MathsCBSE Class 11 · Chapter 11

Introduction to Three Dimensional Geometry Class 11: Notes and Solved Questions

Updated 14 October 2026 · Fact-checked

Introduction to Three Dimensional Geometry is Chapter 11 of CBSE Class 11 Maths. For 2026-27 it covers the three coordinate axes and the coordinate planes, the eight octants, the coordinates (x, y, z) of a point in space, and the distance between two points, with uses such as collinearity, types of triangles and simple loci.

What Introduction to Three Dimensional Geometry is about

A point on a sheet of paper needs two numbers to fix it; a point in a room needs three. To say exactly where a ceiling fan or a flying drone is, you need its distance east, its distance north and its height. This chapter extends the familiar xy-plane to three-dimensional space.

Three mutually perpendicular lines through an origin O, the x-, y- and z-axes, determine three coordinate planes (XY, YZ and ZX). These planes divide space into eight octants. Each point P gets an ordered triplet $(x, y, z)$ : x is its signed distance from the YZ-plane, y from the ZX-plane and z from the XY-plane. Points on the x-axis look like $(x, 0, 0)$ ; points in the XY-plane like $(x, y, 0)$ .

The main tool is the distance formula
$P Q = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2} + (z_{2} - z_{1})^{2}$
proved by applying Pythagoras twice inside a box. With it you can test whether points are collinear, whether a triangle is isosceles or right-angled, and find simple loci such as all points equidistant from two given points. For 2026-27, the section formula in three dimensions is assessed only formatively and is not in the annual exam.

Weightage: This chapter is part of Unit III, Coordinate Geometry, which the syllabus gives 12 of the 80 theory marks in the school annual exam (CBSE adds that there is no fixed chapter-wise weightage).Source: CBSE Mathematics (041) syllabus 2026-27, Class XI course structure, Unit III Coordinate Geometry

Key concepts

Coordinate axes and coordinate planes

NCERT section 11.1 and 11.2

Take three mutually perpendicular lines meeting at a point O, the origin: the x-axis (XOX′), y-axis (YOY′) and z-axis (ZOZ′). Usually x and y are drawn in the horizontal page and z points up.

Each pair of axes fixes a plane:

XY-plane (x- and y-axes): every point has z = 0.
YZ-plane (y- and z-axes): every point has x = 0.
ZX-plane (z- and x-axes): every point has y = 0.

These three coordinate planes are perpendicular to each other. A room corner is a good model: the floor is the XY-plane, and the two walls are the YZ- and ZX-planes.

Worked example

Fill in: (a) A point on the z-axis has x = __ and y = __. (b) A point in the YZ-plane has x = __. (c) The y- and z-axes together determine the __ plane. [1 mark]

Try it first, then show the solutionSolution

(a) Both are 0: the point is (0, 0, z)
(b) 0. (c) YZ-plane

Answer: (a) 0, 0 (b) 0 (c) YZ

Octants

NCERT section 11.2

The three coordinate planes split space into eight octants, named by the signs of the coordinates:

Octant	I	II	III	IV	V	VI	VII	VIII
x	+	−	−	+	+	−	−	+
y	+	+	−	−	+	+	−	−
z	+	+	+	+	−	−	−	−

Octants I to IV lie above the XY-plane (z > 0) and follow the four quadrants of the plane; V to VIII lie below (z < 0) in the same order.

A point with a zero coordinate lies on a coordinate plane, not inside an octant.

Worked example

Name the octants of (−5, 2, 1), (−5, 2, −1), (3, −6, −2) and (−1, −3, −8). [2 marks]

Try it first, then show the solutionSolution

(−5, 2, 1): signs (−, +, +): II; (−5, 2, −1): (−, +, −): VI
(3, −6, −2): (+, −, −): VIII; (−1, −3, −8): (−, −, −): VII

Answer: II, VI, VIII, VII

Coordinates of a point in space

NCERT section 11.3

Through a point P draw planes parallel to the three coordinate planes; they meet the axes at A(x, 0, 0), B(0, y, 0) and C(0, 0, z). Then P has coordinates (x, y, z):

x = signed distance of P from the YZ-plane
y = signed distance from the ZX-plane
z = signed distance from the XY-plane

Conversely, every ordered triplet of real numbers gives exactly one point in space.

