1. Chapter at a Glance
"The car is moving at 60 km/h" is incomplete — in which direction? Quantities like displacement, velocity, force and acceleration need both a magnitude and a direction. These are vectors. Quantities like mass, temperature and time need only a size; they are scalars. This chapter introduces vectors, their addition, components in three dimensions, and two kinds of products — the scalar (dot) product, used to find angles and projections, and the vector (cross) product, used to find areas and perpendicular directions.
- Scalars and vectors; magnitude, direction cosines and ratios.
- Types: zero, unit, coinitial, collinear, equal, negative vectors.
- Addition (triangle and parallelogram laws); scalar multiplication.
- Components: a = a₁î + a₂ĵ + a₃k̂; position vectors; section formula.
- Scalar (dot) product: angle, projection, perpendicularity.
- Vector (cross) product: area of parallelogram and triangle.
2. Vectors and Their Types
A vector is a directed line segment AB⃗ with magnitude |AB⃗| (its length) and a direction. The position vector of P(x, y, z) is OP⃗ = xî + yĵ + zk̂, with |OP⃗| = √(x² + y² + z²).
| Type | Meaning |
|---|---|
| Zero vector 0⃗ | Magnitude 0, direction undefined |
| Unit vector â | Magnitude 1: â = a⃗/|a⃗| |
| Coinitial | Same initial point |
| Collinear (parallel) | Along the same or parallel lines: b⃗ = λa⃗ |
| Equal | Same magnitude and direction |
| Negative of a⃗ | Same magnitude, opposite direction |
If r⃗ makes angles α, β, γ with the x, y, z axes, then l = cos α, m = cos β, n = cos γ are its direction cosines, with l² + m² + n² = 1. For r⃗ = xî + yĵ + zk̂: l = x/r, m = y/r, n = z/r. The numbers x, y, z are direction ratios.
3. Addition and Scalar Multiplication
- Addition is commutative and associative: a⃗ + b⃗ = b⃗ + a⃗.
- λa⃗ has magnitude |λ||a⃗|; same direction if λ > 0, opposite if λ < 0.
- In components: (a₁ + b₁)î + (a₂ + b₂)ĵ + (a₃ + b₃)k̂.
- AB⃗ = (position vector of B) − (position vector of A).
The point dividing AB (position vectors a⃗, b⃗) internally in the ratio m : n has position vector (m b⃗ + n a⃗)/(m + n); midpoint (a⃗ + b⃗)/2.
Find the unit vector in the direction of a⃗ = 2î + 3ĵ + k̂, and its direction cosines.
Solution: |a⃗| = √14. â = (2î + 3ĵ + k̂)/√14; direction cosines 2/√14, 3/√14, 1/√14.
4. Scalar (Dot) Product
a⃗ · b⃗ = |a⃗||b⃗| cos θ, where θ (0 ≤ θ ≤ π) is the angle between them. In components: a⃗ · b⃗ = a₁b₁ + a₂b₂ + a₃b₃.
- a⃗ · b⃗ = 0 ⇔ a⃗ ⊥ b⃗ (for non-zero vectors).
- î·î = ĵ·ĵ = k̂·k̂ = 1; î·ĵ = ĵ·k̂ = k̂·î = 0.
- a⃗ · a⃗ = |a⃗|²; dot product is commutative and distributive.
- Projection of a⃗ on b⃗ = (a⃗ · b⃗)/|b⃗|.
Find the angle between a⃗ = î + ĵ − k̂ and b⃗ = î − ĵ + k̂.
Solution: a⃗·b⃗ = 1 − 1 − 1 = −1; |a⃗| = |b⃗| = √3. cos θ = −1/3 → θ = cos⁻¹(−1/3).
Find the projection of a⃗ = 2î + 3ĵ + 2k̂ on b⃗ = î + 2ĵ + k̂.
Solution: (2 + 6 + 2)/√6 = 10/√6.
5. Vector (Cross) Product
a⃗ × b⃗ = |a⃗||b⃗| sin θ n̂, where n̂ is the unit vector perpendicular to both, by the right-hand rule.
- a⃗ × b⃗ = −(b⃗ × a⃗) (not commutative).
- a⃗ × b⃗ = 0⃗ ⇔ a⃗ ∥ b⃗ (non-zero vectors).
- î × ĵ = k̂, ĵ × k̂ = î, k̂ × î = ĵ; î × î = 0⃗.
