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The determinant is $z\cdot z^3-(-i)(-i)=z^4+1$. Thus $z^4=-1=e^{i(\pi +2\pi n)}$, and its four polynomial roots areIn Cartesian form these are $(1+i)/\sqrt{2}$, $(-1+i)/\sqrt{2}$, $(-1-i)/\sqrt{2}$ and $(1-i)/\sqrt{2}$. Their moduli are one and their complex arguments are equally spaced by $\pi /2$. The Argand diagram is therefore a square centered at the origin, with sides parallel to the coordinate axes.

The three vectors are coplanar precisely when their scalar triple product vanishes:This includes linearly dependent or zero vectors. It expresses that the determinant with those vectors as columns is zero.

For the two moving particles, distinguish closest approach at a common time from the geometric distance between their supporting lines. For fixed reference times, putTheir relative position is ${\mathbf{d}}_0+t\mathbf{v}$. If $\mathbf{v}\ne 0$, the closest approach of two uniformly moving particles occurs atIf $\mathbf{v}=0$, the distance is constant. If only $t\ge 0$ is permitted, replace $t_{\ast }$ by $\max (0,t_{\ast })$ before evaluating the distance.

Now suppose the velocities are nonparallel, and allow the relative starting-time offset to be chosen so that the particles reach the closest points of the two lines simultaneously. Write $\Delta ={\mathbf{a}}_2-{\mathbf{a}}_1$, $A=|{\mathbf{b}}_1{|}^2$, $B={\mathbf{b}}_1\cdot {\mathbf{b}}_2$, $C=|{\mathbf{b}}_2{|}^2$ and $D=AC-B^2>0$. The independent line parameters $s_1,s_2$ at the closest points satisfyThese equations make the connecting vector perpendicular to both velocities. Solving the two-by-two linear system givesThe corresponding distance between skew lines is $|\Delta \cdot ({\mathbf{b}}_1\times {\mathbf{b}}_2)|/|{\mathbf{b}}_1\times {\mathbf{b}}_2|$. Simultaneous arrival requires $t_1+s_1=t_2+s_2$, so $t_1-t_2=s_2-s_1$. Since the vector triple product givesthis condition is preciselyThus the printed formula has a valid interpretation as a synchronization condition for attaining the line-distance lower bound. It is not an identity that must hold for arbitrary fixed reference times at the time of closest approach.

For a counterexample to that stronger reading, take ${\mathbf{a}}_1=0$, ${\mathbf{a}}_2=(1,0,1)$, ${\mathbf{b}}_1=(1,0,0)$, ${\mathbf{b}}_2=(0,1,0)$ and $t_1=t_2=0$. The actual squared distance is $2(t-1/2{)}^2+3/2$, minimized at $t=1/2$. The printed right-hand side is $-1$, whereas the fixed difference $t_1-t_2$ is zero. The line-distance bound is one, but it is not attained by that synchronized motion. Parallel velocities also require separate treatment because the displayed denominator then vanishes.

Eliminate $y$ using $y=1-x^{\prime }$ to get $x^{\prime \prime }+x=-1$. The initial data give $x(0)=1$ and $x^{\prime }(0)=0$, so the simple harmonic motion about the equilibrium $x=-1$ isIndeed, $x^{\prime }+y=1$ and $y^{\prime }-x=1$ identically, and both initial coordinates equal one. This also describes a circle of radius two around $(-1,1)$ in the phase plane.

Work on an interval with $x\ne 0$ and $\cos (y/x)\ne 0$, as required by the original equation. Set $v=y/x$, so $y^{\prime }=v+x{v}^{\prime }$, and obtain $x{v}^{\prime }=3\tan v$. Where $\sin v\ne 0$, separation of variables gives $\cot vdv=3dx/x$ and thereforeThe solutions lost by division by $\tan v$ are $v=k\pi$, and they are included by $C=0$. On a fixed local branch one can writeThe integer $k$ is fixed along a connected regular branch. Implicit differentiation recovers the original equation whenever $x\cos (y/x)\ne 0$. The boundary $|C{x}^3|=1$ is excluded because the original tangent function has a pole there.

The last equation gives $x=1/2-y$, and the first then gives $y+3z=3/2$. Together with the middle equation this reduces the consistency question to $(1-3\alpha )z=\beta -3\alpha /2$.

If $\alpha \ne 1/3$, the linear system has the unique solutionIf $\alpha =1/3$ and $\beta =1/2$, the middle equation repeats one third of $y+3z=3/2$, giving the affine line of solutionsIf $\alpha =1/3$ and $\beta \ne 1/2$, the two equations contradict each other and there is no solution. These are three possible types of solution set, rather than three individual solutions for one fixed choice of parameters.