Home › NCERT Solutions › Class 11 Maths › Chapter 11

NCERT Solutions · Class 11 Maths · Chapter 11

Chapter 11: Introduction to Three Dimensional Geometry (Coordinate Geometry)

Step-by-step solutions to all exercises of Chapter 11, Introduction to Three Dimensional Geometry (NCERT Class 11 Maths, 2026-27): coordinate axes and octants, distance formula, collinearity and the Miscellaneous Exercise. All 13 questions are answered, with the key answer highlighted.

NCERT textbook Chapter 11 (PDF)

Exam revision? Chapter 11 notes with key points, diagrams and practice questions: see the free sample · ₹199 for all 14 chapters

Free NCERT solutions by Notes Bazar · www.notesbazar.in/ncert-solutions/class-11-maths/chapter-11-introduction-to-three-dimensional-geometry

Distance $P Q = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2} + (z_{2} - z_{1})^{2}$ . A point on the x-axis is (x, 0, 0); in the XY-plane, z = 0. Octant signs follow the signs of x, y, z.

Exercise 11.1

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

Solution

On the x-axis a point is at zero distance from the ZX- and XY-planes: y = 0 and z = 0; it is (x, 0, 0).

y = 0 and z = 0

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

Solution

Its distance from the ZX-plane is zero: y = 0; it is (x, 0, z).

y = 0

Name the octants of (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

Read the signs (x, y, z) from Table 11.1: (1, 2, 3): (+, +, +) I; (4, −2, 3): (+, −, +) IV; (4, −2, −5): (+, −, −) VIII; (4, 2, −5): (+, +, −) V; (−4, 2, −5): (−, +, −) VI; (−4, 2, 5): (−, +, +) II; (−3, −1, 6): (−, −, +) III; (−2, −4, −7): (−, −, −) VII.

I; IV; VIII; V; VI; II; III; VII

Fill in the blanks: (i) The x-axis and y-axis together determine the ______. (ii) Points in the XY-plane have the form ______. (iii) Coordinate planes divide space into ______ octants.

Solution

(i) XY-plane (ii) (x, y, 0) (iii) eight.

(i) XY-plane; (ii) (x, y, 0); (iii) eight

Exercise 11.2

Find the distance between (i) (2, 3, 5), (4, 3, 1) (ii) (−3, 7, 2), (2, 4, −1) (iii) (−1, 3, −4), (1, −3, 4) (iv) (2, −1, 3), (−2, 1, 3).

Solution

(i) √(2² + 0² + (−4)²) = √20 = 2√5. (ii) √(5² + (−3)² + (−3)²) = √(25 + 9 + 9) = √43. (iii) √(2² + (−6)² + 8²) = √(4 + 36 + 64) = √104 = 2√26. (iv) √((−4)² + 2² + 0²) = √20 = 2√5.

(i) 2√5; (ii) √43; (iii) 2√26; (iv) 2√5

Show that (−2, 3, 5), (1, 2, 3) and (7, 0, −1) are collinear.

Solution

P(−2, 3, 5), Q(1, 2, 3), R(7, 0, −1): PQ = √(9 + 1 + 4) = √14; QR = √(36 + 4 + 16) = √56 = 2√14; PR = √(81 + 9 + 36) = √126 = 3√14. Since PQ + QR = √14 + 2√14 = 3√14 = PR, P, Q, R are collinear.

P, Q, R are collinear

Verify: (i) (0, 7, −10), (1, 6, −6), (4, 9, −6) form an isosceles triangle. (ii) (0, 7, 10), (−1, 6, 6), (−4, 9, 6) form a right-angled triangle. (iii) (−1, 2, 1), (1, −2, 5), (4, −7, 8), (2, −3, 4) form a parallelogram.

