JEE Mains Maths · Three Dimensional Geometry

Foot, Image and Distance from a Line

Dropping a perpendicular from a point to a line in space: where it lands (the foot), the mirror image of the point in the line, how far the point is from the line, and triangles that have one side on the line.

All Three Dimensional Geometry notes JEE Mains Maths strategy

Why this matters

Fifty-seven PYQs, the second-largest page in the chapter, and almost all of them start the same way: write the line's general point, then force it to be perpendicular. Four ideas cover the page.

Concept 1 of 4: The foot of the perpendicular on a line

Every point of the line is

A+t\vec d

for one number

t

. The foot

M

of the perpendicular from

P

is the one for which

\vec{PM}

is perpendicular to

\vec d

. That single dot-product equation is linear in

t

, so the foot always comes out in one step.

Definition

General point of the line: $M=A+t\vec d$ .
Condition: $(M-P)\cdot\vec d=0$ .
Solving: $t=\frac{(P-A)\cdot\vec d}{|\vec d|^2}$ .
The same equation, read backwards, finds an unknown in $P$ when the foot is given.

Parameter of the foot

t=\frac{(P-A)\cdot\vec d}{|\vec d|^2},\qquad M=A+t\vec d

Worked example

Find the foot of the perpendicular from

P(1,2,3)

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

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2024 · 30 Jan 2024 · Q70Moderate

Example 1 · Three Dimensional Geometry · Foot, Image and Distance from a Line

Let

(\alpha,\beta,\gamma)

be the foot of perpendicular from the point

(1,2,3)

on the line

\frac{x+3}{5}=\frac{y-1}{2}=\frac{z+4}{3}

. Then

19(\alpha+\beta+\gamma)

is equal to:

Divide by $|\vec d|^2$ , not $|\vec d|$

t

multiplies

\vec d

itself, so the projection must be divided by the SQUARED length. Using

|\vec d|

gives a point that is off the perpendicular.

Drill 15 more on the foot of the perpendicular on a line

Concept 2 of 4: The image of a point in a line

The image

Q

P

in a line is as far beyond the line as

P

is in front of it, along the same perpendicular. So the foot

M

is the midpoint of

PQ

, and

Q=2M-P

. Find the foot, then double it.

Definition

$M$ = foot from $P$ ; image $Q=2M-P$ .
$PQ$ is perpendicular to the line and its midpoint lies on the line.
When the image is given with unknowns, use both facts: the midpoint is on the line, and $\vec{PQ}\cdot\vec d=0$ .
A point on the line is its own image.

Image from the foot

Q=2M-P

Worked example

Find the image of

P(1,2,3)

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

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2025 · 28 Jan 2025 · Q62Moderate

Example 2 · Three Dimensional Geometry · Foot, Image and Distance from a Line

If the image of the point

(4,4,3)

in the line

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

(\alpha,\beta,\gamma)

, then

\alpha + \beta + \gamma

is equal to

$2M-P$ , not $M+P$

The foot is the MIDPOINT of

P

and its image, so

Q=2M-P

. Adding

M

and

P

gives a point nowhere near the line.

Drill 21 more on the image of a point in a line

Concept 3 of 4: Distance of a point from a line

The distance is the length

PM

to the foot. A faster route skips the foot: the parallelogram on

\vec{AP}

and

\vec d

has area

|\vec{AP}\times\vec d|

, and dividing by the base

|\vec d|

leaves its height, which is the distance. When the line is given as two planes, its direction is the cross product of their normals.

Definition

$d=\frac{|\vec{AP}\times\vec d|}{|\vec d|}$ , with $A$ any point of the line.
Or $d^2=|\vec{AP}|^2-\frac{(\vec{AP}\cdot\vec d)^2}{|\vec d|^2}$ .
Line given as two planes: direction $\vec n_1\times\vec n_2$ ; find one common point by setting one coordinate to $0$ .
From the axes: distance of $(x,y,z)$ from the $x$ -axis is $\sqrt{y^2+z^2}$ .

Distance from a line

d=\frac{|\vec{AP}\times\vec d|}{|\vec d|}

Worked example

Find the distance of

(1,2,3)

from the line

x=y=z

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 5 Apr 2026 Shift 1 · Q63Moderate

Example 3 · Three Dimensional Geometry · Foot, Image and Distance from a Line

The square of the distance of the point

P(5,6,7)

from the line

\frac{x- 2}{2}=\frac{y- 5}{3}=\frac{z- 2}{4}

is equal to :

Divide by $|\vec d|$

|\vec{AP}\times\vec d|

is a parallelogram's AREA. Only after dividing by the base length

|\vec d|

is it a distance.

Drill 8 more on distance of a point from a line

Concept 4 of 4: Triangles with a side on the line

When two vertices of a triangle lie on a line and the third,

P

, is off it, the height of the triangle is the distance

PM

to the foot. Points on the line at a given distance

k

from

P

sit symmetrically about

M

, at

\sqrt{k^2-PM^2}

either side. That symmetry also puts their midpoint at

M

, which makes centroids easy.

Definition

Height = $PM$ ; area $=\frac12\cdot$ base $\cdot PM$ .
Points on the line at distance $k$ from $P$ : $M\pm\sqrt{k^2-PM^2}\,\hat d$ .
Those two points have midpoint $M$ , so the centroid with $P$ is $\frac{2M+P}{3}$ .
A right angle at the foot: $PM^2+MQ^2=PQ^2$ .

Half-chord and area

MQ=\sqrt{k^2-PM^2},\qquad [PQR]=\tfrac12\,QR\cdot PM

Worked example

Q

and

R

are the points on the

z

-axis at distance

13

from

P(3,4,0)

. Find the area of

\triangle PQR

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2025 · 24 Jan 2025 · Q58Moderate

Example 4 · Three Dimensional Geometry · Foot, Image and Distance from a Line

Let in a

\bigtriangleup ABC

, the length of the side AC be 6 , the vertex

B

(1,2,3)

and the vertices

A,C

lie on the line

\frac{x - 6}{3}=\frac{y - 7}{2}=\frac{z - 7}{- 2}

. Then the area (in sq. units) of

\bigtriangleup ABC

The height is $PM$ , not $PQ$

The height of the triangle is the perpendicular distance to the line.

PQ

is a slanted side; using it overstates the area.

Drill 9 more on triangles with a side on the line

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 (4)

The foot of the perpendicular on a line
Parameter of the foot
$t=\frac{(P-A)\cdot\vec d}{|\vec d|^2},\qquad M=A+t\vec d$
The image of a point in a line
Image from the foot
$Q=2M-P$
Distance of a point from a line
Distance from a line
$d=\frac{|\vec{AP}\times\vec d|}{|\vec d|}$
Triangles with a side on the line
Half-chord and area
$MQ=\sqrt{k^2-PM^2},\qquad [PQR]=\tfrac12\,QR\cdot PM$

Watch out for (4)

Divide by $|\vec d|^2$ , not $|\vec d|$
→ The foot of the perpendicular on a line
$2M-P$ , not $M+P$
→ The image of a point in a line
Divide by $|\vec d|$
→ Distance of a point from a line
The height is $PM$ , not $PQ$
→ Triangles with a side on the line

Test yourself on Three Dimensional Geometry

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