Class 11 Maths NCERT Solutions

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Introduction to Three Dimensional Geometry Class 11 Maths NCERT Solutions

The complete NCERT exercise solutions for Chapter 11, Introduction to Three Dimensional Geometry — 9 questions from Ex 11.1 to Ex 11.2, each worked through step by step in the CBSE marking pattern. Coordinates in space, the distance formula, the section formula, direction cosines and the angle between two lines.

Class:11Subject:MathsChapter:11

4 Key Formulas

DWritten byDeep Narayan

UpdatedSeptember 2026

Key Concept Summary

How many questions are in NCERT Class 11 Maths Chapter 11?

Chapter 11 carries 2 exercise questions, numbered Ex 11.1 to Ex 11.2. All of them are solved step by step on this page, along with the chapter's key formulas and exam pointers.

Chapter Overview

Three axes — x, y and z — cut space into eight octants. A point is located by its perpendicular distances from the three coordinate planes, written as (x, y, z). In this chapter, the key skill is reading signs to name the octant, and the key formula is the distance between two points in space, which extends the two-dimensional distance formula by one more squared difference.

Board pattern

The distance between (x₁, y₁, z₁) and (x₂, y₂, z₂) is √((x₂ − x₁)² + (y₂ − y₁)² + (z₂ − z₁)²). Signs decide the octant: all three positive is octant I, then the octants run anti-clockwise on top and bottom, so (+,+,−) is V, (−,+,−) is VI, (−,−,−) is VII and (+,−,−) is VIII. Locus questions are solved by writing the given distance condition with squares, then simplifying until x, y and z appear in one clean equation.

$Ex 11.1 — Planes, octants, coordinates (4 Q)$
$Ex 11.2 — Distance between points (5 Q)$
$Misc — Triangles, parallelograms, locus (4 Q)$

Exercise 11.1 — Coordinates, Planes and Octants

4Exercise questions

Step-by-step solution

1On the x-axis, the point can move only along x; y and z stay zero.
2Any point on the x-axis has the form (x, 0, 0).

Final answer

y-coordinate = 0 and z-coordinate = 0.

Step-by-step solution

1The XZ-plane is the set of points where the y-coordinate is zero.
2A point (x, y, z) lies in the XZ-plane exactly when y = 0.

Final answer

The y-coordinate is 0.

Step-by-step solution

1Sign pattern (+,+,+) is octant I.
2(4, −2, 3): (+, −, +) is octant IV.
3(4, −2, −5): (+, −, −) is octant VIII.
4(4, 2, −5): (+, +, −) is octant V.
5(−4, 2, −5): (−, +, −) is octant VI.
6(−4, 2, 5): (−, +, +) is octant II.
7(−3, −1, 6): (−, −, +) is octant III.
8(−2, −4, −7): (−, −, −) is octant VII.

Final answer

I, IV, VIII, V, VI, II, III, VII respectively.

Step-by-step solution

1The x-axis and y-axis together determine the XY-plane.
2A point in the XY-plane has zero z-coordinate, so its form is (x, y, 0).
3Three mutually perpendicular planes divide space into 2 × 2 × 2 = 8 octants.

Final answer

(i) XY-plane, (ii) (x, y, 0), (iii) eight (8).

Exercise 11.2 — Distance Between Two Points

5Exercise questions

Step-by-step solution

1(i) d = √((4−2)² + (3−3)² + (1−5)²) = √(4 + 0 + 16) = √20 = 2√5.
2(ii) d = √((2+3)² + (4−7)² + (−1−2)²) = √(25 + 9 + 9) = √43.
3(iii) d = √((1+1)² + (−3−3)² + (4+4)²) = √(4 + 36 + 64) = √104 = 2√26.
4(iv) d = √((−2−2)² + (1+1)² + (3−3)²) = √(16 + 4 + 0) = √20 = 2√5.

Final answer

(i) 2√5, (ii) √43, (iii) 2√26, (iv) 2√5.

