Scalar triple product invariance under circular shift proof

[pastoreerrante]

Homework Statement

Prove that for any three vectors $\hat a, \hat b$ and $\hat c$, $\hat a \cdot (\hat b \times \hat c)$ = $(\hat a \times \hat b) \cdot \hat c$

Homework Equations

[/B]
$\hat i \cdot \hat i = \hat j \cdot \hat j = \hat k \cdot \hat k = (1)(1)\cos(0) = 1$
$\hat i \cdot \hat j = \hat i \cdot \hat k = \hat j \cdot \hat k = (1)(1)\cos(90) = 0$

The Attempt at a Solution

I tried this procedure:

1. calculate the inner cross product $(\hat b \times \hat c)$ and $(\hat a \times \hat b)$
2. dotting the resulting vector with the third remaining vector ($\hat a$ and $\hat c$ respectively)
3. I expected that the resulting vectors were identical in terms of their components, but they aren't.

These are the calculations for the LHS of my thesis:

1. $(\hat b \times \hat c)$= $(b_y c_z - b_z c_y)\hat i$ + $(b_z c_x - b_x c_z)\hat j$ + $(b_x c_y - b_y c_x)\hat k$

2. dotting with $\hat a$:

$(a_x \hat i + a_y \hat j + a_z \hat k)$ $\cdot [(b_y c_z - b_z c_y)\hat i$ + $(b_z c_x - b_x c_z)\hat j$ + $(b_x c_y - b_y c_x)\hat k]$ =
$(a_x b_y c_z - a_x b_z c_y)\hat i$ + $(a_y b_z c_x - a_y b_x c_z)\hat j$ + $(a_z b_x c_y - a_z b_y c_x)\hat k$

Now the calculations for the RHS of my thesis:

1. $(\hat a \times \hat b)$= $(a_y b_z - a_z b_y)\hat i$ + $(a_z b_x - a_x b_z)\hat j$ + $(a_x b_y - a_y b_x)\hat k$

2. dotting with $\hat c$:

$(c_x \hat i + c_y \hat j + c_z \hat k)$ $\cdot [(a_y b_z - a_z b_y)\hat i$ + $(a_z b_x - a_x b_z)\hat j$ + $(a_x b_y - a_y b_x)\hat k]$ =
$(a_y b_z c_x - a_z b_y c_x)\hat i$ + $(a_z b_x c_y - a_x b_z c_y)\hat j$ + $(a_x b_y c_z - a_y b_x c_z)\hat k$

Clearly,

$(a_x b_y c_z - a_x b_z c_y)\hat i$ + $(a_y b_z c_x - a_y b_x c_z)\hat j$ + $(a_z b_x c_y - a_z b_y c_x)\hat k$

is not equal to:

$(a_y b_z c_x - a_z b_y c_x)\hat i$ + $(a_z b_x c_y - a_x b_z c_y)\hat j$ + $(a_x b_y c_z - a_y b_x c_z)\hat k$

What am I missing here?

Thank you in advance

 

[Orodruin]

You can never get a vector as a result from a scalar product. Your result should be a number.

 

[pastoreerrante]

Thank you, I got it! If I don't consider the results as vectors but as normal algebraic expressions, they are perfectly equivalent.

 

[Orodruin]

If I don't consider the results as vectors but as normal algebraic expressions, they are perfectly equivalent.

This rings a warning bell for me. It is not a matter of considering the result as a vector or not, the result of the triple product is a scalar and not a vector.