Special positions:

On the x-axis: (x, 0, 0); y-axis: (0, y, 0); z-axis: (0, 0, z).
In the XY-plane: (x, y, 0); YZ-plane: (0, y, z); ZX-plane: (x, 0, z).

Feet of perpendiculars from P(x, y, z): on the XY-plane, (x, y, 0); on the x-axis, (x, 0, 0); and so on. The distance of P from the z-axis is $x^{2} + y^{2}$ .

Worked example

For P(2, −3, 5), write the feet of the perpendiculars on the XY-plane and on the y-axis, and its distance from the XY-plane. [2 marks]

Try it first, then show the solutionSolution

On the XY-plane: set z = 0: (2, −3, 0). On the y-axis: keep only y: (0, −3, 0)
Distance from the XY-plane = |z| = 5

Answer: (2, −3, 0); (0, −3, 0); 5 units

Distance between two points

NCERT section 11.4

For $P (x_{1}, y_{1}, z_{1})$ and $Q (x_{2}, y_{2}, z_{2})$ :
$P Q = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2} + (z_{2} - z_{1})^{2}$

Proof idea: draw planes through P and Q parallel to the coordinate planes to form a rectangular box with PQ as a diagonal. Its edges have lengths $∣ x_{2} - x_{1} ∣$ , $∣ y_{2} - y_{1} ∣$ , $∣ z_{2} - z_{1} ∣$ . Pythagoras on the base gives the square of the base diagonal; Pythagoras again with the vertical edge gives $P Q^{2}$ .

From the origin: $O P = x^{2} + y^{2} + z^{2}$ .

It is the 2D formula with one more square added.

Worked example

Find the distance between (1, 2, 3) and (4, −2, 15), and the distance of (3, 4, 12) from the origin. [2 marks]

Try it first, then show the solutionSolution

$3^{2} + (- 4)^{2} + 1 2^{2} = 9 + 16 + 144 = 169 = 13$
$9 + 16 + 144 = 13$

Answer: 13 and 13

Using distances: collinearity and shapes

NCERT section 11.4 (Examples)

With the three side lengths of a triangle you can decide many things:

Collinear: points P, Q, R lie on one line if the largest distance equals the sum of the other two (e.g. PQ + QR = PR).
Isosceles: two sides equal. Equilateral: all three equal.
Right-angled: the square of the longest side equals the sum of squares of the other two.
Parallelogram: the diagonals have the same mid-point (in space, equal opposite sides alone do not guarantee a flat parallelogram, so check the mid-points).
Fourth vertex of parallelogram ABCD: $D = A + C - B$ (coordinate by coordinate), since the diagonals bisect each other.

Tip: compare squares of distances to avoid square roots until the end.

Worked example

Show that A(1, 2, 3), B(3, 4, 7) and C(6, 7, 13) are collinear. [3 marks]

Try it first, then show the solutionSolution

$A B = 4 + 4 + 16 = 24 = 26$ ; $B C = 9 + 9 + 36 = 54 = 36$
$A C = 25 + 25 + 100 = 150 = 56$
$A B + B C = 56 = A C$ , so A, B, C are collinear (B lies between A and C)

Answer: Collinear.

Loci in space

NCERT section 11.4 (Examples 6 and 8)

A locus is the set of all points satisfying a condition. Let P = (x, y, z), write the condition with the distance formula, square to remove roots, and simplify.

Equidistant from two points A and B: $P A^{2} = P B^{2}$ . The squared terms cancel and you get a linear equation: a plane (the perpendicular bisector plane of AB).
Fixed distance r from a point C: $P C^{2} = r^{2}$ : a sphere.
** $P A^{2} + P B^{2} = k^{2}$ :** a sphere-type equation.
Sum of distances from two points constant: after squaring twice you get an equation like $16 x^{2} + 25 y^{2} + 25 z^{2} = 400$ (an ellipsoid).

