JEE Mains Maths · Three Dimensional Geometry

Shortest Distance, Intersection and Coplanar Lines

Two lines in space: the shortest distance between skew or parallel lines, the common perpendicular, where two lines meet and when they are coplanar, and a third line drawn to meet both.

All Three Dimensional Geometry notes JEE Mains Maths strategy

Why this matters

Sixty-nine PYQs, the largest page in the chapter; more than half are one formula, the shortest distance between skew lines. Five ideas cover the page.

Concept 1 of 5: Shortest distance between skew lines

Two skew lines neither meet nor run parallel. Their common perpendicular has direction

\vec d_1\times\vec d_2

, and the shortest distance is the length of the projection of any joining vector

\vec a_2-\vec a_1

onto that direction. It is zero exactly when the lines meet.

Definition

$SD=\frac{|(\vec a_2-\vec a_1)\cdot(\vec d_1\times\vec d_2)|}{|\vec d_1\times\vec d_2|}$ .
$SD=0$ : the lines intersect. $\vec d_1\times\vec d_2=\vec 0$ : they are parallel (use the next concept).
Rewrite odd forms first: $x+1=2y=-12z$ is $\frac{x+1}{1}=\frac{y}{1/2}=\frac{z}{-1/12}$ , direction $(12,6,-1)$ .
With an unknown in a point or direction, the formula gives an equation in it; an absolute value usually gives two roots.

Shortest distance between skew lines

SD=\frac{\left|(\vec a_2-\vec a_1)\cdot(\vec d_1\times\vec d_2)\right|}{|\vec d_1\times\vec d_2|}

Worked example

Find the shortest distance between

\vec r=(1,0,0)+\lambda(1,1,0)

and

\vec r=(0,0,2)+\mu(0,1,1)

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 6 Apr 2026 Shift 2 · Q64Moderate

Example 1 · Three Dimensional Geometry · Shortest Distance, Intersection and Coplanar Lines

The shortest distance between the lines

\frac{x - 4}{1}=\frac{y - 3}{2}=\frac{z - 2}{- 3}

and

\frac{x + 2}{2}=\frac{y - 6}{4}=\frac{z - 5}{- 5}

is:

Absolute value, and two answers

The formula has

|\dots|

. With an unknown inside,

|expression|=c

gives two cases; questions asking for 'the sum of all values' need both.

Drill 36 more on shortest distance between skew lines

Concept 2 of 5: Distance between parallel lines

When the directions are parallel,

\vec d_1\times\vec d_2=\vec 0

and the skew-line formula breaks. The distance is then just the distance of a point of one line from the other line.

Definition

Parallel lines share a direction $\vec d$ .
$d=\frac{|(\vec a_2-\vec a_1)\times\vec d|}{|\vec d|}$ .

Distance between parallel lines

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

Worked example

Find the distance between

\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{6}

and

\frac{x-3}{2}=\frac{y-2}{3}=\frac{z-3}{6}

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2024 · 8 April 2024 · Q158Moderate

Example 2 · Three Dimensional Geometry · Shortest Distance, Intersection and Coplanar Lines

If the shortest distance between the lines

\frac{x - \lambda}{2}=\frac{y - 4}{3}=\frac{z - 3}{4}

and

\frac{x - 2}{4}=\frac{y - 4}{6}=\frac{z - 7}{8}

\frac{13}{\sqrt{29}}

, then a value of

\lambda

is :

Check for parallel directions first

(2,3,4)

and

(4,6,8)

are parallel. Plugging them into the skew-line formula divides by zero; spot it before computing.

Drill 1 more on distance between parallel lines

Concept 3 of 5: The line of shortest distance: its feet on the two lines

Sometimes the question wants the ends of the common perpendicular, not just its length. Take a general point

P

on the first line and

Q

on the second.

\vec{PQ}

must be perpendicular to both directions: two linear equations in the two parameters. Solve them and you have both feet.

Definition

$P=\vec a_1+\lambda\vec d_1$ , $Q=\vec a_2+\mu\vec d_2$ .
$\vec{PQ}\cdot\vec d_1=0$ and $\vec{PQ}\cdot\vec d_2=0$ .
$PQ$ is the shortest distance, and $\vec{PQ}$ is parallel to $\vec d_1\times\vec d_2$ .
A point lies on the line of shortest distance if it is $P+s\,\vec{PQ}$ for some $s$ .

Conditions for the feet

\vec{PQ}\cdot\vec d_1=0,\qquad \vec{PQ}\cdot\vec d_2=0

Worked example

Find the feet of the common perpendicular of the

x

-axis and the line through

(1,0,1)

with direction

(0,1,0)

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2024 · 29 January 2024 · Q176Moderate

Example 3 · Three Dimensional Geometry · Shortest Distance, Intersection and Coplanar Lines

Let

O

be the origin, and

M

and

N

be the points on the lines

\frac{x- 5}{4}=\frac{y- 4}{1}=\frac{z- 5}{3}

and

\frac{x + 8}{12}=\frac{y + 2}{5}=\frac{z + 11}{9}

respectively such that

MN

is the shortest distance between the given lines. Then

\overrightarrow{OM}\cdot\overrightarrow{ON}

is equal to

Perpendicular to BOTH lines

Making

\vec{PQ}

perpendicular to only one direction gives the foot from a point, not the common perpendicular. Both dot products must vanish.

