Basic Concepts of Vector Algebra

Q: How do I decide quickly whether three given points are collinear?

Form two vectors from the three points, for example AB and AC, and check whether one is a scalar multiple of the other. Equivalently, look for scalars x, y, z not all zero with x a plus y b plus z c equal to the zero vector and x plus y plus z equal to zero.

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Basic Concepts of Vector Algebra

Maths · Vector Algebra · Concept 1

A vector carries two pieces of information at once: how much, and which way. Once you can add vectors, scale them, break them into components and locate points using position vectors, the whole of vector algebra (dot product, cross product, triple products, lines and planes) becomes routine. This page builds that foundation completely.

Scalars and vectors

A scalar is a quantity specified by magnitude alone, that is, by a real number with a unit. Distance, speed, mass, volume, temperature, work and electric potential are scalars.

A vector is a quantity specified by both a magnitude and a direction, and which obeys the triangle law of addition. Displacement, velocity, acceleration, force, momentum and electric field are vectors.

A vector is drawn as a directed line segment $A B$ , where $A$ is the initial point (tail) and $B$ is the terminal point (head). The line containing the segment is its line of support. The length of $A B$ is its magnitude, written $∣ A B ∣$ or $A B$ . Single letters are written $a$ , with magnitude $∣ a ∣$ or simply $a$ .

Direction matters, not position

Two directed segments drawn at different places in the plane represent the same vector as long as they have equal length and the same direction. This is why a vector can be translated freely without changing it.

Types of vectors

Type	Definition
Zero (null) vector $0$	Magnitude zero, initial and terminal points coincide. Its direction is indeterminate, so it may be taken along any direction.
Unit vector $\overset{a}{^}$	A vector of magnitude $1$ along $a$ , given by $\overset{a}{^} = \frac{a}{∣ a ∣}$ , defined only for $a \neq = 0$ .
Equal vectors	Same magnitude, same direction and representing the same physical quantity.
Negative of a vector	$- a$ has the same magnitude as $a$ but opposite direction. Note $B A = - A B$ .
Collinear (parallel) vectors	Their supports are parallel, irrespective of direction. Same direction gives like vectors, opposite direction gives unlike vectors.
Coplanar vectors	All their supports are parallel to one and the same plane. Any two vectors are always coplanar.
Co-initial vectors	Vectors having the same initial point.
Free vector	Fully described by magnitude and direction alone, with no fixed point of application (displacement, velocity).
Localised (line) vector	Also requires a line of action or point of application (force acting on a rigid body, moment of a force).

Condition for collinearity

$a$ and $b$ (both non-zero) are collinear if and only if $b = λ a$ for some $λ \in R$ .

In component form, if $a = a_{1} \hat{i} + a_{2} \hat{j} + a_{3} \hat{k}$ and $b = b_{1} \hat{i} + b_{2} \hat{j} + b_{3} \hat{k}$ , then they are collinear if

\frac{a _{1}}{b _{1}} = \frac{a _{2}}{b _{2}} = \frac{a _{3}}{b _{3}}

Example 1

Find $x$ and $y$ for which $a = (x + 2) \hat{i} - (x - y) \hat{j} + \hat{k}$ and $b = (x - 1) \hat{i} + (2 x + y) \hat{j} + 2 \hat{k}$ are parallel.

Solution. Parallel means $\frac{x + 2}{x - 1} = \frac{- ( x - y )}{2 x + y} = \frac{1}{2}$ .

From the first and last ratios: $2 (x + 2) = x - 1 \Rightarrow x = - 5$ .

From the last two: $2 (y - x) = 2 x + y \Rightarrow y = 4 x = - 20$ . Hence $x = - 5$ , $y = - 20$ .

Multiplication of a vector by a scalar

For a scalar $m$ and a vector $a$ , the product $m a$ is a vector of magnitude $∣ m ∣ ∣ a ∣$ , parallel to $a$ , pointing along $a$ if $m > 0$ and opposite to $a$ if $m < 0$ . If $m = 0$ or $a = 0$ then $m a = 0$ .