- Area of parallelogram with sides a⃗, b⃗ = |a⃗ × b⃗|; area of triangle = ½|a⃗ × b⃗|.
Find |a⃗ × b⃗| for a⃗ = 2î + ĵ + 3k̂, b⃗ = 3î + 5ĵ − 2k̂.
Solution: a⃗ × b⃗ = î(−2 − 15) − ĵ(−4 − 9) + k̂(10 − 3) = −17î + 13ĵ + 7k̂. |·| = √(289 + 169 + 49) = √507.
Find the area of the triangle with vertices A(1, 1, 1), B(1, 2, 3), C(2, 3, 1).
Solution: AB⃗ = ĵ + 2k̂, AC⃗ = î + 2ĵ. AB⃗ × AC⃗ = î(0 − 4) − ĵ(0 − 2) + k̂(0 − 1) = −4î + 2ĵ − k̂. Area = ½√21.
6. Solved Problems (Board Pattern)
Show that A(2, 6, 3), B(1, 2, 7), C(3, 10, −1) are collinear.
Solution: AB⃗ = −î − 4ĵ + 4k̂, BC⃗ = 2î + 8ĵ − 8k̂ = −2AB⃗. Parallel with common point B → collinear.
Find a unit vector perpendicular to both a⃗ = î + ĵ + k̂ and b⃗ = î + 2ĵ + 3k̂.
Solution: a⃗ × b⃗ = î(3 − 2) − ĵ(3 − 1) + k̂(2 − 1) = î − 2ĵ + k̂. Unit vector ±(î − 2ĵ + k̂)/√6.
If |a⃗| = 3, |b⃗| = 4 and a⃗ · b⃗ = 6, find |a⃗ + b⃗|.
Solution: |a⃗ + b⃗|² = 9 + 16 + 12 = 37 → √37.
7. Master Formula Table
| Result | Formula |
|---|---|
| Magnitude | |a⃗| = √(a₁² + a₂² + a₃²) |
| Unit vector | â = a⃗/|a⃗| |
| Direction cosines | l² + m² + n² = 1 |
| Section formula | (m b⃗ + n a⃗)/(m + n) |
| Dot product | |a||b| cos θ = a₁b₁ + a₂b₂ + a₃b₃ |
| Projection | (a⃗·b⃗)/|b⃗| |
| Cross product | |a||b| sin θ n̂ (determinant) |
| Areas | parallelogram |a⃗ × b⃗|; triangle ½|a⃗ × b⃗| |
| |a⃗ ± b⃗|² | |a⃗|² + |b⃗|² ± 2a⃗·b⃗ |
8. Master Summary Table
| Product | Result type | Zero when | Used for |
|---|---|---|---|
| Dot a⃗·b⃗ | Scalar | Perpendicular | Angles, projections, work |
| Cross a⃗ × b⃗ | Vector | Parallel | Areas, normal directions, torque |
9. Mind Map
- Magnitude, direction
- Unit, collinear
- Triangle law
- Section formula
- cos θ
- Projection, ⊥
- sin θ n̂
- Areas, ∥
10. Tips and Tricks
- To test perpendicularity, use the dot product; to test parallelism, compare ratios or use the cross product.
- Cyclic order î → ĵ → k̂ → î gives positive cross products.
- |a⃗ × b⃗|² + (a⃗·b⃗)² = |a⃗|²|b⃗|² (Lagrange's identity) — a quick check.
- Forgetting the minus sign on the ĵ term in the cross-product determinant.
- Writing a⃗·b⃗ as a vector or a⃗ × b⃗ as a number.
- Using the full cross product magnitude for a triangle (needs ½).
11. Important One-liners
- a⃗·b⃗ = 0 ⇔ perpendicular; a⃗ × b⃗ = 0⃗ ⇔ parallel.
- a⃗ × b⃗ = −b⃗ × a⃗.
- l² + m² + n² = 1 for direction cosines.
- |a⃗ × b⃗| equals the area of the parallelogram on a⃗ and b⃗.
- Projection of a⃗ on b⃗ is a⃗·b̂.
12. Exam-pattern Questions with Detailed Answers
The value of î·(ĵ × k̂) + ĵ·(î × k̂) + k̂·(î × ĵ) is:
- 0
- −1
- 1
- 3
Show answer
î·î + ĵ·(−ĵ) + k̂·k̂ = 1 − 1 + 1 = 1.
Answer: (C)
Find |a⃗| for a⃗ = î + ĵ + k̂.
Show answer
√3.
