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I was reading a book and I found this question :

how to prove the Grassman identity : (a × b) × c=(a ⋅ c) b−(a ⋅ b) c where a,b and c are vectors.

There was a hint that I should begin by expanding the left and right sides of the equation and then comparing the coordinates in cartesian basis.

how?

Martin Sleziak
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Nour
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1 Answers1

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I will state some facts first.

Suppose that $a = (a_1, a_2, a_3)$ and that $b = (b_1, b_2, b_3)$. Then $$ a \times b = \left( \det {\begin{bmatrix} a_2 & b_2 \\ a_3 & b_3 \\ \end{bmatrix}}, -\det {\begin{bmatrix} a_1 & b_1 \\ a_3 & b_3 \\ \end{bmatrix}}, \det {\begin{bmatrix} a_1 & b_1 \\ a_2 & b_2 \\ \end{bmatrix}} \right), $$ and $$ a \cdot b = a_1 b_1 + a_2 b_2 + a_3 c_3. $$

Lemma 1. $$ (a \times b) \cdot (c \times d) = \det {\begin{bmatrix} a \cdot c & a \cdot d \\ b \cdot c & b \cdot d \\ \end{bmatrix}}. $$

Proof. $$ \begin{aligned} & (a \times b) \cdot (c \times d) \\ = {} & \sum_{1 \leq i < j \leq 3} {(a_i b_j - a_j b_i) (c_i d_j - c_j d_i)} \\ = {} & \sum_{1 \leq i < j \leq 3} {(a_i b_j - a_j b_i) (c_i d_j - c_j d_i)} \\ = {} & \sum_{1 \leq i < j \leq 3} {(a_i b_j c_i d_j + a_j b_i c_j d_i - a_i b_j c_j d_i - a_j b_i c_i d_j)} \\ = {} & \sum_{1 \leq i < j \leq 3} {(a_i b_j c_i d_j + a_j b_i c_j d_i)} - \sum_{1 \leq i < j \leq 3} {(a_i b_j c_j d_i + a_j b_i c_i d_j)} \\ = {} & \sum_{1 \leq i < j \leq 3} {(a_i b_j c_i d_j + a_j b_i c_j d_i)} + \sum_{i = 1}^{3} {a_i b_i c_i d_i} \\ & - \sum_{1 \leq i < j \leq 3} {(a_i b_j c_j d_i + a_j b_i c_i d_j)} - \sum_{i = 1}^{3} {a_i b_i c_i d_i} \\ = {} & \sum_{1 \leq i, j \leq 3} {a_i b_j c_i d_j} - \sum_{1 \leq i, j \leq 3} {a_i b_j c_j d_i} \\ = {} & \sum_{i = 1}^{3} \sum_{j = 1}^{3} {a_i b_j c_i d_j} - \sum_{i = 1}^{3} \sum_{j = 1}^{3} {a_i b_j c_j d_i} \\ = {} & \sum_{i = 1}^{3} {a_i c_i} \sum_{j = 1}^{3} {b_j d_j} - \sum_{i = 1}^{3} {a_i d_i} \sum_{j = 1}^{3} {b_j c_j} \\ = {} & (a \cdot c)(b \cdot d) - (a \cdot d)(b \cdot c). \end{aligned} $$

Lemma 2. $$ (a \times b) \cdot c = a \cdot (b \times c). $$

Proof. $$ \begin{aligned} & (a \times b) \cdot c \\ = {} & \det {\begin{bmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \\ \end{bmatrix} } \\ = {} & \det {\begin{bmatrix} b_1 & c_1 & a_1 \\ b_2 & c_2 & a_2 \\ b_3 & c_3 & a_3 \\ \end{bmatrix} } \\ = {} & (b \times c) \cdot a \\ = {} & a \cdot (b \times c). \end{aligned} $$

Theorem. $$ (a \times b) \times c = (c \cdot a) b - (b \cdot c) a. $$

Proof.

For any vector $d$, $$ ((a \times b) \times c) \cdot d = (a \times b) \cdot (c \times d), $$ and $$ ((c \cdot a) b - (b \cdot c) a) \cdot d = (c \cdot a)(b \cdot d) - (b \cdot c)(a \cdot d) = (a \cdot c)(b \cdot d) - (a \cdot d)(b \cdot c). $$ By Lemma 1, we have $$ ((a \times b) \times c) \cdot d = ((c \cdot a) b - (b \cdot c) a) \cdot d $$ for any vector $d$, which is $$ ((a \times b) \times c - ((c \cdot a) b - (b \cdot c) a)) \cdot d = 0 $$ for any vector $d$. Specially, $$ ((a \times b) \times c - ((c \cdot a) b - (b \cdot c) a)) \cdot ((a \times b) \times c - ((c \cdot a) b - (b \cdot c) a)) = 0. $$ Hence we conclude that $$ (a \times b) \times c - ((c \cdot a) b - (b \cdot c) a) = 0, $$ which is just that $$ (a \times b) \times c = (c \cdot a) b - (b \cdot c) a. $$

Juliamisto
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