Problem source

The edges $OA,OB,OC$ of a rigid cube are taken as coordinate axes and $O',A',B',C'$ are the vertices diagonally opposite $O,A,B,C,$ respectively. The four forces acting on the cube are 
	\[
	\begin{pmatrix}\alpha\\
	\beta\\
	\gamma
	\end{pmatrix}\mbox{ at }O\ (0,0,0),\ \begin{pmatrix}\lambda\\
	0\\
	1
	\end{pmatrix}\mbox{ at }O'\ (a,a,a),\ \begin{pmatrix}-1\\
	0\\
	2
	\end{pmatrix}\mbox{ at }B\ (0,a,0),\ \mbox{ and }\begin{pmatrix}1\\
	\mu\\
	\nu
	\end{pmatrix}\mbox{ at }B'\ (a,0,a).
	\]
	The moment of the system about $O$ is zero: find $\lambda,\mu$ and $\nu$.
\begin{questionparts}
\item  Given that $\alpha=\beta=\gamma=0,$ find the system consisting of a single force at $B$ together with a couple which is equivalent to the given system. 
\item Given that $\alpha=2,\beta=3$ and $\gamma=2,$ find the equation of the locus about each point of which the moment of the system is zero. Find the number of units of work done on the cube when it moves (without rotation) a distance in the direction of this line under the action of the given forces only.  
\end{questionparts}

Solution source

\begin{align*}
&&\mathbf{M} &= \begin{pmatrix}\lambda \\ 0\\ 1 \end{pmatrix} \times 
       \begin{pmatrix}a\\ a \\ a \end{pmatrix} 
       + \begin{pmatrix}-1\\ 0\\ 2 \end{pmatrix} \times 
      \begin{pmatrix} 0 \\ a \\ 0 \end{pmatrix} +
      \begin{pmatrix}1\\ \mu\\ \nu \end{pmatrix} \times  \begin{pmatrix} a \\ 0 \\ a \end{pmatrix} \\
&&&= \begin{pmatrix}
-a \\ -a(\lambda -1) \\ \lambda a
\end{pmatrix} + \begin{pmatrix}
-2a \\ 0 \\ -a
\end{pmatrix} + \begin{pmatrix}
\mu a \\ -a(1-\nu) \\ -a \mu
\end{pmatrix} \\
&&&=a \begin{pmatrix}
\mu - 3 \\ \nu - \lambda \\ \lambda-1-\mu
\end{pmatrix} \\
\Rightarrow && \mu &= 3, \lambda = 4, \nu = 4 
\end{align*}

\begin{questionparts}
\item To find the force we add all vectors:

\begin{align*}
\mathbf{F} &= \begin{pmatrix}\lambda \\ 0\\ 1 \end{pmatrix} 
       + \begin{pmatrix}-1\\ 0\\ 2 \end{pmatrix} +
      \begin{pmatrix}1\\ \mu\\ \nu \end{pmatrix} \\
&= \begin{pmatrix}4\\ 0\\ 1 \end{pmatrix} 
       + \begin{pmatrix}-1\\ 0\\ 2 \end{pmatrix} +
      \begin{pmatrix}1\\ 3 \\  4 \end{pmatrix} \\
&= \begin{pmatrix} 4 \\ 3 \\ 7 \end{pmatrix}
\end{align*}
Since the moment about $O$ is $0$, we have the moment about $B$ is:
\begin{align*}
\mathbf{M} &= \begin{pmatrix} 0 \\ a \\ 0 \end{pmatrix} \times \begin{pmatrix} 4 \\ 3 \\ 7\end{pmatrix} \\
&= \begin{pmatrix} 7a \\ 0 \\ -4a\end{pmatrix}
\end{align*}

\item \begin{align*}
\mathbf{0} &= \mathbf{r} \times  \begin{pmatrix} 4 + 2 \\ 3+3 \\ 7+2 \end{pmatrix} \\
&= \mathbf{r} \times  \begin{pmatrix} 6 \\ 6 \\ 9 \end{pmatrix} \\
\end{align*}

Therefore $\mathbf{r} = t\begin{pmatrix} 2 \\ 2 \\ 3 \end{pmatrix}$ (ie a line)


\begin{align*}
\text{Work done} &= \text{Force} \cdot \text{distance} 
\end{align*}

Since they are parallel, it's just the magnitude of the force, which is $3\sqrt{2^2+2^2+3^2} = 3\sqrt{17}$
\end{questionparts}