Properties of scalar multiplication

$m (a) = (a) m$ , $m (n a) = n (m a) = (mn) a$

$(m + n) a = m a + n a$ , $m (a + b) = m a + m b$

Addition of vectors

Triangle law. If $A B = a$ and $B C = b$ (head of the first joined to tail of the second), then $a + b = A C$ .

Parallelogram law. If $a$ and $b$ are drawn as co-initial sides $O A$ and $O B$ of a parallelogram $O A C B$ , then $a + b = O C$ , the diagonal through $O$ .

Polygon law. If several vectors are drawn head to tail, their sum is the vector from the tail of the first to the head of the last. If the polygon closes, the sum is $0$ .

Triangle law and parallelogram law give the same sum. Along the route O to A to C you add head to tail; the diagonal OC is the resultant.

Properties of vector addition

$a + b = b + a$ (commutative)

$(a + b) + c = a + (b + c)$ (associative)

$a + 0 = a$ (additive identity), $a + (- a) = 0$ (additive inverse)

$∣ a + b ∣ \leq ∣ a ∣ + ∣ b ∣$ and $∣ a + b ∣ \geq ∣ a ∣ - ∣ b ∣$

Subtraction is addition of the negative: $a - b = a + (- b)$ . In a parallelogram with co-initial sides $a$ and $b$ , one diagonal represents $a + b$ and the other represents $a - b$ .

Common mistake

Equality in $∣ a + b ∣ = ∣ a ∣ + ∣ b ∣$ holds only when $a$ and $b$ are like parallel vectors. Equality in $∣ a + b ∣ = ∣ a ∣ - ∣ b ∣$ holds only when they are unlike parallel vectors. Students often quote these as always true.

Example 2

If $a = 2 \hat{i} - 3 \hat{j} + \hat{k}$ and $b = 4 \hat{i} + 5 \hat{j} - 2 \hat{k}$ are two adjacent sides of a parallelogram, find unit vectors parallel to the diagonals.

Solution. The diagonals are $a + b = 6 \hat{i} + 2 \hat{j} - \hat{k}$ and $b - a = 2 \hat{i} + 8 \hat{j} - 3 \hat{k}$ .

$∣ a + b ∣ = 36 + 4 + 1 = 41$ and $∣ b - a ∣ = 4 + 64 + 9 = 77$ .

Hence the required unit vectors are $\frac{1}{41} (6 \hat{i} + 2 \hat{j} - \hat{k})$ and $\frac{1}{77} (2 \hat{i} + 8 \hat{j} - 3 \hat{k})$ .

Example 3

$A B C D E$ is a pentagon. Prove that $A B + A E + B C + D C + E D + A C = 3 A C$ .

Solution. Group the terms so that each group telescopes head to tail:

$(A B + B C) + (A E + E D + D C) + A C = A C + A C + A C = 3 A C$ .

Position vector, distance and section formula

Fix an origin $O$ . The position vector of a point $P$ is $r = O P$ . If $A$ and $B$ have position vectors $a$ and $b$ , then

Position vector and distance

A B = b - a (head minus tail), A B = ∣ b - a ∣

Section formula

Internal division. If $C$ divides $A B$ internally in the ratio $m : n$ , then

c = \frac{m b + n a}{m + n}

External division. If $C$ divides $A B$ externally in the ratio $m : n$ (with $m \neq = n$ ), then

c = \frac{m b - n a}{m - n}

Midpoint. $c = \frac{a + b}{2}$ .

C divides AB internally in the ratio m : n. Every point of the line AB is a weighted average of the endpoints.

Example 4

$P$ , $Q$ , $S$ have position vectors $p, q, s$ . $M$ is the midpoint of $P Q$ , and $T$ lies on $S M$ with $S T : T M = 2 : 1$ . Find the position vector of $T$ .

