Rectangular Coordinate System in Space

Q: What are the coordinates of a point in space?

The coordinates (x, y, z) of a point are its signed perpendicular distances from the YZ-plane, ZX-plane and XY-plane respectively. You can read them by completing a cuboid with the origin and the point as opposite corners: the three edge lengths along the axes are x, y and z.

Q: How do you find which octant a point lies in?

Look only at the signs of the coordinates. (+, +, +) is octant I, (-, +, +) is II, (-, -, +) is III and (+, -, +) is IV; the same four patterns with z negative give V to VIII. If any coordinate is zero, the point lies on a coordinate plane or axis instead.

Q: What is the distance of a point from the x-axis?

For P(x, y, z) the foot of the perpendicular on the x-axis is (x, 0, 0), so the distance is √(y² + z²). Similarly the distances from the y-axis and z-axis are √(z² + x²) and √(x² + y²). The distance from the YZ-plane, by contrast, is simply |x|.

Q: How do you prove that three points are collinear in 3D?

There are three quick tests. Show that the sum of two distances equals the third, or show that one point divides the join of the other two in a ratio k : 1 that satisfies all three coordinates, or show that the area of the triangle formed by the points, half of |AB × AC|, is zero.

Q: What is the difference between internal and external division?

In internal division the point lies between A and B and the formula uses m + n in the denominator. In external division the point lies on AB produced and the formula uses m - n, with a minus sign in the numerator too. In the k : 1 method, a negative k signals external division.

Q: How do you find the area of a triangle with vertices in 3D?

The fastest method is area = half of |AB × AC| using a 3 × 3 determinant. Alternatively, find the areas Δx, Δy, Δz of the projections on the three coordinate planes and use Δ² = Δx² + Δy² + Δz². Both give the same answer.

Q: How important is this topic for JEE Main?

Distance formula, section formula and centroid are the first step in most JEE Main 3D geometry questions, including line and shortest-distance problems. Direct questions on collinearity, ratio of division and the third vertex from a centroid also appear, so aim to solve them accurately in under two minutes.

Q: What should JEE Advanced aspirants focus on in this concept?

JEE Advanced rarely asks these formulas directly. They appear inside multi-step problems such as a locus, a ratio found from a variable point, or an area or volume found through vectors. Become fluent with the k : 1 method and the cross-product area formula, since both return in lines and planes.

MathsThree Dimensional GeometryFor JEE aspirants

Preview: Rectangular Coordinate System in Space

Rectangular Coordinate System in Space

The rectangular coordinate system in space fixes the position of every point using three mutually perpendicular axes, $O X$ , $O Y$ and $O Z$ , that meet at the origin $O$ . A point is written as an ordered triple $(x, y, z)$ , where each number is its signed perpendicular distance from one coordinate plane. The rectangular coordinate system in space is the base for the distance formula, section formula, centroid and area results of 3D geometry, and it feeds directly into lines and planes for JEE Main and JEE Advanced.

Key Formulas: Quick Reference

Position vector of $P (x, y, z)$ : $r = x \hat{i} + y \hat{j} + z \hat{k}$ and $O P = x^{2} + y^{2} + z^{2}$
Distance: $P Q = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2} + (z_{2} - z_{1})^{2}$
Distance of $P (x, y, z)$ from the x, y and z axes: $y^{2} + z^{2}$ , $z^{2} + x^{2}$ , $x^{2} + y^{2}$
Origin shifted to $O^{'} (x^{'}, y^{'}, z^{'})$ : $x_{1} = x - x^{'}$ , $y_{1} = y - y^{'}$ , $z_{1} = z - z^{'}$
Internal division in $m : n$ : $x = \frac{m x _{2} + n x _{1}}{m + n}$ ; external division: $x = \frac{m x _{2} - n x _{1}}{m - n}$ (same pattern for $y$ and $z$ )
Midpoint: $x = \frac{x _{1} + x _{2}}{2}$ (same pattern for $y$ and $z$ )
Centroid of a triangle: $x = \frac{x _{1} + x _{2} + x _{3}}{3}$ ; of a tetrahedron: $x = \frac{x _{1} + x _{2} + x _{3} + x _{4}}{4}$ (same pattern for $y$ and $z$ )
Area of a triangle: $Δ = \frac{1}{2} ∣ A B \times A C ∣$ and $Δ^{2} = Δ_{x}^{2} + Δ_{y}^{2} + Δ_{z}^{2}$
Volume of a tetrahedron: $V = \frac{1}{6} [A B A C A D]$

1. Coordinate Axes and Coordinate Planes

Let $O$ be any point in space and let $X^{'} O X$ , $Y^{'} O Y$ and $Z^{'} O Z$ be three lines through $O$ that are perpendicular to each other. These lines are called the coordinate axes and $O$ is called the origin. The planes $X Y$ , $Y Z$ and $Z X$ , each containing two of the axes, are called the coordinate planes.

