Introduction to Three Dimensional Geometry: Class 11 Maths · Chapter 11

Three numbers for every point of space, and the distance between any two

Mathematics · Companion notes, simulations and practice · Official NCERT chapter PDF ↗

Published 18 Oct 2026 · Updated 18 Oct 2026

55 min
1 simulation
20 practice questions
17 MCQs

What you will learn in Introduction to Three Dimensional Geometry

Describe the rectangular coordinate system of space: the origin, the three mutually perpendicular axes and the three coordinate planes.
Name the eight octants and decide from the signs of its coordinates whether a point lies in an octant, on a coordinate plane or on an axis.
Find the coordinates of a point in space by dropping perpendiculars, and locate the point described by a given ordered triplet.
Explain why x, y and z are the signed distances of a point from the YZ-, ZX- and XY-planes, and write the general form of points on each axis and each coordinate plane.
Derive the distance formula in space by applying Pythagoras' theorem twice, and use it to find distances between points.
Use distances to test collinearity, to classify triangles and parallelograms, and to find the equation of a set of points defined by a distance condition.
Use the midpoint and centroid of points in space to find missing vertices and the lengths of medians.

Original companion text aligned to NCERT Mathematics Class 11, Chapter 11, pp. 208–216. Not affiliated with or endorsed by NCERT.

11.1 Introduction

Ask a friend at a cricket match where she is sitting and two facts settle it: the stand and the seat number. A drone filming the same match is another story. Saying that it is above the pitch is not enough; you must also say how high it is hovering. Points of a flat surface need two numbers, but points of space need three.

You already know the two-number story. A plane gets two perpendicular number lines, the axes, and every point of it is described by the pair $(x, y)$ of its signed distances from them. In this chapter we add a third axis, perpendicular to both, and rebuild the same tools for space: coordinates, signs, regions and distances.

A small example shows why three numbers are exactly enough. At Diwali a paper lantern (कंदील) hangs near the corner of a room. The floor and the two walls that meet at that corner are three flat surfaces, each perpendicular to the other two. Measure the lantern's distance from each of them, as in Fig. 11.1, and you have pinned it down: no other point of the room has the same three distances.

4/4It hangs 2 m above the floor. The triplet $(1.5, 2.5, 2)$ fixes it.

Fig. 11.1Three perpendicular surfaces, three distances, one position.

A corner of a room drawn at a slant: a back wall, a side wall on the left and the floor, all meeting at one corner. A diamond-shaped lantern hangs on a string. Step 1 shows only the room and the lantern. Step 2 adds a dashed line from the lantern straight to the back wall, labelled 1.5 m. Step 3 adds a dotted line straight to the side wall, labelled 2.5 m. Step 4 adds a solid line straight down to the floor, labelled 2 m, and the triplet (1.5, 2.5, 2) appears at the top.

Mathematics replaces the walls and floor by three mutually perpendicular planes and calls the three distances the coordinates of the point. The rest of the chapter makes this precise: setting up the axes and planes (Section 11.2), reading and writing coordinates (Section 11.3) and measuring the distance between two points (Section 11.4). In Class 12 you will use the same coordinates to describe lines in space.

Quick check

1 of 2

A drone hovers above a playground. Which information fixes its position completely?

Its distances east and north of a corner postIts distances east and north of the post, and its heightIts height above the ground and its speedIts straight-line distance from the corner post

Show answer

Answer (B) Its distances east and north of the post, and its height

Two horizontal distances fix only the spot on the ground below the drone; the height is the third number. A single distance from the post leaves a whole sphere of possible positions, and speed says nothing about position.

11.2 Coordinate Axes and Coordinate Planes in Three Dimensional Space

Begin with a plane, think of it as the floor, and draw two perpendicular lines in it that cross at a point $O$ . Now stand a third line upright at $O$ , perpendicular to the floor. These three lines, each at right angles to the other two, are the coordinate axes: $X^{'} O X$ is the $x$ -axis, $Y^{'} O Y$ the $y$ -axis and $Z^{'} O Z$ the $z$ -axis. Their common point $O$ is the origin (Fig. 11.2).

4/4Along $O X$ , $O Y$ , $O Z$ distances are positive; along $O X^{'}$ , $O Y^{'}$ , $O Z^{'}$ negative.

Fig. 11.2The axes and the three coordinate planes; the positive half of each axis is solid, the negative half dashed.NCERT Fig. 11.1 · p. 209

Step 1 shows a horizontal plane, drawn as a shaded parallelogram, with two perpendicular lines through the point O: X′OX running from the back right to the front left and Y′OY running from left to right. Step 2 adds the vertical line Z′OZ through O, with small right-angle marks. Step 3 shades two vertical planes, labelled YZ-plane and ZX-plane, alongside the horizontal XY-plane. Step 4 puts a plus sign at the solid end of each axis, OX, OY and OZ, and a minus sign at the dashed ends OX′, OY′ and OZ′.

Any two of the axes lie in one plane, so the axes come with three planes, each containing two axes and perpendicular to the third. Seen the other way round, these three planes cut one another in pairs exactly along the axes. The origin with its axes and planes is a rectangular coordinate system for space.

Definition

Coordinate planes

The three planes that each contain two of the axes: the XY-plane (containing the $x$ - and $y$ -axes), the YZ-plane and the ZX-plane. Each is perpendicular to the remaining axis.

ExampleIn a room, take the floor as the XY-plane and two walls that meet it at a corner as the YZ- and ZX-planes.

On paper we draw the XY-plane as a floor seen at a slant, so the $z$ -axis is vertical. Each axis also needs a direction, because every formula in this chapter depends on signs. The convention is fixed once and for all:This makes the system right-handed: curl the fingers of your right hand from $O X$ towards $O Y$ and your thumb points along $O Z$ . Class 12 vectors rely on it.