Drill 3 more on the line of shortest distance: its feet on the two lines

Concept 4 of 5: Where two lines meet, and when they are coplanar

To find where two lines meet, set their general points equal: three equations, two unknowns. Two of them give

\lambda

and

\mu

; the third either checks out (they meet) or fails (they are skew). Equivalently, two lines are coplanar exactly when the joining vector,

\vec d_1

and

\vec d_2

lie in one plane: their scalar triple product is zero.

Definition

Meeting point: solve $\vec a_1+\lambda\vec d_1=\vec a_2+\mu\vec d_2$ using two coordinates; check the third.
Coplanar: $(\vec a_2-\vec a_1)\cdot(\vec d_1\times\vec d_2)=0$ .
Non-parallel coplanar lines meet; parallel lines are always coplanar.
A line given as two planes: find its direction and one point first.

Condition for coplanar lines

(\vec a_2-\vec a_1)\cdot(\vec d_1\times\vec d_2)=0

Worked example

Where do

\frac{x-1}{1}=\frac{y-2}{2}=\frac{z-3}{3}

and

\frac{x-2}{1}=\frac{y-3}{1}=\frac{z-4}{1}

meet?

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2024 · 5 Apr 2024 · Q61Moderate

Example 4 · Three Dimensional Geometry · Shortest Distance, Intersection and Coplanar Lines

Let

d

be the distance of the point of intersection of the lines

\frac{x + 6}{3}=\frac{y}{2}=\frac{z + 1}{1}

and

\frac{x- 7}{4}=\frac{y- 9}{3}=\frac{z- 4}{2}

from the point

(7,8,9)

. Then

d^{2}+ 6

is equal to:

Always check the third equation

Any two coordinates can be solved for

\lambda

and

\mu

. Only if the third coordinate also agrees do the lines actually meet.

Drill 15 more on where two lines meet, and when they are coplanar

Concept 5 of 5: A line that meets two given lines

A third line that meets two given lines is fixed by its two meeting points

P

(on the first) and

Q

(on the second). Write both as general points. If the third line has a given direction,

\vec{PQ}

must be parallel to it; if it passes through a given point

R

, then

R

P

Q

must be collinear. Either way you get equations in the two parameters.

Definition

$P=\vec a_1+\lambda\vec d_1$ , $Q=\vec a_2+\mu\vec d_2$ .
Given direction $\vec v$ : $\vec{PQ}=k\vec v$ (compare ratios).
Through a given point $R$ : $\vec{RP}\parallel\vec{RQ}$ .
Then $PQ$ , the points themselves, or a point on the line follow directly.

Condition for a given direction

Q-P=k\,\vec v

Worked example

A line with direction

(1,1,1)

meets the

x

-axis at

P

and the line through

(0,0,1)

with direction

(0,1,0)

Q

. Find

P

Q

and

PQ

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2024 · 29 January 2024 · Q90Moderate

Example 5 · Three Dimensional Geometry · Shortest Distance, Intersection and Coplanar Lines

A line with direction ratios

2,1,2

meets the lines

x = y + 2 = z

and

x + 2 = 2y = 2z

respectively at the point

P

and

Q

. if the length of the perpendicular from the point

(1,2,12)

to the line

PQ

l

, then

l^{2}

Two parameters, not one

Using the same letter for the parameters on both lines forces the meeting points to correspond, which they do not. Give each line its own parameter.

Drill 9 more on a line that meets two given lines

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (5)

Shortest distance between skew lines
Shortest distance between skew lines
$SD=\frac{\left|(\vec a_2-\vec a_1)\cdot(\vec d_1\times\vec d_2)\right|}{|\vec d_1\times\vec d_2|}$
Distance between parallel lines
Distance between parallel lines
$d=\frac{|(\vec a_2-\vec a_1)\times\vec d|}{|\vec d|}$
The line of shortest distance: its feet on the two lines
Conditions for the feet
$\vec{PQ}\cdot\vec d_1=0,\qquad \vec{PQ}\cdot\vec d_2=0$
Where two lines meet, and when they are coplanar
Condition for coplanar lines
$(\vec a_2-\vec a_1)\cdot(\vec d_1\times\vec d_2)=0$
A line that meets two given lines
Condition for a given direction
$Q-P=k\,\vec v$

Watch out for (5)

Absolute value, and two answers
→ Shortest distance between skew lines
Check for parallel directions first
→ Distance between parallel lines
Perpendicular to BOTH lines
→ The line of shortest distance: its feet on the two lines
Always check the third equation
→ Where two lines meet, and when they are coplanar
Two parameters, not one
→ A line that meets two given lines

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