The axes are taken in right-handed order: if the fingers of your right hand curl from

O X

towards

O Y

, the thumb points along

O Z

. Following NCERT, every figure on this page draws the x-axis coming towards you, the y-axis to the right and the z-axis upwards.

Figure 1: The rectangular coordinate system in space. The axes

X^{'} O X

Y^{'} O Y

Z^{'} O Z

meet at the origin

O

and pairs of axes form the XY, YZ and ZX coordinate planes.

2. Octants and the Sign Convention

The three coordinate planes divide space into eight octants. The signs of the three coordinates of a point decide which octant it lies in.

Figure 2: The three coordinate planes split space into eight octants. The sign pattern

(\pm, \pm, \pm)

of a point's coordinates fixes its octant.

Octant	Sign of x	Sign of y	Sign of z	Example point
I	+	+	+	$(2, 3, 1)$
II	−	+	+	$(- 2, 3, 1)$
III	−	−	+	$(- 2, - 3, 1)$
IV	+	−	+	$(2, - 3, 1)$
V	+	+	−	$(2, 3, - 1)$
VI	−	+	−	$(- 2, 3, - 1)$
VII	−	−	−	$(- 2, - 3, - 1)$
VIII	+	−	−	$(2, - 3, - 1)$

A point with a zero coordinate is not inside any octant. With one zero it lies on a coordinate plane, for example

(0, 3, - 2)

is on the YZ-plane; with two zeros it lies on an axis, for example

(0, 0, 5)

is on the z-axis.

3. Coordinates of a Point in Space

Consider a point $P$ in space. Its position is given by the triad $(x, y, z)$ , where $x$ , $y$ and $z$ are the perpendicular distances of $P$ from the YZ-plane, ZX-plane and XY-plane respectively, each taken with a sign.

How to read the coordinates of P

Drop a perpendicular $P L$ from $P$ to the XY-plane. Its algebraic length is the z-coordinate.
From the foot $L$ , drop perpendiculars to the x-axis and the y-axis. Their algebraic lengths give the y-coordinate and the x-coordinate, so $O A = x$ and $O B = y$ .
Equivalently, complete the cuboid with $O$ and $P$ as opposite corners. Then $P M = x$ , $P N = y$ and $P L = z$ .

Figure 3: Coordinates of a point in space.

P (x, y, z)

and

O

are opposite corners of a cuboid, so

P M = x

P N = y

and

P L = z

are the perpendicular distances from the three coordinate planes.

Position vector of a point

If $\hat{i}$ , $\hat{j}$ and $\hat{k}$ are unit vectors along $O X$ , $O Y$ and $O Z$ , the position vector of $P (x, y, z)$ with respect to $O$ is

O P = r = x \hat{i} + y \hat{j} + z \hat{k}, O P = ∣ r ∣ = x^{2} + y^{2} + z^{2}

In short, the point is often written simply as $(x, y, z)$ .

Figure 4: Position vector

r = O P = x \hat{i} + y \hat{j} + z \hat{k}

built head to tail from its three components along the coordinate axes.

Points on the axes and coordinate planes

Set of points	Set notation	General point
x-axis	${(x, y, z) ∣ y = z = 0}$	$(x, 0, 0)$
y-axis	${(x, y, z) ∣ x = z = 0}$	$(0, y, 0)$
z-axis	${(x, y, z) ∣ x = y = 0}$	$(0, 0, z)$
XY-plane	${(x, y, z) ∣ z = 0}$	$(x, y, 0)$
YZ-plane	${(x, y, z) ∣ x = 0}$	$(0, y, z)$
ZX-plane	${(x, y, z) ∣ y = 0}$	$(x, 0, z)$

4. Distance Formula

The distance between the points $P (x_{1}, y_{1}, z_{1})$ and $Q (x_{2}, y_{2}, z_{2})$ is

P Q = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2} + (z_{2} - z_{1})^{2}

Why it works

Build the cuboid whose edges are parallel to the axes and whose diagonal is $P Q$ (Figure 5). In the horizontal face, $P S = ∣ x_{2} - x_{1} ∣$ and $S R = ∣ y_{2} - y_{1} ∣$ , so $P R^{2} = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}$ . The edge $R Q = ∣ z_{2} - z_{1} ∣$ is vertical, so $△ P R Q$ is right-angled at $R$ and $P Q^{2} = P R^{2} + R Q^{2}$ .