Front or back ( $x$ ). A point on your side of the YZ-plane, the $O X$ side, gets a positive $x$ ; a point behind it, on the $O X^{'}$ side, a negative one.
Right or left ( $y$ ). The $O Y$ side of the ZX-plane carries positive $y$ , the $O Y^{'}$ side negative $y$ .
Up or down ( $z$ ). Points above the floor, the XY-plane, have $z > 0$ ; points below it, towards $O Z^{'}$ , have $z < 0$ .

Try this

Axes inside a carton

You’ll need

an empty carton or shoebox with one side open
a marker
a ruler
a small eraser

Predict first

The eraser rests on the bottom of the carton. Which of its coordinates must be

0

x

y

z

Turn the open side towards you. Call the inside corner at the back, bottom and left $O$ .
Mark the bottom edge that runs from $O$ towards you as $O X$ , the bottom edge running right as $O Y$ and the upright back edge as $O Z$ .
Label the bottom panel XY, the back panel YZ and the left panel ZX.
Drop the eraser inside. Measure its distance from each panel and write the three numbers as a triplet.

The three coordinate planes slice space into eight pieces, much as two upright cuts and one level cut divide a cake into eight. Each piece is an octant (अष्टांश). An octant is named by the three half-axes along its edges: $X O Y Z$ has edges $O X$ , $O Y$ and $O Z$ , while $X^{'} O Y Z^{'}$ has edges $O X^{'}$ , $O Y$ and $O Z^{'}$ . Octants also carry Roman numerals (Fig. 11.3). Octants I to IV lie above the XY-plane and follow the four quadrants of that plane in their usual order. Octants V to VIII lie below, each directly under the octant numbered four less.

3/3Each octant has its own pattern of signs for $(x, y, z)$ .

Fig. 11.3The three coordinate planes cut space into eight octants: I–IV above the XY-plane, V–VIII below.

A block of eight cubes, separated by gaps along the three coordinate planes, with the x-axis pointing towards the viewer, the y-axis to the right and the z-axis up. Step 1 shows the four upper cubes clearly, labelled I, II, III and IV, while the four lower cubes are faint. Step 2 shows the lower cubes too, labelled V, VI, VII and VIII, each directly below the cube numbered four less. Step 3 adds the signs of x, y and z on every cube, from (+,+,+) on I to (+,−,−) on VIII.

Definition

Octant

Each of the eight regions into which the three coordinate planes divide space. Octants I–IV lie above the XY-plane and V–VIII below it, with octant $n + 4$ directly below octant $n$ .

11.3 Coordinates of a Point in Space

Two questions drive this section. Which three numbers belong to a given point? And which point belongs to three given numbers? Fig. 11.4 answers both for one point $P$ .

Let $M$ be the point of the floor (the XY-plane) straight below or above $P$ , so that $P M$ is perpendicular to the floor, and let $L$ be the point of the $x$ -axis nearest to $M$ , so that $M L$ meets the axis at a right angle. Read off three signed lengths: $O L$ , then $L M$ , then $M P$ . They are the coordinates of $P$ , written $P (x, y, z)$ . Signed means each length carries the sign of its direction: $M P$ is negative when $P$ lies below the floor, $L M$ is negative when $M$ is on the $O Y^{'}$ side, and $O L$ is negative when $L$ is behind $O$ .

4/4 $O L = 3$ , $L M = 4$ , $M P = 5$ , so $P = (3, 4, 5)$ . Walking $O \to L \to M \to P$ reverses the process.

Fig. 11.4Down to the floor, then across to the

x

-axis: the three legs are

x

y

and

z

.NCERT Fig. 11.2 · p. 209

Axes OX, OY and OZ, with the XY-plane drawn as a tiled floor. Step 1 shows the point P floating above the floor. Step 2 drops a vertical line from P to its foot M on the floor, with a right-angle mark. Step 3 draws a dashed line from M, parallel to the y-axis, to the point L on the x-axis, again with a right-angle mark. Step 4 adds the segment OL and labels the three legs x = 3, y = 4 and z = 5; M is marked (3, 4, 0) and the heading reads P(3, 4, 5).

Going from numbers to a point is a walk in three legs. Starting at $O$ , walk $x$ units along the $x$ -axis (backwards if $x < 0$ ) to reach $L$ ; turn and walk $y$ units parallel to the $y$ -axis to reach $M$ ; then climb $z$ units, or descend if $z < 0$ , to reach $P$ . Every point of space has exactly one triplet and every triplet leads to exactly one point. Mathematicians say the points of space and the ordered triplets of real numbers are in one-to-one correspondence.

The word ordered matters. $(2, 5, - 1)$ and $(5, 2, - 1)$ use the same three numbers but name different points: the first is 2 units in front of the YZ-plane and 5 units to the right of the ZX-plane, the second the other way round.

Coordinates as distances from the planes

A second picture treats the three coordinates alike. Imagine the box, a rectangular parallelepiped, that has $O$ and $P$ as opposite corners and its edges along the axes (Fig. 11.5). Three of its faces lie in the coordinate planes; the other three pass through $P$ , each parallel to one coordinate plane. Three corners of the box sit on the axes; call them $A$ on the $x$ -axis, $B$ on the $y$ -axis and $C$ on the $z$ -axis. The edges from $O$ to them are exactly $x$ , $y$ and $z$ : $O A = x$ , $O B = y$ , $O C = z$ .

4/4 $O A = 3$ , $O B = 5$ , $O C = 4$ . Every corner is $P$ with some coordinates replaced by $0$ .

Fig. 11.5Three planes through

P

make a box; its edges along the axes are

x

y

and

z

.NCERT Fig. 11.3 · p. 210

Axes OX, OY and OZ with the point P(3, 5, 4). Step 1 shades a plane through P parallel to the YZ-plane; it meets the x-axis at A. Step 2 adds a plane through P parallel to the ZX-plane, meeting the y-axis at B. Step 3 adds a plane through P parallel to the XY-plane, meeting the z-axis at C. Step 4 labels OA as x = 3, OB as y = 5 and OC as z = 4, and names three corners of the box: (3, 5, 0) on the floor below P, (3, 0, 4) on the ZX-plane and (0, 5, 4) on the YZ-plane.