Solution. $m = \frac{p + q}{2}$ . Since $T$ divides $S M$ internally in the ratio $2 : 1$ ,

$t = \frac{2 m + 1 \cdot s}{2 + 1} = \frac{( p + q ) + s}{3} = \frac{1}{3} (p + q + s)$ .

This is the centroid of triangle $P QS$ , which is the point where the medians meet.

Components, the orthogonal system and magnitude

Take three mutually perpendicular unit vectors $\hat{i}, \hat{j}, \hat{k}$ along the positive $x$ , $y$ , $z$ axes of a right handed system. Every vector in space has a unique expression

Component form

a = a_{1} \hat{i} + a_{2} \hat{j} + a_{3} \hat{k}, ∣ a ∣ = a_{1}^{2} + a_{2}^{2} + a_{3}^{2}

If $P (x, y, z)$ is a point, its position vector is $O P = x \hat{i} + y \hat{j} + z \hat{k}$ and $O P = x^{2} + y^{2} + z^{2}$ .

If $A (x_{1}, y_{1}, z_{1})$ and $B (x_{2}, y_{2}, z_{2})$ , then $A B = (x_{2} - x_{1}) \hat{i} + (y_{2} - y_{1}) \hat{j} + (z_{2} - z_{1}) \hat{k}$ .

Addition and scalar multiplication act componentwise: $a \pm b = (a_{1} \pm b_{1}) \hat{i} + (a_{2} \pm b_{2}) \hat{j} + (a_{3} \pm b_{3}) \hat{k}$ and $m a = m a_{1} \hat{i} + m a_{2} \hat{j} + m a_{3} \hat{k}$ . Two vectors are equal exactly when all three components match.

Resolution in a plane

A vector $a$ in the $x y$ plane making an angle $θ$ with the positive $x$ axis has components

Plane resolution

a = ∣ a ∣ cos θ \hat{i} + ∣ a ∣ sin θ \hat{j}

The pieces $∣ a ∣ cos θ$ and $∣ a ∣ sin θ$ are called the rectangular components of $a$ along the two axes.

Direction cosines and direction ratios

Let $a = a_{1} \hat{i} + a_{2} \hat{j} + a_{3} \hat{k}$ make angles $α$ , $β$ , $γ$ with the positive $x$ , $y$ , $z$ axes respectively. These are the direction angles, and their cosines are the direction cosines of $a$ , written $l, m, n$ .

Direction cosines

l = cos α = \frac{a _{1}}{∣ a ∣}, m = cos β = \frac{a _{2}}{∣ a ∣}, n = cos γ = \frac{a _{3}}{∣ a ∣}

l^{2} + m^{2} + n^{2} = 1, \overset{a}{^} = l \hat{i} + m \hat{j} + n \hat{k}

Equivalently $cos^{2} α + cos^{2} β + cos^{2} γ = 1$ , which also gives $sin^{2} α + sin^{2} β + sin^{2} γ = 2$ .

Any three numbers proportional to $l, m, n$ are called direction ratios of $a$ . The components $a_{1}, a_{2}, a_{3}$ themselves are a set of direction ratios. Direction cosines are unique up to an overall sign (the two senses along the line), but direction ratios are not unique at all: if $a, b, c$ are direction ratios then so are $k a, k b, k c$ for any $k \neq = 0$ .

Direction ratios to direction cosines

l = \pm \frac{a}{a ^{2} + b ^{2} + c ^{2}}, m = \pm \frac{b}{a ^{2} + b ^{2} + c ^{2}}, n = \pm \frac{c}{a ^{2} + b ^{2} + c ^{2}}

The direction angles of OP with the three axes. Their cosines l, m, n always satisfy l squared plus m squared plus n squared equals 1.

Example 5

Find the direction cosines of $a = 2 \hat{i} - \hat{j} + 2 \hat{k}$ and the angle it makes with the $y$ axis.

