The force vector $\mathbf{F}$ points along the straight line from point $A$ to point $B$. Use Eqs. (2.28)-(2.31) to prove that

$\mathbf{r}_B \times \mathbf{F}=\mathbf{r}_A \times \mathbf{F} .$

Strategy: Let $\mathbf{r}_{A B}$ be the position vector from point $A$ to point B. Express $\mathbf{r}_B$ in terms of $\mathbf{r}_A$ and $\mathbf{r}_{A B}$. Notice that the vectors $\mathbf{r}_{A B}$ and $\mathbf{F}$ are parallel.

Solution

VerifiedLet $r_{AB}$ the position vector from point A to B.

Express $r_{B}$ in terms of $r_{A} \ and \ r_{AB}.$

We have:

$r_{B}=r_{A} + r_{AB}$

Therefore,

$r_{B} \times F = (r_{A} + r_{AB}) \times F = (r_{A} \times F)+(r_{AB} \times F)$

The last term is zero since $r_{AB}$ is in parallel with F

Therefore,

$\boxed{r_{B} \times F = r_{A} \times F}$

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