Each face of the box through $P$ is parallel to a coordinate plane, and the gap between two parallel planes can be measured along any common perpendicular. So $∣ x ∣$ is the distance of $P$ from the YZ-plane, $∣ y ∣$ its distance from the ZX-plane and $∣ z ∣$ its distance from the XY-plane. The corners of the box are just as easy: replace some of the coordinates of $P$ by zero.

Zeros tell you where a point sits (Fig. 11.6). The origin is $(0, 0, 0)$ . A point of the $x$ -axis is at zero distance from the ZX- and XY-planes, so it has the form $(x, 0, 0)$ ; points of the other axes are $(0, y, 0)$ and $(0, 0, z)$ . A point of the XY-plane has the form $(x, y, 0)$ , a point of the YZ-plane $(0, y, z)$ and a point of the ZX-plane $(x, 0, z)$ .

3/3One coordinate is zero: the one whose axis is missing from the plane.

Fig. 11.6One zero coordinate: the point is on a coordinate plane. Two zeros: it is on an axis.

Axes OX, OY and OZ. Step 1 marks the origin O(0, 0, 0). Step 2 adds one point on each axis, labelled (x, 0, 0) on the x-axis, (0, y, 0) on the y-axis and (0, 0, z) on the z-axis. Step 3 shades the three coordinate planes and adds one point in each, labelled (x, y, 0) in the XY-plane, (0, y, z) in the YZ-plane and (x, 0, z) in the ZX-plane.

Away from the planes, the signs decide the octant; Table 11.1 lists them. It reads like the four quadrants of the plane twice over: octants I–IV repeat the quadrant signs of $(x, y)$ with $z > 0$ , and V–VIII repeat them with $z < 0$ . A point with a zero coordinate lies on a coordinate plane and so belongs to no octant.

Table 11.1Signs of the coordinates in the eight octantsNCERT Table 11.1 · p. 210

Octant	Edges	$x$	$y$	$z$	Sample point
I	$X O Y Z$	$+$	$+$	$+$	$(2, 1, 3)$
II	$X^{'} O Y Z$	$-$	$+$	$+$	$(- 2, 1, 3)$
III	$X^{'} O Y^{'} Z$	$-$	$-$	$+$	$(- 2, - 1, 3)$
IV	$X O Y^{'} Z$	$+$	$-$	$+$	$(2, - 1, 3)$
V	$X O Y Z^{'}$	$+$	$+$	$-$	$(2, 1, - 3)$
VI	$X^{'} O Y Z^{'}$	$-$	$+$	$-$	$(- 2, 1, - 3)$
VII	$X^{'} O Y^{'} Z^{'}$	$-$	$-$	$-$	$(- 2, - 1, - 3)$
VIII	$X O Y^{'} Z^{'}$	$+$	$-$	$-$	$(2, - 1, - 3)$

Example 11.1NCERT Example 1 · p. 211

Corners of a cupboard

A steel cupboard stands in the corner of a room. Take the corner of the floor as $O$ , the floor as the XY-plane and the two walls behind the cupboard as the YZ- and ZX-planes, with lengths in metres. The cupboard is a box with one corner at $O$ and the opposite corner at $P (1.2, 0.6, 2)$ . Find the coordinates of (a) the corner on the floor directly below $P$ , (b) the top corner that touches both walls, (c) the corner on the floor that lies on the $y$ -axis.

Try it yourself first. Then reveal the solution one step at a time.

Example 11.2NCERT Example 2 · p. 211

Reading the octant

Where do the points $A (4, - 2, - 3)$ , $B (- 5, - 1, 6)$ and $C (0, 2, - 1)$ lie?

Try it yourself first. Then reveal the solution one step at a time.

InteractivePoint-in-space lab

A three-dimensional view of the axes, positive halves solid and negative halves dashed. In Point mode three sliders set x, y and z from −5 to 5; the point P is drawn with its coordinate box and the route from O along the x-axis to L, parallel to the y-axis to the foot M, and up or down to P. A panel names the octant, or the plane or axis the point lies on, and lists the distances from the three coordinate planes, from the three axes and from the origin. In Two points mode a toggle chooses whether the sliders move P or Q; the box with diagonal PQ is drawn with its three edge directions coloured, and the panel shows the differences of the coordinates, their squares, their sum and the distance PQ. Dragging the picture or two view sliders turns and tilts it. Nothing moves until the learner acts.

Fig. 11.7Set

x

y

and

z

with the sliders and drag the picture to turn it. Switch to Two points to watch the distance formula of Section 11.4 at work.

