Worked examples — Skew lines — shortest distance

This page is a complete gym. The parent note built the one formula $d=\frac{|({\vec{a}}_2-{\vec{a}}_1)\cdot ({\vec{b}}_1\times {\vec{b}}_2)|}{|{\vec{b}}_1\times {\vec{b}}_2|}.$ Here we throw every kind of input at it — clean skew lines, lines that secretly intersect, parallel lines where the formula explodes, one direction vector that is a coordinate axis, a degenerate zero-gap case, a real-world word problem, and an exam twist with an unknown constant. Before we start, two pieces of vocabulary so nothing is used before it is earned.

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The scenario matrix

Every problem this topic can throw is exactly one row below. The examples that follow are labelled with the cell they hit, so together they cover the whole table.

#	Case class	What is special about the input	What could go wrong
A	Generic skew	${\vec{b}}_1\nparallel {\vec{b}}_2$, triple product $\ne 0$	pick wrong points → panic (you shouldn't)
B	Secretly intersecting	triple product $=0$, not parallel	miscalling it "skew"
C	Parallel (distinct) lines	${\vec{b}}_2=k{\vec{b}}_1$ ⇒ cross $=\vec{0}$	dividing $0/0$
C'	Coincident lines	parallel AND share every point	both top and bottom vanish, $0/0$
D	Axis-aligned direction	a $\vec{b}$ is $\hat{i},\hat{j}$, or $\hat{k}$	forgetting a component is $0$, not "missing"
E	Degenerate gap (${\vec{a}}_1={\vec{a}}_2$)	gap arrow is $\vec{0}$ ⇒ they touch	is $d=0$ automatic? yes
F	Sign of triple product negative	numerator comes out $<0$	forgetting the $
G	Real-world word problem	two flight paths / girders	translating words → vectors
H	Exam twist: unknown constant	find $k$ so lines intersect	solving triple product $=0$

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A — Generic skew (with the perpendicular picture)

Forecast: guess whether $d$ is bigger or smaller than $1$ before computing. (The lines aren't far apart, so it's small.)

The figure below is worth reading carefully before the algebra. Notice three features: the magenta line is $L_1$, the violet line is $L_2$, and the short orange stick is the one segment joining them at a right angle to both. The dot $P$ (magenta) is the closest point on $L_1$; the dot $Q$ (violet) is the closest point on $L_2$. The printed label on the orange stick is the very number $d$ we are about to compute by hand — so once you finish step 5, come back and confirm the two agree.

1. Gap arrow ${\vec{a}}_2-{\vec{a}}_1=(1-2)\hat{i}+(-1-1)\hat{j}+(1-(-1))\hat{k}=-\hat{i}-2\hat{j}+2\hat{k}$. Why this step? This is the arrow (from a point of $L_1$ to a point of $L_2$) that we will project onto the perpendicular direction.
2. Cross product (perpendicular-to-both direction): =\hat i((-1)(-1)-2\cdot2)-\hat j(3\cdot(-1)-2\cdot1)+\hat k(3\cdot2-(-1)\cdot1).$$ $=\hat i(1-4)-\hat j(-3-2)+\hat k(6+1)=-3\hat i+5\hat j+7\hat k.$ *Why this step?* By definition this arrow is $\perp$ both wires — the direction of the orange stick in the figure.
3. Length of the cross $|{\vec{b}}_1\times {\vec{b}}_2|=\sqrt{9+25+49}=\sqrt{83}$. Why this step? We divide by this to turn a raw projection into a true length.
4. Triple product (dot the gap into the cross): $(-1)(-3)+(-2)(5)+(2)(7)=3-10+14=7$. Why this step? The dot picks out how much of the gap lies along the perpendicular.
5. Distance $d=\frac{|7|}{\sqrt{83}}=\frac{7}{\sqrt{83}}\approx 0.768$.

Verify: $0.768<1$ — matches the forecast, matches the orange-stick label in the figure, and $d>0$ so the lines truly don't meet. Sanity: units are the same as the coordinates (pure length). ✓

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B — Secretly intersecting (triple product $=0$)

Forecast: the directions differ, so if $d\ne 0$ they're skew. Watch the numerator.

