Skew lines — shortest distance — Maths notes

WHAT are we computing?

HOW: Derivation from first principles

We want the shortest segment joining a point $P$ on $L_1$ to a point $Q$ on $L_2$ .

Step 1 — Geometric insight (WHY perpendicular?). Let $\vec{PQ}$ be the joining vector. If $\vec{PQ}$ were not perpendicular to $L_1$ , we could slide $P$ a tiny bit along $L_1$ and shorten $\vec{PQ}$ (drop a perpendicular — the hypotenuse is always longer than a leg). So at the minimum, $\vec{PQ}\perp \vec b_1$ . By the same argument $\vec{PQ}\perp \vec b_2$ . Why this step? The minimum of a distance occurs where moving along either line gives no first-order change — that's exactly perpendicularity.

Step 2 — Direction of the common perpendicular. A vector perpendicular to both $\vec b_1$ and $\vec b_2$ must point along $\vec n = \vec b_1 \times \vec b_2.$ Why this step? The cross product is by definition perpendicular to both inputs. (It's nonzero precisely because the lines aren't parallel — that's where "skew needs $\vec b_1\nparallel\vec b_2$ " is used.)

Step 3 — Project the gap onto $\vec n$ . Take ANY point on each line: $\vec a_1$ and $\vec a_2$ . The vector between them is $\vec a_2 - \vec a_1$ . The shortest distance is the length of the component of this gap along the common perpendicular direction $\hat n = \vec n/|\vec n|$ : $d = \left|\,(\vec a_2 - \vec a_1)\cdot \hat n\,\right| = \frac{\left|(\vec a_2 - \vec a_1)\cdot(\vec b_1\times\vec b_2)\right|}{|\vec b_1\times\vec b_2|}.$ Why this step? No matter which points $\vec a_1,\vec a_2$ we pick, the along- $\hat n$ component is identical, because moving along a line adds only $\vec b$ -vectors, which are $\perp \hat n$ and contribute nothing.

Worked examples

Worked example Example 1 — Standard skew lines

$L_1: \vec r = (\hat i + 2\hat j + \hat k) + \lambda(\hat i - \hat j + \hat k)$ $L_2: \vec r = (2\hat i - \hat j - \hat k) + \mu(2\hat i + \hat j + 2\hat k)$

Step 1. $\vec a_2 - \vec a_1 = (2-1)\hat i + (-1-2)\hat j + (-1-1)\hat k = \hat i - 3\hat j - 2\hat k.$ Why: this is the gap vector we project.

Step 2. $\vec b_1\times\vec b_2 = \begin{vmatrix}\hat i & \hat j & \hat k\\ 1 & -1 & 1\\ 2 & 1 & 2\end{vmatrix} = \hat i(-2-1) - \hat j(2-2) + \hat k(1+2) = -3\hat i + 0\hat j + 3\hat k.$ Why: this is the common perpendicular direction $\vec n$ .

Step 3. $|\vec n| = \sqrt{9+0+9}=\sqrt{18}=3\sqrt2.$ Numerator: $(\hat i - 3\hat j -2\hat k)\cdot(-3\hat i + 0\hat j +3\hat k) = -3 + 0 -6 = -9.$ Why: dot product picks out the component along $\vec n$ .

Result: $d = \dfrac{|-9|}{3\sqrt2} = \dfrac{9}{3\sqrt2} = \dfrac{3}{\sqrt2} = \dfrac{3\sqrt2}{2}.$

Worked example Example 2 — Detecting intersection (d = 0)

$L_1: \vec r = (\hat i + \hat j) + \lambda(2\hat i + \hat j + \hat k)$ , $L_2: \vec r = (2\hat i + \hat j + \hat k) + \mu(\hat i + 2\hat j)$ .

$\vec a_2-\vec a_1 = \hat i + 0\hat j + \hat k.$ $\vec b_1\times\vec b_2 = \begin{vmatrix}\hat i&\hat j&\hat k\\2&1&1\\1&2&0\end{vmatrix}=\hat i(0-2)-\hat j(0-1)+\hat k(4-1)= -2\hat i + \hat j + 3\hat k.$ Triple product $= (1)(-2)+(0)(1)+(1)(3) = 1 \ne 0$ . So $d = \dfrac{|1|}{\sqrt{4+1+9}} = \dfrac{1}{\sqrt{14}}$ — they are skew (not intersecting). Why this check matters: a zero numerator would have meant the four points are coplanar → intersecting.

Worked example Example 3 — Parallel lines

$L_1:\vec r=(\hat i+2\hat j-4\hat k)+\lambda(2\hat i+3\hat j+6\hat k)$ , $L_2:\vec r=(3\hat i+3\hat j-5\hat k)+\mu(4\hat i+6\hat j+12\hat k)$ . Note $\vec b_2 = 2\vec b_1$ → parallel, so use the special formula with $\vec b=(2,3,6)$ . $\vec a_2-\vec a_1 = 2\hat i + \hat j - \hat k.$ $\vec b\times(\vec a_2-\vec a_1)=\begin{vmatrix}\hat i&\hat j&\hat k\\2&3&6\\2&1&-1\end{vmatrix}=\hat i(-3-6)-\hat j(-2-12)+\hat k(2-6)=-9\hat i+14\hat j-4\hat k.$ $|\cdot|=\sqrt{81+196+16}=\sqrt{293}$ , $|\vec b|=\sqrt{49}=7$ . $d=\dfrac{\sqrt{293}}{7}.$

Cartesian form

For $\dfrac{x-x_1}{a_1}=\dfrac{y-y_1}{b_1}=\dfrac{z-z_1}{c_1}$ and $\dfrac{x-x_2}{a_2}=\dfrac{y-y_2}{b_2}=\dfrac{z-z_2}{c_2}$ : $d = \frac{\left|\begin{matrix} x_2-x_1 & y_2-y_1 & z_2-z_1\\ a_1 & b_1 & c_1\\ a_2 & b_2 & c_2\end{matrix}\right|}{\sqrt{(b_1c_2-b_2c_1)^2+(c_1a_2-c_2a_1)^2+(a_1b_2-a_2b_1)^2}}.$ Same formula, just written with components.