Practice · 6 questions

Practice 11.1: questions with answers

Short answerEasyNCERT Exercise 11.1, Q1
A point lies on the $z$ -axis, 4 units below the XY-plane. What are its $x$ - and $y$ -coordinates, and what is the point?
Show solution
Answer $x = 0$ , $y = 0$ ; the point is $(0, 0, - 4)$ .
The $z$ -axis lies in both the YZ-plane and the ZX-plane, so the point is at zero distance from each: $x = 0$ and $y = 0$ . Below the XY-plane the $z$ -coordinate is negative, so $z = - 4$ and the point is $(0, 0, - 4)$ .
2
Short answerEasyNCERT Exercise 11.1, Q2
What can you say about
1. the $x$ -coordinate of every point of the YZ-plane?
  Show solution
  AnswerIt is $0$ .
  The $x$ -coordinate is the signed distance from the YZ-plane, and a point of that plane is at distance $0$ from it.
2. the $z$ -coordinate of every point of the XY-plane?
  Show solution
  AnswerIt is $0$ .
  The $z$ -coordinate is the signed distance from the XY-plane, which is $0$ for its own points. Such points have the form $(x, y, 0)$ .
3
Quick answerEasyNCERT Exercise 11.1, Q3
Match each point with the octant it lies in.
Tap an item on the left, then its match on the right.
Show answer
Answer
$(2, 5, 1)$ : I
$(- 3, 4, 7)$ : II
$(- 1, - 6, 2)$ : III
$(3, 1, - 4)$ : V
$(- 4, - 4, - 4)$ : VII
$(6, - 3, - 2)$ : VIII
Sign patterns: $(+, +, +)$ I, $(-, +, +)$ II, $(-, -, +)$ III, $(+, +, -)$ V, $(-, -, -)$ VII, $(+, -, -)$ VIII.
Quick answerEasyNCERT Exercise 11.1, Q4
Fill in the blanks. The $y$ -axis and the $z$ -axis together determine the -plane. Every point of the ZX-plane has the form . The $x$ -coordinate is negative in of the eight octants.
Show answer
Answer (1) YZ, (2) (x, 0, z), (3) 4
The $y$ - and $z$ -axes span the YZ-plane. On the ZX-plane $y = 0$ , so points are $(x, 0, z)$ . Octants II, III, VI and VII have $x < 0$ : four of them.
Quick answerMedium
How far is the point $(5, - 12, 7)$ from the $z$ -axis?
units
Show answer
Answer 13 units
The nearest point of the $z$ -axis is $(0, 0, 7)$ , so the distance is $\sqrt{5^{2} + 12^{2}} = 13$ . Its distance from the XY-plane, $7$ , is a different quantity.
Quick answerMedium
A point has $x \neq 0$ , $y = 0$ and $z < 0$ . Which statement is true?
It lies in octant V or VIIIIt lies on the ZX-plane, below the $x$ -axisIt lies on the $z$ -axisIt lies on the XY-plane
Show answer
Answer (B) It lies on the ZX-plane, below the $x$ -axis
$y = 0$ puts the point on the ZX-plane, and $z < 0$ puts it below the XY-plane, which in the ZX-plane means below the $x$ -axis. A point on a coordinate plane is in no octant.

11.4 Distance between Two Points

In the plane, the distance between $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ is $\sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}}$ , which is Pythagoras' theorem in disguise. In space a single right triangle is not enough, but two are: the first lies flat in a horizontal plane, and the second stands upright on it.

DerivationThe distance formula in space

Fix any two points: $P$ with coordinates $(x_{1}, y_{1}, z_{1})$ and $Q$ with coordinates $(x_{2}, y_{2}, z_{2})$ . Change one coordinate at a time to get from $P$ to $Q$ , and the three moves trace three edges of a box with edges parallel to the axes and $P Q$ as its long diagonal (Fig. 11.8). The edge lengths are $∣ x_{2} - x_{1} ∣$ , $∣ y_{2} - y_{1} ∣$ and $∣ z_{2} - z_{1} ∣$ .

4/4Right angle at $S$ : $P Q^{2} = P S^{2} + S Q^{2}$ .

Fig. 11.8Pythagoras twice: once across the floor of the box, once up its side.NCERT Fig. 11.4 · p. 211

Step 1 shows two points P and Q joined by a dashed segment, with the coordinates of the two points written at the top left and the question PQ = ?. Step 2 draws the box with edges parallel to the axes that has PQ as a diagonal. Step 3 shades the triangle PRS in the bottom face: PR, parallel to the x-axis, has length |x2 − x1|, RS, parallel to the y-axis, has length |y2 − y1|, there is a right angle at R and the note reads PS² = PR² + RS². Step 4 shades the upright triangle PSQ: the vertical edge SQ has length |z2 − z1|, there is a right angle at S and the note reads PQ² = PS² + SQ².

In order, the moves go parallel to the $x$ -axis to $R (x_{2}, y_{1}, z_{1})$ , parallel to the $y$ -axis to $S (x_{2}, y_{2}, z_{1})$ , and parallel to the $z$ -axis up to $Q$ . The points $P$ , $R$ and $S$ all have height $z_{1}$ , so triangle $P R S$ lies in a horizontal plane. It has a right angle at $R$ , because $P R$ and $R S$ are parallel to two perpendicular axes, with $P R = ∣ x_{2} - x_{1} ∣$ and $R S = ∣ y_{2} - y_{1} ∣$ :

\begin{aligned} P S^{2} & = P R^{2} + R S^{2} \\ = (x_{2} - x_{1})^{2} \\ + (y_{2} - y_{1})^{2} \end{aligned}

(11.1)

Next, $S Q$ is parallel to the $z$ -axis, so it is perpendicular to the horizontal plane of triangle $P R S$ and in particular to $P S$ . Triangle $P S Q$ therefore has a right angle at $S$ , and $S Q = ∣ z_{2} - z_{1} ∣$ :

P Q^{2} = P S^{2} + S Q^{2}

(11.2)

Put (11.1) into (11.2). Since $∣ a ∣^{2} = a^{2}$ , the absolute values disappear and we obtain the distance formula; $P Q$ is the positive square root of the right side.

Distance formula

\begin{aligned} P Q^{2} = (x_{2} - x_{1})^{2} \\ + (y_{2} - y_{1})^{2} \\ + (z_{2} - z_{1})^{2} \end{aligned}

(11.3)

Taking the first point at the origin, with the second point written $P (x, y, z)$ , gives the distance of any point from $O$ :

Distance from the origin

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

(11.4)

Three remarks prevent most slips. The order of the points does not matter, since $(x_{1} - x_{2})^{2} = (x_{2} - x_{1})^{2}$ . Subtract matching coordinates and keep their signs: from $- 3$ to $4$ is a change of $7$ , not $1$ . And the plane formula sits inside the new one: if both points lie in the XY-plane, then $z_{1} = z_{2} = 0$ and the third term vanishes.

Example 11.3NCERT Example 3 · p. 212

Two points, one distance

Find the distance between $P (2, - 1, 5)$ and $Q (- 4, 1, 2)$ .

Try it yourself first. Then reveal the solution one step at a time.