Worked example

Find the equation of the set of points equidistant from A(1, 0, 2) and B(3, 4, 0). [3 marks]

Try it first, then show the solutionSolution

$(x - 1)^{2} + y^{2} + (z - 2)^{2} = (x - 3)^{2} + (y - 4)^{2} + z^{2}$
Expanding, the squares cancel: $- 2 x + 1 - 4 z + 4 = - 6 x + 9 - 8 y + 16$
$4 x + 8 y - 4 z = 20$ , i.e. $x + 2 y - z = 5$

Answer: $x + 2 y - z = 5$ (a plane)

Maths in real life: Introduction to Three Dimensional Geometry

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Distance in a room

Suppose a room is 5 m long, 4 m wide and 3 m high, and a fly travels in a straight line from one floor corner to the opposite ceiling corner.

Taking the starting corner as the origin, find the straight-line distance.

Try it first, then show the solutionSolution

The opposite top corner is (5, 4, 3)
Distance $= 25 + 16 + 9 = 50 = 52 \approx 7.07$ m

Answer: About 7.07 m

Memory hook: Box diagonal = square root of (length² + width² + height²).

Coordinates of a drone

Imagine a drone that is 30 m east, 40 m north and 120 m above its launch point.

Write its coordinates and find (a) its straight-line distance from the launch point (b) the distance along the ground from the launch point to the spot directly below it.

Try it first, then show the solutionSolution

Coordinates (30, 40, 120) with the launch point at the origin
(a) $900 + 1600 + 14400 = 16900 = 130$ m
(b) The spot below is (30, 40, 0): $900 + 1600 = 50$ m

Answer: 130 m; 50 m

Memory hook: Drop z to get the point directly below.

Longest rod in a box

Suppose a delivery box measures 12 cm × 4 cm × 3 cm inside.

What is the longest thin rod that fits inside?

Try it first, then show the solutionSolution

The longest segment joins opposite corners (0, 0, 0) and (12, 4, 3)
$144 + 16 + 9 = 169 = 13$ cm

Answer: 13 cm

Memory hook: Longest fit = space diagonal.

Topics in this chapter

Introduction to Three Dimensional Geometry formula sheet

Distance between two points: $P Q = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2} + (z_{2} - z_{1})^{2}$
Distance from the origin: $O P = x^{2} + y^{2} + z^{2}$
Points on axes: $x -axis : (x, 0, 0), y -axis : (0, y, 0), z -axis : (0, 0, z)$
Points on coordinate planes: $X Y : z = 0, Y Z : x = 0, Z X : y = 0$
Distances from planes and axes: $from XY-plane = ∣ z ∣, from z-axis = x^{2} + y^{2}$
Fourth vertex of a parallelogram ABCD: $D = (x_{A} + x_{C} - x_{B}, y_{A} + y_{C} - y_{B}, z_{A} + z_{C} - z_{B})$

How the school exam asks Introduction to Three Dimensional Geometry

The question patterns to expect, and the method that works for each.

Question type	How it is asked	Method
Axes, planes and octants	Fill in the blanks; coordinates of points on axes or planes; name the octant (Exercise 11.1).	Use the sign table; zero coordinate means on a plane.
Distance between points	Find distances between given pairs; find an unknown coordinate from a given distance.	Apply the formula carefully with signs; square both sides for unknowns.
Collinearity and triangle types	Show points are collinear, or vertices of an isosceles, right-angled or equilateral triangle, or a parallelogram (Exercise 11.2).	Find all needed distances (or their squares) and compare.
Locus	Equation of points equidistant from two points; sum of distances constant; $P A^{2} + P B^{2} = k^{2}$ .	Let P = (x, y, z), write the condition, square, expand, simplify.
Mixed (Miscellaneous)	Fourth vertex of a parallelogram, lengths of medians, unknowns from a given centroid.	Mid-points and centroids work coordinate by coordinate, exactly as in 2D.