Common mistake Steel-manned common errors

(1) Using $\vec a_2-\vec a_1$ with the wrong points... then panicking. Why it feels right: "Surely the answer changes if I start from a different point on the line." Fix: It doesn't — adding a multiple of $\vec b$ to the gap changes nothing after dotting with $\hat n$ (since $\vec b\cdot\hat n=0$ ). You may pick any convenient points.

(2) Forgetting the absolute value / dividing by the wrong magnitude. Why it feels right: triple product looks like a final answer. Fix: distance $\ge 0$ ⇒ take $|\cdot|$ , and divide by $|\vec b_1\times\vec b_2|$ , NOT by $|\vec b_1||\vec b_2|$ .

(3) Applying the cross-product formula to parallel lines. Why it feels right: it's "the formula." Fix: $\vec b_1\times\vec b_2=\vec0$ → $0/0$ . Switch to the parallel-line formula.

(4) Concluding skew just because lines don't share an obvious point. Fix: compute the triple product. $=0$ (and not parallel) ⇒ they intersect, not skew.

Recall Feynman: explain to a 12-year-old

Imagine two straight pieces of wire floating in a room that never touch and aren't pointing the same way. There's exactly one shortest stick you could place connecting them — and that stick pokes both wires at a perfect right angle. To find its length: pick any point on each wire, draw an arrow between them, then measure only the part of that arrow pointing in the "right-angle-to-both" direction. The cross product tells you that direction; the dot product measures how much of the arrow points that way.

Flashcards

When are two lines skew?

Neither parallel nor intersecting (different directions AND no common point).

Shortest-distance formula for skew lines?

d=\dfrac{|(\vec a_2-\vec a_1)\cdot(\vec b_1\times\vec b_2)|}{|\vec b_1\times\vec b_2|}

Why is the shortest segment perpendicular to both lines?

Otherwise you could slide a point along a line and shorten it; min distance ⇒ no first-order change ⇒ perpendicular.

Why does the cross product appear?

\vec b_1\times\vec b_2

is, by definition, perpendicular to both direction vectors — the common-perpendicular direction.

What does numerator

=0

(lines not parallel) imply?

Lines are coplanar and therefore intersect.

Why can we pick any point on each line for

\vec a_1,\vec a_2

Adding a multiple of

\vec b

to the gap is

\perp \hat n

, so it doesn't change the projection onto

\hat n

Formula for shortest distance between PARALLEL lines?

d=\dfrac{|\vec b\times(\vec a_2-\vec a_1)|}{|\vec b|}

The numerator scalar triple product geometrically equals?

The signed volume of the parallelepiped formed by

\vec a_2-\vec a_1,\vec b_1,\vec b_2

Connections

Vector Cross Product — supplies the perpendicular direction.
Scalar Triple Product — numerator; zero ⇔ coplanar.
Distance from a Point to a Line (3D) — the parallel-line case.
Equation of a Line in 3D — vector & Cartesian forms.
Coplanarity of Two Lines — condition $[\vec a_2-\vec a_1,\vec b_1,\vec b_2]=0$ .
Plane containing a line — the common perpendicular lies in related planes.

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, 3D space me do lines teen tarah se behave karti hain: ya to intersect karti hain, ya parallel hoti hain, ya phir skew — matlab na parallel na milti hain, bilkul alag-alag planes me. Skew lines ke beech ki "shortest distance" wo chhota sa segment hai jo dono lines ko ekdum perpendicular (90°) par touch karta hai. Yahi unique stick sabse chhoti hoti hai.

Formula yaad rakhne ka logic simple hai. Dono lines par koi bhi ek-ek point pakdo: $\vec a_1$ aur $\vec a_2$ . Inke beech ka gap $\vec a_2-\vec a_1$ banao. Ab common-perpendicular ki direction chahiye — wo milti hai $\vec b_1\times\vec b_2$ se, kyunki cross product hamesha dono vectors ke perpendicular hota hai. Phir gap ko us perpendicular direction par project kar do (dot product), aur magnitude se divide kar do. Bas: $d=\dfrac{|(\vec a_2-\vec a_1)\cdot(\vec b_1\times\vec b_2)|}{|\vec b_1\times\vec b_2|}$ .

Ek important baat: chahe tum koi bhi point lo line par, answer same aayega, kyunki line ke along chalna means $\vec b$ add karna, aur $\vec b$ to perpendicular direction par zero contribute karta hai. Agar numerator (triple product) zero aa jaye aur lines parallel na ho, to lines actually intersect kar rahi hain — skew nahi. Aur agar lines parallel ho jayein, to cross product zero ho jata hai, tab alag formula use karo: $d=\dfrac{|\vec b\times(\vec a_2-\vec a_1)|}{|\vec b|}$ .

Exam tip: "Cross gives the way, dot gives the stretch, divide to make it pure length" — isi mantra se pura formula recall ho jayega. Absolute value lagana mat bhoolna, distance kabhi negative nahi hoti!

Skew lines — shortest distance

WHAT are we computing?

HOW: Derivation from first principles

Worked examples

Cartesian form

Flashcards

Connections

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — 3D Geometry

Connections