SimulationThe floor plan of Example 11.3

A square grid from −6 to 6 on both axes, read as the XY-plane seen from above. Two draggable points, P′ at (2, −1) and Q′ at (−4, 1), are the feet of the perpendiculars from P(2, −1, 5) and Q(−4, 1, 2) of Example 11.3. The segment P′Q′ is drawn and its length is measured; this is the horizontal part of the distance, and the vertical part is the difference of the heights, 3.

Fig. 11.9

P^{'}

and

Q^{'}

are the feet of

P (2, - 1, 5)

and

Q (- 4, 1, 2)

on the XY-plane, seen from above. Their distance is the floor diagonal

P S

of the derivation.

Try this

Read the length of $P^{'} Q^{'}$ and square it. Add $(5 - 2)^{2} = 9$ and take the square root: you should get $7$ , the answer of Example 11.3.
Drag $Q^{'}$ on top of $P^{'}$ . Now $P$ and $Q$ lie on one vertical line: what does the distance formula reduce to?
Find a position of $Q^{'}$ with $P^{'} Q^{'} = 5$ . If the heights stay $5$ and $2$ , what is $P Q$ now?

With the distance formula, many geometric questions turn into arithmetic:

Collinear points. Three points are collinear exactly when the longest of the three distances between them equals the sum of the other two; otherwise they form a triangle.
Kinds of triangle. Compare side lengths to spot isosceles or equilateral triangles, and test $a^{2} + b^{2} = c^{2}$ for the longest side $c$ to detect a right angle (the converse of Pythagoras' theorem).
Sets of points. A condition on distances, such as being equally far from two fixed points, becomes an equation in $x$ , $y$ , $z$ once the distances are squared. The set of points it describes is called a locus.

Example 11.4NCERT Example 4 · p. 212

Three points on one line

Show that $A (- 1, 2, 0)$ , $B (1, 5, 6)$ and $C (5, 11, 18)$ are collinear.

Try it yourself first. Then reveal the solution one step at a time.

Example 11.5NCERT Example 5 · p. 212

Is there a right angle?

Does the triangle with corners $A (1, 2, 3)$ , $B (3, 3, 5)$ and $C (3, 0, 2)$ have a right angle?

Try it yourself first. Then reveal the solution one step at a time.

Example 11.6NCERT Example 6 · p. 213

A condition becomes a sphere

$A$ and $B$ are the points $(2, - 1, 3)$ and $(0, 3, - 1)$ . A point $P$ moves so that $P A^{2} + P B^{2} = 30$ . Find the equation it satisfies, and describe its path.

Try it yourself first. Then reveal the solution one step at a time.

Distance conditions in space describe surfaces. Fig. 11.10 shows the three you meet in this chapter's examples and exercises.

2/4 $P A = P B$ : the plane through $M$ at right angles to $A B$ .

Fig. 11.10Fixed distance conditions in space give surfaces, not curves.

Step 1 shows two points A and B joined by a segment, with its midpoint M. Step 2 shows a plane standing across the segment at right angles, through M, with a point P on it joined to A and B by dashed lines and the label PA = PB. Step 3 replaces the plane by a sphere centred at M, with a point P on it joined to A and B and the label PA² + PB² = constant. Step 4 replaces the sphere by an elongated, rugby-ball shaped surface around A and B, with a point P on it and the label PA + PB = constant.

3/3 $G$ divides each median in the ratio $2 : 1$ from the vertex.

Fig. 11.11The midpoint averages two points; the centroid averages three.

Step 1 shows a segment AB with equal tick marks on its two halves and its midpoint M, with the midpoint formula written at the top. Step 2 adds a third point C, shades the triangle ABC and draws the median from C to M as a dashed line. Step 3 draws the other two medians, marks their common point G and labels the two parts of the median CM as 2 and 1, with the note CG : GM = 2 : 1.