Figure 5: Proof of the distance formula. Pythagoras in

△ P S R

and then in

△ P R Q

gives

P Q^{2} = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2} + (z_{2} - z_{1})^{2}

Vector method

If the position vectors of $A$ and $B$ are $O A = x_{1} \hat{i} + y_{1} \hat{j} + z_{1} \hat{k}$ and $O B = x_{2} \hat{i} + y_{2} \hat{j} + z_{2} \hat{k}$ , then $A B = O B - O A$ , so

A B = ∣ O B - O A ∣ = ∣ (x_{2} - x_{1}) \hat{i} + (y_{2} - y_{1}) \hat{j} + (z_{2} - z_{1}) \hat{k} ∣

\Rightarrow A B = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2} + (z_{2} - z_{1})^{2}

Putting

x_{1} = y_{1} = z_{1} = 0

gives the distance of a point from the origin,

O P = x^{2} + y^{2} + z^{2}

Solved Example 1

Show that the points

(0, 7, 10)

(- 1, 6, 6)

and

(- 4, 9, 6)

form a right-angled isosceles triangle.

Solution:

Let $A (0, 7, 10)$ , $B (- 1, 6, 6)$ and $C (- 4, 9, 6)$ .

$A B^{2} = (0 + 1)^{2} + (7 - 6)^{2} + (10 - 6)^{2} = 1 + 1 + 16 = 18$ , so $A B = 32$ .

$B C^{2} = (- 1 + 4)^{2} + (6 - 9)^{2} + (6 - 6)^{2} = 9 + 9 + 0 = 18$ , so $B C = 32$ .

$A C^{2} = (0 + 4)^{2} + (7 - 9)^{2} + (10 - 6)^{2} = 16 + 4 + 16 = 36$ , so $A C = 6$ .

Since $A B^{2} + B C^{2} = 36 = A C^{2}$ , $∠ A B C = 9 0^{\circ}$ . Also $A B = B C$ . Hence $△ A B C$ is right-angled and isosceles.

Solved Example 2

Show, using the distance formula, that the points

(4, 5, - 5)

(0, - 11, 3)

and

(2, - 3, - 1)

are collinear.

Solution:

Let $A (4, 5, - 5)$ , $B (0, - 11, 3)$ and $C (2, - 3, - 1)$ .

$A B = (4 - 0)^{2} + (5 + 11)^{2} + (- 5 - 3)^{2} = 16 + 256 + 64 = 336 = 421$

$B C = (0 - 2)^{2} + (- 11 + 3)^{2} + (3 + 1)^{2} = 4 + 64 + 16 = 84 = 221$

$A C = (4 - 2)^{2} + (5 + 3)^{2} + (- 5 + 1)^{2} = 4 + 64 + 16 = 84 = 221$

Since $B C + A C = 421 = A B$ , the points $A$ , $B$ , $C$ are collinear and $C$ lies between $A$ and $B$ . In fact $B C = A C$ , so $C$ is the midpoint of $A B$ .

Solved Example 3

Find the locus of a point which moves so that the sum of its distances from

A (0, 0, - α)

and

B (0, 0, α)

is constant.

Solution:

Let the moving point be $P (x, y, z)$ and let $P A + P B = 2 a$ (a constant, with $a > α$ ).

x^{2} + y^{2} + (z + α)^{2} = 2 a - x^{2} + y^{2} + (z - α)^{2}

Squaring both sides and cancelling $x^{2} + y^{2} + z^{2} + α^{2}$ :

4 α z - 4 a^{2} = - 4 a x^{2} + y^{2} + (z - α)^{2} \Rightarrow a - \frac{α z}{a} = x^{2} + y^{2} + (z - α)^{2}

Squaring again:

a^{2} - 2 α z + \frac{α ^{2} z ^{2}}{a ^{2}} = x^{2} + y^{2} + z^{2} + α^{2} - 2 α z

\Rightarrow x^{2} + y^{2} + z^{2} (1 - \frac{α ^{2}}{a ^{2}}) = a^{2} - α^{2}

\Rightarrow \frac{x ^{2}}{a ^{2} - α ^{2}} + \frac{y ^{2}}{a ^{2} - α ^{2}} + \frac{z ^{2}}{a ^{2}} = 1

This is the required locus, an ellipsoid with $A$ and $B$ as foci on the z-axis.

5. Distance of a Point from the Coordinate Axes and Planes

Let $P A$ , $P B$ and $P C$ be the perpendiculars from $P (x, y, z)$ to the x-axis, y-axis and z-axis. The feet are $A (x, 0, 0)$ , $B (0, y, 0)$ and $C (0, 0, z)$ , so by the distance formula

P A = y^{2} + z^{2}, P B = z^{2} + x^{2}, P C = x^{2} + y^{2}

The distance from an axis uses the other two coordinates. The distance from a coordinate plane uses only the missing coordinate.

