NCERT Solutions for Class 11 Maths Chapter 11: Introduction to Three Dimensional Geometry

These Class 11 Maths Chapter 11 solutions cover Introduction to Three Dimensional Geometry from the latest NCERT textbook (Reprint 2026–27). Every question of Exercise 11.1, Exercise 11.2 and the Miscellaneous Exercise is solved step by step — coordinate axes and planes, octants and the distance formula in 3D — with each answer cross-checked against the book’s answer key.

Class: 11 Subject: Mathematics Chapter: 11 Chapter Name: Introduction to Three Dimensional Geometry Exercises: 11.1, 11.2 & Miscellaneous Session: 2026–27
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Chapter 11 Overview

To fix the position of a point in a plane we need two perpendicular axes, but in real space — a ball thrown in the air, the tip of a hanging bulb — we need three numbers. Chapter 11, Introduction to Three Dimensional Geometry, sets up the rectangular coordinate system in space: the three mutually perpendicular x, y and z axes, the three coordinate planes (XY, YZ, ZX), the eight octants they create, and how every point corresponds to an ordered triplet (x, y, z). It then extends the plane distance formula to the distance between two points in space and applies it to collinearity, types of triangles, parallelograms and loci. The Class 11 Maths Chapter 11 solutions below work through every exercise question and the Miscellaneous Exercise in full.

Key Concepts & Definitions

Coordinate axes: three mutually perpendicular lines X′OX, Y′OY, Z′OZ meeting at the origin O, called the x, y and z-axes.

Coordinate planes: the planes XY, YZ and ZX determined by pairs of axes. The XY-plane is z = 0, the YZ-plane is x = 0 and the ZX-plane is y = 0.

Coordinates of a point: a point P in space is written P(x, y, z), where x, y, z are the perpendicular distances of P from the YZ, ZX and XY planes respectively.

Octants: the three coordinate planes divide space into eight regions called octants (I–VIII). The signs of (x, y, z) decide the octant.

Points on axes/planes: a point on the x-axis is (x, 0, 0); on the y-axis (0, y, 0); on the z-axis (0, 0, z); a point in the XY-plane is (x, y, 0).

Collinear points: three points are collinear if the sum of the two shorter distances between them equals the longest distance (PQ + QR = PR).

Important Formulas (Chapter 11)

Distance between two points P(x1, y1, z1) and Q(x2, y2, z2):
PQ = √[(x2 − x1)2 + (y2 − y1)2 + (z2 − z1)2].

Distance from the origin O to Q(x, y, z): OQ = √(x2 + y2 + z2).

Signs in octants (I–VIII): the sign pattern of (x, y, z) is what determines the octant of a point.

Centroid of a triangle with vertices (x1, y1, z1), (x2, y2, z2), (x3, y3, z3):
G = ((x1 + x2 + x3)/3, (y1 + y2 + y3)/3, (z1 + z2 + z3)/3).

(The section / midpoint formula in 3D is introduced in the next chapter set; only the distance formula and the centroid idea are needed here.)

Exercise 11.1 Solutions

Questions are reproduced verbatim from the NCERT textbook; the worked solutions are original and verified against the answers given at the back of the book.

1. A point is on the x-axis. What are its y-coordinate and z-coordinates?

SOLUTION Any point lying on the x-axis has no displacement along the y-axis or the z-axis. So such a point is of the form (x, 0, 0). ∴ its y-coordinate and z-coordinate are both 0.

2. A point is in the XZ-plane. What can you say about its y-coordinate?

SOLUTION The XZ-plane (also called the ZX-plane) is the plane y = 0. Every point in this plane is at zero perpendicular distance from the ZX-plane along the y-direction. ∴ its y-coordinate is 0.

3. Name the octants in which the following points lie: (1, 2, 3), (4, –2, 3), (4, –2, –5), (4, 2, –5), (–4, 2, –5), (–4, 2, 5), (–3, –1, 6), (–2, –4, –7).

