Show that if $Q'$ is any point on the line of action of $F$, then $PQ × F$ = $PQ'× F$

Question

If a force $F$ is applied to an object at a point $Q$, then the line through $Q$ parallel to $F$ is called the line of action of the force. We defined the vector moment of $F$ about a point $P$ to be $PQ × F$

Show that if $Q'$ is any point on the line of action of $F$, then $PQ × F$ = $PQ'× F$

I'm a little confused with this question. I thought that the torque exerted on an object must be calculated from the pivot point, which in this case is $P$, to the point of application, which is $Q$.

The question gives the following as a hint,

Write $PQ'=PQ+QQ'$ and use properties of the cross product.

Could anyone elaborate on how you could use the hint to arrive at $PQ × F$ = $PQ'× F$?

Answer 1

Write $PQ'=PQ+QQ'$

Then $\left(PQ+QQ'\right) \times F = PQ'\times F$

Then $PQ \times F +QQ'\times F = PQ'\times F$

Since $QQ'$ is parallel to $F$ we have $QQ'\times F = ...$ and it follows.