NDA Maths · Vectors

Foundations: Vectors, Operations, and Position

What a vector is, how to add and scale them, the standard î-ĵ-k̂ basis that turns vectors into numbers, and how anchoring at an origin turns geometry into algebra.

All Vectors notes NDA Maths strategy

Why this matters

Start here. The rest of the chapter — magnitude, dot product, cross product, vector geometry — is built on the eight ideas below. The first six are pure FOUNDATIONS (what a vector is, addition, scalar multiplication, components, types of vectors); the last two are the chapter's first geometric payoff — a collinearity test and the section formula for dividing a segment. The PYQ bank has 6 questions tagged on this subtopic (collinearity + section, three of them HARD) — but you can't drill those without the foundations underneath them. Read top-to-bottom.

Concept 1 of 8

What is a vector? (Scalars vs vectors)

Intuition

Mathematics splits quantities into two families. A scalar — temperature, mass, speed, electric charge — has only magnitude: one number tells you everything. A vector has both magnitude AND direction; one number is not enough. The classical contrast is speed vs velocity: \"60 km/h\" is a scalar (the speed); \"60 km/h east\" is a vector (the velocity). Pictorially we draw a vector as a directed arrow — the length encodes magnitude, the arrowhead encodes direction.

Definition

A scalar is a real number — magnitude only. A vector is an entity with both magnitude (a non-negative real number) and direction (a way of pointing in space). Two vectors are equal if and only if they have the same magnitude AND the same direction — equality is independent of where the arrow is drawn on the page. We write vectors with an over-arrow, like $\vec{v}$ , or in bold, like $\mathbf{v}$ ; their magnitude is written $|\vec{v}|$ .

Worked example

Classify each as scalar or vector: (a) mass of an apple, (b) velocity of a falling stone, (c) temperature of a room, (d) electric current in a wire, (e) force pushing a box, (f) displacement from school to home.

(a) Mass — a single positive number (grams). Direction is meaningless. Scalar.
(b) Velocity — specifies both how fast AND in which direction. Vector.
(c) Temperature — a single number (degrees). No direction. Scalar.
(d) Electric current — a magnitude (amperes) with only a sign convention for flow along the wire, not a spatial direction. Scalar. (Current density IS a vector — but plain current is treated as scalar in school physics.)
(e) Force — specified by both how strong and in what direction it pushes. Vector.
(f) Displacement — \"5 km north-east\" needs both magnitude and direction. Vector.

Answer:Scalars: (a), (c), (d). Vectors: (b), (e), (f).

An arrow drawn anywhere on the page represents the same vector

Two arrows of the same length and direction, drawn in different places, denote the SAME vector — vectors are not tied to a starting point unless we explicitly anchor them. We'll see in concept 6 (Types of Vectors) when that distinction matters and the term \"localized vector\" applies.

Concept 2 of 8

Position vectors and displacement vectors

Intuition

To do geometry with vectors we need to anchor them. Pick any point in space and call it the origin

O

. Every other point

P

is then represented by its position vector

\vec{OP}

— the arrow from

O

P

. Once points have position vectors, geometry becomes algebra: the arrow from

A

B

(its displacement vector) is just the difference of their position vectors:

\vec{AB} = \vec{b} - \vec{a}

. This single idea — point

\leftrightarrow

position vector — is the bridge that the entire rest of the chapter walks over.

Definition

Fix an origin $O$ . The position vector of a point $P$ (often written $\vec{p}$ ) is the vector $\vec{OP}$ from $O$ to $P$ ; its magnitude is the distance $OP$ , its direction is from $O$ towards $P$ . The displacement vector from $A$ to $B$ is $\vec{AB} = \vec{OB} - \vec{OA} = \vec{b} - \vec{a}$ (head minus tail). Its magnitude $|\vec{AB}|$ is the distance from $A$ to $B$ .

Position vector → displacement

\vec{AB} = \vec{b} - \vec{a} \qquad AB = |\vec{b} - \vec{a}|

$\vec{a}, \vec{b}$ position vectors of $A, B$ from the chosen origin
$\vec{AB}$ displacement vector from $A$ to $B$ — head minus tail

Diagram · position vectors & displacement

From the origin O, the position vectors a and b locate points A and B. The displacement from A to B is AB = b − a — move it anywhere and shift the origin: the difference, and so AB, is unchanged.

