Class 12 Maths Vector Algebra – NCERT Formulas & Examples

Class 12 Maths — Chapter 10: Vector Algebra

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Introduction

Vector Algebra deals with quantities that have both magnitude and direction, essential in physics, engineering, and advanced mathematics. This post provides a concise introduction, key formulas, worked examples, 30 practice questions from basic to advanced levels, 20 labelled diagrams for note-taking, and 15 FAQs addressing common doubts. Try solving each question independently, then use the toggle to check solutions.

Key Formulas (Quick Reference)

• Position / component form: vector a = a₁i + a₂j + a₃k
• Magnitude: |a| = √(a₁² + a₂² + a₃²)
• Unit vector: â = a / |a|
• Dot product: a·b = a₁b₁ + a₂b₂ + a₃b₃ = |a||b|cosθ
• Cross product: a × b = determinant form; |a × b| = |a||b|sinθ (direction via right-hand rule)
• Scalar triple product (volume): a·(b×c)
• Projection of a on b (vector): proj_ba = (a·b / |b|²) b
• Line through point r₀ with direction d: r = r₀ + t d
• Plane with normal n through r₀: n·(r − r₀) = 0

Worked Examples

Example 1: Find the unit vector in the direction of a = 3i + 4j.
Solution: |a| = √(3² + 4²) = 5. Unit vector = (3/5)i + (4/5)j.

Example 2: Find the angle between a = i + j + k and b = 2i − j + k.
Solution: a·b = 1·2 + 1·(−1) + 1·1 = 2. |a| = √3, |b| = √6. cosθ = 2/(√3·√6) = 2/√18 = 1/√3. θ = cos⁻¹(1/√3).

Practice — 30 Questions (Basic to Advanced)

Solutions are hidden by default. Use the buttons to expand or collapse all.

Q1. Find the magnitude of vector a = 3i + 4j.

|a| = √(3² + 4²) = √25 = 5.

Q2. Find the unit vector of a = 6i + 8j.

|a| = √(6² + 8²) = 10. Unit vector = (6/10)i + (8/10)j = (3/5)i + (4/5)j.

Q3. Compute a·b for a = 2i + j − k, b = i + 2j + k.

a·b = 2·1 + 1·2 + (−1)·1 = 2 + 2 − 1 = 3.

Q4. Find the angle between a = i + j and b = j + k.

a·b = 1·0 + 1·1 = 1. |a| = √2, |b| = √2. cosθ = 1/(√2·√2) = 1/2. θ = 60°.

Q5. Compute a × b for a = i + j, b = j + k.

a × b = (1·1 − 1·0)i − (1·1 − 0·1)j + (1·1 − 1·0)k = i − j + k.

Q6. Find the area of the parallelogram formed by a = 2i + j, b = i + 2j.

a × b = (2·2 − 1·1)k = 3k. Area = |a × b| = 3.

Q7. Find |a| for a = 5i − 12j + 13k.

|a| = √(5² + (−12)² + 13²) = √(25 + 144 + 169) = √338 = 13√2.

Q8. Show a = 2i + 2j and b = −i + j are perpendicular.

a·b = 2·(−1) + 2·1 = −2 + 2 = 0 → perpendicular.

Q9. If a = 3i − j and b = λi + 2j are perpendicular, find λ.

a·b = 3λ + (−1)·2 = 3λ − 2 = 0. λ = 2/3.

Q10. Find the vector equation of the line through (2, 1, 3) and (5, −2, 4).

Direction = (5−2, −2−1, 4−3) = (3, −3, 1). r = (2,1,3) + t(3,−3,1).

Q11. Compute the scalar triple product a·(b×c) for a = i + j + k, b = 2i − j, c = i + k.

b×c = (0·1 − (−1)·1)i − (2·1 − 0·1)j + (2·1 − (−1)·0)k = i − 2j + 2k. a·(b×c) = 1·1 + 1·(−2) + 1·2 = 1 − 2 + 2 = 1.

Q12. Are vectors a = i + j, b = 2i − j + k, c = 3i + k coplanar?

Scalar triple product = det([1,1,0; 2,−1,1; 3,0,1]) = 1·(−1·1 − 0·1) − 1·(2·1 − 1·3) + 0 = −1 − (−1) = 0 → coplanar.

Q13. If a·b = 0, |a| = 5, |b| = 12, find |a + b|.