Figure 6: Distance of

P (x, y, z)

from the axes:

P A = y^{2} + z^{2}

to the x-axis,

P B = z^{2} + x^{2}

to the y-axis and

P C = x^{2} + y^{2}

to the z-axis.

Distance of $P (x, y, z)$ from	Value
origin	$x^{2} + y^{2} + z^{2}$
x-axis	$y^{2} + z^{2}$
y-axis	$z^{2} + x^{2}$
z-axis	$x^{2} + y^{2}$
YZ-plane	$∣ x ∣$
ZX-plane	$∣ y ∣$
XY-plane	$∣ z ∣$

Solved Example 4

Find the distance of the point

P (3, - 4, 12)

from the origin, from each coordinate axis and from each coordinate plane.

Solution:

From the origin: $O P = 9 + 16 + 144 = 169 = 13$ .

From the x-axis: $(- 4)^{2} + 1 2^{2} = 160 = 410$ . From the y-axis: $1 2^{2} + 3^{2} = 153 = 317$ . From the z-axis: $3^{2} + (- 4)^{2} = 5$ .

From the YZ-plane: $∣3∣ = 3$ . From the ZX-plane: $∣ - 4∣ = 4$ . From the XY-plane: $∣12∣ = 12$ .

6. Shifting the Origin (Translation of Axes)

Shifting the origin to another point without changing the directions of the axes is called the translation of axes.

Let the origin $O$ be shifted to $O^{'} (x^{'}, y^{'}, z^{'})$ without changing the direction of the axes, and let the new coordinate frame be $O^{'} X^{'} Y^{'} Z^{'}$ . If $P (x, y, z)$ is a point with respect to the frame $O X Y Z$ , then its coordinates with respect to $O^{'} X^{'} Y^{'} Z^{'}$ are $(x_{1}, y_{1}, z_{1})$ , where

x_{1} = x - x^{'}, y_{1} = y - y^{'}, z_{1} = z - z^{'}

Figure 7: Translation of axes. When the origin moves to

O^{'} (x^{'}, y^{'}, z^{'})

with the same axis directions, the new coordinates of

P

are

x_{1} = x - x^{'}

y_{1} = y - y^{'}

z_{1} = z - z^{'}

New coordinates = old coordinates of the point minus the coordinates of the new origin. A shift of origin changes coordinates but never changes distances, angles or areas.

Solved Example 5

If the origin is shifted to

(1, 2, - 3)

without changing the directions of the axes, find the new coordinates of the point

(0, 4, 5)

with respect to the new frame.

Solution:

Here the new origin is $(x^{'}, y^{'}, z^{'}) = (1, 2, - 3)$ and the point is $(x, y, z) = (0, 4, 5)$ .

$x_{1} = x - x^{'} = 0 - 1 = - 1$

$y_{1} = y - y^{'} = 4 - 2 = 2$

$z_{1} = z - z^{'} = 5 + 3 = 8$

Therefore the coordinates of the point with respect to the new frame are $(- 1, 2, 8)$ .

7. Section Formula

Internal division

If $P$ divides the join of $A (x_{1}, y_{1}, z_{1})$ and $B (x_{2}, y_{2}, z_{2})$ internally in the ratio $m : n$ , that is $A P : P B = m : n$ , then

P = (\frac{m x _{2} + n x _{1}}{m + n}, \frac{m y _{2} + n y _{1}}{m + n}, \frac{m z _{2} + n z _{1}}{m + n}), m + n \neq = 0

External division

If $Q$ divides $A B$ externally in the ratio $m : n$ , that is $Q$ lies on $A B$ produced with $A Q : QB = m : n$ , then

Q = (\frac{m x _{2} - n x _{1}}{m - n}, \frac{m y _{2} - n y _{1}}{m - n}, \frac{m z _{2} - n z _{1}}{m - n}), m \neq = n

Figure 8: Section formula.

P

divides

A B

internally and

Q

divides it externally, both in the ratio

m : n

Midpoint and the $k : 1$ form

Putting $m = n$ gives the midpoint of $A B$ :

(\frac{x _{1} + x _{2}}{2}, \frac{y _{1} + y _{2}}{2}, \frac{z _{1} + z _{2}}{2})

When the ratio is unknown, take it as $k : 1$ . The dividing point is

(\frac{k x _{2} + x _{1}}{k + 1}, \frac{k y _{2} + y _{1}}{k + 1}, \frac{k z _{2} + z _{1}}{k + 1})

If $k > 0$ the division is internal; if $k < 0$ it is external in the ratio $∣ k ∣ : 1$ .