Step-by-step solution

1Label A(−2, 3, 5), B(1, 2, 3), C(7, 0, −1).
2AB = √((1+2)² + (2−3)² + (3−5)²) = √(9 + 1 + 4) = √14.
3BC = √((7−1)² + (0−2)² + (−1−3)²) = √(36 + 4 + 16) = √56 = 2√14.
4AC = √((7+2)² + (0−3)² + (−1−5)²) = √(81 + 9 + 36) = √126 = 3√14.
5AB + BC = √14 + 2√14 = 3√14 = AC, so B lies between A and C.

Final answer

Collinear — shown by AB + BC = AC.

Step-by-step solution

1(i) A(0,7,−10), B(1,6,−6), C(4,9,−6): AB = √(1+1+16) = √18 = 3√2; BC = √(9+9+0) = √18 = 3√2; AC = √(16+4+16) = 6.
2(i) AB = BC ≠ AC → isosceles triangle. ▲
3(ii) A(0,7,10), B(−1,6,6), C(−4,9,6): AB = √18 = 3√2; BC = √(9+9+0) = √18 = 3√2; AC = √(16+4+16) = 6.
4(ii) (3√2)² + (3√2)² = 18 + 18 = 36 = 6² → right angled at B.
5(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 = √((2−4)² + (−3+7)² + (4−8)²) = √36 = 6; DA = √((−1−2)² + (2+3)² + (1−4)²) = √43.
6(iii) AB = CD and BC = DA → opposite sides equal → parallelogram.

Final answer

(i) isosceles, (ii) right angled, (iii) parallelogram — all verified.

Step-by-step solution

1Let P(x, y, z) be equidistant from (1,2,3) and (3,2,−1).
2(x−1)² + (y−2)² + (z−3)² = (x−3)² + (y−2)² + (z+1)².
3The (y−2)² terms cancel. Expand x and z terms: x² −2x +1 + z² −6z +9 = x² −6x +9 + z² +2z +1.
4−2x −6z +10 = −6x +2z +10 → 4x − 8z = 0 → x = 2z.

Final answer

x − 2z = 0 (the perpendicular bisector plane).

Step-by-step solution

1Let P(x, y, z) with PA + PB = 10.
2PA² = (x−4)² + y² + z² and PB² = (x+4)² + y² + z². Write r² = x² + y² + z² + 16, so PA² = r² − 8x, PB² = r² + 8x.
3(PA + PB)² = 100 gives PA² + PB² + 2·PA·PB = 100 → 2r² + 2·PA·PB = 100 → PA·PB = 50 − r².
4Square: (r² − 8x)(r² + 8x) = (50 − r²)² → r⁴ − 64x² = 2500 − 100r² + r⁴.
5100r² = 2500 + 64x² → 100(x² + y² + z² + 16) = 2500 + 64x² → 36x² + 100y² + 100z² = 900.
6Divide by 4 and multiply out: 9x² + 25y² + 25z² = 225.

Final answer

9x² + 25y² + 25z² = 225.

Quick Revision

Key formulas at a glance

Memorise these equations — direct application numericals and derivations in CBSE & JEE frequently hinge on these.

Distance between two points

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

Section formula

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})

Direction cosines

l^{2} + m^{2} + n^{2} = 1

Angle between two lines

cos θ = l_{1} l_{2} + m_{1} m_{2} + n_{1} n_{2}

Exam Strategy

How this chapter is asked

High-yield question patterns observed across CBSE boards, JEE Main & Advanced, and NEET.

Direction cosines must satisfy l² + m² + n² = 1, so ratios of direction ratios have to be normalised before any angle can be computed.
The distance formula in space is the plane formula with a third squared term — nothing else changes.

FAQ

Frequently asked questions

How many questions are in NCERT Class 11 Maths Chapter 11 (Introduction to Three Dimensional Geometry)?

There are 2 exercise questions in this chapter, numbered Ex 11.1 to Ex 11.2. Every one is solved step by step on this page in the official NCERT numbering.

Which formulas come up in Introduction to Three Dimensional Geometry Class 11 Maths?

The formulas this chapter's questions actually turn on are: Distance between two points, Section formula, Direction cosines, Angle between two lines. They are listed with their expressions in the key formulas section below, and the solved questions show where each one is used.

Is Introduction to Three Dimensional Geometry important for JEE Main?

Moderate — short and formulaic, worth one or two marks in boards, but a stated prerequisite for the line and plane geometry of Class 12.

Next Chapters

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