Prove Vector Problem: PQ∙v = ∫a b dγ/dt ∙v dt

[ballzac]

Homework Statement

Prove $$\mathbf{PQ}\cdot \mathbf v=\int_a^b\frac{\textup d\gamma}{\textup d t}(t)\cdot\mathbf v\textup d t$$
where $$\mathbf P=\gamma(a)$$ and $$\mathbf Q=\gamma(b)$$

The Attempt at a Solution

I get
$$\int_a^b\frac{\textup d\gamma}{\textup d t}(t)\cdot\mathbf v\textup d t=v_x\int_a^b{\frac{\textup d\gamma_x}{\textup d t}\textup d t}+v_y\int_a^b{\frac{\textup d\gamma_y}{\textup d t}\textup d t}+v_z\int_a^b{\frac{\textup d\gamma_z}{\textup d t}\textup d t}\\<br /> =v_x(\gamma_x(b)-\gamma_x(a))+v_y(\gamma_y(b)-\gamma_y(a))+v_y(\gamma_y(b)-\gamma_y(a))\\<br /> =\mathbf v\cdot(\mathbf{Q-P})$$

I'm not sure how to turn this into what is given. I'm not even sure I know what the left hand side of the given identity means. Is it the same thing as
$$\mathbf{P}(\mathbf Q}\cdot \mathbf v)$$
Any help would be appreciated. Thanks in advance :)

 

[ballzac]

By the way, $$v$$ is assumed to be a unit vector.

 

[iamalexalright]

hopefully somebody with more authority can correct me but...

I'm very certain that PQ is defined as (Q - P) so you are correct (didn't look through the whole derivation though)

 

[ballzac]

Oh, I get it. It's the vector from P to Q. Thanks heaps for that. I was so busy trying to figure out where I'd gone wrong that it never occurred to me that I may have it right. :)