1. Directions not parallel? $(2,-1,1)$ is not a multiple of $(1,1,-1)$ (ratios $2,-1,-1$ differ). Good, not parallel. Why this step? Parallel would send us to the C-formula; we must rule it out first.
2. Gap arrow ${\vec{a}}_2-{\vec{a}}_1=(3-1,0-1,2-1)=(2,-1,1)$. Why this step? Same as always — the arrow whose along-perpendicular component becomes the distance.
3. Cross product $\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& -1& 1\\ 1& 1& -1\end{array}\right|=\hat{i}(1-1)-\hat{j}(-2-1)+\hat{k}(2+1)=0\hat{i}+3\hat{j}+3\hat{k}.$ Why this step? This is the common-perpendicular direction; it is nonzero here, confirming the lines aren't parallel so the skew formula is legal.
4. Triple product $(2)(0)+(-1)(3)+(1)(3)=0-3+3=0$. Why this step? A zero numerator (with nonzero denominator) forces $d=0$: the three vectors are coplanar ⇒ the lines intersect.
5. Find the meeting point. Set $(1+2\lambda ,1-\lambda ,1+\lambda )=(3+\mu ,\mu ,2-\mu )$. Why this step? $d=0$ tells us they meet; to find where, we equate the two parametrised points and solve for the dials $\lambda ,\mu$. From $x$: $1+2\lambda =3+\mu$. From $y$: $1-\lambda =\mu$. Substitute: $1+2\lambda =3+1-\lambda \Rightarrow 3\lambda =3\Rightarrow \lambda =1$, $\mu =0$. Check $z$: $1+1=2$ and $2-0=2$. ✓ Point $=(3,0,2)$.

Verify: put $\lambda =1$ into $L_1$: $(1+2,1-1,1+1)=(3,0,2)$; put $\mu =0$ into $L_2$: $(3,0,2)$. Identical → they intersect, $d=0$, not skew. ✓

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C — Parallel (distinct) lines (the formula that doesn't explode)

Forecast: notice ${\vec{b}}_2=-2{\vec{b}}_1$ — parallel. The skew formula would give $0/0$, so use the point-to-line version.

1. Confirm parallel ${\vec{b}}_2=(-4,-2,2)=-2(2,1,-1)=-2{\vec{b}}_1$. Take $\vec{b}=(2,1,-1)$. Why this step? If the directions are proportional the cross of directions is $\vec{0}$, breaking the main formula; distance then = perpendicular distance from a point of $L_2$ to $L_1$, using Distance from a Point to a Line (3D).
2. Gap arrow ${\vec{a}}_2-{\vec{a}}_1=(2-1,1-(-2),0-3)=(1,3,-3)$. Why this step? We need one arrow from $L_1$ to $L_2$; crossing it with $\vec{b}$ next will extract the part perpendicular to the common direction.
3. Cross $\vec{b}\times ({\vec{a}}_2-{\vec{a}}_1)=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& 1& -1\\ 1& 3& -3\end{array}\right|=\hat{i}(-3+3)-\hat{j}(-6+1)+\hat{k}(6-1)=0\hat{i}+5\hat{j}+5\hat{k}.$ Why this step? $|\vec{b}\times ({\vec{a}}_2-{\vec{a}}_1)|$ equals $|\vec{b}|$ times the perpendicular gap, so dividing by $|\vec{b}|$ leaves exactly that gap.
4. Distance $d=\frac{|\vec{b}\times ({\vec{a}}_2-{\vec{a}}_1)|}{|\vec{b}|}=\frac{\sqrt{0+25+25}}{\sqrt{4+1+1}}=\frac{\sqrt{50}}{\sqrt{6}}=\frac{5\sqrt{2}}{\sqrt{6}}=\frac{5}{\sqrt{3}}\approx 2.887$. Why this step? This is the final arithmetic that turns the two magnitudes into one clean length.

Verify: $\frac{\sqrt{50}}{\sqrt{6}}=\sqrt{50/6}=\sqrt{8.333}\approx 2.887$. ✓ Positive length, as required.

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C' — Coincident lines (the pure $0/0$ trap)

Forecast: the directions are proportional, so this looks like case C. But watch what happens to the gap arrow — if it also lies along $\vec{b}$, the lines are literally the same line, and $d=0$.

1. Directions proportional? ${\vec{b}}_2=(4,2,-2)=2(2,1,-1)=2{\vec{b}}_1$. Parallel — take $\vec{b}=(2,1,-1)$. Why this step? Proportional directions rule out the main skew formula (cross of directions $=\vec{0}$); we must next check whether the two lines are merely parallel (case C) or the same line.
2. Gap arrow ${\vec{a}}_2-{\vec{a}}_1=(3-1,2-1,0-1)=(2,1,-1)$. Why this step? If this gap is itself a multiple of $\vec{b}$, then ${\vec{a}}_2$ already lies on $L_1$, so the two "lines" trace the identical set of points.
3. Is the gap along $\vec{b}$? $(2,1,-1)=1\cdot (2,1,-1)=\vec{b}$. Yes — proportional. Why this step? Parallel direction and a gap along that direction together mean the lines coincide.
4. Point-to-line cross $\vec{b}\times ({\vec{a}}_2-{\vec{a}}_1)=\vec{b}\times \vec{b}=\vec{0}$. Why this step? The cross of any vector with a parallel vector is zero — so both the main formula ($0/0$) and the parallel formula ($0/|\vec{b}|$) collapse to $0$.
5. Conclusion $d=0$: the lines are coincident (one and the same). Every point of $L_1$ is a point of $L_2$.

