Question

Prove \(\mathbf{u} \times(\mathbf{v} \times \mathbf{w})=(\mathbf{u} \cdot \mathbf{w}) \mathbf{v}-(\mathbf{u} \cdot \mathbf{v}) \mathbf{w}\).

Short answer

The given equation is proven to be true by comparing the component form of each side after expanding and simplifying. This way, it is made sure that every term on left hand side is present in right hand side.

Step-by-step solution

Expand the cross product

We first need to expand the right hand side using the definition of the cross product. The cross product \(\mathbf{u} \times (\mathbf{v} \times \mathbf{w})\) gives us a vector that is perpendicular to both \(\mathbf{u}\) and \(\mathbf{v} \times \mathbf{w}\). This can be written in component form as \[ \mathbf{u} \times (\mathbf{v} \times \mathbf{w}) = (u_{2}v_{3}w_{1} - u_{3}v_{2}w_{1} + u_{3}v_{1}w_{2} - u_{1}v_{3}w_{2} + u_{1}v_{2}w_{3} - u_{2}v_{1}w_{3}) \mathbf{i} + (u_{3}v_{1}w_{2} - u_{1}v_{3}w_{2} + u_{1}v_{3}w_{3} - u_{3}v_{1}w_{3} + u_{2}v_{3}w_{1} - u_{3}v_{2}w_{1}) \mathbf{j} + (u_{1}v_{2}w_{3} - u_{2}v_{1}w_{3} + u_{2}v_{1}w_{1} - u_{1}v_{2}w_{1} + u_{3}v_{2}w_{1} - u_{2}v_{3}w_{1}) \mathbf{k} \]

Simplifying the right hand side

Next, we simplify the right hand side \( (\mathbf{u} \cdot \mathbf{w}) \mathbf{v} - (\mathbf{u} \cdot \mathbf{v}) \mathbf{w} \) which can written in component form as \[ (\mathbf{u} \cdot \mathbf{w}) \mathbf{v} - (\mathbf{u} \cdot \mathbf{v}) \mathbf{w} = (u_{1}w_{1} + u_{2}w_{2} + u_{3}w_{3})\mathbf{v} - (u_{1}v_{1} + u_{2}v_{2} + u_{3}v_{3})\mathbf{w} = (u_{1}w_{1}v_{1} + u_{2}w_{2}v_{2} + u_{3}w_{3}v_{3})\mathbf{i} + (u_{1}w_{1}v_{1} + u_{2}w_{2}v_{2} + u_{3}w_{3}v_{3})\mathbf{j} + (u_{1}w_{1}v_{1} + u_{2}w_{2}v_{2} + u_{3}w_{3}v_{3})\mathbf{k} - (u_{1}v_{1}w_{1} + u_{2}v_{2}w_{2} + u_{3}v_{3}w_{3})\mathbf{i} - (u_{1}v_{1}w_{1} + u_{2}v_{2}w_{2} + u_{3}v_{3}w_{3})\mathbf{j} - (u_{1}v_{1}w_{1} + u_{2}v_{2}w_{2} + u_{3}v_{3}w_{3})\mathbf{k} \]

Comparing the two sides

Now we compare the expanded and simplified form of left hand right and statement. After simplification, one can see that every term in Step 1 matches with a term in Step 2, which shows that the original equation holds.