Vector Algebra

Physical quantities are broadly divided into two kinds. A scalar quantity, such as mass, length, time or temperature, is completely described by a magnitude (a number with an appropriate unit). A vector quantity, such as displacement, velocity, force or acceleration, needs both a magnitude and a direction to be fully described.

Geometrically, a vector is represented by a directed line segment. If a vector starts at a point A (the initial point) and ends at a point B (the terminal point), it is written as vector AB, and its magnitude is the length of the segment AB, written |AB|. The direction of the arrowhead shows the direction of the vector.

Several special types of vectors appear repeatedly in this chapter:

• Zero (null) vector: a vector whose initial and terminal points coincide, so its magnitude is 0. It is written as 0; its direction is not defined.
• Unit vector: a vector whose magnitude is exactly 1. A unit vector in the direction of a is written â ("a cap") and â = a / |a|.
• Coinitial vectors: two or more vectors having the same initial point.
• Collinear vectors: vectors that are parallel to the same line, irrespective of their magnitudes or direction.
• Equal vectors: two vectors are equal if they have the same magnitude and the same direction, regardless of where their initial points are.
• Negative of a vector: a vector with the same magnitude as a given vector but the opposite direction; the negative of a is written −a.

Worked Example: Find the unit vector in the direction of a = 3i − 4k.

Magnitude: |a| = √(3² + 0² + (−4)²) = √(9+16) = √25 = 5. So the required unit vector is â = a/|a| = (3i − 4k)/5 = (3/5)i − (4/5)k. As a check, its magnitude is √((3/5)² + (4/5)²) = √(9/25+16/25) = √(25/25) = 1, confirming it is indeed a unit vector.

Every point P(x, y, z) in space can be associated with a unique vector called its position vector, obtained by joining the origin O to P: OP = x i + y j + z k, where i, j, k are the standard unit vectors along the x, y and z axes. The magnitude of the position vector is |OP| = √(x² + y² + z²).

If a vector makes angles α, β, γ with the positive x, y and z axes respectively, then cos α, cos β, cos γ are called its direction cosines, usually denoted l, m, n. Direction cosines always satisfy l² + m² + n² = 1. Any three numbers a, b, c that are proportional to the direction cosines (l : m : n = a : b : c) are called the direction ratios of the vector.

Given direction ratios a, b, c of a vector, the actual direction cosines are recovered by dividing each by the magnitude √(a²+b²+c²): l = a/√(a²+b²+c²), m = b/√(a²+b²+c²), n = c/√(a²+b²+c²) (all three signs reverse together if the opposite direction is required).

Worked Example: Find the direction cosines of the vector a = 2i − 3j + 6k.

|a| = √(2² + (−3)² + 6²) = √(4+9+36) = √49 = 7. So l = 2/7, m = −3/7, n = 6/7. Check: l² + m² + n² = 4/49 + 9/49 + 36/49 = 49/49 = 1, as required.

Two vectors are added geometrically using the triangle law of addition: if AB represents vector a and BC (coinitial with the terminal point of a) represents vector b, then the third side AC of the triangle represents the sum a + b. Equivalently, the parallelogram law states that if two vectors are represented as the two adjacent sides of a parallelogram drawn from the same point, their sum is the diagonal of the parallelogram from that point.

Vector addition obeys the same familiar algebraic rules as addition of numbers: it is commutative (a + b = b + a) and associative ((a + b) + c = a + (b + c)). Adding the zero vector leaves a vector unchanged (a + 0 = a), and a + (−a) = 0, which is how subtraction a − b is defined, namely as a + (−b).

Multiplication by a scalar: If λ is a real number (scalar) and a is a vector, then λa is a vector whose magnitude is |λ||a| and whose direction is the same as a if λ > 0, opposite to a if λ < 0, and the zero vector if λ = 0. Scalar multiplication is distributive over vector addition: λ(a + b) = λa + λb, and over scalar addition: (λ + μ)a = λa + μa.

Worked Example: If a = 2i + 3j − k and b = i − 2j + 3k, find (i) a + b, (ii) a − b, (iii) 2a − 3b.

(i) a + b = (2+1)i + (3−2)j + (−1+3)k = 3i + j + 2k.
(ii) a − b = (2−1)i + (3−(−2))j + (−1−3)k = i + 5j − 4k.
(iii) 2a = 4i + 6j − 2k and 3b = 3i − 6j + 9k, so 2a − 3b = (4−3)i + (6−(−6))j + (−2−9)k = i + 12j − 11k.

Any vector a in three-dimensional space can be written in terms of the standard unit vectors i, j, k as a = a₁i + a₂j + a₃k, where a₁, a₂, a₃ are called the scalar components (or simply components) of a along the x, y and z axes. The magnitude is then |a| = √(a₁² + a₂² + a₃²).