SOLUTION Use the sign table of the eight octants: I(+,+,+), II(−,+,+), III(−,−,+), IV(+,−,+), V(+,+,−), VI(−,+,−), VII(−,−,−), VIII(+,−,−). (1, 2, 3): signs (+,+,+) → Octant I. (4, –2, 3): signs (+,−,+) → Octant IV. (4, –2, –5): signs (+,−,−) → Octant VIII. (4, 2, –5): signs (+,+,−) → Octant V. (–4, 2, –5): signs (−,+,−) → Octant VI. (–4, 2, 5): signs (−,+,+) → Octant II. (–3, –1, 6): signs (−,−,+) → Octant III. (–2, –4, –7): signs (−,−,−) → Octant VII. ∴ the octants are I, IV, VIII, V, VI, II, III, VII respectively.

4. Fill in the blanks: (i) The x-axis and y-axis taken together determine a plane known as_______. (ii) The coordinates of points in the XY-plane are of the form _______. (iii) Coordinate planes divide the space into ______ octants.

SOLUTION (i) The plane containing the x-axis and y-axis is the XY-plane. (ii) A point in the XY-plane has z = 0, so its coordinates are of the form (x, y, 0). (iii) The three mutually perpendicular coordinate planes split space into eight (8) octants.

Exercise 11.2 Solutions

1. Find the distance between the following pairs of points: (i) (2, 3, 5) and (4, 3, 1) (ii) (–3, 7, 2) and (2, 4, –1) (iii) (–1, 3, –4) and (1, –3, 4) (iv) (2, –1, 3) and (–2, 1, 3).

SOLUTION Use PQ = √[(x2 − x1)2 + (y2 − y1)2 + (z2 − z1)2]. (i) = √[(4 − 2)2 + (3 − 3)2 + (1 − 5)2] = √(4 + 0 + 16) = √20 = 2√5 units. (ii) = √[(2 + 3)2 + (4 − 7)2 + (−1 − 2)2] = √(25 + 9 + 9) = √43 units. (iii) = √[(1 + 1)2 + (−3 − 3)2 + (4 + 4)2] = √(4 + 36 + 64) = √104 = 2√26 units. (iv) = √[(−2 − 2)2 + (1 + 1)2 + (3 − 3)2] = √(16 + 4 + 0) = √20 = 2√5 units.

2. Show that the points (–2, 3, 5), (1, 2, 3) and (7, 0, –1) are collinear.

SOLUTION Let A(−2, 3, 5), B(1, 2, 3), C(7, 0, −1). Find the three distances. AB = √[(1 + 2)2 + (2 − 3)2 + (3 − 5)2] = √(9 + 1 + 4) = √14. BC = √[(7 − 1)2 + (0 − 2)2 + (−1 − 3)2] = √(36 + 4 + 16) = √56 = 2√14. AC = √[(7 + 2)2 + (0 − 3)2 + (−1 − 5)2] = √(81 + 9 + 36) = √126 = 3√14. Since AB + BC = √14 + 2√14 = 3√14 = AC, the three points lie on one line. ∴ the points are collinear. (Here B lies between A and C.)

3. Verify the following: (i) (0, 7, –10), (1, 6, –6) and (4, 9, –6) are the vertices of an isosceles triangle. (ii) (0, 7, 10), (–1, 6, 6) and (–4, 9, 6) are the vertices of a right angled triangle. (iii) (–1, 2, 1), (1, –2, 5), (4, –7, 8) and (2, –3, 4) are the vertices of a parallelogram.

