Laplace expansion of the inner product (Geometric Algebra)

[NoPhysicsGenius]

Homework Statement

Prove that $\vec {a} \cdot (\vec {b} \wedge \vec {C_r}) = \vec {a} \cdot \vec {b} \vec {C_r} - \vec {b} \wedge (\vec {a} \cdot \vec {C_r})$.

Note that $\vec {a}$ is a vector, $\vec {b}$ is a vector, and $\vec {C_r}$ is an r-blade with $r > 0$.

Also, the dot denotes the inner product, the wedge denotes the outer product, and no operator between vectors (or blades) denotes the geometric product.

Finally, the identity to be proven can be called the "Laplace expansion of the inner product".

Homework Equations

Equation (1a): $\vec {a} \cdot \vec {b} = \frac {1}{2} (\vec {a} \vec {b} + \vec {b} \vec {a})$
$\Rightarrow$
Equation (1b): $\vec {a} \vec {b} = - \vec {b} \vec {a} + 2 \vec {a} \cdot \vec {b}$

Equation (2): $\vec {a} \cdot \vec {A_r} = \frac {1}{2} (\vec {a} \vec {A_r} - (-1)^r \vec {A_r} \vec {a})$

Equation (3): $\vec {b} \vec {C_r} = \vec {b} \cdot \vec {C_r} + \vec {b} \wedge \vec {C_r}$

The Attempt at a Solution

By Equation (1a), we have $\vec {a} \vec {b} \vec {C_r} = (- \vec {b} \vec {a} + 2 \vec {a} \cdot \vec {b}) \vec {C_r} = - \vec {b} \vec {a} \vec {C_r} + 2 \vec {a} \cdot \vec {b} \vec {C_r}$

Note that Equation (2) implies $2 \vec {a} \cdot \vec {C_r} = \vec {a} \vec {C_r} - (-1)^r \vec {C_r} \vec {a} \Rightarrow \vec {a} \vec {C_r} = (-1)^r \vec {C_r} \vec {a} + 2 \vec {a} \cdot \vec {C_r}$

Therefore, $\vec {a} \vec {b} \vec {C_r} = - \vec {b} [(-1)^r \vec {C_r} \vec {a} + 2 \vec {a} \cdot \vec {C_r}] + 2 \vec {a} \cdot \vec {b} \vec {C_r}$
$\Rightarrow \frac {1}{2} \vec {a} \vec {b} \vec {C_r} = - \frac {1}{2} \vec {b} [(-1)^r \vec {C_r} \vec {a} + 2 \vec {a} \cdot \vec {C_r}] + \vec {a} \cdot \vec {b} \vec {C_r}$
$\Rightarrow \vec {a} \cdot \vec {b} \vec {C_r} - \vec {b} \vec {a} \cdot \vec {C_r} = \frac {1}{2} [\vec {a} (\vec {b} \vec {C_r}) - (-1)^r (\vec {b} \vec {C_r}) \vec {a}]$

Using the reverse of Equation (2), we get $\vec {a} \cdot \vec {b} \vec {C_r} - \vec {b} \vec {a} \cdot \vec {C_r} = \vec {a} \cdot (\vec {b} \vec {C_r})$

Applying Equation (3), we have $\vec {a} \cdot \vec {b} \vec {C_r} - \vec {b} \vec {a} \cdot \vec {C_r} = \vec {a} \cdot (\vec {b} \cdot \vec {C_r} + \vec {b} \wedge \vec {C_r}) =$
$\vec {a} \cdot (\vec {b} \cdot \vec {C_r}) + \vec {a} \cdot (\vec {b} \wedge \vec {C_r})$

$\Rightarrow \vec {a} \cdot (\vec {b} \wedge \vec {C_r}) = \vec {a} \cdot \vec {b} \vec {C_r} - \vec {b} \vec {a} \cdot \vec {C_r} - \vec {a} \cdot (\vec {b} \cdot \vec {C_r})$

This is where I got stuck. Somehow, the r-vector part of this last equation gives the desired identity; but I don't know how...

 

[fresh_42]

With equation (3) you directly get, that the statement is equivalent to $\vec{a}\cdot (\vec{b}\cdot \vec{C_r})=\vec{b}\wedge (\vec{a}\cdot\vec{C_r})\,$?

 

[NoPhysicsGenius]

With equation (3) you directly get, that the statement is equivalent to $\vec{a}\cdot (\vec{b}\cdot \vec{C_r})=\vec{b}\wedge (\vec{a}\cdot\vec{C_r})\,$?

I am confused. To finish my proof, I must show that $- \vec {b} \wedge ( \vec {a} \cdot \vec{C_r} ) = - \vec {b} \vec {a} \cdot \vec {C_r} - \vec {a} \cdot ( \vec {b} \cdot \vec {C_r} )$
$\Rightarrow \vec {b} \wedge ( \vec {a} \cdot \vec{C_r} ) = \vec {b} \vec {a} \cdot \vec {C_r} + \vec {a} \cdot ( \vec {b} \cdot \vec {C_r} )$.

Does $\vec {b} \vec {a} \cdot \vec {C_r}$ equal zero?

 

[fresh_42]

I am confused. To finish my proof, I must show that $- \vec {b} \wedge ( \vec {a} \cdot \vec{C_r} ) = - \vec {b} \vec {a} \cdot \vec {C_r} - \vec {a} \cdot ( \vec {b} \cdot \vec {C_r} )$
$\Rightarrow \vec {b} \wedge ( \vec {a} \cdot \vec{C_r} ) = \vec {b} \vec {a} \cdot \vec {C_r} + \vec {a} \cdot ( \vec {b} \cdot \vec {C_r} )$.

Does $\vec {b} \vec {a} \cdot \vec {C_r}$ equal zero?

I don't know. I simply used your equation (3) to get rid of the geometric product in your assertion and ended up with one which is likely wrong. The only other thing I used was distributivity of the dot product.