If P(x₁, y₁, z₁) and Q(x₂, y₂, z₂) are any two points, the vector joining them, directed from P to Q, is PQ = (x₂ − x₁)i + (y₂ − y₁)j + (z₂ − z₁)k, and its magnitude |PQ| gives the distance between P and Q.

The section formula gives the position vector of a point R that divides the line segment joining two points A and B (with position vectors a and b) in a given ratio. If R divides AB internally in the ratio m : n, then r = (mb + na)/(m + n). If R divides AB externally in the ratio m : n, then r = (mb − na)/(m − n). When R is the midpoint of AB, this reduces to r = (a + b)/2.

Worked Example: Find the position vector of the point R that divides the join of A and B, with position vectors a = i − 2j + k and b = 4i + 3j − k, internally in the ratio 2 : 1.

Here m = 2, n = 1. r = (2b + 1·a)/(2+1) = [2(4i+3j−k) + (i−2j+k)]/3 = (8i+6j−2k + i−2j+k)/3 = (9i + 4j − k)/3 = 3i + (4/3)j − (1/3)k.

The scalar product (or dot product) of two nonzero vectors a and b, inclined at angle θ (0 ≤ θ ≤ π) to each other, is the scalar defined by a · b = |a| |b| cos θ. When either vector is the zero vector, the dot product is taken to be 0.

The dot product has several important consequences. Since cos θ = 1 when θ = 0, a · a = |a|². Since cos(π/2) = 0, two nonzero vectors are perpendicular exactly when a · b = 0. For the standard unit vectors, i · i = j · j = k · k = 1 and i · j = j · k = k · i = 0. The dot product is commutative (a · b = b · a) and distributes over addition (a · (b + c) = a · b + a · c). In component form, if a = a₁i+a₂j+a₃k and b = b₁i+b₂j+b₃k, then a · b = a₁b₁ + a₂b₂ + a₃b₃.

Once a·b is known, the angle between the vectors follows from cos θ = (a · b)/(|a||b|). A related idea is the projection of a on b, which is the scalar (a · b)/|b|; geometrically it is the length of the "shadow" of a cast on the line along b.

Worked Example: Find the angle between a = i + j − k and b = i − j + k.

a · b = (1)(1) + (1)(−1) + (−1)(1) = 1 − 1 − 1 = −1. |a| = √(1+1+1) = √3, |b| = √(1+1+1) = √3. So cos θ = −1/(√3·√3) = −1/3, giving θ = cos⁻¹(−1/3), an obtuse angle since the dot product is negative.

The vector product (or cross product) of two nonzero, non-parallel vectors a and b inclined at angle θ is itself a vector, defined as a × b = |a| |b| sin θ n̂, where n̂ is the unit vector perpendicular to both a and b, with direction fixed by the right-hand rule (curl the fingers of the right hand from a to b; the thumb points along n̂). If a and b are parallel (θ = 0 or π) or either is the zero vector, a × b = 0.

The cross product is anti-commutative: b × a = −(a × b), because reversing the order reverses the direction given by the right-hand rule (though the magnitude stays the same). It distributes over addition: a × (b + c) = a × b + a × c, and a × a = 0 for every vector a. For the standard unit vectors, i × i = j × j = k × k = 0, while i × j = k, j × k = i, and k × i = j (cyclic order); reversing any pair flips the sign, e.g. j × i = −k.

In component form, if a = a₁i+a₂j+a₃k and b = b₁i+b₂j+b₃k, the cross product is computed from the determinant of i, j, k over the two rows of components, which expands to a × b = (a₂b₃ − a₃b₂)i − (a₁b₃ − a₃b₁)j + (a₁b₂ − a₂b₁)k. Two very useful geometric applications follow directly from |a × b| = |a||b| sin θ: the area of a triangle with adjacent sides a and b is (1/2)|a × b|, and the area of a parallelogram with adjacent sides a and b is |a × b|.

Worked Example: Find a × b for a = i + 2j + 3k and b = 2i − j + k, and hence find the area of the triangle with adjacent sides a and b.

a × b = (a₂b₃ − a₃b₂)i − (a₁b₃ − a₃b₁)j + (a₁b₂ − a₂b₁)k = ((2)(1)−(3)(−1))i − ((1)(1)−(3)(2))j + ((1)(−1)−(2)(2))k = (2+3)i − (1−6)j + (−1−4)k = 5i + 5j − 5k. Its magnitude is |a × b| = √(25+25+25) = √75 = 5√3. So the area of the triangle is (1/2)(5√3) = (5√3)/2 square units.