Answer: √3
Find λ if 2î + 3ĵ + λk̂ is perpendicular to î − ĵ + k̂.
Show answer
2 − 3 + λ = 0 → λ = 1.
Answer: λ = 1
Find the position vector of the point dividing the join of P(î + 2ĵ − k̂) and Q(−î + ĵ + k̂) in the ratio 2 : 1 internally.
Show answer
[2(−î + ĵ + k̂) + (î + 2ĵ − k̂)]/3 = (−î + 4ĵ + k̂)/3.
Answer: (−î + 4ĵ + k̂)/3
Find the direction cosines of the vector joining A(1, 2, −3) and B(−1, −2, 1), directed from A to B.
Show answer
AB⃗ = −2î − 4ĵ + 4k̂, |AB⃗| = 6. DCs −1/3, −2/3, 2/3.
Answer: −1/3, −2/3, 2/3
If a⃗ is a unit vector and (x⃗ − a⃗)·(x⃗ + a⃗) = 8, find |x⃗|.
Show answer
|x⃗|² − 1 = 8 → |x⃗| = 3.
Answer: 3
Find the area of the parallelogram with adjacent sides 3î + ĵ + 4k̂ and î − ĵ + k̂.
Show answer
Cross = î(1 + 4) − ĵ(3 − 4) + k̂(−3 − 1) = 5î + ĵ − 4k̂. Area = √42.
Answer: √42
Show that the points with position vectors 2î − ĵ + k̂, î − 3ĵ − 5k̂, 3î − 4ĵ − 4k̂ form a right triangle.
Show answer
AB⃗ = −î − 2ĵ − 6k̂, BC⃗ = 2î − ĵ + k̂, CA⃗ = −î + 3ĵ + 5k̂. AB⃗·BC⃗ = −2 + 2 − 6 = −6; BC⃗·CA⃗ = −2 − 3 + 5 = 0 → right angle at C.
Answer: Right-angled at C
If a⃗ + b⃗ + c⃗ = 0⃗ with |a⃗| = 3, |b⃗| = 5, |c⃗| = 7, find the angle between a⃗ and b⃗.
Show answer
a⃗ + b⃗ = −c⃗ → 9 + 25 + 2a⃗·b⃗ = 49 → a⃗·b⃗ = 7.5 = 15 cos θ → cos θ = ½ → θ = 60°.
Answer: 60°
Prove that (a⃗ × b⃗)² = |a⃗|²|b⃗|² − (a⃗·b⃗)².
Show answer
|a⃗ × b⃗|² = |a|²|b|² sin²θ = |a|²|b|²(1 − cos²θ) = |a|²|b|² − (|a||b| cos θ)². ∎
Answer: Proved
Find a vector of magnitude 5 units parallel to the resultant of a⃗ = 2î + 3ĵ − k̂ and b⃗ = î − 2ĵ + k̂.
Show answer
a⃗ + b⃗ = 3î + ĵ, magnitude √10. Required: 5(3î + ĵ)/√10 = (√10/2)(3î + ĵ).
Answer: (3√10/2)î + (√10/2)ĵ
Let a⃗ = î + 4ĵ + 2k̂, b⃗ = 3î − 2ĵ + 7k̂, c⃗ = 2î − ĵ + 4k̂. Find d⃗ perpendicular to both a⃗ and b⃗ with c⃗·d⃗ = 15.
Show answer
a⃗ × b⃗ = î(28 + 4) − ĵ(7 − 6) + k̂(−2 − 12) = 32î − ĵ − 14k̂. d⃗ = λ(32î − ĵ − 14k̂); c⃗·d⃗ = λ(64 + 1 − 56) = 9λ = 15 → λ = 5/3. d⃗ = (1/3)(160î − 5ĵ − 70k̂).
Answer: d⃗ = (160î − 5ĵ − 70k̂)/3
13. Quick Revision Summary
- Vector = magnitude + direction; unit vector â = a⃗/|a⃗|; DCs satisfy l² + m² + n² = 1.
- Add by triangle/parallelogram law; AB⃗ = b⃗ − a⃗; section formula (mb⃗ + na⃗)/(m + n).
- Dot: |a||b| cos θ = Σaᵢbᵢ; zero ⇔ ⊥; projection a⃗·b⃗/|b⃗|.
- Cross: |a||b| sin θ n̂ (determinant); zero ⇔ ∥.
- Areas: parallelogram |a⃗ × b⃗|, triangle ½|a⃗ × b⃗|.