Worked example

Points

A

and

B

have position vectors

\vec{a} = 2\hat{i} + \hat{j}

and

\vec{b} = 5\hat{i} - 3\hat{j}

. Find (i) the displacement

\vec{AB}

, (ii) the displacement

\vec{BA}

, and (iii) the distance

AB

(i) Head minus tail: $\vec{AB} = \vec{b} - \vec{a} = (5-2)\hat{i} + (-3-1)\hat{j} = 3\hat{i} - 4\hat{j}$ .
(ii) Reverse the direction by negating: $\vec{BA} = -\vec{AB} = -3\hat{i} + 4\hat{j}$ . Equivalently $\vec{BA} = \vec{a} - \vec{b}$ .
(iii) Distance = magnitude: $|\vec{AB}| = \sqrt{3^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5$ .

Answer:

\vec{AB} = 3\hat{i} - 4\hat{j}

;

\vec{BA} = -3\hat{i} + 4\hat{j}

;

AB = 5

units.

Practice this concept4 quick reps

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
$\vec{a} = \hat{i} + \hat{j}$ , $\vec{b} = 4\hat{i} + 5\hat{j}$ . Find $\vec{AB}$ .
2.
For the same points, find the distance $AB$ .
3.
$\vec{a} = 2\hat{i}$ , $\vec{b} = 2\hat{i} + 3\hat{j}$ . Find $\vec{BA}$ .
4.
Find $\vec{AB}$ for $A(1,2,0)$ , $B(1,2,5)$ .

Head minus tail — $\vec{AB} = \vec{b} - \vec{a}$ , not $\vec{a} - \vec{b}$

Reverse the subtraction and you compute

\vec{BA}

instead. The magnitudes match (

|\vec{AB}| = |\vec{BA}|

), but the directions are opposite. Direction matters whenever the result feeds into a dot product, an angle, or a cross product downstream.

Position vectors depend on the choice of origin; displacement vectors do NOT

If you move the origin from

O

O'

, every position vector changes (they all shift by the same fixed amount), but the displacement

\vec{AB}

is unchanged — that's why displacements are the more \"physical\" quantity and most theorems are stated in terms of them.

Concept 3 of 8

Addition of vectors (triangle, parallelogram, polygon laws)

Intuition

Two vectors add by joining them tip-to-tail. Place the second arrow's tail at the first arrow's head, then draw a new arrow from the very first tail to the very last head — that resultant arrow is the sum. Equivalently, if you draw both vectors from a common tail, the sum is the diagonal of the parallelogram they span. Subtraction is just addition of the negative:

\vec{a} - \vec{b} = \vec{a} + (-\vec{b})

Definition

Triangle law: place $\vec{b}$ so its tail starts where $\vec{a}$ 's head ends; then $\vec{a} + \vec{b}$ is the arrow from $\vec{a}$ 's tail to $\vec{b}$ 's head.
Parallelogram law (equivalent): if $\vec{a}$ and $\vec{b}$ share a tail, $\vec{a} + \vec{b}$ is the diagonal of the parallelogram on $\vec{a}, \vec{b}$ from that shared tail.
Polygon law (generalisation): the sum of any number of vectors placed tip-to-tail is the arrow from the very first tail to the very last head.
Properties: addition is commutative ( $\vec{a} + \vec{b} = \vec{b} + \vec{a}$ ), associative, has identity $\vec{a} + \vec{0} = \vec{a}$ , and inverse $\vec{a} + (-\vec{a}) = \vec{0}$ .

Vector addition properties

\vec{a} + \vec{b} = \vec{b} + \vec{a} \qquad \vec{a} - \vec{b} = \vec{a} + (-\vec{b})

$\vec{a} + \vec{b}$ tip-to-tail sum (a vector, not a number)
$\vec{0}$ zero vector — the additive identity
$-\vec{a}$ same length as $\vec{a}$ , opposite direction

Visualization · add two vectors tip-to-tail

a₁: 4a₂: 1b₁: 1b₂: 3

a + b = (5, 4)|a + b| = 6.40|a| + |b| = 7.29

The dashed amber arrow is b moved to the head of a — the resultant (indigo) runs from the shared tail to that head, which is also the diagonal of the parallelogram. Notice |a + b| equals |a| + |b| only when a and b point the same way.

Worked example

A particle is displaced first by

\vec{a} = 3\hat{i} + 2\hat{j}

, then by

\vec{b} = -\hat{i} + 4\hat{j}

. Find its total displacement from the starting point and its distance from the start.

Total displacement = sum of the two displacement vectors (tip-to-tail): $\vec{R} = \vec{a} + \vec{b}$ .
Add componentwise: $\vec{R} = (3 + (-1))\hat{i} + (2 + 4)\hat{j} = 2\hat{i} + 6\hat{j}$ .
Distance from start = magnitude of the resultant: $|\vec{R}| = \sqrt{2^2 + 6^2} = \sqrt{4 + 36} = \sqrt{40} = 2\sqrt{10}$ .

Answer:Total displacement

\vec{R} = 2\hat{i} + 6\hat{j}

; distance from start

= 2\sqrt{10}

units.