Tricks to remember Introduction to Three Dimensional Geometry

Short memory aids. Say each one aloud once, then use it on a question the same day.

Zero tells the plane
z = 0 → XY-plane; x = 0 → YZ-plane; y = 0 → ZX-plane. The missing letter is the zero coordinate.
Use it when: Points on planes.
Upstairs, downstairs
Octants I-IV are 'upstairs' (z > 0) and copy the four quadrants; V-VIII are the same rooms 'downstairs' (z < 0).
Use it when: Naming octants.
2D formula plus one square
Add $(z_{2} - z_{1})^{2}$ under the root.
Use it when: Distance formula.
Equidistant gives a plane
$P A^{2} = P B^{2}$ always loses the squared terms and leaves a linear equation.
Use it when: Locus questions.
Distance from a plane is the missing coordinate
Distance from the XY-plane is |z|, from YZ is |x|, from ZX is |y|.
Use it when: Quick answers.

Introduction to Three Dimensional Geometry: exam-level solved questions

Ordered from exam level to HOTS. Cover the solution, write your own, then compare step by step.

Example 1: Points in a plane · 1 mark · MCQExam level

The point (0, −3, 7) lies:
(a) on the y-axis
(b) in the YZ-plane
(c) in the XY-plane
(d) in octant II

Try it first, then show the solutionSolution

x = 0, so the point is in the YZ-plane.

Answer: (b)

Example 2: Octants · 1 mark · Assertion-reasonExam level

Assertion (A): The point (−4, 2, −5) lies in octant VI.
Reason (R): Octant VI is the region where x < 0, y > 0 and z < 0.
(a) Both true, R explains A. (b) Both true, R does not explain A. (c) A true, R false. (d) A false, R true.

Try it first, then show the solutionSolution

Signs (−, +, −) match octant VI.

Answer: (a)

Example 3: Point on the x-axis equidistant from two points · 2 marks · Very short answerHarder exam level

Find the point on the x-axis equidistant from (1, 2, 3) and (3, −1, 2). [2 marks]

Try it first, then show the solutionSolution

Let it be (x, 0, 0): $(x - 1)^{2} + 4 + 9 = (x - 3)^{2} + 1 + 4$
$- 2 x + 14 = - 6 x + 14 \Rightarrow x = 0$ : the origin (both distances are $14$ )

Answer: (0, 0, 0)

Example 4: Equilateral triangle · 2 marks · Very short answerHarder exam level

Show that (2, 0, 0), (0, 2, 0) and (0, 0, 2) are the vertices of an equilateral triangle and find its side. [2 marks]

Try it first, then show the solutionSolution

Each pair differs by 2 in two coordinates and 0 in the third: distance $4 + 4 = 22$
All three sides equal: equilateral, side $22$

Answer: Side $22$

Example 5: Median length · 2 marks · Very short answerHarder exam level

Find the length of the median from A in the triangle with vertices A(0, 0, 4), B(4, 0, 0) and C(0, 4, 0). [2 marks]

Try it first, then show the solutionSolution

Mid-point of BC: (2, 2, 0)
Median $= 4 + 4 + 16 = 24 = 26$

Answer: $26$

Example 6: Unknowns from a centroid · 3 marks · Short answerHarder exam level

The centroid of the triangle with vertices (a, 1, 4), (2, b, 0) and (−1, 3, c) is (1, 2, 3). Find a, b and c. [3 marks]

Try it first, then show the solutionSolution

Centroid = average of the vertices: $\frac{a + 2 - 1}{3} = 1 \Rightarrow a = 2$
$\frac{1 + b + 3}{3} = 2 \Rightarrow b = 2$
$\frac{4 + 0 + c}{3} = 3 \Rightarrow c = 5$

Answer: a = 2, b = 2, c = 5

Example 7: Locus with a sum of squares · 3 marks · Short answerHOTS / competency

Find the locus of P such that $P A^{2} + P B^{2} = k^{2}$ , where A(1, 0, 0) and B(−1, 0, 0). For which k is the locus non-empty? [3 marks]