SOLUTION (i) Let A(0, 7, −10), B(1, 6, −6), C(4, 9, −6). AB = √[(1)2 + (−1)2 + (4)2] = √(1 + 1 + 16) = √18. BC = √[(3)2 + (3)2 + (0)2] = √(9 + 9 + 0) = √18. CA = √[(−4)2 + (−2)2 + (−4)2] = √(16 + 4 + 16) = √36 = 6. Since AB = BC = √18, two sides are equal, so the triangle is isosceles. Verified. ✓ (ii) Let A(0, 7, 10), B(−1, 6, 6), C(−4, 9, 6). AB2 = (−1)2 + (−1)2 + (−4)2 = 1 + 1 + 16 = 18. BC2 = (−3)2 + (3)2 + (0)2 = 9 + 9 + 0 = 18. AC2 = (−4)2 + (2)2 + (−4)2 = 16 + 4 + 16 = 36. Since AB2 + BC2 = 18 + 18 = 36 = AC2, the angle at B is 90°, so it is a right angled triangle. Verified. ✓ (iii) Let A(−1, 2, 1), B(1, −2, 5), C(4, −7, 8), D(2, −3, 4). AB = √[(2)2 + (−4)2 + (4)2] = √(4 + 16 + 16) = √36 = 6. BC = √[(3)2 + (−5)2 + (3)2] = √(9 + 25 + 9) = √43. CD = √[(−2)2 + (4)2 + (−4)2] = √(4 + 16 + 16) = √36 = 6. DA = √[(−3)2 + (5)2 + (−3)2] = √(9 + 25 + 9) = √43. Opposite sides are equal: AB = CD = 6 and BC = DA = √43. So ABCD is a parallelogram. Verified. ✓ (Diagonals: AC = √[(5)2 + (−9)2 + (7)2] = √155 and BD = √[(1)2 + (−1)2 + (−1)2] = √3 are unequal, so it is not a rectangle — a genuine parallelogram.)

4. Find the equation of the set of points which are equidistant from the points (1, 2, 3) and (3, 2, –1).

SOLUTION Let P(x, y, z) be equidistant from A(1, 2, 3) and B(3, 2, −1), so PA = PB, hence PA2 = PB2. (x − 1)2 + (y − 2)2 + (z − 3)2 = (x − 3)2 + (y − 2)2 + (z + 1)2. Expand: (x2 − 2x + 1) + (y − 2)2 + (z2 − 6z + 9) = (x2 − 6x + 9) + (y − 2)2 + (z2 + 2z + 1). Cancel x2, y, z2 terms: −2x + 1 − 6z + 9 = −6x + 9 + 2z + 1. −2x − 6z + 10 = −6x + 2z + 10 ⇒ 4x − 8z = 0. ∴ the required equation is x − 2z = 0 (a plane — the perpendicular bisector plane of AB).

5. Find the equation of the set of points P, the sum of whose distances from A (4, 0, 0) and B (–4, 0, 0) is equal to 10.

SOLUTION Let P(x, y, z). The condition is PA + PB = 10. √[(x − 4)2 + y2 + z2] + √[(x + 4)2 + y2 + z2] = 10. Shift one root and square: (x − 4)2 + y2 + z2 = 100 − 20√[(x + 4)2 + y2 + z2] + (x + 4)2 + y2 + z2. Since (x − 4)2 − (x + 4)2 = −16x, this gives 20√[(x + 4)2 + y2 + z2] = 100 + 16x, i.e. 5√[(x + 4)2 + y2 + z2] = 25 + 4x. Square again: 25[(x + 4)2 + y2 + z2] = (25 + 4x)2. 25x2 + 200x + 400 + 25y2 + 25z2 = 625 + 200x + 16x2. 9x2 + 25y2 + 25z2 − 225 = 0. ∴ the required equation is 9x2 + 25y2 + 25z2 − 225 = 0 (an ellipsoid of revolution).

Miscellaneous Exercise on Chapter 11 — Solutions

1. Three vertices of a parallelogram ABCD are A(3, –1, 2), B (1, 2, –4) and C (–1, 1, 2). Find the coordinates of the fourth vertex.

SOLUTION In a parallelogram ABCD the diagonals AC and BD bisect each other, so the midpoint of AC equals the midpoint of BD. Midpoint of AC = ((3 + (−1))/2, (−1 + 1)/2, (2 + 2)/2) = (1, 0, 2). Let D(x, y, z). Midpoint of BD = ((1 + x)/2, (2 + y)/2, (−4 + z)/2). Equate: (1 + x)/2 = 1 ⇒ x = 1; (2 + y)/2 = 0 ⇒ y = −2; (−4 + z)/2 = 2 ⇒ z = 8. ∴ the fourth vertex is D(1, –2, 8).