Practice this concept4 quick reps

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
$\vec{a} = 2\hat{i} + \hat{j}$ , $\vec{b} = \hat{i} + 3\hat{j}$ . Find $\vec{a} + \vec{b}$ .
2.
$\vec{a} = 3\hat{i} - \hat{j}$ , $\vec{b} = -3\hat{i} + \hat{j}$ . Find $\vec{a} + \vec{b}$ .
3.
Find $\vec{a} - \vec{b}$ for $\vec{a} = 5\hat{i} + 2\hat{j}$ , $\vec{b} = \hat{i} + 2\hat{j}$ .
4.
$|\vec{a}| = 3$ , $|\vec{b}| = 4$ , same direction. Find $|\vec{a} + \vec{b}|$ .

Closed-polygon identity: if vectors form a closed loop, they sum to $\vec{0}$

Tip-to-tail vectors that return to the starting point span a closed polygon, so their sum is the zero vector. The triangle identity

\vec{AB} + \vec{BC} + \vec{CA} = \vec{0}

is exactly this rule for a triangle — used heavily in the Vector Geometry subtopic later.

Magnitudes don't add: $|\vec{a} + \vec{b}| \neq |\vec{a}| + |\vec{b}|$ in general

Two arrows of length 3 don't always combine to length 6 — they combine to length 6 only if perfectly aligned, length 0 if perfectly opposed, and anything in between otherwise. The triangle inequality

|\vec{a} + \vec{b}| \leq |\vec{a}| + |\vec{b}|

is the correct bound.

Concept 4 of 8

Scalar multiplication

Intuition

Multiplying a vector by a real number scales its length. If

k > 0

k\vec{v}

points the SAME way as

\vec{v}

but is

k

times as long. If

k < 0

k\vec{v}

points the OPPOSITE way and has length

|k|

times the original. If

k = 0

, you get the zero vector. Combined with addition, this single operation is the engine behind every vector expression in the chapter — every formula you'll meet is some combination of

\vec{a} + \vec{b}

and

k\vec{v}

Definition

For a scalar $k \in \mathbb{R}$ and a vector $\vec{v}$ , the product $k\vec{v}$ is the vector with magnitude $|k|\,|\vec{v}|$ and direction the same as $\vec{v}$ when $k > 0$ , opposite when $k < 0$ , and $\vec{0}$ when $k = 0$ . It is distributive over both vector and scalar addition: $k(\vec{a} + \vec{b}) = k\vec{a} + k\vec{b}$ and $(k + l)\vec{a} = k\vec{a} + l\vec{a}$ . It is also associative with ordinary multiplication: $k(l\vec{a}) = (kl)\vec{a}$ .

Scalar multiplication

|k\vec{v}| = |k|\,|\vec{v}| \qquad k(\vec{a} + \vec{b}) = k\vec{a} + k\vec{b}

$k$ a real number (positive, negative, or zero)
$|k|$ absolute value of $k$ (gives the magnitude-scaling factor)
sign of $k$ controls whether the direction is preserved or flipped

Visualization · slide k, scale the vector

k = 2.0

k·v = (4.0, 2.0)|k·v| = |k|·|v| = 4.47

Multiplying by k scales the length by |k| and keeps the same line. k > 1 stretches, 0 < k < 1 shrinks, k < 0 flips to the opposite direction, and k = 0 collapses it to the zero vector.

Worked example

Given

\vec{a} = 4\hat{i} + 3\hat{j}

, compute

2\vec{a}

and

-\tfrac{1}{2}\vec{a}

, and verify both magnitudes satisfy

|k\vec{v}| = |k|\,|\vec{v}|

Scale each component: $2\vec{a} = 8\hat{i} + 6\hat{j}$ ; $-\tfrac{1}{2}\vec{a} = -2\hat{i} - \tfrac{3}{2}\hat{j}$ .
Original magnitude: $|\vec{a}| = \sqrt{4^2 + 3^2} = \sqrt{25} = 5$ .
Check $|2\vec{a}| = \sqrt{8^2 + 6^2} = \sqrt{100} = 10 = 2 \cdot 5$ . $\checkmark$
Check $|-\tfrac{1}{2}\vec{a}| = \sqrt{(-2)^2 + (-3/2)^2} = \sqrt{4 + 9/4} = \sqrt{25/4} = 5/2 = \tfrac{1}{2} \cdot 5$ . $\checkmark$

Answer:

2\vec{a} = 8\hat{i} + 6\hat{j}

;

-\tfrac{1}{2}\vec{a} = -2\hat{i} - \tfrac{3}{2}\hat{j}

. Both magnitudes confirm the scaling rule.

