Prove that $\sum_i \vec r_i \times \vec uF_i = (\sum_i \vec r_iF_i) \times \vec u$

Question

Given an expression $$\sum_i \vec r_i \times \vec uF_i $$ where $\vec r_i$ are position vectors from the origin, $\vec u$ is a constant unit vector and $F_i$ are constants, it is true that$$\sum_i \vec r_i \times \vec uF_i = (\sum_i \vec r_iF_i) \times \vec u.$$

How to prove that the above equation is true?

Answer 1

Let's keep it simple. You need just the following two properties of the cross product:

$$\vec{a}\times(k\vec{b})=(k\vec{a})\times\vec{b}\tag{1}$$

$$(\vec{a}+\vec{b})\times\vec{c}=\vec{a}\times\vec{c}+\vec{b}\times\vec{c}\tag{2}$$

Both can be easily proved if you use the fact that the cross product of vectors $\vec{a}=a_x\vec{i}+a_y\vec{j}+a_z\vec{k}$ and $\vec{b}=b_x\vec{i}+b_y\vec{j}+b_z\vec{k}$ is:

$$\vec{a}\times\vec{b}=(a_yb_z-a_zb_y)\vec{i}+(a_zb_x-a_xb_z)\vec{j}+(a_xb_y-a_yb_x)\vec{k}$$

All this is explained in excellent detail on Wikipedia.

The proof is now simple:

$$\sum_i \vec r_i \times (F_i\vec u)=$$

$$\sum_i (F_i\vec r_i) \times \vec u=\tag{(1) applied}$$

$$(\sum_i F_i\vec r_i) \times \vec u\tag{(2) applied}$$

Answer 2

I will use superscript for the index of the vector, so the subscript can be the element of a vector. $$\left(\sum_{i} \vec{r}^i \times \vec{u}F^i\right)_j=$$ $$\sum_{i}\left(\vec{r}^i \times \vec{u}F^i\right)_j=$$ $$\sum_{i}\left(\sum_{k,l}\epsilon_{jkl} r^i_k u_lF^i\right)=$$ $$\sum_{k,l}\left(\sum_{i}\epsilon_{jkl} r^i_k u_lF^i\right)=$$ $$\sum_{k,l}\epsilon_{jkl}\left(\sum_{i} r^i_k F^i\right)u_l=$$ $$\sum_{k,l}\epsilon_{jkl}\left(\sum_{i} \vec{r}^i F^i\right)_ku_l=$$ $$\left(\left(\sum_{i} \vec{r}^i F^i\right)\times\vec{u}\right)_j$$ So we have that $$\sum_{i} \vec{r}^i \times \vec{u}F^i=\left(\sum_{i} \vec{r}^i F^i\right)\times\vec{u}$$

Answer 3

I know this already has an accepted answer, however there is a much more obvious proof. First note that $\vec{r}_i\times\vec{u}$ is just another vector, call it $\vec{q}_i$ so that \begin{equation} \sum_i \vec{r}_i\times\vec{u} F_i = \sum_i \vec{q}_i F_i \end{equation} but because scalar multiplication of a vector by a scalar is commutative, this becomes \begin{eqnarray} \sum_i \vec{r}_i\times\vec{u} F_i &=& \sum_i \vec{q}_i F_i = \sum_i F_i \vec{q}_i \\ &=& \sum_i F_i (\vec{r}_i\times\vec{u}) \\ &=& \left( \sum_i F_i \vec{r}_I \right)\times\vec{u} \end{eqnarray} where we have factored the constant vector $\vec{u}$ outside of the sum in the last step.