Skew Lines
Skew lines are the genuinely 3-dimensional case: two lines that never meet and yet are not parallel. Between any two skew lines there is a unique shortest distance, and finding it requires the projection idea you already know — project the connector vector $\mathbf{a} - \mathbf{c}$ onto a direction $\mathbf{n}$ that is perpendicular to both lines.
The scalar projection of a vector $\mathbf{u}$ onto a unit vector $\hat{\mathbf{n}}$ is $\mathbf{u} \cdot \hat{\mathbf{n}}$. Why is this the right tool for measuring the perpendicular distance between two skew lines? Sketch the geometry before reading on.
Solving the shortest-distance problem rests on two moves: find a vector $\mathbf{n}$ perpendicular to both direction vectors $\mathbf{d}_1$ and $\mathbf{d}_2$, then project the connector $\mathbf{a} - \mathbf{c}$ onto $\hat{\mathbf{n}}$. The absolute value of that scalar projection is the distance — positive because distance is unsigned.
The perpendicular-then-project reading: (1) construct $\mathbf{n}$ with $\mathbf{n} \cdot \mathbf{d}_1 = 0$ and $\mathbf{n} \cdot \mathbf{d}_2 = 0$, (2) normalise $\hat{\mathbf{n}} = \mathbf{n} / |\mathbf{n}|$, (3) take the absolute scalar projection of the connector.
$d = |(\mathbf{a} - \mathbf{c}) \cdot \hat{\mathbf{n}}| = \dfrac{|(\mathbf{a} - \mathbf{c}) \cdot \mathbf{n}|}{|\mathbf{n}|}$
Key facts
- Skew lines are non-parallel and non-intersecting (3D only)
- Shortest distance: $d = \dfrac{|(\mathbf{a} - \mathbf{c}) \cdot \mathbf{n}|}{|\mathbf{n}|}$, where $\mathbf{n} \perp \mathbf{d}_1$ and $\mathbf{n} \perp \mathbf{d}_2$
- Distance is realised along the unique common perpendicular
- NESA MEX-V1: vectors in three dimensions and lines
Concepts
- Why $\mathbf{n}$ must be perpendicular to both directions
- Why the absolute scalar projection is the right measure
- Why the connector $\mathbf{a} - \mathbf{c}$ may be chosen between any pair of points
Skills
- Identify skew lines from parametric form
- Construct $\mathbf{n}$ from two dot-product equations
- Compute the shortest distance via projection
From Lesson 07, two lines are skew iff their direction vectors $\mathbf{d}_1, \mathbf{d}_2$ are non-parallel and the simultaneous parametric system is inconsistent. Once skewness is established, build $\mathbf{n}$ as follows:
- Write $\mathbf{n} = (a, b, c)$ — three unknowns.
- Impose $\mathbf{n} \cdot \mathbf{d}_1 = 0$ and $\mathbf{n} \cdot \mathbf{d}_2 = 0$. That's two linear equations in three unknowns.
- Two equations in three unknowns have a one-parameter family of solutions — pick one component freely (say $c = 1$ or whichever clears fractions) and solve for the other two.
Any non-zero $\mathbf{n}$ from this family points along the unique common perpendicular. The formula $d = |(\mathbf{a} - \mathbf{c}) \cdot \mathbf{n}| / |\mathbf{n}|$ is scale-invariant in $\mathbf{n}$, so the particular choice does not matter.
Skew $\Leftrightarrow$ non-parallel directions $+$ no common point · $\mathbf{n}$ from $\mathbf{n} \cdot \mathbf{d}_1 = 0$ and $\mathbf{n} \cdot \mathbf{d}_2 = 0$ · Two equations, three unknowns — one free parameter · Skew lines lie in parallel planes separated by exactly $d$
Pause — copy the skew-lines criterion (non-parallel directions + no common point), the two dot-product conditions for $\mathbf{n}$, and the geometric picture of parallel planes into your book.
