What this chapter is about
Vector Algebra introduces quantities that have both magnitude and direction, distinguishing them from scalars which have only magnitude. In physics and engineering, displacement, velocity, force and acceleration are vectors; temperature, mass and time are scalars. This chapter builds the algebraic machinery to add, subtract and multiply vectors in two and three dimensions.
A Class 12 student meets this chapter because many results in mechanics, electromagnetism and three-dimensional geometry depend on vector operations. The dot product helps find angles and projections; the cross product gives a vector perpendicular to two given vectors and measures area. Together they form the toolkit for the chapters on three-dimensional geometry and for physics problems involving torque, work and magnetic force.
After working through this chapter you should be able to represent a vector in component form, compute its magnitude, add and subtract vectors, find the scalar (dot) and vector (cross) products, and interpret these products geometrically.
Key ideas
- A vector is written as an ordered triple (a₁, a₂, a₃) or as a₁ i + a₂ j + a₃ k, where i, j, k are unit vectors along the x-, y- and z-axes.
- The magnitude (or modulus) of vector a = a₁ i + a₂ j + a₃ k is |a| = √(a₁² + a₂² + a₃²).
- Two vectors are equal if and only if their corresponding components are equal; a zero vector has all components zero and no definite direction.
- Scalar multiplication: k a stretches or shrinks the vector; if k < 0 the direction reverses.
- The dot product a · b = |a| |b| cos θ, where θ is the angle between the vectors; in component form a · b = a₁b₁ + a₂b₂ + a₃b₃.
- The cross product a × b is a vector perpendicular to both a and b, with magnitude |a| |b| sin θ and direction given by the right-hand rule.
- Cross product in component form: a × b = (a₂b₃ − a₃b₂) i − (a₁b₃ − a₃b₁) j + (a₁b₂ − a₂b₁) k.
- Scalar triple product [a b c] = a · (b × c) equals the volume of the parallelepiped formed by the three vectors; if it is zero the vectors are coplanar.
Formulas and facts to remember
- Formula / Rule: |a| = √(a₁² + a₂² + a₃²) · Meaning: Magnitude of a vector.
- Formula / Rule: Unit vector â = a / |a| · Meaning: A vector of magnitude 1 in the direction of a.
- Formula / Rule: a · b = a₁b₁ + a₂b₂ + a₃b₃ · Meaning: Dot product via components.
- Formula / Rule: a · b = |a| |b| cos θ · Meaning: Dot product via magnitudes and angle.
- Formula / Rule: a × b = (a₂b₃ − a₃b₂) i − (a₁b₃ − a₃b₁) j + (a₁b₂ − a₂b₁) k · Meaning: Cross product in components.
- Formula / Rule: |a × b| = |a| |b| sin θ · Meaning: Magnitude of cross product.
- Formula / Rule: a · a = |a|² · Meaning: Useful identity for magnitude.
- Formula / Rule: a × a = 0 · Meaning: Cross product of a vector with itself is the zero vector.
- Formula / Rule: a · b = 0 ⇔ a ⊥ b (if both non-zero) · Meaning: Perpendicularity test.
- Formula / Rule: a × b = 0 ⇔ a ∥ b (if both non-zero) · Meaning: Parallelism test.
- Formula / Rule: [a b c] = a · (b × c) · Meaning: Scalar triple product (gives volume).
Worked examples
Example 1 — Finding magnitude and unit vector
Problem. Find the magnitude and unit vector in the direction of p = 2 i − 3 j + 6 k.
Solution.
Magnitude |p| = √(2² + (−3)² + 6²) = √(4 + 9 + 36) = √49 = 7.
Unit vector p̂ = p / |p| = (2/7) i − (3/7) j + (6/7) k.
Example 2 — Dot product and angle between vectors
Problem. Find the angle between a = i + 2 j + 2 k and b = 3 i − 4 j.
Solution.
a · b = (1)(3) + (2)(−4) + (2)(0) = 3 − 8 + 0 = −5.
a| = √(1 + 4 + 4) = 3.
b| = √(9 + 16 + 0) = 5.
cos θ = (a · b) / (|a| |b|) = −5 / 15 = −1/3.
Therefore θ = cos⁻¹(−1/3).
Example 3 — Cross product and area of a triangle
Problem. Points A, B, C have position vectors a = i + j, b = 2 i + 3 j + k, c = i + 2 j + 3 k. Find the area of triangle ABC.
Solution.
AB = b − a = (2 − 1) i + (3 − 1) j + (1 − 0) k = i + 2 j + k.
AC = c − a = (1 − 1) i + (2 − 1) j + (3 − 0) k = 0 i + j + 3 k.
AB × AC: Using the determinant expansion,
= (2 × 3 − 1 × 1) i − (1 × 3 − 1 × 0) j + (1 × 1 − 2 × 0) k
= 5 i − 3 j + k.
AB × AC| = √(25 + 9 + 1) = √35.
Area of triangle = ½ |AB × AC| = √35 / 2 square units.
Common mistakes
- Confusing dot and cross products → remember: dot gives a scalar, cross gives a vector.
- Forgetting the minus sign in the j-component when expanding a cross product determinant → write out the 3 × 3 determinant carefully every time.
- Using sin θ in the dot product formula or cos θ in the cross product magnitude → dot uses cos, cross uses sin.
- Assuming a × b = b × a → cross product is anti-commutative: a × b = −(b × a).
- Treating the scalar triple product as a vector → [a b c] is a scalar (the determinant value), not a vector.
Quick revision
- Magnitude of a = a₁ i + a₂ j + a₃ k is √(a₁² + a₂² + a₃²).
- a · b = 0 means the vectors are perpendicular (if both non-zero).
- a × b = 0 means the vectors are parallel (if both non-zero).
- Area of parallelogram with adjacent sides a, b equals |a × b|.
- Volume of parallelepiped = |a · (b × c)|.
- Cross product is anti-commutative; dot product is commutative.