NCERT Solutions for Class 12 Maths Chapter 11: Three Dimensional Geometry (NCERT 2026–27)

These Class 12 Maths Chapter 11 solutions cover Three Dimensional Geometry from the NCERT textbook (Reprint 2026–27). Every question of Exercise 11.1, Exercise 11.2 and the Miscellaneous Exercise is reproduced verbatim and solved step by step — direction cosines, direction ratios, equations of lines in space, angle between two lines and shortest distance between skew and parallel lines — with every answer cross-checked against the NCERT answer key.

Class: 12 Subject: Mathematics Chapter: 11 Chapter name: Three Dimensional Geometry Exercises: 11.1, 11.2, Miscellaneous Session: 2026–27
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Chapter 11 Overview

Chapter 11, Three Dimensional Geometry, extends vector algebra to lines in space. It begins with the direction cosines (l, m, n) and direction ratios (a, b, c) of a directed line, and the relation l2 + m2 + n2 = 1. It then derives the vector and Cartesian equations of a line — through a point parallel to a vector, and through two points — before moving to the angle between two lines using direction ratios or direction cosines. The chapter closes with the shortest distance between two skew lines (the scalar triple product formula) and the distance between parallel lines. The Class 12 Maths Chapter 11 solutions below solve every textbook exercise in order.

Key Concepts & Definitions

Direction cosines: if a directed line makes angles α, β, γ with the positive x, y, z-axes, then l = cosα, m = cosβ, n = cosγ are its direction cosines, and l2 + m2 + n2 = 1.

Direction ratios: any three numbers a, b, c proportional to l, m, n. From them l = ±a/√(a2+b2+c2), and similarly for m and n.

Direction cosines of a join: for P(x1, y1, z1) and Q(x2, y2, z2), direction ratios are x2−x1, y2−y1, z2−z1.

Skew lines: lines in space that are neither parallel nor intersecting; they lie in different planes.

Collinear / parallel test: three points are collinear, or two lines parallel, when their direction ratios are proportional; two lines are perpendicular when a1a2 + b1b2 + c1c2 = 0.

Important Formulas (Chapter 11)

Direction cosines: l2 + m2 + n2 = 1.

Vector equation of a line: r = a + λb (through point a, parallel to b).

Cartesian equation of a line: (x−x1)/a = (y−y1)/b = (z−z1)/c.

Angle between two lines (d.r.): cosθ = |a1a2 + b1b2 + c1c2| ÷ (√(a12+b12+c12) · √(a22+b22+c22)).

Shortest distance (skew): d = |(b1 × b2) · (a2 − a1)| ÷ |b1 × b2|.

Distance (parallel): d = |b × (a2 − a1)| ÷ |b|.

Exercise 11.1 Solutions

Questions are reproduced verbatim from the NCERT textbook; the worked solutions are original and verified against the book’s answer key.

1. If a line makes angles 90°, 135°, 45° with the x, y and z-axes respectively, find its direction cosines.

SOLUTION l = cos 90° = 0. m = cos 135° = −1/√2. n = cos 45° = 1/√2. Check: 0 + 1/2 + 1/2 = 1. ✓ ∴ direction cosines are 0, −1/√2, 1/√2.

2. Find the direction cosines of a line which makes equal angles with the coordinate axes.

SOLUTION Equal angles ⇒ l = m = n = cosα. Using l2 + m2 + n2 = 1: 3l2 = 1, so l = ±1/√3. ∴ direction cosines are ±1/√3, ±1/√3, ±1/√3.

3. If a line has the direction ratios −18, 12, −4, then what are its direction cosines?

SOLUTION √(a2+b2+c2) = √((−18)2 + 122 + (−4)2) = √(324 + 144 + 16) = √484 = 22. Divide each ratio by 22: −18/22, 12/22, −4/22. ∴ direction cosines are −9/11, 6/11, −2/11.

4. Show that the points (2, 3, 4), (−1, −2, 1), (5, 8, 7) are collinear.

SOLUTION Let A(2, 3, 4), B(−1, −2, 1), C(5, 8, 7). Direction ratios of AB: (−1−2, −2−3, 1−4) = (−3, −5, −3). Direction ratios of AC: (5−2, 8−3, 7−4) = (3, 5, 3). AC = −1 × AB, i.e. the ratios are proportional and A is common. ∴ A, B, C are collinear.

