벡터의 연산

정의에 의해,

$$\begin{aligned} \mathbf{b}\times\mathbf{c} \ =\ &(b_2c_3-b_3c_2)\hat{\boldsymbol{\imath}}-(b_1c_3-b_3c_1)\hat{\boldsymbol{\jmath}} +(b_1c_2-b_2c_1)\hat{\mathbf{k}} \\ \\ \text{따라서 좌변과 우변은 다음과 같다} \\ \mathbf{a}\times(\mathbf{b}\times\mathbf{c}) \ =\ &\det\left[\begin{matrix}\hat{\boldsymbol{\imath}}&\hat{\boldsymbol{\jmath}} &\hat{\mathbf{k}}\\ a_1&a_2&a_3\\ b_2c_3-b_3c_2&-b_1c_3+b_3c_1&b_1c_2-b_2c_1\end{matrix}\right]\\ \ =\ &\hat{\boldsymbol{\imath}}\,[a_2(b_1c_2-b_2c_1)-a_3(-b_1c_3+b_3c_1)]\\ {-}&\hat{\boldsymbol{\jmath}}\,[a_1(b_1c_2-b_2c_1)-a_3(b_2c_3-b_3c_2)]\\ {+}&\hat{\mathbf{k}}\,[a_1(-b_1c_3+b_3c_1)-a_2(b_2c_3-b_3c_2)]&{\rm (lhs)}\\ (\mathbf{a}\cdot\mathbf{c})\mathbf{b}-(\mathbf{a}\cdot\mathbf{b})\mathbf{c} \ =\ &(a_1c_1+a_2c_2+a_3c_3)(b_1\hat{\boldsymbol{\imath}}+b_2\hat{\boldsymbol{\jmath}}+b_3\hat{\mathbf{k}}) -(a_1b_1+a_2b_2+a_3b_3)(c_1\hat{\boldsymbol{\imath}}+c_2\hat{\boldsymbol{\jmath}}+c_3\hat{\mathbf{k}})\\ {=}\ & \hat{\boldsymbol{\imath}}\,[a_1b_1c_1+a_2b_1c_2+a_3b_1c_3-a_1b_1c_1-a_2b_2c_1-a_3b_3c_1] \\ {+}&\hat{\boldsymbol{\jmath}}\,[a_1b_2c_1+a_2b_2c_2+a_3b_2c_3-a_1b_1c_2-a_2b_2c_2-a_3b_3c_2] \\ {+}&\hat{\mathbf{k}}\,[a_1b_3c_1+a_2b_3c_2+a_3b_3c_3-a_1b_1c_3-a_2b_2c_3-a_3b_3c_3]\\ {=}\ & \hat{\boldsymbol{\imath}}\,[a_2b_1c_2+a_3b_1c_3-a_2b_2c_1-a_3b_3c_1] \\ {+}&\hat{\boldsymbol{\jmath}}\,[a_1b_2c_1+a_3b_2c_3-a_1b_1c_2-a_3b_3c_2] \\ {+}&\hat{\mathbf{k}}\,[a_1b_3c_1+a_2b_3c_2-a_1b_1c_3-a_2b_2c_3] &{\rm (rhs)} \end{aligned}$$

(lhs)와 (rhs)는 같다.