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I encounter this triple product property in wikipedia But I can't find proof for $$[\vec{a}\cdot (\vec{b} \times \vec{c})]\vec{a}=(\vec{a}\times\vec{b})\times(\vec{a}\times\vec{c})$$
The RHS cross product produce vector while the LHS produce scalar.
So this got me stumble on working out this equation.
How do I get scalar equals to vector?
Does anyone know proof for this?

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I'm not sure what "The RHS cross product produce vector while the LHS produce dot product" means. Both sides of the equation are in fact vectors. – JavaMan Oct 30 '12 at 2:49
Sorry,edited the text.Btw,I though dot product with vector always result in scalar. – kypronite Oct 30 '12 at 2:50
The dot product of two vectors is a scalar. However you are taking the scalar $a \cdot (b \times c)$ and multiplying the vector $a$ by this scalar. – JavaMan Oct 30 '12 at 2:51
sorry, you are correct.I overlooked that. – kypronite Oct 30 '12 at 2:53
up vote 4 down vote accepted

All quantities below are vectors. I will use the following properties of cross-products and dot-products:

$$ (x \times y) \times z = (x \cdot z) y - (y \cdot z)x \\ x \cdot ( y \times z) = y \cdot (z \times x) = z \cdot (x \times y) \\ x \cdot (x \times y) = 0 $$

We start with the righthand side. For convenience, denote $a \times c = v$. Then

\begin{align} (a \times b) \times (a \times c) = (a \times b) \times v &= (a \cdot v)b - (b \cdot v) a \\ &= (a \cdot(a \times c) )b - (b \cdot(a \times c)) a \\ &= 0 - (- a \cdot (b \times c))a \\ &= (a \cdot (b \times c) )a \end{align}

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thanks for the workout.I had some geometry intuition after your initial commend but my vector algebra is sill elementary, difficulty piecing vector properties. – kypronite Oct 30 '12 at 3:24

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