Cross pairing:

m

multiplies the coordinates of the far end

B

and

n

multiplies the coordinates of the near end

A

. Swapping them gives the point that divides

A B

n : m

instead.

Solved Example 6

Find the coordinates of the point which divides the line joining the points

(2, 3, 4)

and

(3, - 4, 7)

in the ratio

3 : 5

Solution:

Let the required point be $(x, y, z)$ . Here $m = 3$ , $n = 5$ , $(x_{1}, y_{1}, z_{1}) = (2, 3, 4)$ and $(x_{2}, y_{2}, z_{2}) = (3, - 4, 7)$ .

$x = \frac{3 ( 3 ) + 5 ( 2 )}{3 + 5} = \frac{19}{8}$

$y = \frac{3 ( - 4 ) + 5 ( 3 )}{3 + 5} = \frac{3}{8}$

$z = \frac{3 ( 7 ) + 5 ( 4 )}{3 + 5} = \frac{41}{8}$

Hence the required point is $(\frac{19}{8}, \frac{3}{8}, \frac{41}{8})$ .

Solved Example 7

Prove that the three points

A (3, - 2, 4)

B (1, 1, 1)

and

C (- 1, 4, - 2)

are collinear.

Solution:

The general coordinates of a point $R$ which divides the join of $A (3, - 2, 4)$ and $B (1, 1, 1)$ in the ratio $μ : 1$ are

R = (\frac{μ + 3}{μ + 1}, \frac{μ - 2}{μ + 1}, \frac{μ + 4}{μ + 1}) ...(1)

If $C (- 1, 4, - 2)$ lies on the line $A B$ , then for some value of $μ$ the point $R$ coincides with $C$ . Equating the x-coordinates:

$\frac{μ + 3}{μ + 1} = - 1 \Rightarrow μ + 3 = - μ - 1 \Rightarrow μ = - 2$

Putting $μ = - 2$ in (1), $R = (\frac{1}{- 1}, \frac{- 4}{- 1}, \frac{2}{- 1}) = (- 1, 4, - 2)$ , which are the coordinates of $C$ .

Hence $A$ , $B$ , $C$ are collinear. Since $μ < 0$ , $C$ divides $A B$ externally in the ratio $2 : 1$ .

Solved Example 8

Show that the points

A (2, 3, 4)

B (- 1, 2, - 3)

and

C (- 4, 1, - 10)

are collinear. Also find the ratio in which

C

divides

A B

Solution:

Let $C$ divide $A B$ in the ratio $k : 1$ . Then

C = (\frac{- k + 2}{k + 1}, \frac{2 k + 3}{k + 1}, \frac{- 3 k + 4}{k + 1})

Equating the x-coordinate to $- 4$ : $- k + 2 = - 4 k - 4 \Rightarrow 3 k = - 6 \Rightarrow k = - 2$ .

For $k = - 2$ : $\frac{2 k + 3}{k + 1} = \frac{- 1}{- 1} = 1$ and $\frac{- 3 k + 4}{k + 1} = \frac{10}{- 1} = - 10$ , which match the y and z coordinates of $C$ .

So one value of $k$ satisfies all three coordinates, and $A$ , $B$ , $C$ are collinear. Since $k < 0$ , $C$ divides $A B$ externally in the ratio $2 : 1$ .

Solved Example 9

The vertices of a triangle are

A (5, 4, 6)

B (1, - 1, 3)

and

C (4, 3, 2)

. The internal bisector of

∠ B A C

meets

B C

D

. Find

A D

Solution:

$A B = 4^{2} + 5^{2} + 3^{2} = 50 = 52$ and $A C = 1^{2} + 1^{2} + 4^{2} = 18 = 32$ .

By the angle bisector theorem, $\frac{B D}{D C} = \frac{A B}{A C} = \frac{5}{3}$ , so $D$ divides $B C$ internally in the ratio $5 : 3$ .

D = (\frac{5 ( 4 ) + 3 ( 1 )}{8}, \frac{5 ( 3 ) + 3 ( - 1 )}{8}, \frac{5 ( 2 ) + 3 ( 3 )}{8}) = (\frac{23}{8}, \frac{12}{8}, \frac{19}{8})

A D = (\frac{23}{8} - 5)^{2} + (\frac{12}{8} - 4)^{2} + (\frac{19}{8} - 6)^{2} = \frac{289 + 400 + 841}{64} = \frac{1530}{8}

Hence $A D = \frac{1530}{8} = \frac{3 170}{8}$ units.

Solved Example 10

If the points

P

Q

R

S

are

(4, 7, 8)

(- 1, - 2, 1)

(2, 3, 4)

and

(1, 2, 5)

respectively, show that

P Q

and

R S

intersect. Also find the point of intersection.