Practice · 7 questions

Practice 11.2: questions with answers

1
NumericalEasyNCERT Exercise 11.2, Q1
Find the distance between each pair of points.
1. $(3, - 1, 2)$ and $(5, 2, 8)$
  Show answer
  Answer 7
  Differences $2, 3, 6$ : $\sqrt{4 + 9 + 36} = \sqrt{49} = 7$ .
2. $(- 2, 0, 5)$ and $(1, - 4, 5)$
  Show answer
  Answer 5
  Differences $3, - 4, 0$ : $\sqrt{9 + 16 + 0} = 5$ . Both points are at height $5$ , so this is really a plane problem.
3. $(0, - 3, 1)$ and $(2, 1, - 3)$
  Show answer
  Answer 6
  Differences $2, 4, - 4$ : $\sqrt{4 + 16 + 16} = 6$ .
4. $(\sqrt{2}, 1, - 1)$ and $(0, 2, 1)$
  Show answer
  Answer 2.6458
  Differences $- \sqrt{2}, 1, 2$ : $\sqrt{2 + 1 + 4} = \sqrt{7} \approx 2.646$ .
ProofEasyNCERT Exercise 11.2, Q2
Show that $(2, - 1, 4)$ , $(4, 0, 2)$ and $(8, 2, - 2)$ are collinear.
Show solution
AnswerThe distances are $3$ , $6$ and $9$ , and $3 + 6 = 9$ .
Call the points $A$ , $B$ , $C$ . Then $A B = \sqrt{2^{2} + 1^{2} + (- 2)^{2}} = 3$ , $B C = \sqrt{4^{2} + 2^{2} + (- 4)^{2}} = 6$ and $A C = \sqrt{6^{2} + 3^{2} + (- 6)^{2}} = 9$ .
Since $A B + B C = A C$ , the points are collinear, and $B$ lies between $A$ and $C$ .
3
ProofMediumNCERT Exercise 11.2, Q3
Verify each statement.
1. $A (2, 0, 1)$ , $B (4, 1, 3)$ and $C (2, 3, 1)$ are the vertices of an isosceles triangle.
  Show solution
  $A B = \sqrt{4 + 1 + 4} = 3$ , $A C = \sqrt{0 + 9 + 0} = 3$ and $B C = \sqrt{4 + 4 + 4} = 2 \sqrt{3}$ . Two sides are equal, and $3 + 3 > 2 \sqrt{3}$ , so the three points do form a triangle and it is isosceles.
2. Triangle $P Q R$ with $P (1, - 1, 2)$ , $Q (3, 1, 1)$ and $R (7, - 1, 5)$ has a right angle.
  Show solution
  $P Q^{2} = 4 + 4 + 1 = 9$ , $Q R^{2} = 16 + 4 + 16 = 36$ and $P R^{2} = 36 + 0 + 9 = 45$ . Since $P Q^{2} + Q R^{2} = 45 = P R^{2}$ , the angle at $Q$ is a right angle.
3. $A (0, 1, 2)$ , $B (1, 3, 4)$ , $C (4, 2, 5)$ and $D (3, 0, 3)$ are the vertices of a parallelogram $A B C D$ .
  Show solution
  The midpoint of $A C$ is $(2, \frac{3}{2}, \frac{7}{2})$ and the midpoint of $B D$ is also $(2, \frac{3}{2}, \frac{7}{2})$ . The diagonals bisect each other, so $A B C D$ is a parallelogram.
  As a check, $A B = C D = 3$ and $B C = D A = \sqrt{11}$ .
Short answerMediumNCERT Exercise 11.2, Q4
Write the equation of the locus of points equidistant from $(2, - 1, 1)$ and $(0, 3, - 3)$ .
Show solution
Answer $x - 2 y + 2 z + 3 = 0$
For $P (x, y, z)$ , equal distances mean equal squared distances: $(x - 2)^{2} + (y + 1)^{2} + (z - 1)^{2} = x^{2} + (y - 3)^{2} + (z + 3)^{2}$ .
Expanding, the left side is $x^{2} + y^{2} + z^{2} - 4 x + 2 y - 2 z + 6$ and the right side is $x^{2} + y^{2} + z^{2} - 6 y + 6 z + 18$ .
Cancelling and collecting: $- 4 x + 8 y - 8 z - 12 = 0$ , that is $x - 2 y + 2 z + 3 = 0$ . Check: the midpoint $(1, 1, - 1)$ gives $1 - 2 - 2 + 3 = 0$ .
Long answerHardNCERT Exercise 11.2, Q5
A point $P$ moves so that its distances from $A (0, 0, 2)$ and $B (0, 0, - 2)$ always add up to $6$ . Find the equation of its locus.
Show solution
Answer $9 x^{2} + 9 y^{2} + 5 z^{2} = 45$
$P A^{2} = x^{2} + y^{2} + (z - 2)^{2}$ and $P B^{2} = x^{2} + y^{2} + (z + 2)^{2}$ , so $P B^{2} - P A^{2} = 8 z$ .
Since $P B^{2} - P A^{2} = (P B + P A) (P B - P A) = 6 (P B - P A)$ , we get $P B - P A = \frac{4 z}{3}$ . Adding this to $P B + P A = 6$ gives $P B = 3 + \frac{2 z}{3}$ .
Square: $x^{2} + y^{2} + z^{2} + 4 z + 4 = 9 + 4 z + \frac{4 z^{2}}{9}$ , so $x^{2} + y^{2} + \frac{5}{9} z^{2} = 5$ .
Multiplying by $9$ : $9 x^{2} + 9 y^{2} + 5 z^{2} = 45$ , an ellipsoid stretched along the $z$ -axis, reaching $z = \pm 3$ on it.
Quick answerMedium
The point of the $x$ -axis that is equidistant from $(1, 2, 3)$ and $(3, 5, - 2)$ is
$(6, 0, 0)$ $(- 6, 0, 0)$ $(4, 0, 0)$ $(0, 6, 0)$
Show answer
Answer (A) $(6, 0, 0)$
With $(a, 0, 0)$ : $(a - 1)^{2} + 4 + 9 = (a - 3)^{2} + 25 + 4$ , so $- 2 a + 14 = - 6 a + 38$ and $a = 6$ . $(0, 6, 0)$ is not on the $x$ -axis at all.
Quick answerMedium
The distance between $(2, k, 1)$ and $(- 1, 3, 5)$ is $\sqrt{34}$ . Find the sum of all possible values of $k$ .
Show answer
Answer 6
$9 + (k - 3)^{2} + 16 = 34$ gives $(k - 3)^{2} = 9$ , so $k = 0$ or $k = 6$ . Their sum is $6$ .

Miscellaneous Examples

Example 11.7NCERT Example 7 · p. 213

A parallelogram that is not a rectangle

Show that $A (1, 0, 2)$ , $B (3, 3, 3)$ , $C (4, 5, 1)$ and $D (2, 2, 0)$ are the vertices of a parallelogram $A B C D$ , and that it is not a rectangle.

Try it yourself first. Then reveal the solution one step at a time.

2/4In a parallelogram the diagonals bisect each other.

Fig. 11.12In space, test a parallelogram with its diagonals: equal opposite sides are not enough.

Step 1 shows a parallelogram ABCD with single tick marks on AB and DC and double tick marks on BC and AD. Step 2 draws the diagonals AC and BD, which cross at one point marked same midpoint. Step 3 adds the note that AC is not equal to BD, so the figure is not a rectangle. Step 4 replaces it by a tetrahedron whose edges AB, BC, CD and DA form a closed path with four equal sides; the diagonals AC and BD, drawn dashed, have different midpoints joined by a short dotted segment, and the note reads four equal sides, not flat.

Example 11.8NCERT Example 8 · p. 214

Equally far from two points

Which points $P$ are as far from $A (1, - 2, 4)$ as from $B (3, 2, 0)$ ? Find the equation they satisfy.

Try it yourself first. Then reveal the solution one step at a time.

Example 11.9NCERT Example 9 · p. 214

The missing vertex

The centroid of triangle $A B C$ is $G (2, - 1, 3)$ . If $A = (4, 1, - 2)$ and $B = (- 3, 0, 5)$ , find $C$ .