Verify: the point ${\vec{a}}_2=(3,2,0)$ should sit on $L_1$. Solve $(1+2\lambda ,1+\lambda ,1-\lambda )=(3,2,0)$: from the first, $\lambda =1$; check $1+1=2$ ✓ and $1-1=0$ ✓. So ${\vec{a}}_2$ is on $L_1$ ⇒ coincident ⇒ $d=0$. The genuine $0/0$ is resolved not by arithmetic but by recognising "same line." ✓

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D — Axis-aligned direction (a component is exactly zero)

Forecast: these are like a lamp-post and a floor rail one metre apart — expect $d=1$.

1. Directions ${\vec{b}}_1=(0,0,1)$, ${\vec{b}}_2=(1,0,0)$. Not parallel (different axes). Treat the zeros as real components, not "blanks." Why this step? A common slip is writing ${\vec{b}}_1=(1)$; it is genuinely $(0,0,1)$, and the zeros must ride through every dot and cross that follows.
2. Gap ${\vec{a}}_2-{\vec{a}}_1=(0-1,1-0,0-0)=(-1,1,0)$. Why this step? Same routine — the arrow from a point of $L_1$ to a point of $L_2$ that we will project.
3. Cross $\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 0& 0& 1\\ 1& 0& 0\end{array}\right|=\hat{i}(0-0)-\hat{j}(0-1)+\hat{k}(0-0)=(0,1,0)$. Why this step? This is the common-perpendicular direction; here it comes out along $\hat{j}$, i.e. the horizontal direction linking the post and the rail.
4. Triple product $(-1)(0)+(1)(1)+(0)(0)=1$; $|{\vec{b}}_1\times {\vec{b}}_2|=1$. Why this step? The dot extracts the along-$\hat{j}$ part of the gap (which is $1$), and the unit-length cross means no rescaling is needed.
5. $d=\frac{|1|}{1}=1$. Why this step? Final division of numerator by denominator gives the pure length.

Verify: the two lines are the $z$-axis-parallel line at $x=1,y=0$ and the $x$-axis-parallel line at $y=1,z=0$; their nearest points $(1,0,0)$ and $(1,1,0)$ are exactly $1$ apart. Matches forecast. ✓

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E — Degenerate gap: the lines share a point

Forecast: both lines start at the same point $(1,1,1)$, so they obviously meet there — $d$ must be $0$.

1. Gap ${\vec{a}}_2-{\vec{a}}_1=(0,0,0)$. Why this step? Whatever we later dot this zero arrow with is $0$, so the numerator vanishes before we even compute the cross — worth spotting early to save work.
2. Triple product $(0,0,0)\cdot (\text{anything})=0$. Why this step? Confirms the numerator is zero regardless of the perpendicular direction; the distance can only be $0$ provided the denominator is nonzero, which we check next.
3. Cross (nonzero, they aren't parallel): $\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 2& 3\\ 2& -1& 1\end{array}\right|=\hat{i}(2+3)-\hat{j}(1-6)+\hat{k}(-1-4)=(5,5,-5)$, length $5\sqrt{3}\ne 0$. Why this step? A nonzero denominator distinguishes this genuine "$0/\text{nonzero}=0$" from the coincident-line "$0/0$" of case C'.
4. $d=\frac{0}{5\sqrt{3}}=0$. Why this step? Clean division confirms the lines touch at exactly one shared point — intersecting, not skew.

Verify: shared point $(1,1,1)$ lies on both lines ($\lambda =0$ and $\mu =0$), so distance is genuinely $0$ — not the $0/0$ trap, because the denominator is nonzero. ✓ This is intersecting, not skew.

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F — Negative triple product (why the absolute value matters)

Forecast: you'll get a negative number in the middle — do not report a negative distance.