Solution:

Let $P Q$ and $R S$ meet at $A$ . If $A$ divides $P Q$ in the ratio $λ : 1$ and $R S$ in the ratio $k : 1$ , then

A = (\frac{- λ + 4}{λ + 1}, \frac{- 2 λ + 7}{λ + 1}, \frac{λ + 8}{λ + 1})

A = (\frac{k + 2}{k + 1}, \frac{2 k + 3}{k + 1}, \frac{5 k + 4}{k + 1})

Equating x-coordinates: $(- λ + 4) (k + 1) = (k + 2) (λ + 1)$ , which simplifies to $2 λk + 3 λ - 3 k - 2 = 0$ ...(i)

Equating y-coordinates: $(- 2 λ + 7) (k + 1) = (2 k + 3) (λ + 1)$ , which simplifies to $4 λk + 5 λ - 5 k - 4 = 0$ ...(ii)

(ii) minus 2 times (i) gives $- λ + k = 0$ , so $λ = k$ . Putting this in (i): $2 λ^{2} - 2 = 0$ , so $λ = \pm 1$ .

$λ = - 1$ is rejected because it makes the denominators zero. So $λ = k = 1$ .

Check with z-coordinates: $\frac{λ + 8}{λ + 1} = \frac{9}{2}$ and $\frac{5 k + 4}{k + 1} = \frac{9}{2}$ . They agree, so the lines do intersect.

With $λ = 1$ , $A = (\frac{3}{2}, \frac{5}{2}, \frac{9}{2})$ . Since $λ = k = 1$ , this point is the common midpoint of $P Q$ and $R S$ .

8. Centroid, Incentre and Centroid of a Tetrahedron

Centroid of a triangle

The centroid $G$ of the triangle with vertices $A (x_{1}, y_{1}, z_{1})$ , $B (x_{2}, y_{2}, z_{2})$ and $C (x_{3}, y_{3}, z_{3})$ is

G = (\frac{x _{1} + x _{2} + x _{3}}{3}, \frac{y _{1} + y _{2} + y _{3}}{3}, \frac{z _{1} + z _{2} + z _{3}}{3})

It divides each median in the ratio $2 : 1$ from the vertex.

Incentre of a triangle

With side lengths $a = B C$ , $b = C A$ and $c = A B$ , the incentre is

I = (\frac{a x _{1} + b x _{2} + c x _{3}}{a + b + c}, \frac{a y _{1} + b y _{2} + c y _{3}}{a + b + c}, \frac{a z _{1} + b z _{2} + c z _{3}}{a + b + c})

Centroid of a tetrahedron

If $A (x_{1}, y_{1}, z_{1})$ , $B (x_{2}, y_{2}, z_{2})$ , $C (x_{3}, y_{3}, z_{3})$ and $D (x_{4}, y_{4}, z_{4})$ are the vertices of a tetrahedron, its centroid is

G = (\frac{x _{1} + x _{2} + x _{3} + x _{4}}{4}, \frac{y _{1} + y _{2} + y _{3} + y _{4}}{4}, \frac{z _{1} + z _{2} + z _{3} + z _{4}}{4})

It lies on the line joining a vertex to the centroid of the opposite face and divides it in the ratio $3 : 1$ .

Figure 9: Centroid of a triangle divides each median in

2 : 1

; centroid of a tetrahedron divides the line from a vertex to the opposite face centroid in

3 : 1

Solved Example 11

Two vertices of a triangle are

(4, - 6, 3)

and

(2, - 2, 1)

and its centroid is

(\frac{8}{3}, - 1, 2)

. Find the third vertex.

Solution:

Let the third vertex be $(x, y, z)$ . Using the centroid formula:

$\frac{4 + 2 + x}{3} = \frac{8}{3} \Rightarrow x = 2$

$\frac{- 6 - 2 + y}{3} = - 1 \Rightarrow y = 5$

$\frac{3 + 1 + z}{3} = 2 \Rightarrow z = 2$

Hence the third vertex is $(2, 5, 2)$ .

Solved Example 12

The centroid of the tetrahedron

O A B C

, where

O

is the origin and

A (a, 2, 3)

B (1, b, 2)

C (2, 1, c)

, is

(1, 2, 3)

. Find the distance of the point

(a, b, c)

from the origin.

Solution:

$\frac{0 + a + 1 + 2}{4} = 1 \Rightarrow a = 1$

$\frac{0 + 2 + b + 1}{4} = 2 \Rightarrow b = 5$

$\frac{0 + 3 + 2 + c}{4} = 3 \Rightarrow c = 7$

Distance of $(1, 5, 7)$ from the origin $= 1 + 25 + 49 = 75 = 53$ units.

