Skew Lines and Shortest Distance

Master skew lines, shortest distance between skew lines, and shortest distance between parallel lines for JEE Main and Advanced

8 min read

Real-World Hook: Bridges and Power Lines

Look at two highways on different levels of a flyover - they don’t intersect, and they’re not parallel. They’re skew lines! Similarly, power transmission lines crossing at different heights, railway tracks at different elevations in hilly terrain - these are all examples of skew lines. Engineers need to calculate the shortest distance between them for safety clearances.

What are Skew Lines?

Definition

Two lines in 3D space are called skew lines if they:

  1. Do NOT intersect (no common point)
  2. Are NOT parallel
  3. Lie in different planes (not coplanar)

Types of Line Pairs in 3D

TypeIntersectingParallelCoplanar
Intersectingโœ“โœ—โœ“
Parallelโœ—โœ“โœ“
Skewโœ—โœ—โœ—

Memory Trick ๐Ÿง 

“Neither Together, Nor Same Direction”

  • NOT intersecting (not together)
  • NOT parallel (not same direction)
  • Therefore, they must be SKEW!

Shortest Distance Between Two Lines

The shortest distance (SD) between two lines is the length of the common perpendicular - a line segment perpendicular to both lines.

Formula (Vector Form)

If two lines are:

$$\vec{r} = \vec{a_1} + \lambda\vec{b_1} \quad \text{and} \quad \vec{r} = \vec{a_2} + \mu\vec{b_2}$$

Then shortest distance:

$$\boxed{SD = \frac{|(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})|}{|\vec{b_1} \times \vec{b_2}|}}$$

Interactive Demo: Visualize Skew Lines

See how skew lines exist in different planes with shortest distance.

Formula (Cartesian Form)

If two lines are:

$$\frac{x-x_1}{a_1} = \frac{y-y_1}{b_1} = \frac{z-z_1}{c_1} \quad \text{and} \quad \frac{x-x_2}{a_2} = \frac{y-y_2}{b_2} = \frac{z-z_2}{c_2}$$

Then:

$$\boxed{SD = \frac{\begin{vmatrix} x_2-x_1 & y_2-y_1 & z_2-z_1 \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \end{vmatrix}}{\sqrt{(b_1c_2-b_2c_1)^2 + (c_1a_2-c_2a_1)^2 + (a_1b_2-a_2b_1)^2}}}$$

Memory Trick ๐Ÿง 

“Box Product Over Cross Product Magnitude”

  • Numerator: Scalar triple product = (aโ‚‚ - aโ‚) ยท (bโ‚ ร— bโ‚‚)
    • Represents volume of parallelepiped
  • Denominator: |bโ‚ ร— bโ‚‚|
    • Magnitude of cross product
    • Represents area of base

The height of the parallelepiped = Volume/Base Area = SD!

Step-by-Step Method (Cartesian)

Given: Two lines in symmetric form

Step 1: Identify points and direction ratios

  • Line 1: Point (xโ‚, yโ‚, zโ‚), DR (aโ‚, bโ‚, cโ‚)
  • Line 2: Point (xโ‚‚, yโ‚‚, zโ‚‚), DR (aโ‚‚, bโ‚‚, cโ‚‚)

Step 2: Form the numerator determinant

$$\begin{vmatrix} x_2-x_1 & y_2-y_1 & z_2-z_1 \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \end{vmatrix}$$

Step 3: Calculate denominator

$$\sqrt{(b_1c_2-b_2c_1)^2 + (c_1a_2-c_2a_1)^2 + (a_1b_2-a_2b_1)^2}$$

This is |bโ‚ ร— bโ‚‚| where bโ‚ = (aโ‚, bโ‚, cโ‚) and bโ‚‚ = (aโ‚‚, bโ‚‚, cโ‚‚).

Step 4: Calculate SD = |Numerator|/Denominator

Special Cases

Case 1: Parallel Lines

If lines are parallel: bโ‚ ร— bโ‚‚ = 0 (direction vectors parallel)

Formula becomes: 0/0 (indeterminate)

Instead, use:

$$\boxed{SD = \frac{|(\vec{a_2} - \vec{a_1}) \times \vec{b_1}|}{|\vec{b_1}|}}$$

Or equivalently: Distance from a point on line 2 to line 1.

Case 2: Intersecting Lines

If lines intersect: SD = 0

This happens when:

$$(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2}) = 0$$

The numerator of the SD formula becomes zero.

Common Mistake โš ๏ธ

Forgetting absolute value in determinant:

Determinant might be negative, but distance is always positive!

