Magnitude and direction of a vector
A vector is a quantity with a size and a direction, like a push, a velocity or a move from one place to another. A scalar has only a size, like mass or time.
We draw a vector as an arrow from a start point (tail) to an end point (head). The magnitude is the length of the arrow, written |a|. For a = xî + yĵ + zk̂ the magnitude is |a| = √(x² + y² + z²). This is just Pythagoras used twice.
The direction is told by the angles the arrow makes with the positive x, y and z axes.
Direction cosines and direction ratios
Let the arrow make angles α, β, γ with the x, y and z axes. Then l = cos α, m = cos β, n = cos γ are its direction cosines.
- For a = xî + yĵ + zk̂: l = x/|a|, m = y/|a|, n = z/|a|.
- Always l² + m² + n² = 1 (because x² + y² + z² = |a|²).
Any three numbers in the same proportion as l, m, n are called direction ratios. The components x, y, z themselves are direction ratios. To go from ratios a, b, c to cosines, divide each by √(a² + b² + c²).
Why it matters: the unit vector in the direction of a is simply lî + mĵ + nk̂.
Types of vectors
- Zero (null) vector 0: length 0, start and end are the same point, direction not fixed.
- Unit vector â: length 1. â = a/|a|. î, ĵ, k̂ are unit vectors along the axes.
- Co-initial vectors: same starting point.
- Collinear (parallel) vectors: lie along the same or parallel lines; b = k a for some number k.
- Equal vectors: same length and same direction, wherever they are drawn (free vectors).
- Negative of a vector −a: same length, opposite direction.
Position vector and components
The position vector of a point P(x, y, z) is the arrow from the origin O to P: OP = xî + yĵ + zk̂. Its length is the distance OP.
Here x î, y ĵ and z k̂ are the vector components along the axes, and x, y, z are the scalar components.
Vector joining two points
From P₁(x₁, y₁, z₁) to P₂(x₂, y₂, z₂): P₁P₂ = (x₂ − x₁)î + (y₂ − y₁)ĵ + (z₂ − z₁)k̂ — always "head minus tail".
Addition of vectors and multiplication by a scalar
Triangle law: place the tail of b at the head of a; the arrow from the tail of a to the head of b is a + b. Parallelogram law says the same thing: draw a and b from one point; the diagonal is a + b.
- In components: add î parts, ĵ parts and k̂ parts separately.
- Addition is commutative (a + b = b + a) and associative ((a + b) + c = a + (b + c)).
- a + (−a) = 0, and a − b = a + (−b).
Scalar multiple
k a has length |k||a|. If k > 0 it points the same way, if k < 0 it points the opposite way, if k = 0 it is the zero vector. k(a + b) = ka + kb.
Section formula
Let A and B have position vectors a and b.
- Internal division in m : n: r = (m b + n a)/(m + n).
- External division in m : n: r = (m b − n a)/(m − n), m ≠ n.
- Mid-point (m = n): r = (a + b)/2.
Why: AP : PB = m : n means n·AP = m·PB, i.e. n(r − a) = m(b − r). Solve for r. Remember: m multiplies the far end b.
Also: the centroid of a triangle with vertices a, b, c is (a + b + c)/3.
Dot (scalar) product and its uses
a·b = |a||b| cos θ, where θ (0 ≤ θ ≤ π) is the angle between them. The answer is a number (scalar).
- In components: a·b = a₁b₁ + a₂b₂ + a₃b₃.
- î·î = ĵ·ĵ = k̂·k̂ = 1; î·ĵ = ĵ·k̂ = k̂·î = 0.
- a·b = 0 ⇔ a ⊥ b (for non-zero vectors). a·a = |a|².
- Commutative: a·b = b·a; distributive over addition.
Uses
- Angle: cos θ = a·b / (|a||b|).
- Projection of a on b: a·b̂ = a·b/|b| (the "shadow" of a on b).
- Work done by a force: W = F·d.
Cross (vector) product and its uses
a × b = |a||b| sin θ n̂, where n̂ is the unit vector at right angles to both a and b, chosen by the right-hand rule (curl fingers from a to b, thumb gives n̂).
- î × ĵ = k̂, ĵ × k̂ = î, k̂ × î = ĵ; î × î = 0.
- b × a = −(a × b) (not commutative).
- a × b = 0 ⇔ a ∥ b (for non-zero vectors).
- In components use the 3 × 3 determinant with î, ĵ, k̂ in the first row, a₁ a₂ a₃ in the second, b₁ b₂ b₃ in the third.