2. Find the lengths of the medians of the triangle with vertices A (0, 0, 6), B (0, 4, 0) and (6, 0, 0).

SOLUTION Let A(0, 0, 6), B(0, 4, 0), C(6, 0, 0). A median joins a vertex to the midpoint of the opposite side. Median AD, D = midpoint of BC = ((0 + 6)/2, (4 + 0)/2, (0 + 0)/2) = (3, 2, 0). AD = √[(3 − 0)2 + (2 − 0)2 + (0 − 6)2] = √(9 + 4 + 36) = √49 = 7. Median BE, E = midpoint of AC = (3, 0, 3). BE = √[(3 − 0)2 + (0 − 4)2 + (3 − 0)2] = √(9 + 16 + 9) = √34. Median CF, F = midpoint of AB = (0, 2, 3). CF = √[(0 − 6)2 + (2 − 0)2 + (3 − 0)2] = √(36 + 4 + 9) = √49 = 7. ∴ the medians have lengths 7, √34 and 7 units.

3. If the origin is the centroid of the triangle PQR with vertices P (2a, 2, 6), Q (–4, 3b, –10) and R(8, 14, 2c), then find the values of a, b and c.

SOLUTION The centroid is ((2a − 4 + 8)/3, (2 + 3b + 14)/3, (6 − 10 + 2c)/3), and it equals the origin (0, 0, 0). x: (2a + 4)/3 = 0 ⇒ 2a + 4 = 0 ⇒ a = −2. y: (3b + 16)/3 = 0 ⇒ 3b + 16 = 0 ⇒ b = −16/3. z: (2c − 4)/3 = 0 ⇒ 2c − 4 = 0 ⇒ c = 2. ∴ a = −2, b = −16/3, c = 2.

4. If A and B be the points (3, 4, 5) and (–1, 3, –7), respectively, find the equation of the set of points P such that PA2 + PB2 = k2, where k is a constant.

SOLUTION Let P(x, y, z). Then: PA2 = (x − 3)2 + (y − 4)2 + (z − 5)2. PB2 = (x + 1)2 + (y − 3)2 + (z + 7)2. Add: PA2 + PB2 = 2x2 + 2y2 + 2z2 − 4x − 14y + 4z + (9 + 16 + 25 + 1 + 9 + 49). The constant sum is 9 + 16 + 25 + 1 + 9 + 49 = 109, so PA2 + PB2 = 2x2 + 2y2 + 2z2 − 4x − 14y + 4z + 109. Setting this equal to k2: 2x2 + 2y2 + 2z2 − 4x − 14y + 4z + 109 = k2. Divide by 2 and rearrange: ∴ x2 + y2 + z2 − 2x − 7y + 2z = (k2 − 109)/2 (a sphere).

Common Mistakes to Avoid

Watch out for these

  • Confusing which coordinate is zero on a plane: the XY-plane is z = 0, the YZ-plane is x = 0, the ZX-plane is y = 0.
  • Reading the octant sign table wrongly — line up the signs (x, y, z) carefully with octants I–VIII.
  • Dropping a square or a sign while expanding (x − a)2 in distance / locus problems.
  • Calling a parallelogram a rectangle without checking the diagonals — equal diagonals are needed for a rectangle.
  • For collinearity, forgetting that you must show the sum of the two shorter distances equals the longest (PQ + QR = PR), not just compare squares.
  • Forgetting to simplify surds, e.g. leaving √20 instead of 2√5.

Practice MCQs & Assertion–Reason

1. The coordinate planes divide space into how many octants?

(a) 4    (b) 6    (c) 8    (d) 12

2. A point on the y-axis is of the form:

(a) (x, 0, 0)    (b) (0, y, 0)    (c) (0, 0, z)    (d) (x, y, 0)

3. The distance of the point (3, −4, 12) from the origin is:

(a) 11    (b) 12    (c) 13    (d) 14

4. The distance between (2, 3, 5) and (4, 3, 1) is:

(a) √5    (b) 2√5    (c) 4    (d) 20

5. The point (−3, −1, 6) lies in octant:

(a) I    (b) II    (c) III    (d) VII

6. The XY-plane is described by the equation:

(a) x = 0    (b) y = 0    (c) z = 0    (d) x = y

7. The centroid of the triangle with vertices (0, 0, 6), (0, 4, 0) and (6, 0, 0) is:

(a) (2, 2, 2)    (b) (3, 2, 2)    (c) (2, 4/3, 2)    (d) (6, 4, 6)

8. If a point is in the XZ-plane, then its:

(a) x = 0    (b) y = 0    (c) z = 0    (d) x = z

9. The points (−2, 3, 5), (1, 2, 3) and (7, 0, −1) are:

(a) vertices of a right triangle    (b) collinear    (c) vertices of an equilateral triangle    (d) coincident

10. The midpoint of the segment joining (0, 4, 0) and (6, 0, 0) is:

(a) (3, 2, 0)    (b) (6, 4, 0)    (c) (3, 0, 2)    (d) (0, 2, 3)

Answer key: 1-(c), 2-(b), 3-(c), 4-(b), 5-(c), 6-(c), 7-(c), 8-(b), 9-(b), 10-(a).