Try it first, then show the solutionSolution

$(x - 1)^{2} + y^{2} + z^{2} + (x + 1)^{2} + y^{2} + z^{2} = k^{2}$
$2 x^{2} + 2 y^{2} + 2 z^{2} + 2 = k^{2} \Rightarrow x^{2} + y^{2} + z^{2} = \frac{k ^{2} - 2}{2}$
The left side is never negative, so we need $k^{2} \geq 2$ : a sphere centred at the origin (a single point when $k^{2} = 2$ )

Answer: $x^{2} + y^{2} + z^{2} = \frac{k ^{2} - 2}{2}$ , non-empty for $k^{2} \geq 2$

Example 8: Parallelogram by distances · 5 marks · Long answerHOTS / competency

Show that A(1, 2, 3), B(4, 5, 6), C(7, 3, 1) and D(4, 0, −2) form a parallelogram, using distances, and decide whether it is a rectangle. [5 marks]

Try it first, then show the solutionSolution

$A B^{2} = 9 + 9 + 9 = 27$ ; $C D^{2} = 9 + 9 + 9 = 27$
$B C^{2} = 9 + 4 + 25 = 38$ ; $D A^{2} = 9 + 4 + 25 = 38$
Mid-point of AC $= (4, 2.5, 2)$ and mid-point of BD $= (4, 2.5, 2)$ : the diagonals bisect each other, so ABCD is a parallelogram (the equal opposite sides agree with this)
Diagonals: $A C^{2} = 36 + 1 + 4 = 41$ ; $B D^{2} = 0 + 25 + 64 = 89$
The diagonals are unequal, so it is not a rectangle (also $A B^{2} + B C^{2} = 65 \neq = 41$ )

Answer: A parallelogram, not a rectangle.

Case-study practice (Section E style)

Read the situation, then answer every part. In a school exam set on the board pattern this is a 4-mark question split into 1 + 1 + 1 + 1 marks.

Ceiling fan and lamp

Suppose a room has one floor corner at the origin, the floor as the XY-plane and the ceiling at height 3 m. A fan hangs at F(2, 3, 2.5) and a lamp stands at L(5, 7, 1) (metres).

(i) · 1 mark
How far is the fan below the ceiling?
Answer
Answer: 0.5 m
3 − 2.5
(ii) · 1 mark
Find the distance between the fan and the lamp.
Answer
Answer: $27.25 \approx 5.22$ m
$9 + 16 + 2.25$
(iii) · 1 mark
Write the point on the floor directly below the fan.
Answer
Answer: (2, 3, 0)
Set z = 0.
(iv) · 1 mark
In which octant are all points of the room (not on the walls or floor)?
Answer
Answer: Octant I
All coordinates positive.

Two birds

Suppose two birds sit at P(2, −3, 5) and Q(−2, 1, 3) relative to a tree base at the origin (metres, z upward).

(i) · 1 mark
Find PQ.
Answer
Answer: 6 m
$16 + 16 + 4 = 36$
(ii) · 1 mark
Which bird is higher, and by how much?
Answer
Answer: P, by 2 m
Compare z: 4 and 2.
(iii) · 1 mark
How far is P from the vertical tree trunk (the z-axis)?
Answer
Answer: $13 \approx 3.61$ m
$4 + 9$
(iv) · 1 mark
Name the octants of P and Q.
Answer
Answer: IV and II
(+, −, +) and (−, +, +).

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Find the values of k if the distance between (k, 2, 3) and (1, −1, −3) is 7. [2 marks] (2 marks)

Step	Marks
$(k - 1)^{2} + 9 + 36 = 49 \Rightarrow (k - 1)^{2} = 4$	1
$k - 1 = \pm 2 \Rightarrow k = 3$ or $k = - 1$	1

Show that the points A(0, 0, 0), B(1, 2, 2) and C(−2, 2, −1) form a right-angled isosceles triangle. [3 marks] (3 marks)