Try it yourself first. Then reveal the solution one step at a time.

Quick check

1 of 3

M (3, - 1, 4)

is the midpoint of

A B

, and

A = (1, 2, 5)

. What is the

z

-coordinate of

B

Show answer

Answer 3

\frac{5 + z_{B}}{2} = 4

gives

z_{B} = 3

. In full,

B = (5, - 4, 3)

Recap

Introduction to Three Dimensional Geometry: key points and summary

Space needs three coordinates: three mutually perpendicular axes $X^{'} O X$ , $Y^{'} O Y$ and $Z^{'} O Z$ meet at the origin $O$ (Section 11.2).
The XY-, YZ- and ZX-planes each contain two of the axes and stand at right angles to the third (Section 11.2).
Space splits into eight octants along the coordinate planes, and the signs of $(x, y, z)$ tell you which octant holds a point (Table 11.1).
A point $P (x, y, z)$ is reached by moving $x$ along the $x$ -axis, then $y$ parallel to the $y$ -axis, then $z$ parallel to the $z$ -axis; points and ordered triplets correspond one to one (Section 11.3).
$∣ x ∣$ , $∣ y ∣$ and $∣ z ∣$ are the distances of $P$ from the YZ-, ZX- and XY-planes (Coordinates as distances from the planes).
Points of the axes have the forms $(x, 0, 0)$ , $(0, y, 0)$ , $(0, 0, z)$ ; points of the coordinate planes have the forms $(x, y, 0)$ , $(0, y, z)$ , $(x, 0, z)$ (Fig. 11.6).
$P Q^{2} = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2} + (z_{2} - z_{1})^{2}$ ((11.3)), and $O P = \sqrt{x^{2} + y^{2} + z^{2}}$ ((11.4)).
Distances decide collinearity, kinds of triangle and parallelograms, and turn distance conditions into equations of surfaces (Section 11.4).
Midpoints and centroids average the coordinates; a quadrilateral in space is a parallelogram when its diagonals share a midpoint (midpoint and centroid).

Flash cards · 7

Key ideas

1 / 7

Miscellaneous Exercise on Chapter 11

Practice · 7 questions

Practice questions with answers (NCERT “Miscellaneous exercise”)

Short answerMediumNCERT Miscellaneous Exercise, Q1
$A (2, - 3, 1)$ , $B (4, 1, - 2)$ and $C (1, 2, 3)$ are consecutive corners of a parallelogram $A B C D$ . Find the fourth corner $D$ .
Show solution
Answer $D (- 1, - 2, 6)$
The midpoint of $A C$ is $(\frac{3}{2}, - \frac{1}{2}, 2)$ . It must also be the midpoint of $B D$ .
With $D = (x, y, z)$ : $\frac{4 + x}{2} = \frac{3}{2}$ , $\frac{1 + y}{2} = - \frac{1}{2}$ and $\frac{- 2 + z}{2} = 2$ , so $x = - 1$ , $y = - 2$ , $z = 6$ .
2
NumericalMediumNCERT Miscellaneous Exercise, Q2
The triangle has vertices $A (4, 0, 0)$ , $B (0, 2, 0)$ and $C (0, 0, 6)$ . Find the length of
1. the median from $A$
  Show answer
  Answer 5.099
  The midpoint of $B C$ is $(0, 1, 3)$ ; the median is $\sqrt{16 + 1 + 9} = \sqrt{26} \approx 5.10$ .
2. the median from $B$
  Show answer
  Answer 4.123
  The midpoint of $C A$ is $(2, 0, 3)$ ; the median is $\sqrt{4 + 4 + 9} = \sqrt{17} \approx 4.12$ .
3. the median from $C$
  Show answer
  Answer 6.403
  The midpoint of $A B$ is $(2, 1, 0)$ ; the median is $\sqrt{4 + 1 + 36} = \sqrt{41} \approx 6.40$ . Check: $26 + 17 + 41 = 84 = \frac{3}{4} (20 + 40 + 52)$ , as the sum of the squared medians must be.
Short answerMediumNCERT Miscellaneous Exercise, Q3
The centroid of triangle $P Q R$ is $(1, 2, - 1)$ , where $P = (3 a, - 2, 5)$ , $Q = (1, 4 b, - 8)$ and $R = (- 7, 6, c + 1)$ . Find $a$ , $b$ and $c$ .
Show solution
Answer $a = 3$ , $b = \frac{1}{2}$ , $c = - 1$
$x$ : $3 a + 1 - 7 = 3$ , so $a = 3$ . $y$ : $- 2 + 4 b + 6 = 6$ , so $b = \frac{1}{2}$ . $z$ : $5 - 8 + c + 1 = - 3$ , so $c = - 1$ .
Long answerHardNCERT Miscellaneous Exercise, Q4
$A$ and $B$ are $(1, 0, - 2)$ and $(- 3, 2, 4)$ , and $k$ is a constant. Find the equation of the locus of a point $P$ that moves so that $P A^{2} + P B^{2} = k^{2}$ . For which values of $k$ does such a point exist?
Show solution
Answer $2 (x^{2} + y^{2} + z^{2}) + 4 x - 4 y - 4 z + 34 - k^{2} = 0$ ; points exist only when $k^{2} \geq 28$ .
$P A^{2} = (x - 1)^{2} + y^{2} + (z + 2)^{2}$ and $P B^{2} = (x + 3)^{2} + (y - 2)^{2} + (z - 4)^{2}$ .
Adding: $2 x^{2} + 2 y^{2} + 2 z^{2} + 4 x - 4 y - 4 z + 34 = k^{2}$ , since the constants are $1 + 4 + 9 + 4 + 16 = 34$ .
Halve and complete the squares: $(x + 1)^{2} + (y - 1)^{2} + (z - 1)^{2} = \frac{k^{2}}{2} - 14$ . The left side is never negative, so points exist only when $k^{2} \geq 28$ .
For $k^{2} > 28$ the set is a sphere centred at $(- 1, 1, 1)$ , the midpoint of $A B$ . For $k^{2} = 28$ it is that single point.
Quick answerMedium
The distance of the point $(6, - 2, 3)$ from the $y$ -axis is
$7$ $3 \sqrt{5}$ $\sqrt{13}$ $2 \sqrt{10}$
Show answer
Answer (B) $3 \sqrt{5}$
$\sqrt{6^{2} + 3^{2}} = \sqrt{45} = 3 \sqrt{5}$ . The other options are the distances from the origin ( $7$ ), the $x$ -axis ( $\sqrt{13}$ ) and the $z$ -axis ( $2 \sqrt{10}$ ).
Quick answerMedium
Assertion (A): The point $(- 2, 0, 5)$ lies in octant II. Reason (R): In octant II the $x$ -coordinate is negative and the $y$ - and $z$ -coordinates are positive.
Both A and R are true, and R explains ABoth A and R are true, but R does not explain AA is true but R is falseA is false but R is true
Show answer
Answer (D) A is false but R is true
R is the correct sign pattern of octant II. But the point has $y = 0$ , so it lies on the ZX-plane and is in no octant: A is false.
Quick answerHard
For which value of $k$ are $(1, 2, 3)$ , $(3, 6, 7)$ and $(k, 10, 11)$ collinear?
Show answer
Answer 5
From the first point to the second, $y$ and $z$ rise by $4$ while $x$ rises by $2$ . To the third point $y$ and $z$ again rise by $4$ , so $x$ must again rise by $2$ : $k = 5$ . Distances confirm it: $6 + 6 = 12$ .

