10. 向量的外积
10.1 向量外积的定义
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α×β垂直于
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α,β,α×β构成右手系。(用右手,四指朝向和
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|\pmb{\alpha}\times\pmb{\beta}|=|\pmb{\alpha}||\pmb{\beta}|\sin<\pmb{\alpha},\pmb{\beta}>
∣α×β∣=∣α∣∣β∣sin<α,β>(恰好是
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10.2 向量外积的性质
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α × β = 0 ⃗ ⇔ α / / β \pmb{\alpha}\times\pmb{\beta}=\vec{0}\Leftrightarrow\pmb{\alpha} // \pmb{\beta} α×β=0⇔α//β
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α × β = − β × α \pmb{\alpha}\times\pmb{\beta}=-\pmb{\beta}\times\pmb{\alpha} α×β=−β×α(反交换性)
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假定 α \pmb{\alpha} α是单位向量, β ⊥ α \pmb{\beta}\bot\pmb{\alpha} β⊥α,此时 ∣ β × α ∣ = ∣ β ∣ |\pmb{\beta}\times\pmb{\alpha}|=|\pmb{\beta}| ∣β×α∣=∣β∣(直角正弦为1)
故 β × α \pmb{\beta}\times\pmb{\alpha} β×α可看作 β \pmb{\beta} β绕 α \pmb{\alpha} α旋转 9 0 ∘ 90^{\circ} 90∘而得
若 α \pmb{\alpha} α不是单位向量,此时 ∣ β × α ∣ = ∣ α ∣ ∣ β ∣ |\pmb{\beta}\times\pmb{\alpha}|=|\pmb{\alpha}||\pmb{\beta}| ∣β×α∣=∣α∣∣β∣(直角条件没变,正弦值还是1)
记 α 0 = α ∣ α ∣ \pmb{\alpha}_{0}=\frac{\pmb{\alpha}}{|\pmb{\alpha}|} α0=∣α∣α
所以 α × β = ∣ α ∣ α 0 × β \pmb{\alpha}\times\pmb{\beta}=|\pmb{\alpha}|\pmb{\alpha}_{0}\times\pmb{\beta} α×β=∣α∣α0×β
若 β \pmb{\beta} β不与 α \pmb{\alpha} α垂直,此时 α × β \pmb{\alpha}\times\pmb{\beta} α×β可看作 P ˉ α β \bar{P}_{\pmb{\alpha}}\pmb{\beta} Pˉαβ绕 α \pmb{\alpha} α旋转 9 0 ∘ 90^{\circ} 90∘再乘上 ∣ α ∣ |\pmb{\alpha}| ∣α∣
【定理】如果 α ≠ 0 ⃗ , α × β \pmb{\alpha}\ne\vec{0},\pmb{\alpha}\times\pmb{\beta} α=0,α×β可以看成 P ˉ α β \bar{P}_{\pmb{\alpha}}\pmb{\beta} Pˉαβ绕 α \pmb{\alpha} α旋转 9 0 ∘ 90^{\circ} 90∘再乘上 ∣ α ∣ |\pmb{\alpha}| ∣α∣ -
(双线性性)
α × ( λ β ) = λ ( α × β ) \pmb{\alpha}\times(\lambda\pmb{\beta})=\lambda(\pmb{\alpha}\times\pmb{\beta}) α×(λβ)=λ(α×β)
α × ( β + γ ) = α × β + α × γ \pmb{\alpha}\times(\pmb{\beta}+\pmb{\gamma})=\pmb{\alpha}\times\pmb{\beta}+\pmb{\alpha}\times\pmb{\gamma} α×(β+γ)=α×β+α×γ
【证】(1) α × ( λ β ) = ∣ α ∣ α 0 × ( λ β ) = ∣ α ∣ α 0 × P ˉ α ( λ β ) = ∣ α ∣ α 0 × λ P ˉ α β = λ ∣ α ∣ α 0 × P ˉ α β = λ ( α × β ) \pmb{\alpha}\times(\lambda\pmb{\beta})=|\pmb{\alpha}|\pmb{\alpha}_{0}\times(\lambda\pmb{\beta})=|\pmb{\alpha}|\pmb{\alpha}_{0}\times\bar{P}_{\pmb{\alpha}}(\lambda\pmb{\beta})=|\pmb{\alpha}|\pmb{\alpha}_{0}\times\lambda\bar{P}_{\pmb{\alpha}}\pmb{\beta}=\lambda|\pmb{\alpha}|\pmb{\alpha}_{0}\times\bar{P}_{\pmb{\alpha}}\pmb{\beta}=\lambda(\pmb{\alpha}\times\pmb{\beta}) α×(λβ)=∣α∣α0×(λβ)=∣α∣α0×Pˉα(λβ)=∣α∣α0×λPˉαβ=λ∣α∣α0×Pˉαβ=λ(α×β)