5. Find the direction cosines of the sides of the triangle whose vertices are (3, 5, −4), (−1, 1, 2) and (−5, −5, −2).

SOLUTION Let A(3, 5, −4), B(−1, 1, 2), C(−5, −5, −2). Side AB: d.r. = (−1−3, 1−5, 2+4) = (−4, −4, 6); magnitude = √(16+16+36) = √68 = 2√17. d.c. of AB = (−4, −4, 6)/(2√17) = (−2/√17, −2/√17, 3/√17). Side BC: d.r. = (−5+1, −5−1, −2−2) = (−4, −6, −4); magnitude = √(16+36+16) = √68 = 2√17. d.c. of BC = (−4, −6, −4)/(2√17) = (−2/√17, −3/√17, −2/√17). Side CA: d.r. = (3+5, 5+5, −4+2) = (8, 10, −2); magnitude = √(64+100+4) = √168 = 2√42. d.c. of CA = (8, 10, −2)/(2√42) = (4/√42, 5/√42, −1/√42).

Exercise 11.2 Solutions

1. Show that the three lines with direction cosines 12/13, −3/13, −4/13; 4/13, 12/13, 3/13; 3/13, −4/13, 12/13 are mutually perpendicular.

SOLUTION Two lines with d.c. (l1, m1, n1) and (l2, m2, n2) are perpendicular if l1l2 + m1m2 + n1n2 = 0. Lines 1 & 2: (12×4 + (−3)×12 + (−4)×3)/169 = (48 − 36 − 12)/169 = 0. ✓ Lines 2 & 3: (4×3 + 12×(−4) + 3×12)/169 = (12 − 48 + 36)/169 = 0. ✓ Lines 1 & 3: (12×3 + (−3)×(−4) + (−4)×12)/169 = (36 + 12 − 48)/169 = 0. ✓ ∴ the three lines are mutually perpendicular.

2. Show that the line through the points (1, −1, 2), (3, 4, −2) is perpendicular to the line through the points (0, 3, 2) and (3, 5, 6).

SOLUTION d.r. of first line: (3−1, 4+1, −2−2) = (2, 5, −4). d.r. of second line: (3−0, 5−3, 6−2) = (3, 2, 4). a1a2 + b1b2 + c1c2 = 2×3 + 5×2 + (−4)×4 = 6 + 10 − 16 = 0. ∴ the lines are perpendicular.

3. Show that the line through the points (4, 7, 8), (2, 3, 4) is parallel to the line through the points (−1, −2, 1), (1, 2, 5).

SOLUTION d.r. of first line: (2−4, 3−7, 4−8) = (−2, −4, −4). d.r. of second line: (1+1, 2+2, 5−1) = (2, 4, 4). First = −1 × second, so the ratios are proportional. ∴ the lines are parallel.

4. Find the equation of the line which passes through the point (1, 2, 3) and is parallel to the vector 3î + 2ĵ − 2k̂.

SOLUTION Here a = î + 2ĵ + 3k̂ and b = 3î + 2ĵ − 2k̂. Vector form: r = (î + 2ĵ + 3k̂) + λ(3î + 2ĵ − 2k̂), where λ is a real number. Cartesian form: (x−1)/3 = (y−2)/2 = (z−3)/(−2).

5. Find the equation of the line in vector and in cartesian form that passes through the point with position vector 2î − j + 4k̂ and is in the direction î + 2ĵ − k̂.

SOLUTION a = 2î − ĵ + 4k̂,   b = î + 2ĵ − k̂. Vector form: r = (2î − ĵ + 4k̂) + λ(î + 2ĵ − k̂). Cartesian form: (x−2)/1 = (y+1)/2 = (z−4)/(−1).

6. Find the cartesian equation of the line which passes through the point (−2, 4, −5) and parallel to the line given by (x+3)/3 = (y−4)/5 = (z+8)/6.

SOLUTION Parallel lines share direction ratios, so d.r. = 3, 5, 6. Line through (−2, 4, −5): (x+2)/3 = (y−4)/5 = (z+5)/6. ∴ (x+2)/3 = (y−4)/5 = (z+5)/6.

