7.3 Vectors

Why Vectors Matter for PAF CAE Maths

Vector methods appear throughout FSc maths and form the language of later engineering mechanics: force resultants, velocity, and angular quantities all use the same component algebra. On the Pakistan Air Force Aeronautical Engineering initial academic paper, vector questions are usually short: find a magnitude, a unit vector, an angle via the cosine formula, or one component of a cross product. Treat this section as a formula-and-recognition drill under the commonly reported ~50 maths MCQs / ~25 minutes timing.

Scalars, Vectors, and Component Form

A scalar has magnitude only (mass, temperature, time). A vector has magnitude and direction (displacement, force, velocity).

In Cartesian components:

• 2D: a = a₁ î + a₂ ĵ = ⟨a₁, a₂⟩
• 3D: a = a₁ î + a₂ ĵ + a₃ k̂ = ⟨a₁, a₂, a₃⟩

Equal vectors have equal corresponding components. The zero vector 0 has all components zero and undefined direction.

Worked Example 1 — Resultant

If a = ⟨3, −1, 2⟩ and b = ⟨−5, 4, 1⟩, find 2a − b.

Solution: 2a = ⟨6, −2, 4⟩; 2a − b = ⟨6−(−5), −2−4, 4−1⟩ = ⟨11, −6, 3⟩.

Magnitude and Unit Vector

Magnitude (length):

|a| = √(a₁² + a₂²) in 2D, or √(a₁² + a₂² + a₃²) in 3D.

A unit vector has magnitude 1. For a ≠ 0,

â = a / |a|.

The standard basis vectors î = ⟨1,0,0⟩, ĵ = ⟨0,1,0⟩, k̂ = ⟨0,0,1⟩ are already unit vectors and mutually perpendicular.

Worked Example 2 — Unit vector

Find the unit vector in the direction of v = ⟨2, −3, 6⟩.

Solution: |v| = √(4 + 9 + 36) = √49 = 7. Unit vector ⟨2/7, −3/7, 6/7⟩.

Worked Example 3 — Vector of given length

Find the vector of magnitude 10 in the direction of ⟨3, −4⟩.

Solution: Unit vector ⟨3/5, −4/5⟩ (since √(9+16)=5). Required vector: 10⟨3/5, −4/5⟩ = ⟨6, −8⟩.

Direction Cosines (3D Reminder)

If a makes angles α, β, γ with the positive x-, y-, and z-axes,

cos α = a₁/|a|, cos β = a₂/|a|, cos γ = a₃/|a|,

and cos²α + cos²β + cos²γ = 1.

MCQs may ask you to verify this identity or recover one cosine from the other two.

Dot Product (Scalar Product)

Definition: a · b = |a||b| cos θ, where θ is the angle between the directions of a and b when placed tail-to-tail.

Component form: a · b = a₁b₁ + a₂b₂ + a₃b₃.

Consequences:

• a · a = |a|²
• a · b = 0 ⇔ a ⊥ b (for nonzero vectors)
• cos θ = (a · b) / (|a||b|)
• Projection of a onto b: proj_b a = [(a · b)/|b|²] b; scalar projection (a · b)/|b|

Worked Example 4 — Angle between vectors

Find the angle between a = ⟨1, 2, 2⟩ and b = ⟨2, −1, 2⟩.

Solution: a · b = 2 − 2 + 4 = 4. |a| = 3, |b| = 3. cos θ = 4/9 ⇒ θ = cos⁻¹(4/9).

Worked Example 5 — Orthogonality check

Are ⟨2, −1, 1⟩ and ⟨1, 3, 1⟩ perpendicular?

Solution: Dot product = 2 − 3 + 1 = 0. Yes, they are perpendicular.

Cross Product (Vector Product)

In 3D, a × b is a vector such that:

1. a × b is perpendicular to both a and b.
2. |a × b| = |a||b| sin θ (area of the parallelogram spanned by a and b).
3. Direction follows the right-hand rule.

Determinant form:

a × b = | î    ĵ    k̂ |        | a₁  a₂  a₃ |        | b₁  b₂  b₃ |

= î(a₂b₃ − a₃b₂) − ĵ(a₁b₃ − a₃b₁) + k̂(a₁b₂ − a₂b₁).

Key facts:

• a × b = −(b × a) (anti-commutative)
• a × a = 0
• a × b = 0 (with a, b ≠ 0) ⇔ a ∥ b
• î × ĵ = k̂, ĵ × k̂ = î, k̂ × î = ĵ

Worked Example 6 — Cross product

Compute ⟨1, −2, 3⟩ × ⟨2, 0, −1⟩.

Solution:

î[(−2)(−1) − (3)(0)] − ĵ[(1)(−1) − (3)(2)] + k̂[(1)(0) − (−2)(2)]

= î(2 − 0) − ĵ(−1 − 6) + k̂(0 + 4)

= ⟨2, 7, 4⟩.

(The middle component: −(−7) = +7.)

Worked Example 7 — Area of a triangle

Points A(1,0,0), B(2,1,−1), C(0,−1,2) form a triangle. Find its area.

Solution: AB = ⟨1, 1, −1⟩, AC = ⟨−1, −1, 2⟩.

AB × AC = |î ĵ k̂; 1 1 −1; −1 −1 2| = î(2 − 1) − ĵ(2 − 1) + k̂(−1 + 1) = ⟨1, −1, 0⟩.

|AB × AC| = √2. Area = (1/2)|AB × AC| = √2 / 2.

Scalar Triple Product (Brief)

[a, b, c] = a · (b × c) equals the volume of the parallelepiped formed by a, b, c. It equals the determinant of the 3×3 matrix with rows (or columns) a, b, c. Vanishing triple product means the three vectors are coplanar—occasionally tested as a yes/no MCQ.

Applications Framed for Timed MCQs

You do not need physics depth beyond recognising which product is scalar vs vector.

Exam Traps

1. Dot vs cross: dot → scalar; cross → vector. An option that is a number cannot be a cross product answer.
2. Sign of ĵ component: the determinant expansion subtracts the ĵ minor—forgetting the minus sign is the most common arithmetic error.
3. Unit vector check: components must satisfy squares summing to 1; use this to eliminate options quickly.
4. 2D “cross product”: in plane problems, |a₁b₂ − a₂b₁| is the magnitude of the 3D cross product’s k̂ component—still a scalar area measure.

Drill magnitudes, unit vectors, and one clean cross-product expansion until they feel automatic; that is the depth the CAE academic maths block rewards.