For each Assertion–Reason question, choose: (A) Both Assertion and Reason are true and the Reason is the correct explanation of the Assertion; (B) Both are true but the Reason is not the correct explanation; (C) Assertion is true but Reason is false; (D) Assertion is false but Reason is true.

A-R 1. Assertion: A point lying on the x-axis has the form (x, 0, 0).

Reason: On the x-axis the perpendicular distances along the y and z directions are both zero.

A-R 2. Assertion: The three coordinate planes divide space into eight octants.

Reason: The coordinate planes are the XY, YZ and ZX planes.

A-R 3. Assertion: The distance between (2, 3, 5) and (4, 3, 1) is 2√5 units.

Reason: The distance between P(x1, y1, z1) and Q(x2, y2, z2) is √[(x2 − x1)2 + (y2 − y1)2 + (z2 − z1)2].

A-R 4. Assertion: The point (4, −2, −5) lies in octant VIII.

Reason: Octant VIII has the sign pattern (+, +, −).

A-R 5. Assertion: The points (−2, 3, 5), (1, 2, 3) and (7, 0, −1) are collinear.

Reason: For these points the sum of the two smaller distances equals the largest distance.

Answer key: 1-(A), 2-(B), 3-(A), 4-(C), 5-(A).

Quick Revision Summary

  • Space needs three mutually perpendicular axes (x, y, z); a point is written as the triplet (x, y, z).
  • The three coordinate planes are XY (z = 0), YZ (x = 0) and ZX (y = 0); they split space into eight octants.
  • Points on axes: x-axis (x, 0, 0), y-axis (0, y, 0), z-axis (0, 0, z); a point in the XY-plane is (x, y, 0).
  • Distance: PQ = √[(x2 − x1)2 + (y2 − y1)2 + (z2 − z1)2]; from origin OQ = √(x2 + y2 + z2).
  • Three points are collinear if the sum of the two shorter distances equals the longest.
  • Centroid of a triangle is the average of the three vertices’ coordinates.
  • Equidistant-point loci give planes; sum/difference-of-distance loci give spheres, ellipsoids, etc.

How to score full marks in this chapter

Write the distance formula first, then substitute carefully — most marks are lost in arithmetic, not method. For octant questions, state the sign pattern you are matching. In locus problems, set PA2 = PB2 (or the given condition), expand fully, cancel the common square terms, and simplify to the lowest integer coefficients. Always simplify surds (e.g. √20 = 2√5) and label your final answer with units.

Frequently Asked Questions

What is Class 11 Maths Chapter 11 about?

Chapter 11, Introduction to Three Dimensional Geometry, introduces the rectangular coordinate system in space — the x, y and z axes, the three coordinate planes, the eight octants, the coordinates of a point as an ordered triplet (x, y, z), and the distance formula between two points in three dimensions.

How many exercises are there in Class 11 Maths Chapter 11?

There are two exercises — Exercise 11.1 (4 questions on axes, planes and octants) and Exercise 11.2 (5 questions on the distance formula) — plus a Miscellaneous Exercise of 4 questions. All of them are solved step by step on this page.

What is the distance formula in three dimensions?

The distance between P(x1, y1, z1) and Q(x2, y2, z2) is PQ = √[(x2 − x1)2 + (y2 − y1)2 + (z2 − z1)2]. From the origin, the distance to (x, y, z) is √(x2 + y2 + z2).

Are these Class 11 Maths Chapter 11 solutions free?

Yes. All solutions are free and follow the official NCERT Class 11 Mathematics textbook for the 2026–27 session, with every answer verified against the book’s answer key.

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