Show that $\vec{a} \times (\vec{b} \times \vec{a}) = (\vec{a} \times \vec{b}) \times \vec{a}$.

1. Problem Description

Show that $\vec{a} \times (\vec{b} \times \vec{a}) = (\vec{a} \times \vec{b}) \times \vec{a}$.

2. Solution Steps

We will evaluate both sides of the equation separately using the vector triple product identity.

First, consider the left-hand side: $\vec{a} \times (\vec{b} \times \vec{a})$.

Using the vector triple product identity:

$\vec{A} \times (\vec{B} \times \vec{C}) = (\vec{A} \cdot \vec{C})\vec{B} - (\vec{A} \cdot \vec{B})\vec{C}$

Applying this identity to $\vec{a} \times (\vec{b} \times \vec{a})$ with $\vec{A} = \vec{a}$, $\vec{B} = \vec{b}$, and $\vec{C} = \vec{a}$, we get:

$\vec{a} \times (\vec{b} \times \vec{a}) = (\vec{a} \cdot \vec{a})\vec{b} - (\vec{a} \cdot \vec{b})\vec{a} = |\vec{a}|^2 \vec{b} - (\vec{a} \cdot \vec{b})\vec{a}$.

Now consider the right-hand side: $(\vec{a} \times \vec{b}) \times \vec{a}$.

Using the vector triple product identity:

$(\vec{A} \times \vec{B}) \times \vec{C} = - \vec{C} \times (\vec{A} \times \vec{B}) = -[(\vec{C} \cdot \vec{B})\vec{A} - (\vec{C} \cdot \vec{A})\vec{B}] = (\vec{C} \cdot \vec{A})\vec{B} - (\vec{C} \cdot \vec{B})\vec{A}$

Applying this to $(\vec{a} \times \vec{b}) \times \vec{a}$ with $\vec{A} = \vec{a}$, $\vec{B} = \vec{b}$, and $\vec{C} = \vec{a}$, we get:

$(\vec{a} \times \vec{b}) \times \vec{a} = (\vec{a} \cdot \vec{a})\vec{b} - (\vec{a} \cdot \vec{b})\vec{a} = |\vec{a}|^2 \vec{b} - (\vec{a} \cdot \vec{b})\vec{a}$.

Since both sides simplify to the same expression, the equation holds.

3. Final Answer

$\vec{a} \times (\vec{b} \times \vec{a}) = (\vec{a} \cdot \vec{a})\vec{b} - (\vec{a} \cdot \vec{b})\vec{a}$

$(\vec{a} \times \vec{b}) \times \vec{a} = (\vec{a} \cdot \vec{a})\vec{b} - (\vec{a} \cdot \vec{b})\vec{a}$

Therefore, $\vec{a} \times (\vec{b} \times \vec{a}) = (\vec{a} \times \vec{b}) \times \vec{a}$.