Timeline

Coordinates in space

1637
René Descartes
La Géométrie turns plane curves into equations.
1700
Antoine Parent
Writes a surface as an equation in three coordinates.
1731
Alexis Clairaut
Publishes a book on curves in space, written at about sixteen.
1748
Leonhard Euler
The appendix to volume 2 of his Introductio treats surfaces systematically.
1854
Bernhard Riemann
A famous lecture allows spaces of any number of dimensions.
1908
Hermann Minkowski
Adds time as a fourth coordinate: the space-time of Einstein's relativity.

Find your NCERT exercise question

Our practice questions are written for ncert.ai and follow the NCERT exercises. Find the one that matches the question in your textbook; its answer is with it.

In your NCERT textbook	Practice question here
NCERT Exercise 11.1, Q1	Practice 11.1 · Q1
NCERT Exercise 11.1, Q2	Practice 11.1 · Q2
NCERT Exercise 11.1, Q3	Practice 11.1 · Q3
NCERT Exercise 11.1, Q4	Practice 11.1 · Q4
NCERT Exercise 11.2, Q1	Practice 11.2 · Q1
NCERT Exercise 11.2, Q2	Practice 11.2 · Q2
NCERT Exercise 11.2, Q3	Practice 11.2 · Q3
NCERT Exercise 11.2, Q4	Practice 11.2 · Q4
NCERT Exercise 11.2, Q5	Practice 11.2 · Q5
NCERT Miscellaneous Exercise, Q1	Miscellaneous exercise · Q1
NCERT Miscellaneous Exercise, Q2	Miscellaneous exercise · Q2
NCERT Miscellaneous Exercise, Q3	Miscellaneous exercise · Q3
NCERT Miscellaneous Exercise, Q4	Miscellaneous exercise · Q4

Quick check: MCQs on Introduction to Three Dimensional Geometry

11.1 Introduction2 questions
11.2 Coordinate Axes and Coordinate Planes in Three Dimensional Space4 questions
Coordinates as distances from the planes4 questions
11.4 Distance between Two Points4 questions
Miscellaneous Examples3 questions

Glossary · 12

Key terms (English–Hindi)

centroidकेंद्रकkendrak: The common point of a triangle's three medians; its coordinates are the averages of the vertices' coordinates.First seen in the chapter →
collinearसंरेखsanrekh: Lying on one straight line.First seen in the chapter →
coordinate axesनिर्देशांक अक्षnirdeshank aksh: The three mutually perpendicular lines through the origin, the $x$ -, $y$ - and $z$ -axes, along which coordinates are measured.First seen in the chapter →
coordinate planeनिर्देशांक तलnirdeshank tal: A plane containing two of the axes: the XY-, YZ- or ZX-plane.
coordinates of a pointनिर्देशांकnirdeshank: The ordered triplet $(x, y, z)$ of signed distances of a point from the YZ-, ZX- and XY-planes.First seen in the chapter →

distance formulaदूरी सूत्रdoori sootra: The rule $P Q^{2} = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2} + (z_{2} - z_{1})^{2}$ , where $P$ and $Q$ have coordinates $(x_{1}, y_{1}, z_{1})$ and $(x_{2}, y_{2}, z_{2})$ .First seen in the chapter →

locusबिंदुपथbindupath: The set of all points that satisfy a given condition; in space it is usually a surface.First seen in the chapter →

midpointमध्य-बिंदुmadhya-bindu: The point of a segment that is equally far from its two ends; its coordinates are the averages of theirs.First seen in the chapter →

octantअष्टांशashtansh: One of the eight regions into which the three coordinate planes divide space.First seen in the chapter →
originमूल बिंदुmool bindu: The point $O$ where the three axes meet; its coordinates are $(0, 0, 0)$ .First seen in the chapter →

rectangular coordinate system: The origin together with three mutually perpendicular axes and the three coordinate planes they determine.First seen in the chapter →
rectangular parallelepipedघनाभghanabh: A solid bounded by six rectangular faces, like a brick or a closed carton.First seen in the chapter →

11.1 Introduction

11.2 Coordinate Axes and Coordinate Planes in Three Dimensional Space

11.3 Coordinates of a Point in Space

Coordinates as distances from the planes

11.4 Distance between Two Points

Miscellaneous Examples

Miscellaneous Exercise on Chapter 11

Find your NCERT exercise question

Quick check: MCQs on Introduction to Three Dimensional Geometry

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