Step	Marks
$A B^{2} = 1 + 4 + 4 = 9$ ; $A C^{2} = 4 + 4 + 1 = 9$	1
$B C^{2} = 9 + 0 + 9 = 18$	1
AB = AC = 3 (isosceles) and $A B^{2} + A C^{2} = 18 = B C^{2}$ (right angle at A)	1

(a) Find the equation of the set of points P whose distances from A(3, 0, 0) and B(−3, 0, 0) add up to 10. (b) Check that (5, 0, 0) and (0, 4, 0) lie on it. [5 marks] (5 marks)

Step	Marks
$(x - 3)^{2} + y^{2} + z^{2} = 10 - (x + 3)^{2} + y^{2} + z^{2}$	1
Square: $(x - 3)^{2} = 100 - 20 (x + 3)^{2} + y^{2} + z^{2} + (x + 3)^{2}$ , so $20 (x + 3)^{2} + y^{2} + z^{2} = 100 + 12 x$	1
Divide by 4 and square: $25 [(x + 3)^{2} + y^{2} + z^{2}] = (25 + 3 x)^{2}$	1
$25 x^{2} + 150 x + 225 + 25 y^{2} + 25 z^{2} = 625 + 150 x + 9 x^{2} \Rightarrow 16 x^{2} + 25 y^{2} + 25 z^{2} = 400$	1
(b) (5, 0, 0): 400 ✓ (distances 2 + 8 = 10); (0, 4, 0): $25 \times 16 = 400$ ✓ (distances 5 + 5 = 10)	1

Where students lose marks in Introduction to Three Dimensional Geometry

The slips examiners see most often, and what to write instead.

Mistake: Wrong octant numbering below the XY-plane
Guessing the order.
Do this: V to VIII repeat the order of I to IV with z negative: V (+, +, −), VI (−, +, −), VII (−, −, −), VIII (+, −, −).
Usually costs: 1 mark.
Mistake: Sign slips inside the distance formula
Subtracting a negative coordinate.
Wrong working
From (2, −3, 5) to (−2, 1, 3): $(1 - 3)^{2}$ for the y-part
Right working
$(1 - (- 3))^{2} = 16$
Do this: Write each difference in brackets first.
Usually costs: 1 mark.
Mistake: Collinearity by checking only two distances
Stopping after finding AB and BC.
Do this: Find all three and show the biggest equals the sum of the other two.
Usually costs: 1-2 marks.
Mistake: Confusing distance from an axis with distance from a plane
Both 'feel' like one coordinate.
Do this: From the XY-plane: |z|. From the z-axis: $x^{2} + y^{2}$ .
Usually costs: 1 mark.
Mistake: Not simplifying a locus fully
Leaving squares on both sides.
Do this: Expand both sides and cancel $x^{2}$ , $y^{2}$ , $z^{2}$ where possible.
Usually costs: 1 mark.
Mistake: Using the 3D section formula as the main method
Old notes.
Do this: The 3D section formula is formative-only for 2026-27; use distances (and simple mid-points) instead.
Usually costs: Usually none, but the expected method is distances.

NCERT exercise guide: Introduction to Three Dimensional Geometry

What each exercise in the NCERT chapter tests, how to approach it, and a question of the same kind solved step by step. Questions here are written in our own words.

Exercise 11.1 (axes, planes, octants)

What it tests: Coordinates of points on axes and planes, naming octants, fill-in-the-blank facts about the coordinate system.

How to approach it: Use the zero-coordinate rule and the octant sign table.

Same kind as Exercise 11.1

Name the octants of (1, −2, 3), (−1, −2, 3), (1, 2, −3) and (−1, 2, −3).

Show the solutionSolution

(+, −, +): IV; (−, −, +): III
(+, +, −): V; (−, +, −): VI

Answer: IV, III, V, VI

Exercise 11.2 (distance formula)

What it tests: Distances between points, collinearity, isosceles and right triangles, parallelograms, loci from equidistance and from a constant sum of distances.

How to approach it: Distance formula; compare squared distances; square to simplify loci.

Same kind as Exercise 11.2

Find the distance between (2, −1, 4) and (−1, 3, 4).