Solution. $∣ a ∣ = 4 + 1 + 4 = 3$ , so $l = \frac{2}{3}$ , $m = - \frac{1}{3}$ , $n = \frac{2}{3}$ .

Check: $\frac{4}{9} + \frac{1}{9} + \frac{4}{9} = 1$ . The angle with the $y$ axis is $β = cos^{- 1} (- \frac{1}{3})$ , which is obtuse.

Angle between two vectors, and the angle bisector

The angle between two vectors is the smaller angle formed when they are brought to a common initial point, so $0^{\circ} \leq θ \leq 18 0^{\circ}$ .

Bisectors of the angle between two vectors

Internal bisector direction: $λ (\frac{a}{∣ a ∣} + \frac{b}{∣ b ∣}) = λ (\overset{a}{^} + \hat{b}), λ \in R^{+}$

External bisector direction: $λ (\overset{a}{^} - \hat{b}), λ \in R^{+}$

Common mistake

The bisector is $\overset{a}{^} + \hat{b}$ , not $a + b$ . You must normalise both vectors first. $a + b$ bisects the angle only in the special case $∣ a ∣ = ∣ b ∣$ , which is exactly the rhombus case.

Example 6

The vector $\hat{i} + \hat{j} + \hat{k}$ bisects the angle between $c$ and $3 \hat{i} + 4 \hat{j}$ . Find the unit vector along $c$ .

Solution. Write $\overset{c}{^}$ for the unit vector along $c$ . Then for some $λ > 0$ ,

$λ (\hat{i} + \hat{j} + \hat{k}) = \overset{c}{^} + \frac{3 i ^ + 4 j ^}{5}$ , so $\overset{c}{^} = (λ - \frac{3}{5}) \hat{i} + (λ - \frac{4}{5}) \hat{j} + λ \hat{k}$ .

Imposing $∣ \overset{c}{^} ∣ = 1$ : $3 λ^{2} - \frac{14}{5} λ + 1 = 1$ , giving $λ = \frac{14}{15}$ (rejecting $λ = 0$ ).

Hence $\overset{c}{^} = \frac{1}{3} \hat{i} + \frac{2}{15} \hat{j} + \frac{14}{15} \hat{k}$ .

Linear combinations, dependence and independence

Given vectors $a_{1}, a_{2}, \dots, a_{n}$ and scalars $k_{1}, k_{2}, \dots, k_{n}$ , the vector $k_{1} a_{1} + k_{2} a_{2} + \dots + k_{n} a_{n}$ is called a linear combination of them.

The set is linearly independent if

k_{1} a_{1} + k_{2} a_{2} + \dots + k_{n} a_{n} = 0 ⟹ k_{1} = k_{2} = \dots = k_{n} = 0

and linearly dependent if such a relation holds with at least one $k_{r} \neq = 0$ . In that case $a_{r}$ can be written as a linear combination of the others.

Fundamental theorems

In a plane. If $a, b$ are non-zero and non-collinear, every vector $r$ coplanar with them has a unique expression $r = x a + y b$ .

In space. If $a, b, c$ are non-zero and non-coplanar, every vector $r$ in space has a unique expression $r = x a + y b + z c$ .

Uniqueness gives the comparison rule: $x a + y b + z c = x^{'} a + y^{'} b + z^{'} c \Rightarrow x = x^{'}, y = y^{'}, z = z^{'}$ .

Configuration	Status
Two collinear vectors	Linearly dependent
Two non-zero non-collinear vectors	Linearly independent
Three coplanar vectors	Linearly dependent
Three non-zero non-coplanar vectors	Linearly independent
Any four or more vectors in space	Always linearly dependent
$\hat{i}, \hat{j}, \hat{k}$	Linearly independent (they form a basis of space)

Tests you will reuse constantly

Collinearity of three points $A (a), B (b), C (c)$ : there exist scalars $x, y, z$ not all zero with

$x a + y b + z c = 0$ and $x + y + z = 0$ .

