Hooke's Law vector form

I'm studying spring model. There is a suggested equation (Hooke's Law vector form) But, I couldn't understand how to derive that equation.

I'm reading 'Computer Animation Algorithms and Techniques Third Edition. In the page 241, (7.88) The equation is as follows:

($|v_{1}^* - v_{2}^*|$: the rest length)

($|v_{1} - v_{2}|$: the current length)

$F_s = (\frac{k_s|v_1 -v_2| - |v_{1}^* - v_{2}^*|}{{|v_{1}^* - v_{2}^*|}})\frac{(v_1 -v_2)}{|v_1 -v_2|}$ // I think the $k_s$ is typo.

(I referred to http://www2.cs.uregina.ca/~anima/408/Notes/Cloth/Cloth.htm)

So, maybe $F_s = k_s(\frac{|v_1 -v_2| - |v_{1}^* - v_{2}^*|}{\color{red}{|v_{1}^* - v_{2}^*|}})\frac{(v_1 -v_2)}{|v_1 -v_2|}\cdot\cdot$(1) is correct.

However, I wonder why the red part divides the equation?

If I know correctly about Hooke's Law, $F=-k_s(current_{lenght} - rest_{length})\cdot\cdot$(2).

It means that the red part should be removed like that $F_s = k_s(|v_1 -v_2| - |v_{1}^* - v_{2}^*|)\frac{(v_1 -v_2)}{|v_1 -v_2|}\cdot\cdot$(3).

For example, in 3D space, there is a spring that has the spring term as $k_s$.

The rest length is '3', so let's assume the left rest point, $v_2^*$, is fixed at (0,0,0) and the right rest point, $v_1^*$, is (3,0,0).

And then we moved $v_1^*$ to $v_1 $=(5, 0, 0) with some force. We can calculate the force with (2), and the force is $-2k_s = -k_s(|(5,0,0) - (0,0,0)| - |(3,0,0) - (0,0,0)|)$

But if we use equation (1), the result is different.

$F_s = -k_s(\frac{|v_1 -v_2| - |v_{1}^* - v_{2}^*|}{{|v_{1}^* - v_{2}^*|}})\frac{(v_1 -v_2)}{|v_1 -v_2|} \\\quad= -k_s(\frac{|(5,0,0) -(0,0,0)| - |(3,0,0) - (0,0,0)|}{{|(3,0,0) - (0,0,0)|}})\frac{((5,0,0) -(0,0,0))}{|(5,0,0) -(0,0,0)|} \\\quad= -\frac{2}{3}k_s(1,0,0)$

The force amount of $-\frac{2}{3}k_s(1,0,0)$ is totally different from $-2k_s$

Could you explain what I'm wrong?