9. Area of a Triangle and Volume of a Tetrahedron

Area using projections on the coordinate planes

For the triangle with vertices $A (x_{1}, y_{1}, z_{1})$ , $B (x_{2}, y_{2}, z_{2})$ and $C (x_{3}, y_{3}, z_{3})$ , let $Δ_{x}$ , $Δ_{y}$ and $Δ_{z}$ be the areas of its projections on the YZ-plane, ZX-plane and XY-plane. Each projection is a triangle in two coordinates, so

Δ_{x} = \frac{1}{2} y_{1} z_{1} 1 y_{2} z_{2} 1 y_{3} z_{3} 1

Δ_{y} = \frac{1}{2} x_{1} z_{1} 1 x_{2} z_{2} 1 x_{3} z_{3} 1

Δ_{z} = \frac{1}{2} x_{1} y_{1} 1 x_{2} y_{2} 1 x_{3} y_{3} 1

The area $Δ$ of $△ A B C$ is then given by

Δ^{2} = Δ_{x}^{2} + Δ_{y}^{2} + Δ_{z}^{2}

Figure 10: Area of a triangle in 3D.

Δ_{z}

is the area of the projection on the XY-plane; with

Δ_{x}

Δ_{y}

from the other two planes,

Δ^{2} = Δ_{x}^{2} + Δ_{y}^{2} + Δ_{z}^{2}

Area using vectors

Δ = \frac{1}{2} A B \times A C

A B \times A C = \hat{i} \hat{j} \hat{k} x_{2} - x_{1} y_{2} - y_{1} z_{2} - z_{1} x_{3} - x_{1} y_{3} - y_{1} z_{3} - z_{1}

Three points are collinear exactly when this area is zero, which gives a third collinearity test after the distance and section-ratio methods.

Volume of a tetrahedron

The volume of the tetrahedron with vertices $A (x_{1}, y_{1}, z_{1})$ , $B (x_{2}, y_{2}, z_{2})$ , $C (x_{3}, y_{3}, z_{3})$ and $D (x_{4}, y_{4}, z_{4})$ is $V = \frac{1}{6}$ times the modulus of

x_{1} y_{1} z_{1} 1 x_{2} y_{2} z_{2} 1 x_{3} y_{3} z_{3} 1 x_{4} y_{4} z_{4} 1

Equivalently, $V = \frac{1}{6} [A B A C A D]$ . If $V = 0$ , the four points are coplanar.

Solved Example 13

Find the area of the triangle with vertices

A (1, 2, 1)

B (2, 1, 3)

and

C (- 1, 1, 2)

by both methods.

Solution:

Vector method. $A B = \hat{i} - \hat{j} + 2 \hat{k}$ and $A C = - 2 \hat{i} - \hat{j} + \hat{k}$ .

A B \times A C = \hat{i} \hat{j} \hat{k} 1 - 12 - 2 - 11 = \hat{i} - 5 \hat{j} - 3 \hat{k}

So $Δ = \frac{1}{2} 1 + 25 + 9 = \frac{35}{2}$ square units.

Projection method.

Δ_{x} = \frac{1}{2} 211131121 = \frac{1}{2}

Δ_{y} = \frac{1}{2} 111231 - 121 = \frac{5}{2}

Δ_{z} = \frac{1}{2} 121211 - 111 = - \frac{3}{2}

$Δ^{2} = \frac{1}{4} + \frac{25}{4} + \frac{9}{4} = \frac{35}{4}$ , so $Δ = \frac{35}{2}$ square units. Both methods agree; the sign of $Δ_{z}$ does not matter because it is squared.

Solved Example 14

Find the volume of the tetrahedron with vertices

O (0, 0, 0)

A (1, 2, 1)

B (2, 1, 3)

and

C (- 1, 1, 2)

Solution:

With $O$ as the origin, $O A = (1, 2, 1)$ , $O B = (2, 1, 3)$ and $O C = (- 1, 1, 2)$ .

[O A O B O C] = 121213 - 112 = 1 (2 - 3) - 2 (4 + 3) + 1 (2 + 1) = - 12

$V = \frac{1}{6} ∣ - 12∣ = 2$ cubic units.

