• 0 Votes - 0 Average
• 1
• 2
• 3
• 4
• 5
 Lagrange's Vector Identity: $\forall a,b,c,d \in \mathbb{R}^3$, (a×b)×c = (a·c)b-(b·c)a
12-18-2010, 07:33 PM
Post: #1
 elim Moderator Posts: 577 Joined: Feb 2010 Reputation: 0
Lagrange's Vector Identity: $\forall a,b,c,d \in \mathbb{R}^3$, (a×b)×c = (a·c)b-(b·c)a
(1) $(\mathbf{a} \times \mathbf{b}) \times \mathbf{c} = (\mathbf{a}\cdot \mathbf{c}) \mathbf{b} - (\mathbf{b}\cdot \mathbf{c}) \mathbf{a}$ is actually equivalent to

(2) $(\mathbf{a} \times \mathbf{b})\cdot(\mathbf{c} \times \mathbf{d}) = (\mathbf{a} \cdot \mathbf{c})(\mathbf{b}\cdot \mathbf{d})-(\mathbf{b}\cdot\mathbf{c})(\mathbf{a} \cdot \mathbf{d})$

Proof: By the identity $(\mathbf{u} \times \mathbf{v})\cdot \mathbf{w} = \mathbf{u} \cdot (\mathbf{v} \times \mathbf{w})$, dot product with $\mathbf{d}$ both sides of (1) we get (2).

Conversely, from (2) we have
$((\mathbf{a} \times \mathbf{b}) \times \mathbf{c})\cdot \mathbf{d} = (\mathbf{a} \times \mathbf{b})\cdot(\mathbf{c} \times \mathbf{d}) = (\mathbf{a} \cdot \mathbf{c})(\mathbf{b}\cdot\mathbf{d})-(\mathbf{b}\cdot\mathbf{c})(\mathbf{a}\cdot \mathbf{d}) = ((\mathbf{a} \cdot \mathbf{c}) \mathbf{b} - (\mathbf{b}\cdot\mathbf{c}) \mathbf{a}) \cdot \mathbf{d}$.
Since this is true for all $\mathbf{d}$, (1) follows.

Now we prove (2). Let $(a_1,a_2,a_3) = \mathbf{a}$ and likewise for $\mathbf{b}$ and $\mathbf{c}$, then $(\mathbf{a} \cdot \mathbf{c})(\mathbf{b}\cdot \mathbf{d})-(\mathbf{b}\cdot \mathbf{c})(\mathbf{a} \cdot \mathbf{d}) =$

$\displaystyle{\sum_{1\le i,j \le 3} (a_i c_i b_j d_j - b_i c_i a_j d_j) = \sum_{1\le i<j \le 3} ((a_i c_i b_j d_j + a_j c_j b_i d_i)-(b_i c_i a_j d_j + b_j c_j a_i d_i))=}$
$\displaystyle{\sum_{1\le i<j \le 3} (a_i b_j(c_i d_j-c_j d_i)-a_j b_i (c_i d_j - c_j d_i)) = \sum_{1\le i<j \le 3} (a_i b_j - a_j b_i)(c_i d_j - c_j d_i) = (\mathbf{a} \times \mathbf{b})\cdot(\mathbf{c} \times \mathbf{d})}$
 « Next Oldest | Next Newest »

Forum Jump: