Vectors
1. Introduction
Vectors are quantities that have both magnitude and direction. They are essential in physics for describing forces, velocities, and fields, and are fundamental to three-dimensional geometry.
2. Basic Concepts
2.1 Representation
A vector is denoted as →a or a. Its magnitude is |a| or a.
Position Vector of point P(x, y, z): →OP = x̂i + ŷj + ẑk. Magnitude: |→OP| = √(x² + y² + z²).
2.2 Types of Vectors
Zero Vector: Magnitude 0, direction undefined. Unit Vector: |â| = 1. â = a/|a|. Equal Vectors: Same magnitude and direction. Negative Vector: Same magnitude, opposite direction. Collinear Vectors: Parallel or anti-parallel. Coplanar Vectors: Lie in the same plane.
3. Addition of Vectors
3.1 Triangle Law
If the tail of b coincides with the head of a, a + b is the vector from the tail of a to the head of b.
3.2 Parallelogram Law
If a and b share the same tail, a + b is the diagonal of the parallelogram.
3.3 Properties
Commutative: a + b = b + a Associative: (a + b) + c = a + (b + c) Additive identity: 0 Additive inverse: a + (-a) = 0
4. Scalar (Dot) Product
a · b = |a||b| cos θ, where θ is the angle between a and b.
Properties:
- a · b = 0 iff a ⟂ b (a and b are perpendicular)
- a · a = |a|²
- Commutative: a · b = b · a
- Distributive: a · (b + c) = a · b + a · c
In component form: a · b = a₁b₁ + a₂b₂ + a₃b₃.
5. Vector (Cross) Product
a × b = |a||b| sin θ n̂, where n̂ is a unit vector perpendicular to both a and b.
Properties:
- a × b = 0 iff a ∥ b
- Anti-commutative: a × b = -(b × a)
- Not associative: a × (b × c) ≠ (a × b) × c in general
In component form (determinant method): a × b = |[î, ĵ, k̂], [a₁, a₂, a₃], [b₁, b₂, b₃]|.
6. Scalar Triple Product
[a b c] = a · (b × c) represents the volume of the parallelepiped.
Properties:
- [a b c] = [b c a] = [c a b] (cyclic permutation)
- [a b c] = 0 iff a, b, c are coplanar
- [a b c] = |[a₁ a₂ a₃], [b₁ b₂ b₃], [c₁ c₂ c₃]|
7. Worked Problems
Problem 1: Find the angle between a = (î + ĵ - 2k̂) and b = (2î - ĵ + k̂). Solution: a · b = 1×2 + 1×(-1) + (-2)×1 = 2 - 1 - 2 = -1. |a| = √(1 + 1 + 4) = √6, |b| = √(4 + 1 + 1) = √6. cos θ = -1/6 ⇒ θ = cos^{-1}(-1/6).
Problem 2: Find the area of triangle with vertices A(1, 2, 3), B(2, 3, 1), C(3, 1, 2). Solution: →AB = (î + ĵ - 2k̂), →AC = (2î - ĵ - k̂). →AB × →AC = |[î, ĵ, k̂], [1, 1, -2], [2, -1, -1]| = -3î - 3ĵ - 3k̂. Area = (1/2)|→AB × →AC| = (1/2)√27 = 3√3/2.
Problem 3: Show that a = 2î - ĵ + k̂, b = î - 3ĵ - 5k̂, c = 3î - 4ĵ - 4k̂ are coplanar. Solution: [a b c] = |[2, -1, 1], [1, -3, -5], [3, -4, -4]| = 2(12-20) + 1(-4+15) + 1(-4+9) = -16 + 11 + 5 = 0. Hence coplanar.
8. Common Mistakes
'Students often confuse dot and cross products. Remember: dot gives a scalar, cross gives a vector.'
'The cross product is anti-commutative — a × b = -(b × a). Never forget the negative sign.'
9. ISC Exam Focus
| Topic | Theory Marks | Practical Marks |
|---|---|---|
| Dot product | 3 | 1 |
| Cross product | 4 | 2 |
| Scalar triple product | 3 | 2 |
| Applications | 3 | 2 |
10. Self-Test Questions
- Find the projection of a on b if a = 2î - ĵ + 3k̂, b = î + 2ĵ + 2k̂.
- If |a| = 2, |b| = 3, and a · b = 3, find |a × b|.
- Find the volume of the parallelepiped formed by a = 2î - ĵ + 3k̂, b = î + 2ĵ - k̂, c = 3î + ĵ + 2k̂.
- Show that (a - b) × (a + b) = 2(a × b).
- If a · b = a · c and a × b = a × c, prove that b = c (a ≠ 0).
11. Key Formulae at a Glance
| Quantity | Formula |
|---|---|
| Dot product | a·b = |
| Cross product | a×b = |
| Area of parallelogram | |
| Area of triangle | (1/2) |
| Volume of parallelepiped | |
| Condition for coplanarity | a·(b×c) = 0 |
| Unit vector | â = a/ |
| Angle between vectors | cosθ = (a·b)/( |