Practice this concept4 quick reps

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
$\vec{a} = 2\hat{i} - 3\hat{j}$ . Find $3\vec{a}$ .
2.
$\vec{a} = 4\hat{i} + 2\hat{j}$ . Find $-\tfrac{1}{2}\vec{a}$ .
3.
If $|\vec{a}| = 5$ , find $|3\vec{a}|$ .
4.
If $|\vec{a}| = 6$ , find $|-2\vec{a}|$ .

Sign of $k$ controls DIRECTION, not just signs of components

k = -1

does more than negate components arithmetically — geometrically it FLIPS the arrow (

180^\circ

rotation). That's the same operation as

-\vec{v}

, which is why subtraction reads as

\vec{a} - \vec{b} = \vec{a} + (-1)\vec{b}

Concept 5 of 8

Component form: the î, ĵ, k̂ basis

Intuition

Pick three mutually-perpendicular unit vectors

\hat{i}, \hat{j}, \hat{k}

along the positive

x, y, z

axes. Every 3-D vector can be uniquely written as

\vec{v} = v_1\hat{i} + v_2\hat{j} + v_3\hat{k}

— the numbers

v_1, v_2, v_3

are its components along those axes. In component form, vector addition becomes componentwise addition; scalar multiplication becomes scalar-times-each-component; equality becomes \"components match one-by-one\". This is why component form is so powerful — every operation reduces to elementary arithmetic on three numbers.

Definition

The standard basis of 3-D space is $\hat{i} = (1, 0, 0)$ , $\hat{j} = (0, 1, 0)$ , $\hat{k} = (0, 0, 1)$ — unit-length, mutually perpendicular, along the positive $x, y, z$ axes. Any vector has a unique decomposition $\vec{v} = v_1\hat{i} + v_2\hat{j} + v_3\hat{k}$ ; the triple $(v_1, v_2, v_3)$ is its component form. For 2-D vectors just drop the $\hat{k}$ term. Componentwise rules: (equality) $\vec{a} = \vec{b}$ iff $a_i = b_i$ for every $i$ ; (addition) $\vec{a} + \vec{b} = (a_1+b_1)\hat{i} + (a_2+b_2)\hat{j} + (a_3+b_3)\hat{k}$ ; (scalar mult) $k\vec{a} = (ka_1)\hat{i} + (ka_2)\hat{j} + (ka_3)\hat{k}$ .

Component form

\vec{v} = v_1\hat{i} + v_2\hat{j} + v_3\hat{k} \qquad \vec{a} + \vec{b} = (a_1+b_1)\hat{i} + (a_2+b_2)\hat{j} + (a_3+b_3)\hat{k}

$\hat{i}, \hat{j}, \hat{k}$ standard basis — unit vectors along positive $x, y, z$ axes
$v_1, v_2, v_3$ components of $\vec{v}$ — uniquely determined by the basis choice

Diagram · component form (drag to rotate)

Step along x (2.4 î), then y ( 1.6 ĵ), then z ( 2.0 k̂) to reach the tip: v = 2.4 î + 1.6 ĵ + 2.0 k̂. Any vector is the sum of its axis components, and |v| = √(x² + y² + z²).

Worked example

Given

\vec{a} = 2\hat{i} - \hat{j} + 3\hat{k}

and

\vec{b} = -\hat{i} + 4\hat{j} + 2\hat{k}

, compute

3\vec{a} - 2\vec{b}

in component form.

Apply scalar multiplication componentwise: $3\vec{a} = 6\hat{i} - 3\hat{j} + 9\hat{k}$ and $2\vec{b} = -2\hat{i} + 8\hat{j} + 4\hat{k}$ .
Subtract componentwise: $3\vec{a} - 2\vec{b} = (6 - (-2))\hat{i} + (-3 - 8)\hat{j} + (9 - 4)\hat{k}$ .
Simplify: $8\hat{i} - 11\hat{j} + 5\hat{k}$ .

Answer:

3\vec{a} - 2\vec{b} = 8\hat{i} - 11\hat{j} + 5\hat{k}

Practice this concept4 quick reps

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
$\vec{a} = \hat{i} + 2\hat{j} + 3\hat{k}$ , $\vec{b} = 2\hat{i} - \hat{j} + \hat{k}$ . Find $\vec{a} + \vec{b}$ .
2.
Find $2\vec{a}$ for $\vec{a} = \hat{i} - \hat{j} + 2\hat{k}$ .
3.
If $x\hat{i} + 3\hat{j} = 5\hat{i} + y\hat{j}$ , find $x$ and $y$ .
4.
Find $\vec{a} - \vec{b}$ for $\vec{a} = 3\hat{i} + \hat{k}$ , $\vec{b} = \hat{i} + \hat{j} + \hat{k}$ .

