📐

Vectors

Ch. 5Std 12

Easy Overview

Let's talk about vectors. You already know that a quantity like '5 kilograms' just tells you how much. That's a scalar — it has magnitude only. But what about '5 kilometres per hour towards the north'? That's a vector — it has both magnitude AND direction. Vectors are the language of motion, force, and space itself. They're used everywhere — from calculating the trajectory of a rocket to rendering 3D graphics in video games to analysing forces in a bridge. In physics, almost everything interesting is a vector: displacement, velocity, acceleration, force, momentum. In mathematics, vectors let us work in 2D and 3D space in a clean, algebraic way. We start with the basics: what a vector is, how to represent it (as a directed line segment AB or as a position vector using components like 2i + 3j − k), and the different types — zero vector, unit vector, collinear vectors, co-initial vectors, negative vectors, equal vectors, and free vs. localized vectors. Then we learn how to find the magnitude (length) of a vector and its direction cosines. Vector addition is governed by the triangle law and the parallelogram law — both say the same thing in different visual ways. The triangle law: place the tail of the second vector at the head of the first; the resultant goes from the first tail to the second head. The parallelogram law: if two vectors share the same tail, the resultant is the diagonal of the parallelogram they form. Vector addition is commutative (A + B = B + A) and associative. Then there's scalar multiplication — stretching or shrinking a vector by a factor. If the scalar is negative, the direction reverses. The section formula lets you find a point dividing a line segment in a given ratio — both internally and externally. This has direct applications in coordinate geometry. Now, for the really powerful stuff — multiplying vectors. There are two ways to multiply vectors, and they give completely different results. The dot product (scalar product) A·B = |A||B|cos θ gives a scalar. It tells you how much one vector goes in the direction of another. If A and B are perpendicular, A·B = 0. If they're parallel, A·B = |A||B|. The dot product in component form is a₁b₁ + a₂b₂ + a₃b₃. It's used to find the angle between vectors, to check perpendicularity, and to find the projection of one vector onto another. The cross product (vector product) A × B = |A||B|sin θ n̂ gives a VECTOR that is perpendicular to both A and B. Its magnitude is the area of the parallelogram formed by A and B. The direction is given by the right-hand rule — curl your fingers from A to B, and your thumb points the way. In component form, it's the determinant of a 3×3 matrix. If A and B are parallel, A × B = 0. The cross product is anti-commutative: A × B = −(B × A). Then there's the scalar triple product [A B C] = A·(B × C). This gives the volume of the parallelepiped formed by the three vectors. If [A B C] = 0, the three vectors are coplanar — they lie in the same plane. This is a crucial test for coplanarity. There's also the vector triple product A × (B × C) = (A·C)B − (A·B)C, which has its own applications. By the end of this chapter, vectors will feel like second nature. You'll be able to add, subtract, dot, cross, and triple-product your way through any 2D or 3D geometry problem. And you'll appreciate how vectors simplify things that would be incredibly messy with coordinates alone.

What Is a Vector? — Representation and Notation

A vector is a quantity with both magnitude and direction. Represent it as a directed line segment: AB (from A to B, with A as tail and B as head). The magnitude is written as |AB|. A position vector gives the coordinates of a point relative to the origin: OP = xi + yj + zk, where i, j, k are unit vectors along the x, y, z axes. Two vectors are equal if they have the same magnitude AND direction — location doesn't matter.

Types of Vectors

Zero vector (0): magnitude 0, no direction. Unit vector (â): magnitude 1, direction of a. A unit vector in the direction of A is Â = A/|A|. Collinear (parallel) vectors: same or opposite direction. Co-initial vectors: share the same tail. Negative vector of A: same magnitude, opposite direction. Free vector: can be moved parallel to itself (location doesn't matter). Localized vector: fixed at a specific point.

Magnitude and Direction Cosines

The magnitude of a vector A = a₁i + a₂j + a₃k is |A| = √(a₁² + a₂² + a₃²). Direction cosines are the cosines of the angles α, β, γ that the vector makes with the x, y, z axes: l = cos α = a₁/|A|, m = cos β = a₂/|A|, n = cos γ = a₃/|A|. Property: l² + m² + n² = 1. Direction ratios are any three numbers proportional to (l, m, n).

Vector Addition — Triangle and Parallelogram Laws

Triangle law: to add A and B, place B's tail at A's head. The resultant R = A + B goes from A's tail to B's head. Parallelogram law: place A and B tail-to-tail. Complete the parallelogram — the diagonal from the common tail is R = A + B. Properties: commutative (A + B = B + A), associative (A + (B + C) = (A + B) + C), identity (A + 0 = A), inverse (A + (−A) = 0).

Scalar Multiplication of Vectors

Multiplying a vector A by a scalar k gives kA. If k > 0, direction is same as A; if k < 0, direction is opposite. Magnitude becomes |k|·|A|. Properties: k(A + B) = kA + kB, (k₁ + k₂)A = k₁A + k₂A, (k₁k₂)A = k₁(k₂A), 1·A = A, (−1)·A = −A. Scalar multiplication is how we stretch, shrink, or reverse vectors.

Position Vectors and Section Formula

Position vector of point P is OP = p = xi + yj + zk. The vector from A to B is AB = b − a (head − tail). Section formula: if point R divides AB in ratio m:n internally, then r = (mb + na)/(m + n). For external division in ratio m:n, r = (mb − na)/(m − n). If R is the midpoint, r = (a + b)/2. This is the vector version of the coordinate section formula.

