Vector Algebra is a fundamental chapter in Class 12 Maths that introduces you to quantities having both magnitude and direction. From physics forces to computer graphics, vectors are everywhere. This chapter carries 4–6 marks in CBSE board exams and builds the foundation for 3D Geometry.
Key Concepts
1. Scalars and Vectors
| Scalar | Vector |
|---|---|
| Has only magnitude | Has both magnitude and direction |
| Examples: mass, temperature, speed | Examples: force, velocity, displacement |
| Denoted by ordinary letters | Denoted by letters with arrow (→) on top |
2. Types of Vectors
| Type | Definition | Example |
|---|---|---|
| Zero Vector (0→) | Magnitude = 0, arbitrary direction | Force on a stationary object at rest |
| Unit Vector (â) | Magnitude = 1 | î, ĵ, k̂ along x, y, z axes |
| Co-initial | Same starting point | Vectors from origin |
| Collinear | Parallel (same or opposite direction) | a→ = λb→ |
| Equal | Same magnitude and direction | Free vectors that overlap |
| Negative | Same magnitude, opposite direction | −a→ |
3. Position Vector
OP→ = xî + yĵ + zk̂
Magnitude: |OP→| = √(x² + y² + z²)
4. Addition of Vectors
Triangle Law: Place vectors head-to-tail; the resultant goes from the tail of the first to the head of the last.
Parallelogram Law: Place vectors at same point; diagonal of parallelogram = resultant.
a→ + b→ = (a₁+b₁)î + (a₂+b₂)ĵ + (a₃+b₃)k̂
5. Section Formula
Internally in ratio m:n → r→ = (mb→ + na→)/(m+n)
Externally in ratio m:n → r→ = (mb→ − na→)/(m−n)
Midpoint: r→ = (a→ + b→)/2
6. Scalar (Dot) Product
In component form: a→ · b→ = a₁b₁ + a₂b₂ + a₃b₃
cos θ = (a→ · b→)/(|a→||b→|)
Properties of Dot Product
- Commutative: a→ · b→ = b→ · a→
- Distributive: a→ · (b→ + c→) = a→ · b→ + a→ · c→
- î · î = ĵ · ĵ = k̂ · k̂ = 1
- î · ĵ = ĵ · k̂ = k̂ · î = 0
7. Cross (Vector) Product
where n̂ is the unit vector perpendicular to both a→ and b→ (right-hand rule).
In component form:
a→ × b→ = (a₂b₃ − a₃b₂)î − (a₁b₃ − a₃b₁)ĵ + (a₁b₂ − a₂b₁)k̂
|a→ × b→| = Area of parallelogram with adjacent sides a→ and b→
½|a→ × b→| = Area of triangle
Properties of Cross Product
- NOT commutative: a→ × b→ = −(b→ × a→)
- Distributive: a→ × (b→ + c→) = a→ × b→ + a→ × c→
- î × î = ĵ × ĵ = k̂ × k̂ = 0→
- î × ĵ = k̂, ĵ × k̂ = î, k̂ × î = ĵ
Important Definitions
| Term | Definition |
|---|---|
| Vector | A quantity with both magnitude and direction |
| Unit Vector | Vector with magnitude 1; â = a→/|a→| |
| Collinear Vectors | Vectors that lie along the same or parallel lines |
| Dot Product | Scalar product: a→ · b→ = |a→||b→| cos θ |
| Cross Product | Vector product: a→ × b→ = |a→||b→| sin θ · n̂ |
| Projection | Projection of a→ on b→ = (a→ · b→)/|b→| |
Solved Examples — NCERT-Based
Example 1: Find the unit vector in the direction of a→ = 2î − 3ĵ + 6k̂
Solution:
|a→| = √(4 + 9 + 36) = √49 = 7
â = a→/|a→| = (2î − 3ĵ + 6k̂)/7 = (2/7)î − (3/7)ĵ + (6/7)k̂
Example 2: Find the angle between a→ = î + ĵ + k̂ and b→ = î − ĵ + k̂
Solution:
a→ · b→ = (1)(1) + (1)(−1) + (1)(1) = 1
|a→| = √3, |b→| = √3
cos θ = 1/(√3 × √3) = 1/3
θ = cos⁻¹(1/3)
Example 3: Find a→ × b→ if a→ = 3î + 2ĵ − k̂ and b→ = î − ĵ + 2k̂
Solution:
a→ × b→ = [(2)(2) − (−1)(−1)]î − [(3)(2) − (−1)(1)]ĵ + [(3)(−1) − (2)(1)]k̂
= (4 − 1)î − (6 + 1)ĵ + (−3 − 2)k̂
= 3î − 7ĵ − 5k̂
Example 4: Find the area of the triangle with vertices A(1,1,2), B(2,3,5), C(1,5,5)
Solution:
AB→ = (1)î + (2)ĵ + (3)k̂, AC→ = (0)î + (4)ĵ + (3)k̂
AB→ × AC→ = (6−12)î − (3−0)ĵ + (4−0)k̂ = −6î − 3ĵ + 4k̂
|AB→ × AC→| = √(36 + 9 + 16) = √61
Area = ½√61 ≈ 3.9 sq units
Important Questions for Board Exams
1 Mark Questions
- Find the magnitude of vector 3î − 2ĵ + 6k̂.
- If |a→| = 3 and |b→| = 4, find |a→ × b→| when θ = 90°.
- Write the unit vector along î + ĵ.
2 Mark Questions
- Find the projection of a→ = 2î + 3ĵ + 2k̂ on b→ = î + 2ĵ + k̂.
- Show that vectors a→ = 2î − ĵ + k̂ and b→ = î + 2ĵ − k̂ are not parallel.
- Find λ if vectors î − 2ĵ + 3k̂ and 3î − 6ĵ + λk̂ are collinear.
3 Mark Questions
- If a→ + b→ + c→ = 0→, prove that a→ × b→ = b→ × c→ = c→ × a→.
- Find the area of a parallelogram whose adjacent sides are a→ = 3î + ĵ + 4k̂ and b→ = î − ĵ + k̂.
5 Mark Questions
- If a→ = 2î + 2ĵ + 3k̂, b→ = −î + 2ĵ + k̂ and c→ = 3î + ĵ, find a→ + b→, and verify that (a→ + b→) × c→ = a→ × c→ + b→ × c→.
- Prove using vectors that the diagonals of a rhombus bisect each other at right angles.
Quick Revision Points
- Vectors have magnitude AND direction; scalars have only magnitude
- Unit vector: â = a→/|a→| (divide by magnitude)
- Section formula: (mb→ + na→)/(m+n) for internal division
- Dot product → scalar, gives angle: cos θ = (a→·b→)/(|a→||b→|)
- Cross product → vector, gives area: |a→×b→| = area of parallelogram
- Perpendicular ⟹ dot product = 0
- Parallel ⟹ cross product = 0→
- Projection of a→ on b→ = (a→·b→)/|b→|
- Area of triangle = ½|a→ × b→|
Chapter Navigation
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Next: Three Dimensional Geometry Class 12 Notes
Related Chapters in Class 12 Maths
- Three Dimensional Geometry Class 12 Notes
- Linear Programming Class 12 Notes
- Probability Class 12 Notes
Practice What You Learned
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