7. The cartesian equation of a line is (x−5)/3 = (y+4)/7 = (z−6)/2. Write its vector form.

SOLUTION The line passes through (5, −4, 6) with d.r. 3, 7, 2. So a = 5î − 4ĵ + 6k̂ and b = 3î + 7ĵ + 2k̂. Vector form: r = (5î − 4ĵ + 6k̂) + λ(3î + 7ĵ + 2k̂).

8. Find the angle between the following pairs of lines: (i) r = 2î − 5ĵ + k̂ + λ(3î + 2ĵ + 6k̂) and r = 7î − 6k̂ + μ(î + 2ĵ + 2k̂) (ii) r = 3î + ĵ − 2k̂ + λ(î − ĵ − 2k̂) and r = 2î − ĵ − 56k̂ + μ(3î − 5ĵ − 4k̂)

SOLUTION (i) b1 = (3, 2, 6), b2 = (1, 2, 2). b1·b2 = 3 + 4 + 12 = 19; |b1| = √(9+4+36) = 7; |b2| = √(1+4+4) = 3. cosθ = 19/(7×3) = 19/21, so θ = cos−1(19/21). (ii) b1 = (1, −1, −2), b2 = (3, −5, −4). b1·b2 = 3 + 5 + 8 = 16; |b1| = √(1+1+4) = √6; |b2| = √(9+25+16) = √50 = 5√2. cosθ = 16/(√6 × 5√2) = 16/(5√12) = 16/(10√3) = 8/(5√3), so θ = cos−1(8/(5√3)).

9. Find the angle between the following pair of lines: (i) (x−2)/2 = (y−1)/5 = (z+3)/(−3) and (x+2)/(−1) = (y−4)/8 = (z−5)/4 (ii) x/2 = y/2 = z/1 and (x−5)/4 = (y−2)/1 = (z−3)/8

SOLUTION (i) d.r.: (2, 5, −3) and (−1, 8, 4). Dot = −2 + 40 − 12 = 26; magnitudes = √(4+25+9) = √38 and √(1+64+16) = √81 = 9. cosθ = 26/(9√38), so θ = cos−1(26/(9√38)). (ii) d.r.: (2, 2, 1) and (4, 1, 8). Dot = 8 + 2 + 8 = 18; magnitudes = √(4+4+1) = 3 and √(16+1+64) = √81 = 9. cosθ = 18/(3×9) = 18/27 = 2/3, so θ = cos−1(2/3).

10. Find the values of p so that the lines (1−x)/3 = (7y−14)/(2p) = (z−3)/2 and (7−7x)/(3p) = (y−5)/1 = (6−z)/5 are at right angles.

SOLUTION Rewrite each in standard form (x−x1)/a = … Line 1: (1−x)/3 = (x−1)/(−3); (7y−14)/(2p) = (y−2)/(2p/7); (z−3)/2. d.r. of line 1: (−3, 2p/7, 2). Line 2: (7−7x)/(3p) = (x−1)/(−3p/7); (y−5)/1; (6−z)/5 = (z−6)/(−5). d.r. of line 2: (−3p/7, 1, −5). Perpendicular ⇒ (−3)(−3p/7) + (2p/7)(1) + (2)(−5) = 0. 9p/7 + 2p/7 − 10 = 0 ⇒ 11p/7 = 10 ⇒ p = 70/11.

11. Show that the lines (x−5)/7 = (y+2)/(−5) = z/1 and x/1 = y/2 = z/3 are perpendicular to each other.

SOLUTION d.r.: (7, −5, 1) and (1, 2, 3). Dot = 7×1 + (−5)×2 + 1×3 = 7 − 10 + 3 = 0. ∴ the lines are perpendicular.

12. Find the shortest distance between the lines r = (î + 2ĵ + k̂) + λ(î − ĵ + k̂) and r = 2î − ĵ − k̂ + μ(2î + ĵ + 2k̂).

