋅γ
（2） ( λ α ) ⋅ β = P β λ α β 0 ∣ β ∣ = λ p β α β 0 ∣ β ∣ = λ α β (\lambda\pmb{\alpha})\cdot\pmb{\beta}=\frac{P_{\pmb{\beta}}{\pmb{\lambda \pmb{\alpha}}}}{\pmb{\beta}_{0}}|\pmb{\beta}|=\frac{\lambda p_{\pmb{\beta}}\pmb{\alpha}}{\pmb{\beta}_{0}}|\pmb{\beta}|=\lambda \pmb{\alpha}\pmb{\beta} (λα)⋅β=β0Pβλα∣β∣=β0λpβα∣β∣=λαβ
【例】用向量的内积来证明余弦定理。
【例1.12】
9.1 内积的坐标计算公式
有空间仿射坐标系 [ O : e 1 , e 2 , e 3 ] [O:\pmb{e}_{1},\pmb{e}_{2},\pmb{e}_{3}] [O:e1,e2,e3]， α = ( a 1 , a 2 , a 3 ) , β = ( b 1 , b 2 , b 3 ) \pmb{\alpha}=(a_{1},a_{2},a_{3}),\pmb{\beta}=(b_{1},b_{2},b_{3}) α=(a1,a2,a3),β=(b1,b2,b3)即 α = a 1 e 1 + a 2 e 2 + a 3 e 3 , β = b 1 e 1 + b 2 e 2 + b 3 e 3 \pmb{\alpha}=a_{1}\pmb{e}_{1}+a_{2}\pmb{e}_{2}+a_{3}\pmb{e}_{3},\pmb{\beta}=b_{1}\pmb{e}_{1}+b_{2}\pmb{e}_{2}+b_{3}\pmb{e}_{3} α=a1e1+a2e2+a3e3,β=b1e1+b2e2+b3e3
则 α ⋅ β = ( a 1 e 1 + a 2 e 2 + a 3 e 3 ) ( b 1 e 1 + b 2 e 2 + b 3 e 3 ) = a 1 b 1 e 1 2 + a 1 b 1 e 1 e 2 + a 1 b 3 e 1 e 3 + a 2 b 1 e 1 e 2 + a 2 b 2 e 2 2 + a 2 b 3 e 2 e 3 + a 3 b 1 e 1 e 3 + a 3 b 2 e 2 e 3 + a 3 b 3 e 3 2 \pmb{\alpha}\cdot\pmb{\beta}=(a_{1}\pmb{e}_{1}+a_{2}\pmb{e}_{2}+a_{3}\pmb{e}_{3})(b_{1}\pmb{e}_{1}+b_{2}\pmb{e}_{2}+b_{3}\pmb{e}_{3})=a_{1}b_{1}\pmb{e}_{1}^{2}+a_{1}b_{1}\pmb{e}_{1}\pmb{e}_{2}+a_{1}b_{3}\pmb{e}_{1}\pmb{e}_{3}+a_{2}b_{1}\pmb{e}_{1}\pmb{e}_{2}+a_{2}b_{2}\pmb{e}_{2}^{2}+a_{2}b_{3}\pmb{e}_{2}\pmb{e}_{3}+a_{3}b_{1}\pmb{e}_{1}\pmb{e}_{3}+a_{3}b_{2}\pmb{e}_{2}\pmb{e}_{3}+a_{3}b_{3}\pmb{e}_{3}^{2} α⋅β=(a1e1+a2e2+a3e3)(b1e1+b2e2+b3e3)=a1b1e12+a1b1e1e2+a1b3e1e3+a2b1e1e2+a2b2e22+a2b3e2e3+a3b1e1e3+a3b2e2e3+a3b3e32
若 [ O : e 1 , e 2 , e 3 ] [O:\pmb{e}_{1},\pmb{e}_{2},\pmb{e}_{3}] [O:e1,e2,e3]是直角坐标系， e i ⊥ e j ( i ≠ j ) , ∣ e i ∣ = 1 ⇔ e i e j = 0 ( i ≠ j ) , e i 2 = 1 \pmb{e}_{i}\bot\pmb{e}_{j}(i\ne j),|\pmb{e}_{i}|=1\Leftrightarrow \pmb{e}_{i}\pmb{e}_{j}=0(i\ne j),\pmb{e}_{i}^{2}=1 ei⊥ej(i=j),∣ei∣=1⇔eiej=0(i=j),ei2=1，则 α ⋅ β = ( a 1 e 1 + a 2 e 2 + a 3 e 3 ) ( b 1 e 1 + b 2 e 2 + b 3 e 3 ) = a 1 b 1 + a 2 b 2 + a 3 b 3 \pmb{\alpha}\cdot\pmb{\beta}=(a_{1}\pmb{e}_{1}+a_{2}\pmb{e}_{2}+a_{3}\pmb{e}_{3})(b_{1}\pmb{e}_{1}+b_{2}\pmb{e}_{2}+b_{3}\pmb{e}_{3})=a_{1}b_{1}+a_{2}b_{2}+a_{3}b_{3} α⋅β=(a1e1+a2e2+a3e3)(b1e1+b2e2+b3e3)=a1b1+a2b2+a3b3
【例14】有正四面体
