Class 12 Maths Chapter 10: Vector Algebra

Vector Algebra is Chapter 10 of NCERT Class 12 Maths. It introduces quantities that have both magnitude and direction, the different types of vectors, their addition, the section formula, and the two kinds of products of vectors: the scalar (dot) product and the vector (cross) product. Watch the five video lessons below, then go through the notes, formulas and solved examples to revise.

What you will learn

The difference between scalars and vectors, and the main types of vectors
Position vectors, components and direction cosines of a vector
Addition of vectors, multiplication by a scalar and the section formula
The scalar (dot) product, angle between vectors and projection
The vector (cross) product and its use in finding areas

Video Lessons

Vector Algebra - Part 1

Vector Algebra - Part 2

Vector Algebra - Part 3

Vector Algebra - Part 4

Vector Algebra - Part 5

Key Concepts and Formulas

Scalars, Vectors and Their Types

A scalar has only magnitude (like mass or time), while a vector has magnitude as well as direction (like displacement or force). A vector is drawn as a directed line segment from an initial point to a terminal point.

Unit vector along a: â = a/|a|

Zero vector: magnitude 0, direction not defined
Unit vector: magnitude 1
Equal vectors: same magnitude and same direction; collinear vectors are parallel to the same line
Negative of a vector: same magnitude, opposite direction

Components and Direction Cosines

In three dimensions, î, ĵ and k̂ are unit vectors along the x, y and z axes. The position vector of the point P(x, y, z) is written in terms of them.

r = xî + yĵ + zk̂
|r| = √(x² + y² + z²)
l = x/|r|, m = y/|r|, n = z/|r|
l² + m² + n² = 1
A(x₁, y₁, z₁) to B(x₂, y₂, z₂):
AB = (x₂ − x₁)î + (y₂ − y₁)ĵ + (z₂ − z₁)k̂

Addition and Scalar Multiplication

Vectors follow the triangle law and the parallelogram law of addition. Addition is commutative and associative, and multiplying by a scalar λ stretches, shrinks or reverses a vector. Two vectors are collinear when their components are proportional.

Triangle law: AB + BC = AC
Collinear: b = λa for some scalar λ
(a₁î + a₂ĵ + a₃k̂) + (b₁î + b₂ĵ + b₃k̂)
  = (a₁ + b₁)î + (a₂ + b₂)ĵ + (a₃ + b₃)k̂

Triangle law of vector addition showing AB + BC = AC, Class 12 Maths Chapter 10 Vector Algebra — Head to tail: AB + BC = AC

Section Formula

If a point R divides the line joining points with position vectors a and b in the ratio m : n, its position vector is given by these formulas.

Internal: r = (mb + na)/(m + n)
External: r = (mb − na)/(m − n)
Midpoint: r = (a + b)/2

Scalar or Dot Product

The dot product of two vectors gives a scalar. It is the main tool for finding angles and checking perpendicularity of non-zero vectors.

a · b = |a||b| cos θ
a · b = a₁b₁ + a₂b₂ + a₃b₃
cos θ = (a · b)/(|a||b|)
a ⊥ b ⇔ a · b = 0
a · a = |a|²
Projection of a on b = (a · b)/|b|

Vector or Cross Product

The cross product of two vectors gives a vector perpendicular to both of them, with direction fixed by the right-hand rule. In components, a × b is the determinant with rows (î, ĵ, k̂), (a₁, a₂, a₃) and (b₁, b₂, b₃).

a × b = |a||b| sin θ n̂
a × b = −(b × a)
a × b = 0 ⇔ a ∥ b (non-zero vectors)
Area of parallelogram (sides a, b) = |a × b|
Area of triangle (sides a, b) = ½|a × b|
Parallelogram with diagonals d₁, d₂ = ½|d₁ × d₂|

Dot product as projection and cross product as parallelogram area, Class 12 Vector Algebra — Dot product gives a number, cross product gives area

Solved Examples

Example 1: Find the magnitude and unit vector of a = 2î − 3ĵ + 6k̂.

|a| = √(2² + (−3)² + 6²)
|a| = √(4 + 9 + 36) = √49 = 7
∴ |a| = 7, â = (2î − 3ĵ + 6k̂)/7

Example 2: Find the direction cosines of 2î − ĵ + 2k̂.

Magnitude = √(4 + 1 + 4) = 3
l = 2/3, m = −1/3, n = 2/3
∴ Direction cosines are 2/3, −1/3, 2/3

Example 3: Find the point dividing the join of A(1, −2, 3) and B(7, 4, −3) internally in the ratio 2 : 1.

r = (2b + a)/3
x = (14 + 1)/3 = 5
y = (8 − 2)/3 = 2
z = (−6 + 3)/3 = −1
∴ Point is (5, 2, −1), i.e. 5î + 2ĵ − k̂

Example 4: Find λ if 2î + λĵ + k̂ is perpendicular to 3î − 2ĵ + 4k̂.

Dot product = 0
(2)(3) + (λ)(−2) + (1)(4) = 0
6 − 2λ + 4 = 0
∴ λ = 5

Example 5: Find the angle between a = î + 2ĵ + 2k̂ and b = 2î − ĵ + 2k̂.

a · b = 2 − 2 + 4 = 4
|a| = 3, |b| = 3
cos θ = 4/9
∴ θ = cos⁻¹(4/9)

Example 6: Find the projection of a = 3î + ĵ + 2k̂ on b = î + 2ĵ + 2k̂.

a · b = 3 + 2 + 4 = 9
|b| = √(1 + 4 + 4) = 3
Projection = 9/3
∴ Projection = 3

Example 7: Find the area of the triangle with adjacent sides a = î + 2ĵ − k̂ and b = 2î − ĵ + 3k̂.

a × b = (6 − 1)î − (3 + 2)ĵ + (−1 − 4)k̂
a × b = 5î − 5ĵ − 5k̂
|a × b| = √75 = 5√3
Area = ½|a × b|
∴ Area = 5√3/2 square units

Common Mistakes to Avoid

In the cross product determinant, the middle term carries a minus sign; dropping it is the most common error.
Do not confuse the projection (a · b)/|b| with the projection vector, which is that number multiplied by b̂.
Remember a × b = −(b × a); the order matters for the cross product but not for the dot product.
For the area of a triangle, do not forget the factor ½.

Quick Revision

|a| = √(a₁² + a₂² + a₃²),  â = a/|a|
Section (internal): (mb + na)/(m + n)
a · b = a₁b₁ + a₂b₂ + a₃b₃ = |a||b| cos θ
|a × b| = |a||b| sin θ
Area of parallelogram = |a × b|
Area of triangle = ½|a × b|