Equality of vectors = ALL components match — that's 3 equations, not 1

A statement \"

\vec{a} = \vec{b}

\" with unknowns hidden in the components is implicitly giving you a SYSTEM of equations (one per component). PYQs use this often: given

\vec{a} = \vec{b}

and unknowns

x, y, z

in the components, equate

a_1 = b_1

a_2 = b_2

a_3 = b_3

and solve.

Components depend on the basis; the vector itself does not

Rotate the coordinate axes and the components

(v_1, v_2, v_3)

change, but the underlying vector (the arrow in space) is the same. All NDA questions stick with the standard

\hat{i}, \hat{j}, \hat{k}

basis, so this is rarely an issue in practice — but it explains why some identities are \"basis-free\" (magnitudes, dot products, angles).

Concept 6 of 8

Types of vectors (zero, unit, equal, parallel, collinear, coplanar)

Intuition

Before going further, give names to the special kinds of vectors you'll keep meeting. The zero vector has no length and no defined direction. A unit vector has length exactly 1. Two vectors are parallel if their directions agree or are exactly opposite. Three or more vectors are coplanar if you can draw them all in one flat plane. These names are not optional vocabulary — every PYQ uses them as shorthand for an entire concept.

Definition

Zero vector $\vec{0}$ — magnitude 0, direction undefined. Acts as the additive identity.
Unit vector $\hat{u}$ — magnitude exactly 1. Any non-zero $\vec{v}$ has a unique unit vector along it: $\hat{v} = \vec{v}/|\vec{v}|$ .
Equal vectors — same magnitude AND direction; position on the page is irrelevant.
**Negative of $\vec{v}$ ** — same magnitude, opposite direction, written $-\vec{v}$ .
Parallel vectors $\vec{a} \parallel \vec{b}$ — same OR opposite direction; equivalently, one is a non-zero scalar multiple of the other: $\vec{a} = k\vec{b}$ for some $k \neq 0$ .
Collinear points — three or more points lying on one straight line (a stronger condition than just having parallel direction vectors).
Coplanar vectors / points — all lying in one flat plane.
Free vs localized vector — a free vector cares only about magnitude and direction; a localized vector additionally has a fixed application point (e.g. a force at a specific point on a body). Most NDA questions treat vectors as free.

Unit vector and parallelism

\hat{v} = \dfrac{\vec{v}}{|\vec{v}|} \qquad \vec{a} \parallel \vec{b} \iff \vec{a} = k\vec{b} \text{ for some } k \neq 0

$\hat{v}$ unit vector along $\vec{v}$ — pure direction, magnitude 1
$k$ non-zero scalar; sign of $k$ tells whether the parallel vectors agree or oppose

Worked example

Let

\vec{a} = 2\hat{i} - \hat{j}

and

\vec{b} = -6\hat{i} + 3\hat{j}

. (i) Are

\vec{a}, \vec{b}

parallel? (ii) Is either a unit vector? (iii) Find the unit vector along

\vec{a}

(i) Test parallelism by looking for a scalar $k$ with $\vec{b} = k\vec{a}$ . Component 1: $-6 = 2k \Rightarrow k = -3$ . Component 2: $3 = -1 \cdot k = -k$ , so $k = -3$ . Both give the same $k$ ; the vectors are parallel (specifically anti-parallel, since $k < 0$ ).
(ii) Magnitudes: $|\vec{a}| = \sqrt{4 + 1} = \sqrt{5} \neq 1$ ; $|\vec{b}| = \sqrt{36 + 9} = \sqrt{45} \neq 1$ . Neither is a unit vector.
(iii) Divide $\vec{a}$ by its magnitude: $\hat{a} = \vec{a}/|\vec{a}| = (2\hat{i} - \hat{j})/\sqrt{5}$ . Quick check: $|\hat{a}|^2 = (4 + 1)/5 = 1$ . $\checkmark$

Answer:(i) Parallel (anti-parallel,

\vec{b} = -3\vec{a}

). (ii) Neither is a unit vector. (iii)

\hat{a} = (2\hat{i} - \hat{j})/\sqrt{5}

Practice this concept4 quick reps

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
Find the unit vector along $\vec{v} = 3\hat{i} + 4\hat{j}$ .
2.
Are $\vec{a} = \hat{i} - 2\hat{j}$ and $\vec{b} = -2\hat{i} + 4\hat{j}$ parallel?
3.
Is $\vec{v} = \tfrac{1}{\sqrt{2}}(\hat{i} + \hat{j})$ a unit vector?
4.
What is the magnitude of the zero vector $\vec{0}$ ?

