Shortest distance between two skew lines

At a Glance

• Frequency: 4 sub-parts across 4 of 13 years (2016, 2017, 2018, 2025)
• Priority tier: T3
• Marks (count): 10 (4)
• Average solve time: ~7 min
• Difficulty mix: easy 2, medium 2
• Section: A | Dominant type: computation

Why This Chapter Matters

Shortest-distance questions appear in Section A (Q1 or Q2) in four of the last ten years, always for 10 marks. Every question uses exactly one formula — the scalar triple product divided by the cross-product magnitude. The only variation is how the direction vectors are extracted from the given line forms, and whether the lines turn out to intersect (numerator $= 0$). Master the formula and the sign-extraction step, and these questions take under 7 minutes.

Minimum Theory

Skew lines. Two lines in 3D that are neither parallel nor intersecting are skew. They have a unique shortest connecting segment, perpendicular to both line directions.

Formula. Let line $L_1$ pass through $P_1$ with direction $\mathbf{d}_1$, and $L_2$ through $P_2$ with direction $\mathbf{d}_2$: $d = \frac{\bigl|(\mathbf{P}_2 - \mathbf{P}_1)\cdot(\mathbf{d}_1 \times \mathbf{d}_2)\bigr|}{|\mathbf{d}_1 \times \mathbf{d}_2|}.$

Intersection condition. The lines intersect if and only if the numerator equals zero.

Question Archetypes

shortest-distance (4 question(s); 2016, 2017, 2018, 2025)

Compute $\bigl|(\mathbf{P_2}-\mathbf{P_1})\cdot(\mathbf{d_1}\times\mathbf{d_2})\bigr|\,/\,|\mathbf{d_1}\times\mathbf{d_2}|$

Recognition Cues

• Two lines given in symmetric (standard) form; asked for “shortest distance.”
• May also ask for a parameter $m$ making the lines intersect — set the numerator to zero.
• One line may be given as the intersection of two planes (2018) — extract direction as $\mathbf{n}_1\times\mathbf{n}_2$.

Solution Template

1. Extract $P_1$, $\mathbf{d}_1$ and $P_2$, $\mathbf{d}_2$. Rewrite any "$a-y$" ratio as ”$y-a$ over $-1$” to get the correct sign.
2. Cross product $\mathbf{d}_1\times\mathbf{d}_2$ via $3\times3$ determinant; compute its magnitude.
3. Joining vector $\mathbf{P}_2-\mathbf{P}_1$.
4. Scalar triple product $(\mathbf{P}_2-\mathbf{P}_1)\cdot(\mathbf{d}_1\times\mathbf{d}_2)$.
5. Divide and rationalize any surd.
6. Intersection test (if asked): set numerator $= 0$ and solve for the parameter.

Worked Example 1

2017 Paper 1, 2017-P1-Q1e (10 marks)

Find the shortest distance between $\dfrac{x-3}{3}=\dfrac{8-y}{1}=\dfrac{z-3}{1}$ and $\dfrac{x+3}{-3}=\dfrac{y+7}{2}=\dfrac{z-6}{4}$.

Step 1. Rewrite $\dfrac{8-y}{1}=\dfrac{y-8}{-1}$; the $y$-direction ratio is $-1$. $P_1=(3,8,3),\;\mathbf{d}_1=(3,-1,1);\quad P_2=(-3,-7,6),\;\mathbf{d}_2=(-3,2,4).$

Step 2. $\mathbf{d}_1\times\mathbf{d}_2 = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\3&-1&1\\-3&2&4\end{vmatrix} = (-4-2,\;-12-3,\;6-3) = (-6,-15,3),\quad |\cdot| = \sqrt{270}=3\sqrt{30}.$

Step 3. $\mathbf{P}_2-\mathbf{P}_1 = (-6,-15,3)$.

Step 4. $(-6)(-6)+(-15)(-15)+3\cdot3 = 36+225+9 = 270$.

Step 5. $\boxed{d = \frac{270}{3\sqrt{30}} = \frac{90}{\sqrt{30}} = 3\sqrt{30} \approx 16.43.}$

Worked Example 2

2016 Paper 1, 2016-P1-Q1e (10 marks)

Find the shortest distance between $\dfrac{x-1}{2}=\dfrac{y-2}{4}=\dfrac{z-3}{1}$ and $y-mx=z=0$. For what value of $m$ do they intersect?

Step 1. Line $L_2$: set $x=t$ in $z=0,\,y=mx$ to get point $(0,0,0)$, direction $(1,m,0)$. $P_1=(1,2,3),\;\mathbf{d}_1=(2,4,1);\quad P_2=(0,0,0),\;\mathbf{d}_2=(1,m,0).$

Step 2. $\mathbf{d}_1\times\mathbf{d}_2 = (-m,\;1,\;2m-4),\quad |\cdot|=\sqrt{m^2+1+(2m-4)^2}=\sqrt{5m^2-16m+17}.$

Step 3–4. $\mathbf{P}_2-\mathbf{P}_1=(-1,-2,-3)$; triple product $= m-2-6m+12 = 10-5m$.

Step 5. $\boxed{d = \frac{5|2-m|}{\sqrt{5m^2-16m+17}}.}$

Step 6. Intersection: $10-5m=0 \Rightarrow \boxed{m=2}$.

Common Traps

• Sign trap in symmetric form. $\dfrac{8-y}{1}$ means the $y$-direction ratio is $-1$, not $+1$. Rewrite as $\dfrac{y-8}{-1}$ first — this is the single most common error on this topic.
• Lines as intersections of planes. When a line is given by two plane equations $\mathbf{n}_1\cdot\mathbf{r}=d_1$ and $\mathbf{n}_2\cdot\mathbf{r}=d_2$, extract its direction as $\mathbf{n}_1\times\mathbf{n}_2$ and find a point by setting one coordinate to $0$ and solving $2\times2$.
• Forgetting to rationalize. $90/\sqrt{30}$ should be written as $3\sqrt{30}$; an irrational denominator loses marks.
• Intersection $\ne$ parallel. Numerator $= 0$ means coplanar lines that intersect. Denominator $= 0$ means parallel lines — a different (degenerate) case requiring the distance between two parallel lines formula.

Marks-Aware Writing

Write the cross product as a displayed $3\times3$ determinant, state each factor of the formula explicitly, and show the final rationalization. The formula itself earns 2 marks; working out each intermediate vector earns 3–4 more; the final division earns the last marks. Skipping the determinant display and just writing the answer costs 4–5 marks.

Practice Set

• 2018-P1-Q2d (12 m) — — Hint: the first line is the intersection of two planes; its direction is $\mathbf{n}_1\times\mathbf{n}_2$; find a point by setting $z=0$.
• 2025-P1-Q2c-ii (10 m) — — Hint: same line pair as 2017; direction signs are explicit in this version.