Since a·b = 0, |a + b|² = |a|² + |b|² = 5² + 12² = 25 + 144 = 169. |a + b| = 13.

Q14. Find a vector perpendicular to a = 2i + j and b = i − j in 3D.

a×b = det([i,j,k; 2,1,0; 1,−1,0]) = (1·0 − 0·(−1))i − (2·0 − 0·1)j + (2·(−1) − 1·1)k = −3k. Vector = (0,0,−3).

Q15. Find the shortest distance from P(1,2,3) to the line r = (0,0,0) + t(1,1,1).

AP = (1,2,3), d = (1,1,1). Distance = |AP × d| / |d| = |(1,2,3) × (1,1,1)| / √3 = |−i + 2j − k| / √3 = √6 / √3 = √2.

Q16. Find the equation of the plane through (1,0,0) with normal n = i + j + k.

n·(r − r₀) = 0. (1,1,1)·(x−1,y,z) = 0. x + y + z − 1 = 0.

Q17. Find the scalar projection of a = 2i + 3j + k on b = i + j.

a·b = 2·1 + 3·1 = 5. |b| = √(1² + 1²) = √2. Scalar projection = 5/√2.

Q18. Are vectors a = (1,0,1), b = (0,1,1), c = (1,1,0) coplanar?

Scalar triple product = det([1,0,1; 0,1,1; 1,1,0]) = 1·(1·0 − 1·1) − 0 + 1·(0·1 − 1·1) = −1 − 1 = −2 ≠ 0 → not coplanar.

Q19. If |a| = 4, |b| = 3, a·b = 6, find the angle between a and b.

cosθ = a·b / (|a||b|) = 6/(4·3) = 1/2. θ = 60°.

Q20. Find the equation of the plane through points (1,0,0), (0,1,0), (0,0,1).

Vectors AB = (−1,1,0), AC = (−1,0,1). Normal = AB × AC = (1,1,1). Plane: x + y + z − 1 = 0.

Q21. Find a vector c perpendicular to a = (2,1,0) and b = (1,0,1) with |c| = 2.

c = a × b = (1·1 − 0·0)i − (2·1 − 0·1)j + (2·0 − 1·1)k = i − 2j − k. |c| = √6. Unit c = (1/√6, −2/√6, −1/√6). Scale: c = 2·(1/√6, −2/√6, −1/√6) = (2/√6, −4/√6, −2/√6).

Q22. Do lines r = (1,1,1) + t(1,0,1) and r = (0,0,0) + s(0,1,1) intersect?

Equate: (1+t,1,s,s) = (0,0,0). t = −1, s = 0. Check: (0,1,0) ≠ (0,0,0). No intersection (skew lines).

Q23. Find the shortest distance between lines r1 = (1,0,0) + t(1,1,0) and r2 = (0,0,1) + s(0,1,1).

d1 = (1,1,0), d2 = (0,1,1), AB = (−1,0,1). d1×d2 = (1·1 − 0·1)i − (1·1 − 0·0)j + (1·1 − 1·0)k = i − j + k. |d1×d2| = √3. Distance = |AB·(d1×d2)| / |d1×d2| = |−1 − 1 + 1| / √3 = 1/√3.

Q24. If x + b = 0 and x is parallel to a, where a = (1,2,3), b = (2,−1,0), find x.

x = −b = (−2,1,0). Check parallelism: (−2,1,0) = λ(1,2,3) not possible (no λ satisfies). No solution.

Q25. Find λ such that vectors (1,λ,1), (2,1,λ), (λ,1,2) are coplanar.

det([1,λ,1; 2,1,λ; λ,1,2]) = 0. Compute: λ(1·2 − λ·1) − λ(2·2 − λ·λ) + 1(2·1 − 1·λ) = λ(2−λ) − λ(4−λ²) + (2−λ) = λ² − 5λ + 2 = 0. Solve: λ = (5±√17)/2.

Q26. Find the point dividing the line from (1,2,3) to (4,5,6) in the ratio 2:1 internally.

Point = ((2·4 + 1·1)/(2+1), (2·5 + 1·2)/(2+1), (2·6 + 1·3)/(2+1)) = (3,4,5).

Q27. If a·b = 8, |a| = 4, |b| = 4, verify Cauchy-Schwarz and find θ.

Cauchy-Schwarz: |a·b| ≤ |a||b|. 8 ≤ 4·4 = 16, holds. cosθ = 8/(4·4) = 1/2. θ = 60°.

