- #1
NoPhysicsGenius
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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...