1. Gap ${\vec{a}}_2-{\vec{a}}_1=(2-1,1-0,2-0)=(1,1,2)$. Why this step? The usual arrow from a point of $L_1$ to a point of $L_2$, ready to project.
2. Cross $\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 1& 1\\ 1& -1& 1\end{array}\right|=\hat{i}(1+1)-\hat{j}(1-1)+\hat{k}(-1-1)=(2,0,-2)$. Why this step? The common-perpendicular direction; nonzero, so the lines aren't parallel and the main formula applies.
3. Triple product $(1)(2)+(1)(0)+(2)(-2)=2+0-4=-2$. Why this step? Here the gap leans against the cross direction, so the signed projection is negative — a normal, harmless outcome.
4. Length of the cross $|{\vec{b}}_1\times {\vec{b}}_2|=\sqrt{4+0+4}=2\sqrt{2}$. Why this step? We divide by this to convert the projection into a true length.
5. $d=\frac{|-2|}{2\sqrt{2}}=\frac{2}{2\sqrt{2}}=\frac{1}{\sqrt{2}}\approx 0.707$. Why this step? The absolute value discards the sign (orientation), because distance is never negative.

Verify: the sign only told us the gap points opposite to ${\vec{b}}_1\times {\vec{b}}_2$; distance is a length, so we take $|-2|=2$ and get $d=1/\sqrt{2}>0$. ✓

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G — Real-world word problem (two flight paths)

Forecast: the "1 km" vertical offset suggests $d$ is order-1 km.

1. Translate to vectors ${\vec{a}}_1=(0,0,1),{\vec{b}}_1=(1,2,0)$; ${\vec{a}}_2=(3,0,0),{\vec{b}}_2=(0,1,1)$. These come straight from Equation of a Line in 3D. Why this step? A word problem must first become the five ingredients before any formula can act.
2. Gap ${\vec{a}}_2-{\vec{a}}_1=(3,0,-1)$. Why this step? The arrow between the two starting positions, whose perpendicular part is the closest approach.
3. Cross $\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 2& 0\\ 0& 1& 1\end{array}\right|=\hat{i}(2-0)-\hat{j}(1-0)+\hat{k}(1-0)=(2,-1,1)$. Why this step? The direction perpendicular to both flight paths — the line of shortest approach.
4. Triple product $(3)(2)+(0)(-1)+(-1)(1)=6-1=5$; $|{\vec{b}}_1\times {\vec{b}}_2|=\sqrt{4+1+1}=\sqrt{6}$. Why this step? Numerator (signed gap along the perpendicular) over denominator (perpendicular's length) is the whole formula.
5. $d=\frac{5}{\sqrt{6}}\approx 2.041$ km. Why this step? Final division gives the physical closest-approach distance in km.

Verify: $5/\sqrt{6}=\sqrt{25/6}=\sqrt{4.167}\approx 2.04$ km — same order as the offset, and a sensible positive separation for two non-parallel paths. ✓

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H — Exam twist: choose the constant so lines meet

Forecast: "intersect" ⇔ triple product $=0$; solve a linear equation for $k$.

1. Gap ${\vec{a}}_2-{\vec{a}}_1=(1,2,k-1)$. Why this step? Only the gap depends on $k$, so writing it symbolically lets $k$ flow into the triple product.
2. Cross ${\vec{b}}_1\times {\vec{b}}_2=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 2& 3\\ 2& 1& 1\end{array}\right|=\hat{i}(2-3)-\hat{j}(1-6)+\hat{k}(1-4)=(-1,5,-3)$. Why this step? The cross is fixed (independent of $k$), so we compute it once.
3. Set triple product to zero: $(1)(-1)+(2)(5)+(k-1)(-3)=0\Rightarrow -1+10-3k+3=0\Rightarrow 12-3k=0\Rightarrow k=4$. Why this step? Zero triple product + non-parallel directions = coplanar = intersecting (a plane holds both), so this equation pins down $k$.
4. Directions non-parallel? $(1,2,3)$ vs $(2,1,1)$ — yes, not multiples. So $d=0$ genuinely means intersect. Why this step? If they had been parallel, $d=0$ would instead mean coincident (case C'); ruling that out confirms a true crossing.

Verify: with $k=4$, gap $=(1,2,3)$, triple product $=(1)(-1)+(2)(5)+(3)(-3)=-1+10-9=0$. ✓ For any other $k$ the lines are skew.

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Recall Which formula does each case need?

Case C (parallel, distinct) uses the point-to-line formula? ::: Yes — cross of directions is $\vec{0}$, so use $d=\frac{|\vec{b}\times ({\vec{a}}_2-{\vec{a}}_1)|}{|\vec{b}|}$. Case C' (coincident): how do you resolve the $0/0$? ::: Not by arithmetic — recognise that a point of one line lies on the other, so the lines are identical and $d=0$. Zero triple product with non-parallel directions means? ::: The lines intersect (coplanar), $d=0$, not skew. Why report $|-2|$ not $-2$ in case F? ::: Distance is a length $\ge 0$; the sign only records orientation of the gap. In case E, is $0/0$ ever a danger? ::: No — the denominator $|{\vec{b}}_1\times {\vec{b}}_2|$ is nonzero when lines aren't parallel; only the numerator is $0$.