Question

Prove the property of the cross product. $$ \mathbf{u} \cdot(\mathbf{v} \times \mathbf{w})=(\mathbf{u} \times \mathbf{v}) \cdot \mathbf{w} $$

Short answer

The scalar triple product identity \(\mathbf{u} \cdot (\mathbf{v} \times \mathbf{w}) = (\mathbf{u} \times \mathbf{v}) \cdot \mathbf{w}\) has been proven by expressing the vectors in component form, calculating the cross products, substituting back in, and showing that the results of the two sides of the identity are equal.

Step-by-step solution

Apply the coordinates

Consider vectors \(\mathbf{u}=[u_1,u_2,u_3]\), \(\mathbf{v}=[v_1,v_2,v_3]\) and \(\mathbf{w}=[w_1,w_2,w_3]\). Substitute these into the equation to specify the problem in terms of identifiable components.

Calculate cross products

Calculate the cross product \(\mathbf{v} \times \mathbf{w}\) and \(\mathbf{u} \times \mathbf{v}\) using the rule of cross product in 3D. \(\mathbf{v} \times \mathbf{w} = [v_2w_3 - v_3w_2, v_3w_1 - v_1w_3, v_1w_2 - v_2w_1]\) and \(\mathbf{u} \times \mathbf{v} = [u_2v_3 - u_3v_2, u_3v_1 - u_1v_3, u_1v_2 - u_2v_1]\).

Substitute cross products back into the identity

Substitute the calculated cross products back into the identity, and calculate the dot product for each side. \(\mathbf{u} \cdot (\mathbf{v} \times \mathbf{w}) = u_1(v_2w_3 - v_3w_2) + u_2(v_3w_1 - v_1w_3) + u_3(v_1w_2 - v_2w_1)\), and \((\mathbf{u} \times \mathbf{v}) \cdot \mathbf{w} = w_1(u_2v_3 - u_3v_2) + w_2(u_3v_1 - u_1v_3) + w_3(u_1v_2 - u_2v_1)\). After calculations, you should see they result in the same expression, thereby proving the identity.