Dot and Cross Product Proof: $u \times (v \times w) = ( u \cdot w)v - (u \cdot v)w$

Question

How do you prove that: $u \times (v \times w) = ( u \cdot w)v - (u \cdot v)w$ ? The textbook says as a hint to "first do it for $u=i,j$ and $k$; then write $u-xi+yj+zk$ but I am not sure what that means.

Answer 1

I'm going to use the basic definitions of scalar and vector triple products to prove this. Let $\mathbf u= a_1 \mathbf i + a_2 \mathbf j + a_3 \mathbf k, \mathbf v= b_1 \mathbf i + b_2 \mathbf j + b_3 \mathbf k, \mathbf w= c_1 \mathbf i + c_2 \mathbf j + c_3 \mathbf k$,

Then $(\mathbf{v} \times \mathbf{w})= (b_2c_3-b_3c_2) \mathbf i + (b_3c_1-b_1c_3) \mathbf j + (b_1c_2-b_2c_1) \mathbf k$

Hence, $\mathbf{u} \times (\mathbf{v} \times \mathbf{w})= \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k}\\ a_1 & a_2 & a_3 \\ (b_2c_3-b_3c_2) & (b_3c_1-b_1c_3) & (b_1c_2-b_2c_1) \end{vmatrix}$

=$(a_1c_1+a_2c_2+a_3c_3)(b_1 \mathbf{i}+ b_2 \mathbf{j} + b_3 \mathbf{k}) - (a_1b_1+a_2b_2+a_3b_3)(c_1 \mathbf{i}+ c_2 \mathbf{j} + c_3 \mathbf{k})$

=$( \mathbf{u} \cdot \mathbf{w}) \cdot \mathbf{v} - (\mathbf{u} \cdot \mathbf{v}) \cdot \mathbf{w}$