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If the relative position is $\mathbf{d}(t)={\mathbf{d}}_0+t\mathbf{v}$ with $\mathbf{v}\ne 0$, completing the square givesThis proves the minimum distance over all real times. If time is restricted to an interval, project $t_{\ast }$ onto that interval; if $\mathbf{v}=0$, the distance is constant. Both particles are evaluated at the same time, unlike the two independent parameters in the distance between skew lines.

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.

The parametric equation of a straight line through $a$ with unit direction $\hat{t}$ isIf the two lines meet, then $a_1+s{\hat{t}}_1=a_2+t{\hat{t}}_2$ for some real $s,t$, so $a_1-a_2$ lies in the linear span of their direction vectors. Its dot product with their cross product is consequently zero:This is not sufficient without excluding parallel lines. For instance $a_1=(0,0,0)$, $a_2=(0,1,0)$ and ${\hat{t}}_1={\hat{t}}_2=(1,0,0)$ satisfy it, but the lines are distinct and never meet. For nonparallel lines the condition is sufficient, because the two directions span the plane perpendicular to their nonzero cross product.

For nonparallel skew lines, put $n=({\hat{t}}_1\times {\hat{t}}_2)/|{\hat{t}}_1\times {\hat{t}}_2|$. Every joining vector is $a_1-a_2+s{\hat{t}}_1-t{\hat{t}}_2$, with fixed component $(a_1-a_2)\cdot n$ along $n$. Its length is at least the absolute value of that component. Its perpendicular component can be canceled by a unique choice of $s,t$, since the two independent directions span $n^{\perp }$. This lower bound is attained, proving the distance between skew lines:

Take points $p=(3,0,4)$ and $q=(-2,3,3)$ with directions $u=(1,3,-1)$ and $v=(0,1,-1)$. A common normal is $u\times v=(-2,1,1)$, of length $\sqrt{6}$. The planes through the two lines with this common normal have separationThis is a lower bound for the distance between skew lines. It is attained at $s=-1$ and $t=-4$: the two points are $(2,-3,5)$ and $(-2,-1,7)$, whose difference $(4,-2,-2)$ is perpendicular to both directions and has length $2\sqrt{6}$. Thus the shortest distance is $\boxed{2\sqrt{6}}$.

Assume the direction vectors are nonzero. The cross product equation is equivalent toso $\mathbf{x}-\mathbf{a}$ is parallel to $\mathbf{u}$ and the locus is the line $\mathbf{x}=\mathbf{a}+\lambda \mathbf{u}$.

For nonparallel directions, ${\mathbf{u}}_1\times {\mathbf{u}}_2$ is normal to both lines. Projecting their displacement onto the corresponding unit vector gives the distance between skew linesIf the directions are parallel, the corresponding formula is

The nonparallel direction unit vectors have nonzero cross product. LetFor any joining vector $\mathbf{s}=\mathbf{c}-\mathbf{a}+\mu \hat{\mathbf{d}}-\lambda \hat{\mathbf{b}}$, its normal component is fixed:Thus its length is at least the absolute value of that component. Since the two line directions span ${\mathbf{n}}^{\perp }$, suitable $\lambda ,\mu$ cancel the entire component in that plane, attaining the bound. Therefore the distance between skew lines, including the zero-distance intersecting case, isThe numerator is a scalar triple product. The nonparallel hypothesis is essential to the denominator and to the spanning argument.

The cross product of the direction vectors is ${\mathbf{n}}_1\times {\mathbf{n}}_2=(1,-1,0)$. It is nonzero, so a common perpendicular direction exists. Projecting ${\mathbf{a}}_2-{\mathbf{a}}_1=(-1,0,1)$ onto its unit vector gives the distance between skew lines:ThusThe numerator is a scalar triple product; movement along either line does not change this perpendicular projection.

The square $d(t{)}^2=(t+1{)}^2+1/2$ is minimized when its first term vanishes. ThereforeThis agrees with the distance between skew lines found in part (b). The corresponding parameter on the first line is its orthogonal projection valueThe closest pair is $(-1/2,-3/2,0)$ and $(-1,-1,0)$; their connecting vector $(-1/2,1/2,0)$ is perpendicular to both direction vectors.

Let $\mathbf{d}={\mathbf{a}}_2-{\mathbf{a}}_1$ and $\mathbf{w}={\mathbf{v}}_1\times {\mathbf{v}}_2$. The assumed linear independence gives $\mathbf{w}\ne 0$. Every displacement joining one line to the other isIts dot product with $\mathbf{w}$ is always $\mathbf{d}\cdot \mathbf{w}$. The component in that direction is therefore fixed, so orthogonal decomposition gives $\Vert \mathbf{s}\Vert \ge |\mathbf{d}\cdot \mathbf{w}|/\Vert \mathbf{w}\Vert$.

The two linearly independent vectors ${\mathbf{v}}_1,{\mathbf{v}}_2$ span the plane perpendicular to $\mathbf{w}$. Suitable ${\lambda }_1,{\lambda }_2$ cancel the component of $\mathbf{d}$ in that plane, attaining the lower bound. Thus the distance between skew lines formula, also valid when the lines intersect, gives