SOLUTION a1 = (1, 2, 1), b1 = (1, −1, 1); a2 = (2, −1, −1), b2 = (2, 1, 2). a2 − a1 = (1, −3, −2). b1 × b2 = |i j k; 1 −1 1; 2 1 2| = ((−1)(2)−(1)(1))î − ((1)(2)−(1)(2))ĵ + ((1)(1)−(−1)(2))k̂ = −3î + 0ĵ + 3k̂. |b1 × b2| = √(9 + 0 + 9) = √18 = 3√2. (b1 × b2)·(a2−a1) = (−3)(1) + 0(−3) + 3(−2) = −3 + 0 − 6 = −9. d = |−9|/(3√2) = 9/(3√2) = 3/√2 = 3√2/2.

13. Find the shortest distance between the lines (x+1)/7 = (y+1)/(−6) = (z+1)/1 and (x−3)/1 = (y−5)/(−2) = (z−7)/1.

SOLUTION a1 = (−1, −1, −1), b1 = (7, −6, 1); a2 = (3, 5, 7), b2 = (1, −2, 1). a2 − a1 = (4, 6, 8). b1 × b2 = |i j k; 7 −6 1; 1 −2 1| = ((−6)(1)−(1)(−2))î − ((7)(1)−(1)(1))ĵ + ((7)(−2)−(−6)(1))k̂ = −4î − 6ĵ − 8k̂. |b1 × b2| = √(16 + 36 + 64) = √116 = 2√29. (b1 × b2)·(a2−a1) = (−4)(4) + (−6)(6) + (−8)(8) = −16 − 36 − 64 = −116. d = |−116|/(2√29) = 116/(2√29) = 58/√29 = 58√29/29 = 2√29.

14. Find the shortest distance between the lines whose vector equations are r = (î + 2ĵ + 3k̂) + λ(î − 3ĵ + 2k̂) and r = 4î + 5ĵ + 6k̂ + μ(2î + 3ĵ + k̂).

SOLUTION a1 = (1, 2, 3), b1 = (1, −3, 2); a2 = (4, 5, 6), b2 = (2, 3, 1). a2 − a1 = (3, 3, 3). b1 × b2 = |i j k; 1 −3 2; 2 3 1| = ((−3)(1)−(2)(3))î − ((1)(1)−(2)(2))ĵ + ((1)(3)−(−3)(2))k̂ = −9î + 3ĵ + 9k̂. |b1 × b2| = √(81 + 9 + 81) = √171 = 3√19. (b1 × b2)·(a2−a1) = (−9)(3) + (3)(3) + (9)(3) = −27 + 9 + 27 = 9. d = |9|/(3√19) = 3/√19 = 3/√19 (= 3√19/19).

15. Find the shortest distance between the lines whose vector equations are r = (1−t)î + (t−2)ĵ + (3−2t)k̂ and r = (s+1)î + (2s−1)ĵ − (2s+1)k̂.

SOLUTION Write each in a + (param)b form. Line 1: a1 = (1, −2, 3), b1 = (−1, 1, −2). Line 2: a2 = (1, −1, −1), b2 = (1, 2, −2). a2 − a1 = (0, 1, −4). b1 × b2 = |i j k; −1 1 −2; 1 2 −2| = ((1)(−2)−(−2)(2))î − ((−1)(−2)−(−2)(1))ĵ + ((−1)(2)−(1)(1))k̂ = 2î − 4ĵ − 3k̂. |b1 × b2| = √(4 + 16 + 9) = √29. (b1 × b2)·(a2−a1) = (2)(0) + (−4)(1) + (−3)(−4) = 0 − 4 + 12 = 8. d = |8|/√29 = 8/√29 (= 8√29/29).

Miscellaneous Exercise on Chapter 11 Solutions

1. Find the angle between the lines whose direction ratios are a, b, c and b − c, c − a, a − b.

SOLUTION Sum of products = a(b−c) + b(c−a) + c(a−b). = ab − ac + bc − ab + ca − cb = 0. Since the dot product is 0, cosθ = 0. ∴ the angle between the lines is 90°.

2. Find the equation of a line parallel to x-axis and passing through the origin.

SOLUTION The x-axis has direction ratios 1, 0, 0. A line through the origin (0, 0, 0) parallel to it has equation (x−0)/1 = (y−0)/0 = (z−0)/0. ∴ x/1 = y/0 = z/0 (in vector form, r = λî).