(2) α × ( β + γ ) = ∣ α ∣ α 0 × P ˉ α ( β + γ ) = ∣ α ∣ α 0 × ( P ˉ α β + P ˉ α γ ) = ∣ α ∣ ( α 0 × P ˉ α β + α 0 × P ˉ α γ ) = ∣ α ∣ α 0 × P ˉ α β + ∣ α ∣ α 0 × P ˉ α γ = α × β + α × γ \pmb{\alpha}\times(\pmb{\beta}+\pmb{\gamma})=|\pmb{\alpha}|\pmb{\alpha}_{0}\times\bar{P}_{\pmb{\alpha}}(\pmb{\beta}+\pmb{\gamma})=|\pmb{\alpha}|\pmb{\alpha}_{0}\times(\bar{P}_{\pmb{\alpha}}\pmb{\beta}+\bar{P}_{\pmb{\alpha}}\pmb{\gamma})=|\pmb{\alpha}|(\pmb{\alpha}_{0}\times\bar{P}_{\pmb{\alpha}}\pmb{\beta}+\pmb{\alpha}_{0}\times\bar{P}_{\pmb{\alpha}}\pmb{\gamma})=|\pmb{\alpha}|\pmb{\alpha}_{0}\times\bar{P}_{\pmb{\alpha}}\pmb{\beta}+|\pmb{\alpha}|\pmb{\alpha}_{0}\times\bar{P}_{\pmb{\alpha}}\pmb{\gamma}=\pmb{\alpha}\times\pmb{\beta}+\pmb{\alpha}\times\pmb{\gamma} α×(β+γ)=∣α∣α0×Pˉα(β+γ)=∣α∣α0×(Pˉαβ+Pˉαγ)=∣α∣(α0×Pˉαβ+α0×Pˉαγ)=∣α∣α0×Pˉαβ+∣α∣α0×Pˉαγ=α×β+α×γ
10.3 用坐标计算外积
假设有空间仿射坐标系
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[O:\pmb{e}_{1},\pmb{e}_{2},\pmb{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}
[O:e1,e2,e3],α=a1e1+a2e2+a3e3,β=b1e1+b2e2+b3e3
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\pmb{\alpha}\times\pmb{\beta}=(a_{1}\pmb{e}_{1}+a_{2}\pmb{e}_{2}+a_{3}\pmb{e}_{3})\times(b_{1}\pmb{e}_{1}+b_{2}\pmb{e}_{2}+b_{3}\pmb{e}_{3})=a_{1}b_{1}\pmb{e}_{1}\times\pmb{e}_{1}+a_{1}b_{2}\pmb{e}_{1}\times\pmb{e}_{2}+a_{1}b_{3}\pmb{e}_{1}\times\pmb{e}_{3}+a_{2}b_{1}\pmb{e}_{2}\times\pmb{e}_{1}+a_{2}b_{2}\pmb{e}_{2}\times\pmb{e}_{2}+a_{2}b_{3}\pmb{e}_{2}\times\pmb{e}_{3}+a_{3}b_{1}\pmb{e}_{3}\times\pmb{e}_{1}+a_{3}b_{2}\pmb{e}_{3}\times\pmb{e}_{2}+a_{3}b_{3}\pmb{e}_{3}\times\pmb{e}_{3}=a_{1}b_{2}\pmb{e}_{1}\times\pmb{e}_{2}+a_{1}b_{3}\pmb{e}_{1}\times\pmb{e}_{3}+a_{2}b_{1}\pmb{e}_{2}\times\pmb{e}_{1}+a_{2}b_{3}\pmb{e}_{2}\times\pmb{e}_{3}+a_{3}b_{1}\pmb{e}_{3}\times\pmb{e}_{1}+a_{3}b_{2}\pmb{e}_{3}\times\pmb{e}_{2}=a_{1}b_{2}\pmb{e}_{1}\times\pmb{e}_{2}+ a_{1}b_{3}\pmb{e}_{1}\times\pmb{e}_{3}- a_{2}b_{1}\pmb{e}_{1}\times\pmb{e}_{2}+ a_{2}b_{3}\pmb{e}_{2}\times\pmb{e}_{3}- a_{3}b_{1}\pmb{e}_{1}\times\pmb{e}_{3}- a_{3}b_{2}\pmb{e}_{2}\times\pmb{e}_{3}= (a_{1}b_{2}-a_{2}b_{1})\pmb{e}_{1}\times\pmb{e}_{2}+ (a_{1}b_{3}-a_{3}b_{1})\pmb{e}_{1}\times\pmb{e}_{3}+ (a_{2}b_{3}-a_{3}b_{2})\pmb{e}_{2}\times\pmb{e}_{3}=\begin{vmatrix} a_{1}& a_{2}\\ b_{1}&b_{2} \end{vmatrix}\pmb{e}_{1}\times\pmb{e}_{2}+\begin{vmatrix} a_{1}& a_{3}\\ b_{1}&b_{3} \end{vmatrix}\pmb{e}_{1}\times\pmb{e}_{3}+\begin{vmatrix} a_{2}& a_{3}\\ b_{2}&b_{3} \end{vmatrix}\pmb{e}_{2}\times\pmb{e}_{3}
α×β=(a1e1+a2e2+a3e3)×(b1e1+b2e2+b3e3)=a1b1e1×e1+a1b2e1×e2+a1b3e1×e3+a2b1e2×e1+a2b2e2×e2+a2b3e2×e3+a3b1e3×e1+a3b2e3×e2+a3b3e3×e3=a1b2e1×e2+a1b3e1×e3+a2b1e2×e1+a2b3e2×e3+a3b1e3×e1+a3b2e3×e2=a1b2e1×e2+a1b3e1×e3−a2b1e1×e2+a2b3e2×e3−a3b1e1×e3−a3b2e2×e3=(a1b2−a2b1)e1×e2+(a1b3−a3b1)e1×e3+(a2b3−a3b2)e2×e3=
a1b1a2b2
e1×e2+
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e2×e3
若
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\pmb{\alpha}\times\pmb{\beta}=\begin{vmatrix} a_{1}& a_{2}\\ b_{1}&b_{2} \end{vmatrix}\pmb{e}_{3}+\begin{vmatrix} a_{1}& a_{3}\\ b_{1}&b_{3} \end{vmatrix}\pmb{e}_{2}+\begin{vmatrix} a_{2}& a_{3}\\ b_{2}&b_{3} \end{vmatrix}\pmb{e}_{1}
α×β=
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e1