Dot Product (Scalar Product)

A·B = |A||B|cos θ, where θ is the angle between A and B. The result is a SCALAR. In components: A·B = a₁b₁ + a₂b₂ + a₃b₃. Properties: commutative (A·B = B·A), distributive (A·(B + C) = A·B + A·C), k(A·B) = (kA)·B = A·(kB). If A·B = 0, A and B are perpendicular. A·A = |A|². i·i = j·j = k·k = 1, i·j = j·k = k·i = 0.

Projection of a Vector

The scalar projection of A onto B is the length of A's shadow on B: comp_B A = (A·B)/|B|. The vector projection of A onto B is the actual vector component of A along B: proj_B A = ((A·B)/|B|²) B. The component of A perpendicular to B is A − proj_B A. These are used to decompose vectors into parallel and perpendicular parts relative to a direction.

Angle Between Two Vectors Using Dot Product

From the dot product formula: cos θ = (A·B) / (|A||B|). This is the standard way to find the angle between any two vectors in 2D or 3D. If cos θ = 1, vectors are parallel and same direction. cos θ = −1 means opposite directions. cos θ = 0 means perpendicular. This formula works for any number of dimensions.

Cross Product (Vector Product)

A × B = |A||B|sin θ n̂, where n̂ is a unit vector perpendicular to both A and B (direction by right-hand rule). The result is a VECTOR. Magnitude = area of parallelogram with sides A and B. In components: A × B = |i j k; a₁ a₂ a₃; b₁ b₂ b₃| = (a₂b₃ − a₃b₂)i + (a₃b₁ − a₁b₃)j + (a₁b₂ − a₂b₁)k. Properties: anti-commutative (A × B = −B × A), distributive, not associative. A × A = 0. i × j = k, j × k = i, k × i = j.

Area Using Cross Product

Scalar Triple Product

[A B C] = A·(B × C). This is a SCALAR. Geometric meaning: it's the volume of the parallelepiped formed by A, B, C. In components: [A B C] = det([a₁ a₂ a₃; b₁ b₂ b₃; c₁ c₂ c₃]). Properties: cyclic permutations don't change the value: [A B C] = [B C A] = [C A B]. Swapping two vectors flips the sign: [A B C] = −[B A C]. [A B C] = 0 iff A, B, C are coplanar.

Coplanarity Test Using Scalar Triple Product

Three vectors A, B, C are coplanar if and only if [A B C] = 0. For four points P, Q, R, S to be coplanar, the vectors PQ, PR, PS must satisfy [PQ PR PS] = 0. This is the most efficient way to test coplanarity — one scalar calculation instead of trying to find a plane equation and checking if the point lies on it.

Vector Triple Product

A × (B × C) = (A·C)B − (A·B)C. Note the pattern: it's the middle vector (B) times the dot product of the outer two (A·C), minus the third vector (C) times the dot product of the other pair (A·B). The result lies in the plane of B and C. This is used in physics (e.g., deriving centripetal acceleration formula) and in simplifying complex vector expressions.

Linear Combination of Vectors

A vector R can be expressed as a linear combination of vectors A, B, C if R = xA + yB + zC for some scalars x, y, z. If A, B, C are non-coplanar, any vector in space can be expressed uniquely as a linear combination of them. This is the vector equivalent of having a basis — like i, j, k forming a basis for 3D space.

Key Points

•Vector = magnitude + direction; Scalar = magnitude only
•Position vector of point P(x,y,z): p = xi + yj + zk
•AB = b − a (head minus tail); |AB| = distance between A and B
•Unit vector in direction of a: â = a/|a|
•Triangle law: AB + BC = AC; Parallelogram law: diagonal from common tail
•Section formula (internal): r = (mb + na)/(m + n); midpoint: r = (a + b)/2
•Dot product A·B = |A||B|cos θ = a₁b₁ + a₂b₂ + a₃b₃ — result is SCALAR
•A·B = 0 ⇔ A ⟂ B (perpendicular vectors)
•Cross product A×B = |A||B|sin θ n̂ — result is VECTOR perpendicular to both
•A×B = 0 ⇔ A ∥ B (parallel vectors)
•Area of parallelogram = |A×B|; area of triangle = ½|A×B|
•Scalar triple product [A B C] = A·(B×C) = volume of parallelepiped
•[A B C] = 0 ⇔ A, B, C are coplanar
•Vector triple product: A×(B×C) = (A·C)B − (A·B)C
•Direction cosines l, m, n satisfy l² + m² + n² = 1
•i·i = j·j = k·k = 1; i×j = k, j×k = i, k×i = j
•Three non-coplanar vectors form a basis for 3D space

Practice Questions

If A = 2i + 3j − k and B = i − j + 2k, find A·B, A×B, and the angle between A and B.
Find the area of the triangle with vertices P(1,2,3), Q(2,3,1), and R(3,1,2).
Show that the vectors A = i + 2j + 3k, B = 2i + 3j − 4k, and C = 3i + 5j − k are coplanar.
Prove that [A B C] = [B C A] = [C A B].
Find the projection of vector A = 2i + 3j + k on vector B = i + 2j − 2k.
If a = 2i + 3j − k, b = i − 2j + 3k, and c = −i + j + 2k, find the volume of the parallelepiped formed by a, b, c.
Find the vector of magnitude 5 in the direction opposite to A = 2i + 3j − 6k.
Using vectors, prove that the medians of a triangle are concurrent.