If numerator determinant = -5:
โŒ SD = -5/denominator  [Negative distance!]
โœ“  SD = |-5|/denominator = 5/denominator  [Always positive]

Condition for Lines to be Skew

Two lines are skew if and only if:

  1. Not parallel: bโ‚ ร— bโ‚‚ โ‰  0
  2. Not intersecting: (aโ‚‚ - aโ‚) ยท (bโ‚ ร— bโ‚‚) โ‰  0

Both conditions must be satisfied!

Worked Examples

Example 1: Shortest Distance (JEE Main Level)

Problem: Find shortest distance between lines:

$$\frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4} \quad \text{and} \quad \frac{x-2}{3} = \frac{y-4}{4} = \frac{z-5}{5}$$

Solution: Line 1: Point (1, 2, 3), DR (2, 3, 4) Line 2: Point (2, 4, 5), DR (3, 4, 5)

Numerator:

$$\begin{vmatrix} 2-1 & 4-2 & 5-3 \\ 2 & 3 & 4 \\ 3 & 4 & 5 \end{vmatrix} = \begin{vmatrix} 1 & 2 & 2 \\ 2 & 3 & 4 \\ 3 & 4 & 5 \end{vmatrix}$$

Expanding along R1:

$$= 1(15-16) - 2(10-12) + 2(8-9)$$ $$= 1(-1) - 2(-2) + 2(-1)$$ $$= -1 + 4 - 2 = 1$$

Denominator: |bโ‚ ร— bโ‚‚| where bโ‚ = (2, 3, 4), bโ‚‚ = (3, 4, 5)

$$\vec{b_1} \times \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 4 \\ 3 & 4 & 5 \end{vmatrix}$$ $$= \hat{i}(15-16) - \hat{j}(10-12) + \hat{k}(8-9)$$ $$= -\hat{i} + 2\hat{j} - \hat{k}$$ $$|\vec{b_1} \times \vec{b_2}| = \sqrt{1 + 4 + 1} = \sqrt{6}$$

SD = |1|/โˆš6 = 1/โˆš6 = โˆš6/6 units

Example 2: Checking if Lines Intersect (JEE Main Level)

Problem: Check if the following lines intersect:

$$\frac{x-1}{3} = \frac{y-1}{-1} = \frac{z+1}{0} \quad \text{and} \quad \frac{x-4}{2} = \frac{y}{0} = \frac{z+1}{3}$$

Solution: For intersecting lines, SD = 0, which means numerator determinant = 0.

Line 1: (1, 1, -1), DR (3, -1, 0) Line 2: (4, 0, -1), DR (2, 0, 3)

$$\begin{vmatrix} 4-1 & 0-1 & -1-(-1) \\ 3 & -1 & 0 \\ 2 & 0 & 3 \end{vmatrix} = \begin{vmatrix} 3 & -1 & 0 \\ 3 & -1 & 0 \\ 2 & 0 & 3 \end{vmatrix}$$

R1 and R2 are identical โ†’ Determinant = 0

Therefore, lines intersect (or are parallel - need to check).

Check parallel: 3/2 โ‰  -1/0 (can’t be parallel)

Answer: Lines intersect.

To find point of intersection, solve the line equations simultaneously.

Example 3: Parallel Lines Distance (JEE Advanced Level)

Problem: Find shortest distance between parallel lines:

$$\frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4} \quad \text{and} \quad \frac{x-3}{2} = \frac{y-4}{3} = \frac{z-5}{4}$$

Solution: Lines are parallel (same DR: (2, 3, 4)).

Point on line 1: A(1, 2, 3) Point on line 2: B(3, 4, 5) Direction vector: b = (2, 3, 4)

AB = (3-1, 4-2, 5-3) = (2, 2, 2)

$$\vec{AB} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 2 & 2 \\ 2 & 3 & 4 \end{vmatrix}$$ $$= \hat{i}(8-6) - \hat{j}(8-4) + \hat{k}(6-4)$$ $$= 2\hat{i} - 4\hat{j} + 2\hat{k}$$ $$|\vec{AB} \times \vec{b}| = \sqrt{4 + 16 + 4} = \sqrt{24} = 2\sqrt{6}$$ $$|\vec{b}| = \sqrt{4 + 9 + 16} = \sqrt{29}$$

SD = 2โˆš6/โˆš29 = 2โˆš6/โˆš29 = 2โˆš174/29 units

Example 4: Vector Form (JEE Advanced Level)

Problem: Find shortest distance between:

$$\vec{r} = (\hat{i} + 2\hat{j} + 3\hat{k}) + \lambda(\hat{i} + \hat{j} + \hat{k})$$ $$\vec{r} = (2\hat{i} + 3\hat{j} + 4\hat{k}) + \mu(2\hat{i} + 3\hat{j} + 4\hat{k})$$