Parallel VECTORS vs collinear POINTS — different conditions

Two vectors are parallel when they share a DIRECTION (regardless of where they start). Three or more points are collinear when they all lie on one LINE — a stricter condition that requires them to share a line, not just share a direction. Points

A, B, C

are collinear iff

\vec{AB} \parallel \vec{AC}

(the next concept turns this into a usable test).

Zero vector is parallel to everything and to nothing

Because

\vec{0} = 0 \cdot \vec{v}

for any

\vec{v}

, the zero vector technically satisfies the scalar-multiple definition for every direction. PYQs sidestep the ambiguity by implicitly assuming non-zero vectors when talking about parallelism — read the question carefully if the hypothesis is loose.

Concept 7 of 8

Collinearity of three points (and vector relations in regular figures)

Intuition

Three points are collinear when the line through two of them passes through the third. In vector language that means one displacement vector is a scalar multiple of another, or equivalently any third point on the line is a weighted average of the first two where the weights add to one. The same coefficient-sum-to-one identity hides in many disguises — including the famous

(\vec{a}\times\vec{b})+(\vec{b}\times\vec{c})+(\vec{c}\times\vec{a})=\vec{0}

test. Regular polygons obey similar fixed identities — every diagonal and side can be expressed as a known scalar multiple of any other.

Definition

Points $A, B, C$ with position vectors $\vec{a}, \vec{b}, \vec{c}$ are collinear if and only if there exist scalars $\alpha, \beta, \gamma$ (not all zero) with $\alpha + \beta + \gamma = 0$ and $\alpha\vec{a} + \beta\vec{b} + \gamma\vec{c} = \vec{0}$ . Equivalently $\vec{c} = \lambda\vec{a} + \mu\vec{b}$ with $\lambda + \mu = 1$ .

Collinearity test

\alpha\vec{a} + \beta\vec{b} + \gamma\vec{c} = \vec{0} \;\text{ with }\; \alpha + \beta + \gamma = 0

$\vec{a},\vec{b},\vec{c}$ position vectors of the three points
$\alpha,\beta,\gamma$ scalars; both the linear-combo and the sum vanish

Worked example

Three points have position vectors

\vec{a}, \vec{b}, \vec{c}

with

2\vec{a} - 3\vec{b} + \vec{c} = \vec{0}

. Show they are collinear and find the ratio in which

B

divides

AC

The coefficients are $2, -3, 1$ . Check their sum: $2 + (-3) + 1 = 0$ . Since both the linear combination and the coefficient sum vanish, the three points are collinear.
Rewrite to isolate $\vec{b}$ : $3\vec{b} = 2\vec{a} + \vec{c}$ , so $\vec{b} = \dfrac{2\vec{a} + \vec{c}}{3}$ .
Compare with the internal section formula $\vec{b} = \dfrac{m\vec{c} + n\vec{a}}{m + n}$ . Matching gives $m = 1$ , $n = 2$ , so $B$ divides $AC$ internally in the ratio $AB : BC = m : n = 1 : 2$ .

Answer:Collinear;

B

divides

AC

internally in the ratio

1 : 2

Practice this conceptself-check · 4 quick reps

Try it yourself

Position vectors satisfy

\vec{a} + 3\vec{b} - 4\vec{c} = \vec{0}

. Show the points are collinear and find the ratio in which

C

divides

AB

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
Coefficients in $\alpha\vec{a} + \beta\vec{b} + \gamma\vec{c} = \vec{0}$ are $1, -3, 2$ . Collinear?
2.
Coefficients $2, 1, 1$ . Collinear?
3.
If $\vec{c} = \tfrac{1}{2}\vec{a} + \tfrac{1}{2}\vec{b}$ , are $A, B, C$ collinear?
4.
Three points are collinear when one displacement is a ___ of another.

From the bank · past-year question

Example 7VectorsMODERATE

The position vectors of three points A, B and C are

\vec{a}, \vec{b}

and

\vec{c}

respectively such that

3\vec{a} - 4\vec{b} + \vec{c} = \vec{0}

. What is AB : BC equal to?

[Q68 · Apr · 2025]

Coefficient sum must be zero — don't skip the check

If the scalars in

\alpha\vec{a}+\beta\vec{b}+\gamma\vec{c}=\vec{0}

do NOT sum to zero, the three points are coplanar with the origin (i.e.

\vec{a},\vec{b},\vec{c}

are linearly dependent) but generally NOT collinear. The sum-to-zero condition is what forces them onto one line.

