Proof of Triple Scalar or Box product

• Feb 4th 2010, 01:31 PM
joepinsley18
[solved] Proof of Triple Scalar or Box product
Prove that (u cross v) dot w = u dot (w cross v)

This is called the triple scalar or box product

Please try it using a parallelpiped but be sure to consider all cases (the angle between the vector normal to the base and the height vector could be obtuse)

I will gladly accept any other ways of proving this statement.

Thank you.
• Feb 4th 2010, 04:43 PM
Soroban
Hello, joepinsley18!

Quote:

Prove that: .$\displaystyle (\vec u \times \vec v) \cdot \vec w \;=\; \vec u \cdot (\vec w \times \vec v)$ . . . . Not true!

I think you mean: .$\displaystyle (\vec u \times \vec v) \cdot \vec w \;=\;\vec u \cdot({\color{red}\vec v} \times {\color{red}\vec w})$

We can do it in a most primitive way. . $\displaystyle \text{Let: }\:\begin{Bmatrix}\vec u &=& \langle u_1,u_2,u_3 \rangle \\ \vec v &=& \langle v_1,v_2,v_3\rangle \\ \vec w &=& \langle w_1,w_2,w_3\rangle \end{Bmatrix}$

We have: .$\displaystyle \vec u \times \vec v \;=\;\left|\begin{array}{ccc}\vec i& \vec j& \vec k \\ u_1&u_2&u_3 \\ v_1&v_2&v_3\end{array}\right| \;=\;\bigg\langle u_2v_3-u_3v_2,\;-(u_1v_3-u_3v_1),\;u_1v_2-u_2v_1\bigg\rangle$

Hence: .$\displaystyle (\vec u \times \vec v)\cdot \vec w \;=\;\bigg\langle u_2v_3 - u_3v_2,\;-(u_1v_3-u_3v_1),\;u_1v_2-u_2v_1\bigg\rangle \cdot \bigg\langle w_1,w_2,w_3\bigg\rangle$

. . . . . . . . . . . . .$\displaystyle =\;u_1v_2w_3 - u_1v_3w_2 - u_2v_1w_3 + u_2v_3w_1 + u_3v_1w_2 - u_3v_2w_1$

We have: .$\displaystyle \vec v \times \vec w \;=\;\left|\begin{array}{ccc} \vec i& \vec j& \vec k\\v_1&v_2&v_3\\w_1&w_2&w_3\end{array}\right| \;=\;\bigg\langle v_2w_3-v_3w_2,\;-(v_1w_3-v_3w_1,\;v_1w_2 - v_2w_1\bigg\rangle$

Hence: .$\displaystyle \vec u\cdot(\vec v\times \vec w) \;=\;\bigg\langle u_1,u_2,u_3\bigg\rangle \cdot\bigg\langle v_2w_3-v_2w_3,\;-(v_1w_3-v_3w_1),\;v_1w_2-v_2w_1\bigg\rangle$

. . . . . . . . . . . . .$\displaystyle =\; u_1v_2w_3 - u_1v_3w_2 - u_2v_1w_3 + u_2v_3w_1 + u_3v_1w_2 - u_3v_2w_1$

Therefore: .$\displaystyle (\vec u \times \vec v)\cdot \vec w \;=\;\vec u\cdot(\vec v \times \vec w)$

• Feb 4th 2010, 08:54 PM
joepinsley18
I did intend to prove that "u cross v dot w = u dot w cross v". From your proof, I can see that this statement is false! By switching v and w as you did, the determinants of both matrices become equal. Had you tried proving the above statement, you would have found that the determinants are opposites. Thank you. I see it now.