Common Mistakes to Avoid

Watch out

Reading $x$ as the distance of the point from the x-axis. It is the distance from the YZ-plane; the distance from the x-axis is $y^{2} + z^{2}$ .
Pairing the ratio wrongly in the section formula. $\frac{m x _{1} + n x _{2}}{m + n}$ gives the point dividing $A B$ in $n : m$ , not $m : n$ .
Rejecting a negative $k$ in the $k : 1$ method. A negative value is valid and means external division.
Adding the new origin's coordinates while shifting the origin. The rule is subtraction: $x_{1} = x - x^{'}$ .
Dropping the $\frac{1}{2}$ in $Δ_{x}$ , $Δ_{y}$ , $Δ_{z}$ , or the $\frac{1}{6}$ and the modulus in the volume of a tetrahedron.
Giving an octant for a point that has a zero coordinate. Such a point lies on a coordinate plane or an axis.
Proving collinearity by showing two distances are equal. You must show that the sum of two distances equals the third.

Frequently Asked Questions

What are the coordinates of a point in space?

The coordinates $(x, y, z)$ of a point are its signed perpendicular distances from the YZ-plane, ZX-plane and XY-plane respectively. You can read them by completing a cuboid with the origin and the point as opposite corners: the three edge lengths along the axes are $x$ , $y$ and $z$ .

How do you find which octant a point lies in?

Look only at the signs of the coordinates. $(+, +, +)$ is octant I, $(-, +, +)$ is II, $(-, -, +)$ is III and $(+, -, +)$ is IV; the same four patterns with $z$ negative give V to VIII. If any coordinate is zero, the point lies on a coordinate plane or axis instead.

What is the distance of a point from the x-axis?

For $P (x, y, z)$ the foot of the perpendicular on the x-axis is $(x, 0, 0)$ , so the distance is $y^{2} + z^{2}$ . Similarly the distances from the y-axis and z-axis are $z^{2} + x^{2}$ and $x^{2} + y^{2}$ . The distance from the YZ-plane, by contrast, is simply $∣ x ∣$ .

How do you prove that three points are collinear in 3D?

There are three quick tests. Show that the sum of two distances equals the third, or show that one point divides the join of the other two in a ratio $k : 1$ that satisfies all three coordinates, or show that the area of the triangle formed by the points, $\frac{1}{2} ∣ A B \times A C ∣$ , is zero.

What is the difference between internal and external division?

In internal division the point lies between $A$ and $B$ and the formula uses $m + n$ in the denominator. In external division the point lies on $A B$ produced and the formula uses $m - n$ , with a minus sign in the numerator too. In the $k : 1$ method, a negative $k$ signals external division.

How do you find the area of a triangle with vertices in 3D?

The fastest method is $Δ = \frac{1}{2} ∣ A B \times A C ∣$ using a $3 \times 3$ determinant. Alternatively, find the areas $Δ_{x}$ , $Δ_{y}$ , $Δ_{z}$ of the projections on the three coordinate planes and use $Δ^{2} = Δ_{x}^{2} + Δ_{y}^{2} + Δ_{z}^{2}$ . Both give the same answer.

How important is this topic for JEE Main?

Distance formula, section formula and centroid are the first step in most JEE Main 3D geometry questions, including line and shortest-distance problems. Direct questions on collinearity, ratio of division and the third vertex from a centroid also appear, so aim to solve them accurately in under two minutes.

What should JEE Advanced aspirants focus on in this concept?

JEE Advanced rarely asks these formulas directly. They appear inside multi-step problems such as a locus, a ratio found from a variable point, or an area or volume found through vectors. Become fluent with the $k : 1$ method and the cross-product area formula, since both return in lines and planes.

Previous year questions on Rectangular Coordinate System in Space

1 question from past papers, each with a step-by-step solution.

JEE Main 2025 Jan 28 Shift 1, Mathematics Q6

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Rectangular Coordinate System in Space

1. Coordinate Axes and Coordinate Planes

2. Octants and the Sign Convention

3. Coordinates of a Point in Space

How to read the coordinates of P

Position vector of a point

Points on the axes and coordinate planes

4. Distance Formula

Why it works

Vector method

5. Distance of a Point from the Coordinate Axes and Planes

6. Shifting the Origin (Translation of Axes)

7. Section Formula

Internal division

External division

Midpoint and the k:1 form

8. Centroid, Incentre and Centroid of a Tetrahedron

Centroid of a triangle

Incentre of a triangle

Centroid of a tetrahedron

9. Area of a Triangle and Volume of a Tetrahedron

Area using projections on the coordinate planes

Area using vectors

Volume of a tetrahedron

Common Mistakes to Avoid

Frequently Asked Questions

What are the coordinates of a point in space?

How do you find which octant a point lies in?

What is the distance of a point from the x-axis?

How do you prove that three points are collinear in 3D?

What is the difference between internal and external division?

How do you find the area of a triangle with vertices in 3D?

How important is this topic for JEE Main?

What should JEE Advanced aspirants focus on in this concept?

Previous year questions on Rectangular Coordinate System in Space

Ready to master Three Dimensional Geometry?

Midpoint and the $k : 1$ form