$(\vec{a}\times\vec{b})+(\vec{b}\times\vec{c})+(\vec{c}\times\vec{a})=\vec{0}$ means collinear, not coplanar

A common HARD-paper trap: this cross-product identity vanishes precisely when the three points are collinear. If a question gives

\vec{c} = \cos^2\theta\,\vec{a} + \sin^2\theta\,\vec{b}

the coefficients sum to

\cos^2\theta + \sin^2\theta = 1

, so

C

lies on line

AB

and the cross-product sum is forced to zero.

Drill 4 more on collinearity of three points (and vector relations in regular figures)

Concept 8 of 8

Section Formula — Internal and External Division

Intuition

A point that divides a segment

AB

in a ratio

m : n

is a weighted average of the endpoints. When the dividing point lies between

A

and

B

, the division is internal and the weights add normally. When it lies on the extension outside, the division is external and one weight gets a negative sign — that single sign-flip is the most-tested trap in this subtopic.

Definition

If $P$ divides $AB$ internally in ratio $m : n$ , then $\vec{p} = \dfrac{m\vec{b} + n\vec{a}}{m + n}$ . If $P$ divides externally in ratio $m : n$ , then $\vec{p} = \dfrac{m\vec{b} - n\vec{a}}{m - n}$ . Midpoint is the special case $m = n$ : $\vec{p} = (\vec{a} + \vec{b})/2$ .

Section formula (internal / external)

\vec{p}_{\text{int}} = \dfrac{m\vec{b} + n\vec{a}}{m + n} \qquad \vec{p}_{\text{ext}} = \dfrac{m\vec{b} - n\vec{a}}{m - n}

$\vec{a}, \vec{b}$ position vectors of the endpoints $A, B$
$m : n$ ratio in which $P$ divides $AB$
$\vec{p}$ position vector of the dividing point

Diagram · section formula (internal vs external), m : n = 2 : 1

Internal: P = (m·b + n·a)/(m + n) sits between A and B. External: Q = (m·b − n·a)/(m − n) sits beyond B — the minus sign is what pushes it outside. The midpoint is the m = n case, (a + b)/2.

Worked example

Points

A

and

B

have position vectors

\vec{a} = 2\hat{i} + \hat{j}

and

\vec{b} = 4\hat{i} + 5\hat{j}

. Find the position vector of the point

P

that divides

AB

externally in the ratio

3 : 1

Identify the ratio: $m = 3$ (towards $B$ ), $n = 1$ (towards $A$ ). External division so use the minus-sign formula.
Apply $\vec{p} = \dfrac{m\vec{b} - n\vec{a}}{m - n} = \dfrac{3(4\hat{i}+5\hat{j}) - 1(2\hat{i}+\hat{j})}{3 - 1}$ .
Numerator: $12\hat{i} + 15\hat{j} - 2\hat{i} - \hat{j} = 10\hat{i} + 14\hat{j}$ . Denominator: $2$ .
Divide: $\vec{p} = 5\hat{i} + 7\hat{j}$ .

Answer:

\vec{p} = 5\hat{i} + 7\hat{j}

Practice this conceptself-check · 4 quick reps

Try it yourself

Find the midpoint of the segment joining

A(3\hat{i} - \hat{j})

and

B(5\hat{i} + 3\hat{j})

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
Midpoint of $A(2\hat{i})$ and $B(4\hat{i})$ ?
2.
$P$ divides $A(\vec{0})$ , $B(6\hat{i})$ internally $1:2$ . Find $\vec{p}$ .
3.
Internal-division denominator for ratio $m:n$ ?
4.
External-division denominator for ratio $m:n$ ?

From the bank · past-year question

Example 8VectorsHARD

Let

\vec{p}

\vec{q}

be position vectors of P,Q. R and S divide PQ internally/externally in ratio 2:3. If OR and OS are perpendicular, which is correct?

[Q67 · Apr · 2018]

External division: denominator is $m - n$ , not $m + n$

The most common bug. The external-section formula reverses one sign in the numerator AND swaps the denominator's plus for a minus. If

m = n

, external division is undefined (the point is at infinity) — another way to spot you've mis-set up an internal problem as external.

Watch the ratio order — $m : n$ means $AP : PB$ , not $AP : AB$

PYQs often phrase it as \"divides

AB

in ratio

2 : 3

\" — that is

AP : PB = 2 : 3

, so

m = 2

(the part nearer

B

) and

n = 3

(the part nearer

A

). Reversing them gives the wrong answer.