Solution: aโ‚ = (1, 2, 3), bโ‚ = (1, 1, 1) aโ‚‚ = (2, 3, 4), bโ‚‚ = (2, 3, 4)

aโ‚‚ - aโ‚ = (1, 1, 1)

$$\vec{b_1} \times \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 2 & 3 & 4 \end{vmatrix}$$ $$= \hat{i}(4-3) - \hat{j}(4-2) + \hat{k}(3-2)$$ $$= \hat{i} - 2\hat{j} + \hat{k}$$ $$(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2}) = (1, 1, 1) \cdot (1, -2, 1)$$ $$= 1 - 2 + 1 = 0$$

Since numerator = 0, SD = 0

Answer: Lines intersect (or are parallel - let’s check).

Check parallel: bโ‚ and bโ‚‚ not proportional (1/2 โ‰  1/3), so not parallel.

Therefore, lines intersect.

Practice Problems

Level 1: JEE Main Basics

  1. Find shortest distance between:

    • (x-1)/1 = (y-2)/2 = (z-3)/3
    • (x-2)/2 = (y-3)/3 = (z-4)/4

  2. Check if lines are skew, intersecting, or parallel:

    • (x-1)/2 = (y-2)/3 = (z-3)/4
    • (x-2)/4 = (y-3)/6 = (z-4)/8

  3. Find distance between parallel lines:

    • (x-1)/1 = (y-1)/1 = (z-1)/1
    • (x-2)/1 = (y-3)/1 = (z-4)/1

Level 2: JEE Main Standard

  1. Find shortest distance between lines:

    • r = (3i + 5j + 7k) + ฮป(i + 2j + k)
    • r = (-i - j - k) + ฮผ(7i - 6j + k)

  2. Show that the lines (x+1)/3 = (y+3)/5 = (z+5)/7 and (x-2)/1 = (y-4)/3 = (z-6)/5 are skew.

  3. Find the shortest distance between the lines:

    • x/1 = y/2 = z/3
    • (x-2)/2 = (y-3)/3 = (z-4)/4

Level 3: JEE Advanced

  1. Find the equation of the line of shortest distance between:

    • (x-1)/1 = (y-2)/2 = (z-3)/3
    • (x-2)/3 = (y-4)/4 = (z-5)/5

  2. Prove that the shortest distance between the lines:

    • r = aโ‚ + ฮปb
    • r = aโ‚‚ + ฮผb (parallel lines) is |(aโ‚‚ - aโ‚) ร— b|/|b|.

  3. Find the points on the two lines where the shortest distance occurs:

    • x/1 = (y-1)/2 = (z-2)/3
    • (x-1)/2 = (y-2)/3 = (z-3)/4

  4. Two skew lines have direction ratios (1, 2, 3) and (2, 3, 4). If they pass through (1, 0, 0) and (0, 1, 0) respectively, find the shortest distance.

Quick Reference Formulas

CaseFormula
Skew lines (vector)SD = |(aโ‚‚-aโ‚)ยท(bโ‚ร—bโ‚‚)|/|bโ‚ร—bโ‚‚|
Skew lines (Cartesian)SD = |Det(xโ‚‚-xโ‚, yโ‚‚-yโ‚, zโ‚‚-zโ‚; aโ‚,bโ‚,cโ‚; aโ‚‚,bโ‚‚,cโ‚‚)|/|bโ‚ร—bโ‚‚|
Parallel linesSD = |(aโ‚‚-aโ‚)ร—b|/|b|
Intersecting linesSD = 0 (numerator = 0)

Algorithm: Finding Shortest Distance

Step 1: Identify line type

  • Check if parallel: DR proportional?
  • Check if intersecting: Numerator determinant = 0?
  • Otherwise: Skew lines

Step 2: Choose appropriate formula

  • Skew: Use full formula
  • Parallel: Use simplified formula
  • Intersecting: SD = 0

Step 3: Calculate

  • Extract points and DRs
  • Form determinant (numerator)
  • Calculate cross product magnitude (denominator)
  • Divide: SD = |Numerator|/Denominator

Step 4: Verify units and reasonableness

Common Exam Patterns

JEE Main typically asks:

  • Direct SD calculation (3-4 marks)
  • Checking if lines are skew (2 marks)
  • SD for parallel lines (2-3 marks)
  • Verifying intersection (2 marks)

JEE Advanced patterns:

  • Finding equation of line of shortest distance
  • Finding points where SD occurs
  • Locus problems involving SD
  • Combined with planes and spheres

Pro Tip:

  • Always check for parallel/intersecting cases first - simpler formulas!
  • For skew lines, vector form is often cleaner than Cartesian
  • Remember absolute value in final answer - distance is always positive!

Last updated: November 2025