Solution

(i) A(0, 7, −10), B(1, 6, −6), C(4, 9, −6): AB = √(1 + 1 + 16) = √18 = 3√2; BC = √(9 + 9 + 0) = 3√2; CA = √(16 + 4 + 16) = 6. AB = BC: isosceles. (ii) A(0, 7, 10), B(−1, 6, 6), C(−4, 9, 6): AB² = 1 + 1 + 16 = 18; BC² = 9 + 9 + 0 = 18; AC² = 16 + 4 + 16 = 36. AB² + BC² = 36 = AC²: right angle at B. (iii) A(−1, 2, 1), B(1, −2, 5), C(4, −7, 8), D(2, −3, 4): AB = √(4 + 16 + 16) = 6; BC = √(9 + 25 + 9) = √43; CD = √(4 + 16 + 16) = 6; DA = √(9 + 25 + 9) = √43. Opposite sides equal: a parallelogram.

(i) isosceles; (ii) right angle at B; (iii) a parallelogram

Find the equation of the set of points equidistant from (1, 2, 3) and (3, 2, −1).

Solution

(x − 1)² + (y − 2)² + (z − 3)² = (x − 3)² + (y − 2)² + (z + 1)² ⇒ −2x + 1 − 6z + 9 = −6x + 9 + 2z + 1 ⇒ 4x − 8z = 0 ⇒ x − 2z = 0.

x − 2z = 0

Find the equation of the set of points P with PA + PB = 10, where A(4, 0, 0), B(−4, 0, 0).

Solution

√((x − 4)² + y² + z²) = 10 − √((x + 4)² + y² + z²). Squaring: (x − 4)² = 100 − 20√((x + 4)² + y² + z²) + (x + 4)² ⇒ 20√(…) = 100 + 16x ⇒ 5√(…) = 25 + 4x. Squaring again: 25(x² + 8x + 16 + y² + z²) = 625 + 200x + 16x² ⇒ 9x² + 25y² + 25z² − 225 = 0.

9x² + 25y² + 25z² − 225 = 0

Miscellaneous Exercise on Chapter 11

Three vertices of parallelogram ABCD are A(3, −1, 2), B(1, 2, −4), C(−1, 1, 2). Find D.

Solution

The diagonals AC and BD bisect each other: midpoint of AC = (1, 0, 2) = midpoint of BD = ((1 + x)/2, (2 + y)/2, (−4 + z)/2) ⇒ x = 1, y = −2, z = 8: D(1, −2, 8).

D(1, −2, 8)

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

Solution

Midpoints: BC (3, 2, 0), CA (3, 0, 3), AB (0, 2, 3). Median from A: √(9 + 4 + 36) = 7; from B: √(9 + 16 + 9) = √34; from C: √(36 + 4 + 9) = 7.

7; √34

The origin is the centroid of ΔPQR with P(2a, 2, 6), Q(−4, 3b, −10), R(8, 14, 2c). Find a, b, c.

Solution

(2a − 4 + 8)/3 = 0 ⇒ a = −2; (2 + 3b + 14)/3 = 0 ⇒ b = −16/3; (6 − 10 + 2c)/3 = 0 ⇒ c = 2.

a = −2; b = −16/3; c = 2

A(3, 4, 5), B(−1, 3, −7). Find the set of points P with PA² + PB² = k².

Solution

As in Example 6: 2x² + 2y² + 2z² − 4x − 14y + 4z + 109 = k², i.e. x² + y² + z² − 2x − 7y + 2z = (k² − 109)/2.

x² + y² + z² − 2x − 7y + 2z = (k² − 109)/2

Cover of Complete Maths Notes for Class 11 (NCERT 2026-27)

Done with the answers?Now revise the whole chapter: Class 11 Mathematics notes

For NCERT Mathematics Class 11 (2026-27 edition) - all 14 chapters
Typed concept notes with every formula, theorem and method, with worked examples
Every NCERT exercise and miscellaneous question solved step by step

₹199₹24920% off · all 14 chapters · instant PDF

Read free sample

What is inside the Chapter 11 notes →

← Chapter 10: Conic Sections Chapter 12: Limits and Derivatives →

More free NCERT solutions for Class 11Class 11 Physics · 14 chapters Class 11 Chemistry · 9 chapters Class 11 Biology · 19 chapters All classes →

Found a mistake or need help with a question? Message us on WhatsApp. Free study tips and new solutions: join our WhatsApp Channel.