Uses
- Area of parallelogram with sides a, b = |a × b|; with diagonals d₁, d₂ = ½|d₁ × d₂|.
- Area of triangle ABC = ½|AB × AC|.
- A vector perpendicular to two given vectors; sin θ = |a × b| / (|a||b|).
- Torque τ = r × F in physics.
Board exam: the unit carries 14 marks with 3D geometry; typical questions are angle/projection, area by cross product, unit vector perpendicular to two vectors and section formula.
Key formulas and definitions
- |a| = √(x² + y² + z²)
- l = x/|a|, m = y/|a|, n = z/|a|; l² + m² + n² = 1
- â = a/|a|
- P₁P₂ = (x₂ − x₁)î + (y₂ − y₁)ĵ + (z₂ − z₁)k̂
- Internal: r = (mb + na)/(m + n); external: r = (mb − na)/(m − n)
- a·b = |a||b|cosθ = a₁b₁ + a₂b₂ + a₃b₃
- Projection of a on b = a·b/|b|
- |a × b| = |a||b|sinθ; area of triangle = ½|a × b|
Worked examples
1. Find the magnitude and direction cosines of a = 2î − 3ĵ + 6k̂.
Step 1: |a| = √(4 + 9 + 36) = √49 = 7. Step 2: l = 2/7, m = −3/7, n = 6/7. Check: 4/49 + 9/49 + 36/49 = 1 ✓.
2. Find the unit vector in the direction of a + b, where a = î + 2ĵ − k̂ and b = 2î − ĵ + 3k̂.
Step 1: a + b = 3î + ĵ + 2k̂. Step 2: |a + b| = √(9 + 1 + 4) = √14. Step 3: unit vector = (3î + ĵ + 2k̂)/√14.
3. Find the position vector of the point R dividing P(2, −1, 3) and Q(4, 5, −3) externally in the ratio 2 : 1.
Step 1: r = (2q − 1p)/(2 − 1). Step 2: 2q = (8, 10, −6), p = (2, −1, 3). Step 3: 2q − p = (6, 11, −9). So r = 6î + 11ĵ − 9k̂.
4. Find the angle between a = î + ĵ − k̂ and b = î − ĵ + k̂.
Step 1: a·b = 1 − 1 − 1 = −1. Step 2: |a| = |b| = √3. Step 3: cosθ = −1/3. So θ = cos⁻¹(−1/3) ≈ 109.5°.
5. Find the projection of a = 2î + 3ĵ + 2k̂ on b = î + 2ĵ + k̂.
Step 1: a·b = 2 + 6 + 2 = 10. Step 2: |b| = √6. Step 3: projection = 10/√6 = 5√6/3 ≈ 4.08.
6. Find a unit vector perpendicular to both a = 2î + 3ĵ − k̂ and b = î − ĵ + 2k̂.
Step 1: a × b = î(3·2 − (−1)(−1)) − ĵ(2·2 − (−1)·1) + k̂(2·(−1) − 3·1) = 5î − 5ĵ − 5k̂. Step 2: |a × b| = 5√3. Step 3: unit vector = ±(î − ĵ − k̂)/√3.
7. Find the area of the triangle with vertices A(1, 1, 2), B(2, 3, 5) and C(1, 5, 5).
Step 1: AB = î + 2ĵ + 3k̂, AC = 0î + 4ĵ + 3k̂. Step 2: AB × AC = î(6 − 12) − ĵ(3 − 0) + k̂(4 − 0) = −6î − 3ĵ + 4k̂. Step 3: |AB × AC| = √(36 + 9 + 16) = √61. Area = ½√61 ≈ 3.9 square units.
8. If |a| = 3, |b| = 4 and a·b = 6, find |a × b| and |a + b|.
Step 1: cosθ = 6/12 = 1/2, so θ = 60° and sinθ = √3/2. Step 2: |a × b| = 3 × 4 × √3/2 = 6√3. Step 3: |a + b|² = |a|² + |b|² + 2a·b = 9 + 16 + 12 = 37, so |a + b| = √37.
Common mistakes
- Writing a × b = b × a. The order matters: b × a = −(a × b).
- Forgetting the square root in |a|, or adding the components instead of their squares.
- In the section formula, putting m with a instead of b. m goes with the far end b: (mb + na)/(m + n).
- Using the middle term of the cross-product determinant with a + sign. The ĵ term always carries a minus: −ĵ(a₁b₃ − a₃b₁).