Quick check: A vector $\mathbf{n}$ used in the skew-distance formula must satisfy:
We just saw that skew lines have non-parallel directions and no common point, and that the common perpendicular direction $\mathbf{n}$ satisfies $\mathbf{n}\cdot\mathbf{d}_1 = 0$ and $\mathbf{n}\cdot\mathbf{d}_2 = 0$ (two equations, one free parameter). That raises a question: how do we compute the actual numerical distance between the two skew lines? This card answers it → $d = |(\mathbf{a}-\mathbf{c})\cdot\mathbf{n}|/|\mathbf{n}|$, the projected separation along the common perpendicular.
For skew lines $\ell_1: \mathbf{r}_1 = \mathbf{a} + \lambda \mathbf{d}_1$ and $\ell_2: \mathbf{r}_2 = \mathbf{c} + \mu \mathbf{d}_2$, let $\mathbf{n}$ be any non-zero vector with $\mathbf{n} \perp \mathbf{d}_1$ and $\mathbf{n} \perp \mathbf{d}_2$. Then the shortest distance between the lines is
Why it works: the lines $\ell_1$ and $\ell_2$ lie in parallel planes (both perpendicular to $\mathbf{n}$). The distance between any point on $\ell_1$ and any point on $\ell_2$, when measured along $\hat{\mathbf{n}}$, equals the gap between these planes — independent of which points $\mathbf{a}$ and $\mathbf{c}$ are chosen.
Formula: $d = |(\mathbf{a} - \mathbf{c}) \cdot \mathbf{n}| / |\mathbf{n}|$ · Take absolute value — distance is unsigned · Independent of point choice on each line · $\mathbf{n}$ unique up to scalar — any choice gives the same $d$
Pause — copy the shortest-distance formula $d = |(\mathbf{a}-\mathbf{c})\cdot\mathbf{n}|/|\mathbf{n}|$, the absolute-value rule, and the independence from point choice on each line into your book.
Did you get this? True or false: the value of $d = |(\mathbf{a} - \mathbf{c}) \cdot \mathbf{n}| / |\mathbf{n}|$ depends on which specific points $\mathbf{a}$ and $\mathbf{c}$ on the two skew lines are chosen.
Worked examples · 3 in a row, reveal as you go
Show that $\mathbf{r}_1 = (1,0,2) + \lambda(2,1,-1)$ and $\mathbf{r}_2 = (0,1,3) + \mu(1,-1,2)$ are skew, then find the shortest distance between them.
$\mathbf{n} \cdot (2,1,-1) = 2a + b - c = 0$
$\mathbf{n} \cdot (1,-1,2) = a - b + 2c = 0$.
Adding: $3a + c = 0$, so $c = -3a$. From the second: $b = a + 2c = a - 6a = -5a$. Take $a = 1$: $\mathbf{n} = (1, -5, -3)$.
The line $\ell_1$ is the x-axis, $\mathbf{r}_1 = \lambda(1,0,0)$. The line $\ell_2$ passes through $(0,0,1)$ with direction $(0,1,0)$. Show that they are skew and find the shortest distance.
Find the shortest distance between $\mathbf{r}_1 = (1,2,1) + \lambda(1,0,2)$ and $\mathbf{r}_2 = (0,3,0) + \mu(2,1,0)$, having confirmed the lines are skew.
Fill the gap: The shortest distance between two skew lines equals the scalar projection of the connector $\mathbf{a} - \mathbf{c}$ onto a vector that is to both direction vectors.
Misconceptions to fix · the 3 traps that cost marks
Did you get this? True or false: if $\mathbf{n} = (1, -5, -3)$ satisfies the perpendicularity conditions for two skew lines, then so does $\mathbf{n}' = (-2, 10, 6)$, and using $\mathbf{n}'$ instead gives the same distance.
Activities · practice with the ideas
Show that $\mathbf{r}_1 = (0,0,0) + \lambda(1,0,0)$ and $\mathbf{r}_2 = (0,1,2) + \mu(0,1,0)$ are skew, then find the shortest distance.
Find the shortest distance between $\mathbf{r}_1 = (1,0,0) + \lambda(0,1,1)$ and $\mathbf{r}_2 = (0,1,0) + \mu(1,0,1)$ (skew — confirm).
Identify whether $\mathbf{r}_1 = (1,2,3) + \lambda(1,-1,1)$ and $\mathbf{r}_2 = (2,1,4) + \mu(2,-2,2)$ are skew, parallel, or intersecting. If applicable, give the shortest distance.