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (7)

Position vectors and displacement vectors
Position vector → displacement
$\vec{AB} = \vec{b} - \vec{a} \qquad AB = |\vec{b} - \vec{a}|$
Addition of vectors (triangle, parallelogram, polygon laws)
Vector addition properties
$\vec{a} + \vec{b} = \vec{b} + \vec{a} \qquad \vec{a} - \vec{b} = \vec{a} + (-\vec{b})$
Scalar multiplication
Scalar multiplication
$|k\vec{v}| = |k|\,|\vec{v}| \qquad k(\vec{a} + \vec{b}) = k\vec{a} + k\vec{b}$
Component form: the î, ĵ, k̂ basis
Component form
$\vec{v} = v_1\hat{i} + v_2\hat{j} + v_3\hat{k} \qquad \vec{a} + \vec{b} = (a_1+b_1)\hat{i} + (a_2+b_2)\hat{j} + (a_3+b_3)\hat{k}$
Types of vectors (zero, unit, equal, parallel, collinear, coplanar)
Unit vector and parallelism
$\hat{v} = \dfrac{\vec{v}}{|\vec{v}|} \qquad \vec{a} \parallel \vec{b} \iff \vec{a} = k\vec{b} \text{ for some } k \neq 0$
Collinearity of three points (and vector relations in regular figures)
Collinearity test
$\alpha\vec{a} + \beta\vec{b} + \gamma\vec{c} = \vec{0} \;\text{ with }\; \alpha + \beta + \gamma = 0$
Section Formula — Internal and External Division
Section formula (internal / external)
$\vec{p}_{\text{int}} = \dfrac{m\vec{b} + n\vec{a}}{m + n} \qquad \vec{p}_{\text{ext}} = \dfrac{m\vec{b} - n\vec{a}}{m - n}$

Watch out for (14)

An arrow drawn anywhere on the page represents the same vector
→ What is a vector? (Scalars vs vectors)
Head minus tail — $\vec{AB} = \vec{b} - \vec{a}$ , not $\vec{a} - \vec{b}$
→ Position vectors and displacement vectors
Position vectors depend on the choice of origin; displacement vectors do NOT
→ Position vectors and displacement vectors
Closed-polygon identity: if vectors form a closed loop, they sum to $\vec{0}$
→ Addition of vectors (triangle, parallelogram, polygon laws)
Magnitudes don't add: $|\vec{a} + \vec{b}| \neq |\vec{a}| + |\vec{b}|$ in general
→ Addition of vectors (triangle, parallelogram, polygon laws)
Sign of $k$ controls DIRECTION, not just signs of components
→ Scalar multiplication
Equality of vectors = ALL components match — that's 3 equations, not 1
→ Component form: the î, ĵ, k̂ basis
Components depend on the basis; the vector itself does not
→ Component form: the î, ĵ, k̂ basis
Parallel VECTORS vs collinear POINTS — different conditions
→ Types of vectors (zero, unit, equal, parallel, collinear, coplanar)
Zero vector is parallel to everything and to nothing
→ Types of vectors (zero, unit, equal, parallel, collinear, coplanar)
Coefficient sum must be zero — don't skip the check
→ Collinearity of three points (and vector relations in regular figures)
$(\vec{a}\times\vec{b})+(\vec{b}\times\vec{c})+(\vec{c}\times\vec{a})=\vec{0}$ means collinear, not coplanar
→ Collinearity of three points (and vector relations in regular figures)
External division: denominator is $m - n$ , not $m + n$
→ Section Formula — Internal and External Division
Watch the ratio order — $m : n$ means $AP : PB$ , not $AP : AB$
→ Section Formula — Internal and External Division

Mastery check — 4 interleaved questions

Try each one before clicking. Questions are interleaved across the concepts above, not grouped — interleaving sharpens transfer.

Example 1VectorsEASY

4\hat{i}+\hat{j}-3\hat{k}

and

p\hat{i}+q\hat{j}-2\hat{k}

are collinear vectors, then what are the possible values of

p

and

q

respectively?

[Q96 · Apr · 2022]

Example 2VectorsMODERATE

Let

ABCDEF

be a regular hexagon. If

\overrightarrow{AD}=m\overrightarrow{BC}

and

\overrightarrow{CF}=n\overrightarrow{AB}

, then what is

mn

equal to?

[Q54 · Sep · 2024]

Example 3VectorsHARD

The position vectors of three points A, B and C are

\vec{a}, \vec{b}

and

\vec{c}

respectively, where

\vec{c} = \cos^2\theta\,\vec{a} + \sin^2\theta\,\vec{b}

. What is

(\vec{a}\times\vec{b})+(\vec{b}\times\vec{c})+(\vec{c}\times\vec{a})

equal to?

[Q69 · Apr · 2025]

Example 4VectorsHARD

The vectors

60\hat{i}+3\hat{j}

40\hat{i}-8\hat{j}

and

\beta\hat{i}-52\hat{j}

are collinear if:

[Q94 · Sep · 2023]

Drill every past-year question on this subtopic

6 questions from the bank — paginated, with cart and Word-export support.

Drill the 6 questions

Drill every past-year question on this subtopic

Related notes