3. If the lines (x−1)/(−3) = (y−2)/(2k) = (z−3)/2 and (x−1)/(3k) = (y−1)/1 = (z−6)/(−5) are perpendicular, find the value of k.

SOLUTION d.r.: (−3, 2k, 2) and (3k, 1, −5). Perpendicular ⇒ (−3)(3k) + (2k)(1) + (2)(−5) = 0. −9k + 2k − 10 = 0 ⇒ −7k = 10 ⇒ k = −10/7.

4. Find the shortest distance between lines r = 6î + 2ĵ + 2k̂ + λ(î − 2ĵ + 2k̂) and r = −4î − k̂ + μ(3î − 2ĵ − 2k̂).

SOLUTION a1 = (6, 2, 2), b1 = (1, −2, 2); a2 = (−4, 0, −1), b2 = (3, −2, −2). a2 − a1 = (−10, −2, −3). b1 × b2 = |i j k; 1 −2 2; 3 −2 −2| = ((−2)(−2)−(2)(−2))î − ((1)(−2)−(2)(3))ĵ + ((1)(−2)−(−2)(3))k̂ = 8î + 8ĵ + 4k̂. |b1 × b2| = √(64 + 64 + 16) = √144 = 12. (b1 × b2)·(a2−a1) = (8)(−10) + (8)(−2) + (4)(−3) = −80 − 16 − 12 = −108. d = |−108|/12 = 9.

5. Find the vector equation of the line passing through the point (1, 2, −4) and perpendicular to the two lines: (x−8)/3 = (y+19)/(−16) = (z−10)/7 and (x−15)/3 = (y−29)/8 = (z−5)/(−5).

SOLUTION d.r. of the two given lines: b1 = (3, −16, 7), b2 = (3, 8, −5). The required line is perpendicular to both, so its direction = b1 × b2. b1 × b2 = |i j k; 3 −16 7; 3 8 −5|. i: (−16)(−5) − (7)(8) = 80 − 56 = 24. j: −[(3)(−5) − (7)(3)] = −[−15 − 21] = 36. k: (3)(8) − (−16)(3) = 24 + 48 = 72. Direction = (24, 36, 72) = 12(2, 3, 6). ∴ r = (î + 2ĵ − 4k̂) + λ(2î + 3ĵ + 6k̂).

Common Mistakes to Avoid

Watch out for these

  • Forgetting to make the line equation standard — (1−x)/3 must be rewritten as (x−1)/(−3) before reading off direction ratios (Q10, Q15).
  • Dropping the modulus in the shortest-distance formula; the numerator is |(b1×b2)·(a2−a1)|, so distance is never negative.
  • Using the angle-between-lines formula without the modulus — the acute angle uses |a1a2+b1b2+c1c2|.
  • Confusing direction cosines (l2+m2+n2=1) with direction ratios (any proportional triple) — only d.c. must be normalised.
  • Sign slips when computing the cross product b1 × b2; remember the middle (j) term carries a minus.
  • Treating parallel lines with the skew formula — for parallel lines b1×b2 = 0, so use d = |b×(a2−a1)|/|b|.

Practice MCQs & Assertion–Reason

1. The direction cosines of the x-axis are:

(a) 0, 0, 1    (b) 1, 0, 0    (c) 0, 1, 0    (d) 1, 1, 1

2. If l, m, n are direction cosines of a line, then:

(a) l + m + n = 1    (b) l2 + m2 + n2 = 1    (c) lm + mn + nl = 1    (d) l2 + m2 + n2 = 0

3. The direction cosines of a line with direction ratios −18, 12, −4 are:

(a) −9/11, 6/11, −2/11    (b) −18, 12, −4    (c) 9/11, 6/11, 2/11    (d) −3, 2, −1

4. Two lines with direction ratios (a1, b1, c1) and (a2, b2, c2) are perpendicular if:

(a) a1/a2 = b1/b2 = c1/c2    (b) a1a2 + b1b2 + c1c2 = 0    (c) a1 = a2    (d) a1a2 = b1b2

5. The Cartesian equation of the line through (5, −4, 6) parallel to 3î + 7ĵ + 2k̂ is:

(a) (x−5)/3 = (y+4)/7 = (z−6)/2    (b) (x+5)/3 = (y−4)/7 = (z+6)/2    (c) (x−3)/5 = (y−7)/(−4) = (z−2)/6    (d) (x−5)/7 = (y+4)/3 = (z−6)/2