Show the solutionSolution

$(- 3)^{2} + 4^{2} + 0^{2}$
$= 25 = 5$

Answer: 5 units

Miscellaneous Exercise on Chapter 11

What it tests: Fourth vertex of a parallelogram, lengths of medians, unknown coordinates from a centroid, locus with $P A^{2} + P B^{2} = k^{2}$ .

How to approach it: Mid-points and centroids coordinate by coordinate; distance formula.

Same kind as Miscellaneous Exercise

Three vertices of parallelogram ABCD are A(1, 2, 3), B(4, 5, 6) and C(7, 3, 1). Find D.

Show the solutionSolution

Diagonals AC and BD bisect each other, so A + C = B + D
D = (1 + 7 − 4, 2 + 3 − 5, 3 + 1 − 6) = (4, 0, −2)

Answer: D(4, 0, −2)

Practice questions on Introduction to Three Dimensional Geometry

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQExam level
The distance of (3, 4, 5) from the origin is:
(a) 12
(b) $52$
(c) $12$
(d) 50
Hint
$50$ .
Answer
(b) $52$
Q2 · 1 mark · MCQExam level
The coordinate planes divide space into:
(a) 4 parts
(b) 6 parts
(c) 8 parts
(d) 3 parts
Hint
Octants.
Answer
(c) 8
Q3 · 1 mark · Assertion-reasonExam level
Assertion (A): The distance of (5, −2, 3) from the ZX-plane is 2.
Reason (R): The distance of a point from the ZX-plane is the modulus of its y-coordinate.
(a) Both true, R explains A. (b) Both true, R does not explain A. (c) A true, R false. (d) A false, R true.
Hint
|y|.
Answer
(a)
Q4 · 2 marks · Very short answerExam level
Write the coordinates of the feet of the perpendiculars from (4, −3, 2) to the three coordinate planes.
Hint
Set one coordinate to 0.
Answer
XY: (4, −3, 0); YZ: (0, −3, 2); ZX: (4, 0, 2).
Q5 · 1 mark · Very short answerHarder exam level
Find the distance between (−2, 4, −3) and (2, −4, 5).
Hint
Differences 4, −8, 8.
Answer
$16 + 64 + 64 = 144 = 12$ .
Q6 · 2 marks · Very short answerHarder exam level
Verify that O(0, 0, 0), A(2, 2, 1) and B(1, −2, 2) form an isosceles triangle. Is it also right-angled?
Hint
Find all three sides.
Answer
OA = $4 + 4 + 1 = 3$ , OB = $1 + 4 + 4 = 3$ , AB = $1 + 16 + 1 = 32$ . OA = OB: isosceles. Also $O A^{2} + O B^{2} = 18 = A B^{2}$ : right-angled at O.
Q7 · 2 marks · Very short answerHarder exam level
Find the equation of the set of points equidistant from (2, 1, 0) and (0, 1, 4).
Hint
Squares cancel.
Answer
$(x - 2)^{2} + z^{2} = x^{2} + (z - 4)^{2} \Rightarrow - 4 x + 4 = - 8 z + 16 \Rightarrow x - 2 z + 3 = 0$ .
Q8 · 3 marks · Short answerHarder exam level
Show that A(2, −1, 3), B(4, 3, 1) and C(8, 11, −3) lie on one straight line.
Hint
Three distances.
Answer
AB $= 4 + 16 + 4 = 26$ , BC $= 16 + 64 + 16 = 46$ , AC $= 36 + 144 + 36 = 66$ . AB + BC = AC, so collinear.
Q9 · 3 marks · Short answerHarder exam level
A point P is at distance 13 from the origin, 5 from the z-axis, and lies above the XY-plane with x = 3 and y > 0. Find P.
Hint
Use $x^{2} + y^{2} = 25$ then $z^{2}$ .
Answer
$9 + y^{2} = 25 \Rightarrow y = 4$ ; $25 + z^{2} = 169 \Rightarrow z = 12$ (above the plane). P = (3, 4, 12).
Q10 · 2 marks · Very short answerHOTS / competency
Find the equation of the sphere (set of points) at distance 5 from C(1, −2, 3), and decide whether (4, 2, 3) lies on it.
Hint
$P C^{2} = 25$ .
Answer
$(x - 1)^{2} + (y + 2)^{2} + (z - 3)^{2} = 25$ . For (4, 2, 3): $9 + 16 + 0 = 25$ , so yes.
Q11 · 3 marks · Short answerHOTS / competency
Explain why the set of points equidistant from two different points in space is always a plane, by showing that the equation has no squared terms.
Hint
Expand $P A^{2} = P B^{2}$ in general.
Answer
With $A (a_{1}, a_{2}, a_{3})$ , $B (b_{1}, b_{2}, b_{3})$ : $\sum (x - a_{i})^{2} = \sum (x - b_{i})^{2}$ . The $x^{2}$ , $y^{2}$ , $z^{2}$ terms appear on both sides and cancel, leaving $2 (b_{1} - a_{1}) x + 2 (b_{2} - a_{2}) y + 2 (b_{3} - a_{3}) z = ∣ B ∣^{2} - ∣ A ∣^{2}$ , a linear equation (not all coefficients zero since A ≠ B), i.e. a plane.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.How many octants are there?Show answer
Eight.
2.Form of a point on the z-axis?Show answer
(0, 0, z).
3.Which coordinate is zero in the ZX-plane?Show answer
y.
4.Octant of (4, −2, 3)?Show answer
IV.
5.Distance of (3, 4, 12) from the origin?Show answer
13.
6.Distance of (2, −3, 5) from the XY-plane?Show answer
5.
7.Locus of points equidistant from two fixed points?Show answer
A plane.
8.Is the 3D section formula in the 2026-27 annual exam?Show answer
No, it is formative-only.

