1
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – JavaMan Oct 30 '12 at 2:49
  • $\begingroup$ Sorry,edited the text.Btw,I though dot product with vector always result in scalar. $\endgroup$ – kypronite Oct 30 '12 at 2:50
  • 2
    $\begingroup$ 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. $\endgroup$ – JavaMan Oct 30 '12 at 2:51
  • $\begingroup$ sorry, you are correct.I overlooked that. $\endgroup$ – kypronite Oct 30 '12 at 2:53
4
$\begingroup$

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}

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.