6. The angle between the lines x/2 = y/2 = z/1 and (x−5)/4 = (y−2)/1 = (z−3)/8 is:

(a) cos−1(1/3)    (b) cos−1(2/3)    (c) 90°    (d) 0°

7. Skew lines are lines that are:

(a) parallel    (b) intersecting    (c) neither parallel nor intersecting    (d) coincident

8. The shortest distance between two intersecting lines is:

(a) zero    (b) one    (c) infinite    (d) cannot be found

9. The value of p for which (1−x)/3 = (7y−14)/(2p) = (z−3)/2 and (7−7x)/(3p) = (y−5)/1 = (6−z)/5 are perpendicular is:

(a) 7/11    (b) 11/70    (c) 70/11    (d) 10

10. A line makes equal angles with the coordinate axes; its direction cosines are:

(a) 1, 1, 1    (b) 1/3, 1/3, 1/3    (c) ±1/√3, ±1/√3, ±1/√3    (d) 1/√2, 1/√2, 0

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

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: The lines with direction ratios a, b, c and b−c, c−a, a−b are perpendicular.

Reason: a(b−c) + b(c−a) + c(a−b) = 0.

A-R 2. Assertion: A line making equal angles with the axes has direction cosines ±1/√3, ±1/√3, ±1/√3.

Reason: For direction cosines, l2 + m2 + n2 = 1.

A-R 3. Assertion: The points (2, 3, 4), (−1, −2, 1) and (5, 8, 7) are collinear.

Reason: The direction ratios of the segments joining them are proportional.

A-R 4. Assertion: The shortest distance between two parallel lines is zero.

Reason: Parallel lines lie in the same plane and never meet.

A-R 5. Assertion: The shortest distance between r = (î+2ĵ+k̂) + λ(î−ĵ+k̂) and r = 2î−ĵ−k̂ + μ(2î+ĵ+2k̂) is 3√2/2.

Reason: For skew lines d = |(b1×b2)·(a2−a1)| ÷ |b1×b2|.

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

Quick Revision Summary

  • Direction cosines l, m, n are the cosines of the angles the line makes with the axes; l2 + m2 + n2 = 1.
  • Direction ratios a, b, c are any numbers proportional to l, m, n; d.c. = (a, b, c)/√(a2+b2+c2).
  • Vector equation of a line: r = a + λb; Cartesian: (x−x1)/a = (y−y1)/b = (z−z1)/c.
  • Angle between lines: cosθ = |b1·b2| / (|b1||b2|); perpendicular when the dot product is 0.
  • Shortest distance (skew): d = |(b1×b2)·(a2−a1)| / |b1×b2|.
  • Distance (parallel): d = |b×(a2−a1)| / |b|; for intersecting lines the shortest distance is 0.

How to score full marks in this chapter

Always reduce a line to standard form (x−x1)/a = (y−y1)/b = (z−z1)/c before reading off direction ratios — watch the signs when terms like (1−x) or (6−z) appear. Write the cross product in determinant form and double-check the middle (j) sign. State the formula you use, keep the modulus in distance and angle answers, and verify each result against the NCERT answer key — the marks come from clean, justified steps.

Frequently Asked Questions

What is Class 12 Maths Chapter 11 about?

Chapter 11, Three Dimensional Geometry, deals with direction cosines and direction ratios of a line, the vector and Cartesian equations of a line in space, the angle between two lines, and the shortest distance between skew and parallel lines — all derived using vector algebra.

How many exercises are there in Class 12 Maths Chapter 11?

There are two numbered exercises — Exercise 11.1 (direction cosines and ratios) and Exercise 11.2 (equations of lines, angle and shortest distance) — plus a Miscellaneous Exercise. Every question of all three is solved on this page.

What is the formula for the shortest distance between two skew lines?

For lines r = a1 + λb1 and r = a2 + μb2, the shortest distance is d = |(b1 × b2) · (a2 − a1)| ÷ |b1 × b2|. If the lines intersect, this distance is zero.

Are these Class 12 Maths Chapter 11 solutions free?

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

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