How do I find the minimum magnitude of a force and apply it to get a certain displacement? I was asked this:

A force $F = [3,\,-5,\,2]$ in Newtons pulls a sled through a displacement $s = [5,\,-1,\,3]$ in metres.
  
  
*
  
*How much work is done on the sled by the force?     
  
*What is the minimum
  magnitude of force that could have been applied to the sled to obtain
  the same displacement? Explain your answer.
  

For the first one I have no problem, I figured out the work done is $26$ joules.
For the second I'm a bit unsure though. I get the feeling that the answer is the projection of $\vec F$ on $\vec s$. I'm pretty sure that is the answer, but I can't really explain why.
So I'm wondering if I'm right, and if I am, why that is? If I'm not, what is the right answer; and why is it right?
 A: The problem statement seems to assume that the displacements are
as good as infinitesimal; I will therefore omit some differential symbols
and assume that the sled's motion and the ground are approximately uncurved
in this problem. "Motion" here means displacement as a function of time.
The second question in the problem statement suggests that
different forces $\vec{F}$ can have exactly the same effect
on the motion of the sled.
In other words, there are nonzero force changes
corresponding to zero motion changes.
This implies that the system is constrained.
The constraint that comes to mind for a typical sled on the ground
is that it only follows the direction of its skids.
Since we are assuming that the geometry here is as good as infinitesimal,
this can be simplified to mean that the sled moves only along a fixed axis.
Obviously, the axis determines the direction (up to sign) of $\vec{s}$.
It will be helpful to decompose $$\vec{F} = \vec{F}_s + \vec{F}_o$$
where $\vec{F}_s$ is collinear with $s$ and
$\vec{F}_o$ is orthogonal to $\vec{s}$.
Thus, $\vec{F}_s$ is your projection of $\vec{F}$ on $\vec{s}$.
You can achieve this with the following matrix $\mathbf{S}$:
$$\begin{align}
\text{Let}\quad
\mathbf{S} &= \frac{1}{\vec{s}\cdot\vec{s}}\,\vec{s}\circ\vec{s}
= \frac{1}{35}\begin{pmatrix}5\\-1\\3\end{pmatrix}
\begin{pmatrix}5&-1&3\end{pmatrix}
\\\text{Then}\quad
\vec{F}_s &= \mathbf{S}\cdot\vec{F}
= \frac{\vec{F}\cdot\vec{s}}{\vec{s}\cdot\vec{s}}\,\vec{s}
,\quad\vec{F}_o = \vec{F}-\vec{F}_s = (\mathbf{I}-\mathbf{S})\cdot\vec{F}
\end{align}$$
where $\cdot$ denotes the dot product, $\circ$ denotes the outer product,
and $\mathbf{I}$ is the $3\times3$ identity matrix.
Note that $\mathbf{S}$ is symmetric and positive semidefinite.
Furthermore, $\mathbf{S}$ is idempotent: $\mathbf{S}\cdot\mathbf{S}=\mathbf{S}$.
This allows you to verify that $\vec{F}_s\cdot\vec{F}_o=0$.
Back to the mechanics.
The balance of forces requires that all force components orthogonal to
the sled's acceleration sum to zero.
Due to the constraint imposed,
acceleration can happen only along the axis determined by $\vec{s}$,
therefore $\vec{F}_o$ must be
counteracted by a constraint force
$$\vec{F}_c=-\vec{F}_o$$
The fact that $\vec{F}_o\cdot\vec{s}=0$ also means that $\vec{F}_o$
does not contribute to the work done. The same applies to $\vec{F}_c$
and is consistent with the general fact that constraint forces do not do work:
$$W = \vec{F}\cdot\vec{s} = \vec{F}_s\cdot\vec{s} +
\underbrace{\vec{F}_o\cdot\vec{s}}_0 = \vec{F}_s\cdot\vec{s}$$
In fact, the sled's motion can only be influenced by the sum
$$\vec{F}+\vec{F}_c = \vec{F}_s+
\underbrace{\vec{F}_o+\vec{F}_c}_0 = \vec{F}_s$$
which again points to the conclusion that preserving motion requires
preserving $\vec{F}_s$, whereas $\vec{F}_o$ can be varied arbitrarily
without effect on the sled's motion.
Now, since $\vec{F}_s$ and $\vec{F}_o$ are orthogonal by design,
we can apply the pythagorean theorem:
$$\|\vec{F}\|^2 = \|\vec{F}_s\|^2+\|\vec{F}_o\|^2$$
And this means that minimization of $\|\vec{F}\|$ while preserving $\vec{F}_s$
means minimizing $\|\vec{F}_o\|$. This is achieved with $\vec{F}_o=\vec{0}$.
Thus, the minimal required force is
$$\vec{F} = \vec{F}_s
= \frac{\vec{F}\cdot\vec{s}}{\vec{s}\cdot\vec{s}}\,\vec{s}
= \frac{26\,\mathrm{Nm}}{35\,\mathrm{m}^2}\vec{s}
\quad\text{with}\quad
\|\vec{F}\| = \|\vec{F}_s\| = \frac{26}{\sqrt{35}}\mathrm{N}$$