Best order to study it

Picture the axes using a room corner.
Learn the three coordinate planes and which coordinate is zero on each.
Learn the eight octants with the sign table.
Learn how a point's coordinates are its distances from the planes.
Learn the distance formula and its box proof.
Practise collinearity and triangle-type questions.
Finish with loci and the Miscellaneous Exercise.

Quick revision checklist

The night before the exam: can you do each of these without looking?

Can you draw the three axes and label the coordinate planes?
Can you write the form of points on each axis and each plane?
Can you name the octant of any point?
Can you give the feet of perpendiculars from a point to the planes and axes?
Can you prove the distance formula?
Can you find the distance between two points without sign errors?
Can you show three points are collinear?
Can you test for isosceles, right-angled and equilateral triangles?
Can you find the locus of points equidistant from two points?
Can you find the fourth vertex of a parallelogram?

Introduction to Three Dimensional Geometry: frequently asked questions

Is the section formula in 3D in the 2026-27 exam?

No. The NCERT section on the section formula in three dimensions was removed, and CBSE lists it as assessed only formatively. The annual exam covers axes, planes, octants, coordinates and the distance formula.

Why is this chapter called an 'introduction'?

It sets up coordinates in space and the distance formula. Lines and planes in space, direction cosines and vectors come in Class 12, building on these ideas.

How do I remember the octant numbers?

Octants I to IV are above the XY-plane and follow the quadrant order of the plane (I: + +, II: − +, III: − −, IV: + −); V to VIII are directly below them in the same order.

Can a point lie in more than one octant?

No. A point with all coordinates non-zero lies in exactly one octant. If any coordinate is zero, the point lies on a coordinate plane and in no octant.

How is the Class 11 Maths exam set?

It is a school annual exam of 80 marks (3 hours) plus 20 marks of internal assessment; Unit III, Coordinate Geometry, which includes this chapter, carries 12 marks. CBSE gives no Class 11 sample paper, so schools usually mix MCQs, assertion-reason, short and long answers and case-based questions.

Based on: CBSE Mathematics (041) syllabus 2026-27 (Classes XI-XII); NCERT Mathematics, Textbook for Class XI, Chapter 11 (Reprint 2026-27)