Some Basic
Concepts
Magnitude, unit vectors, direction cosines, components, section formula — the essential building blocks of vector algebra.
The magnitude of vector a is its length. It is always non-negative and is found using the distance formula extended to 3D.
|a| = √(x² + y² + z²)
In 2D — if a = x î + y ĵ: |a| = √(x² + y²)
Find |a| if a = 2î − 3ĵ + 6k̂
|a| = √(4 + 9 + 36) = √49 = 7
Magnitude is always a scalar ≥ 0. |a| = 0 only when a =0 (zero vector).
A unit vector has magnitude = exactly 1. It is used to represent direction alone. The unit vector in the direction of a⃗ is written as â (read "a-hat").
Always: |â| = 1
Find unit vector along b = 3î + 4ĵ
|b| = √(9+16) = 5
b̂ = (3/5)î + (4/5)ĵ
Verify: √((3/5)²+(4/5)²) = √(9/25+16/25) = 1 ✓
î, ĵ, k̂ are themselves unit vectors. They are mutually perpendicular. This means: î·ĵ = ĵ·k̂ = k̂·î = 0 and î·î = ĵ·ĵ = k̂·k̂ = 1
If vector a makes angles α, β, γ with the positive x, y, z axes, then cos α, cos β, cos γ are called the direction cosines (written l, m, n).
l = x/r, m = y/r, n = z/r
★ Golden Identity: l² + m² + n² = 1 (always true)
Direction ratios (a, b, c) are any numbers proportional to direction cosines. From them, we recover direction cosines as:
m = b/√(a²+b²+c²)
n = c/√(a²+b²+c²)
Direction cosines of a = î + 2ĵ + 2k̂
|a| = √(1+4+4) = 3
l = 1/3, m = 2/3, n = 2/3
Check: (1/9)+(4/9)+(4/9) = 9/9 = 1 ✓
| Feature | Direction Cosines | Direction Ratios |
|---|---|---|
| Notation | l, m, n | a, b, c |
| Constraint | l²+m²+n² = 1 | No constraint |
| Uniqueness | Unique (up to sign) | Infinitely many sets |
| Use | Standard form | Easier to write initially |
l² + m² + n² = 1 is used repeatedly in proofs and problems. The direction cosines of axes: x-axis = (1,0,0), y-axis = (0,1,0), z-axis = (0,0,1).
Any vector in 3D is expressed as a linear combination of î, ĵ, k̂. The scalars x, y, z are its components.
a₁ = x-component, a₂ = y-component, a₃ = z-component
Add component-wise:
(a₁+b₁)î + (a₂+b₂)ĵ + (a₃+b₃)k̂
a = b if a₁=b₁, a₂=b₂, a₃=b₃
All three components must match.
Point R divides segment AB (with position vectors a and b) in ratio m : n.
r = (m·b + n·a) / (m + n)
External Division:
r = (m·b − n·a) / (m − n)
Midpoint (m = n):
r = (a + b) / 2
A = (1,2,3), B = (4,5,6). Find midpoint M.
M = ((1+4)/2, (2+5)/2, (3+6)/2) = (5/2, 7/2, 9/2)
Section formula appears in 4-mark questions. Express the result as a position vector: r = (5/2)î + (7/2)ĵ + (9/2)k̂
Two vectors are collinear (parallel) if one is a scalar multiple of the other.
Equivalently: a₁/b₁ = a₂/b₂ = a₃/b₃
Are a = 2î+4ĵ−6k̂ and b = î+2ĵ−3k̂ parallel?
Ratios: 2/1 = 4/2 = −6/−3 = 2 (all equal) → Yes, a = 2b ✓
To prove three points A, B, C are collinear: show AB = λ · AC. This is the fastest method and always works.