Q28. Find a unit vector making equal angles with the axes.

Let vector = (a,a,a). |a| = √(a² + a² + a²) = √3a = 1. a = ±1/√3. Unit vector = (±1/√3, ±1/√3, ±1/√3).

Q29. Are vectors u = (1,2,3), v = (2,4,6), w = (3,6,9) linearly dependent?

v = 2u, w = 3u. det([1,2,3; 2,4,6; 3,6,9]) = 0 (linearly dependent columns). Yes, dependent.

Q30. Show that p = a×(b×c) + b×(c×a) + c×(a×b) = 0.

Use vector triple product: a×(b×c) = (a·c)b − (a·b)c. Sum terms: (a·c)b − (a·b)c + (b·a)c − (b·c)a + (c·b)a − (c·a)b = 0 (terms cancel).

20 Labelled Diagrams (Use in Notes)

Below are 20 compact SVG illustrations you can copy into your notes. Each is labelled for clarity and designed to be simple for web and print use.

1. Vector in 3D (components)

2. Angle between two vectors

3. Parallelogram law

4. Projection of a on b

5. Area via cross product

6. Vector components on axes

7. Skew lines

8. Plane and normal

9. Line-plane intersection

10. Triangle area from two vectors

11. Unit vector with axes

12. Vector addition (triangle)

13. Direction ratios

14. Reflection illustration

15. Orthogonal projection

16. Parametric line points

17. Shortest distance between skew lines

18. Scalar triple product (volume)

19. Orthonormal basis idea

20. Direction cosines

FAQs (Basic to Advanced)

Q1: What is a vector in mathematics?

A: A vector is a quantity with both magnitude and direction, represented as an arrow or in component form (e.g., a = xi + yj + zk). It differs from a scalar, which has only magnitude.

Q2: How do I calculate the magnitude of a vector?

A: For a vector a = xi + yj + zk, the magnitude is |a| = √(x² + y² + z²). This gives the length of the vector.

Q3: What is a unit vector, and how is it found?

A: A unit vector has a magnitude of 1 and points in the same direction as the original vector. Compute it as â = a / |a|.

Q4: When are two vectors perpendicular?

A: Two vectors are perpendicular if their dot product is zero: a·b = 0. This implies the angle between them is 90°.

Q5: What does the dot product represent geometrically?

A: The dot product a·b = |a||b|cosθ measures the projection of one vector onto another and is used to find angles or test perpendicularity.

Q6: When should I use the cross product?

A: Use the cross product to find a vector perpendicular to two given vectors, calculate areas (e.g., parallelogram area = |a × b|), or compute volumes via the scalar triple product.

Q7: How do I determine if vectors are coplanar?

A: Three vectors are coplanar if their scalar triple product a·(b×c) = 0, indicating they lie in the same plane.

Q8: What is the vector equation of a line?

A: The vector equation of a line through point r₀ with direction vector d is r = r₀ + td, where t is a scalar parameter.

Q9: How do I find the equation of a plane?

A: A plane through point r₀ with normal vector n has equation n·(r − r₀) = 0, or in scalar form, ax + by + cz = d.

Q10: How can I improve speed in vector algebra calculations?

A: Practice dot and cross products regularly, memorize determinant shortcuts for 3×3 matrices, and sketch vectors to visualize geometric problems.

Q11: What is the significance of the scalar triple product?

A: The scalar triple product a·(b×c) gives the volume of the parallelepiped formed by vectors a, b, c. If zero, the vectors are coplanar.

Q12: How do I find the shortest distance between a point and a line?

A: For a point P and line r = r₀ + td, the distance is |AP × d| / |d|, where AP is the vector from a point on the line to P.

Q13: How do I handle skew lines in 3D geometry?

A: For skew lines r = r₁ + td₁ and r = r₂ + sd₂, the shortest distance is |AB·(d₁×d₂)| / |d₁×d₂|, where AB is a vector between points on the lines.

Q14: What are direction cosines?

A: Direction cosines are the cosines of the angles a vector makes with the coordinate axes. For a = (x,y,z), they are (x/|a|, y/|a|, z/|a|).

Q15: How do I solve problems involving vector identities?

A: Use identities like a×(b×c) = (a·c)b − (a·b)c and practice algebraic manipulation. Verify with numerical examples for complex identities.

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