For two skew lines, the perpendicular vector you compute has magnitude $|\mathbf{n}| = 7$ and the connector dotted into $\mathbf{n}$ gives $-21$. State the shortest distance and justify the absolute value.
Explain in 2–3 sentences why two parallel lines in $\mathbb{R}^3$ that don't coincide are not classified as skew, even though they never intersect.
Odd one out: Three of these are correct features of $\mathbf{n}$ in the skew-distance formula. Which one is NOT?
Earlier you sketched why the perpendicular segment between two skew lines has the shortest length.
If a connecting segment is not perpendicular to one of the directions, sliding the endpoint along that line will shorten the segment until perpendicularity is achieved. So the global minimum is realised exactly at the common perpendicular — and the projection formula computes its length directly via $d = |(\mathbf{a} - \mathbf{c}) \cdot \hat{\mathbf{n}}|$. The whole technique reduces "3D distance minimisation" to "scalar projection".
Pick your answer, then rate your confidence — that tells the system what to drill next. Each retry pulls a fresh mix from the bank.
Q1. Show that $\mathbf{r}_1 = (0,0,0) + \lambda(1,0,0)$ and $\mathbf{r}_2 = (0,2,3) + \mu(0,1,0)$ are skew, and write down the shortest distance between them (justify briefly). (2 marks)
Q2. Confirm that $\mathbf{r}_1 = (1,2,1) + \lambda(1,0,2)$ and $\mathbf{r}_2 = (0,3,0) + \mu(2,1,0)$ are skew, and find the shortest distance between them. (3 marks)
Q3. Two skew lines have direction vectors $\mathbf{d}_1 = (1, 1, 0)$ and $\mathbf{d}_2 = (1, -1, 1)$, passing through $\mathbf{a} = (0, 0, 0)$ and $\mathbf{c} = (1, 0, 2)$ respectively. Find a perpendicular vector $\mathbf{n}$ and use it to compute the shortest distance between the lines. (3 marks)
Comprehensive answers (click to reveal)
Activity answers:
1. Directions $(1,0,0), (0,1,0)$ not parallel. z gives $0 = 2$ — inconsistent $\Rightarrow$ skew. $\mathbf{n} = (0,0,1)$; connector $(0, -1, -2)$; $d = |-2|/1 = 2$.
2. Directions $(0,1,1), (1,0,1)$ not parallel. Solving: $1 = \mu$ (x), $\lambda = 1$ (y), z: $\lambda = \mu \Rightarrow 1 = 1$ ✓. Actually the lines intersect at $(1, 1, 1)$ — $d = 0$. (Hint to student: a careful skew check before applying the formula prevents this trap.)
3. $(2,-2,2) = 2(1,-1,1)$, parallel. Point $(1,2,3)$ on $\ell_2$: $\mu = -1/2$ in x, check y and z ✓. Coincident; $d = 0$.
4. $d = 21/7 = 3$. Absolute value because scalar projection is signed but distance is unsigned.
5. Skew is defined as non-parallel AND non-intersecting. Parallel lines satisfy non-intersection (when not coincident) but fail non-parallelism, so they form a separate category. The shortest distance is found by the point-to-line formula instead.
Q1 (2 marks): Skew confirmed via inconsistent z [1]. $\mathbf{n} = (0,0,1)$ satisfies both perpendicularity conditions; $d = 3$ [1].
Q2 (3 marks): Skew confirmed (z gives $-5 \neq 0$) [1]. $\mathbf{n} = (-2, 4, 1)$, $|\mathbf{n}| = \sqrt{21}$ [1]. $d = 5/\sqrt{21} = 5\sqrt{21}/21$ [1].
Q3 (3 marks): $\mathbf{n} = (1, -1, -2)$ with $|\mathbf{n}| = \sqrt{6}$ [1]. Connector $(-1, 0, -2)$; dot product $3$ [1]. $d = 3/\sqrt{6} = \sqrt{6}/2$ [1].
Five timed questions on identifying skew lines, constructing the perpendicular vector, and computing the shortest distance. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.
⚔ Enter the arenaClimb platforms by computing skew distances quickly. Lighter alternative to the boss.
Mark lesson as complete
Tick when you've finished the practice and review.