Coplanarity of four points $A, B, C, D$ : there exist $x, y, z, w$ not all zero with

$x a + y b + z c + w d = 0$ and $x + y + z + w = 0$ .

Example 7

Given non-coplanar $a, b, c$ , show that $5 a + 6 b + 7 c$ , $7 a - 8 b + 9 c$ and $3 a + 20 b + 5 c$ are linearly dependent.

Solution. Suppose $5 a + 6 b + 7 c = μ (7 a - 8 b + 9 c) + ν (3 a + 20 b + 5 c)$ .

Comparing coefficients (legal because $a, b, c$ are non-coplanar):

$7 μ + 3 ν = 5$ , $- 8 μ + 20 ν = 6$ , $9 μ + 5 ν = 7$ .

The first and third give $μ = ν = \frac{1}{2}$ , and these values satisfy the second as well. A consistent solution exists, so the three vectors are linearly dependent.

Position vectors of the centres of a triangle

Let $A (a), B (b), C (c)$ be the vertices of a triangle with side lengths $a = B C$ , $b = C A$ , $c = A B$ .

Standard centres

Centroid $G$ : $g = \frac{a + b + c}{3}$ (divides each median in the ratio $2 : 1$ from the vertex)

Incentre $I$ : $i = \frac{a a + b b + c c}{a + b + c}$

Excentre opposite $A$ : $i_{1} = \frac{- a a + b b + c c}{- a + b + c}$

If the circumcentre is taken as origin, the orthocentre is $a + b + c$ , and then $g = \frac{a + b + c}{3}$ lies on $O H$ dividing it in the ratio $1 : 2$ (the Euler line).

JEE Advanced

For a tetrahedron with vertices $a, b, c, d$ , the centroid is $\frac{1}{4} (a + b + c + d)$ . It is the point of concurrency of the four lines joining each vertex to the centroid of the opposite face, and it divides each such line in the ratio $3 : 1$ from the vertex. For a regular tetrahedron it is equidistant from all four vertices and from all four faces.

Example 8

$A B C D$ is a parallelogram and $E$ is the midpoint of $A B$ . Show by vectors that $D E$ trisects $A C$ and is itself trisected by $A C$ .

Solution. Take $A$ as origin, $A B = p$ , $A D = q$ . Then $A C = p + q$ and $A E = \frac{1}{2} p$ .

Let $K$ on $A C$ satisfy $A K : A C = 1 : 3$ , so $A K = \frac{1}{3} (p + q)$ .

Let $M$ on $D E$ satisfy $D M : M E = 2 : 1$ , so $A M = \frac{2 A E + 1 \cdot A D}{3} = \frac{p + q}{3}$ .

Since $A K = A M$ , the points $K$ and $M$ coincide. That single point divides $A C$ in $1 : 3$ from $A$ and divides $D E$ in $2 : 1$ from $D$ , which is exactly the required trisection.

Example 9

The sum of two unit vectors is a unit vector. Show that the magnitude of their difference is $3$ .

Solution. Let $\overset{a}{^}$ and $\hat{b}$ be the unit vectors. Using the parallelogram identity,

∣ \overset{a}{^} + \hat{b} ∣^{2} + ∣ \overset{a}{^} - \hat{b} ∣^{2} = 2 (∣ \overset{a}{^} ∣^{2} + ∣ \hat{b} ∣^{2}) = 4

Given $∣ \overset{a}{^} + \hat{b} ∣ = 1$ , we get $∣ \overset{a}{^} - \hat{b} ∣^{2} = 4 - 1 = 3$ , so $∣ \overset{a}{^} - \hat{b} ∣ = 3$ .

Mistakes that cost marks

Writing $A B = a - b$ . It is always head minus tail, so $A B = b - a$ .
Comparing coefficients of $a, b, c$ without first checking they are non-coplanar. The comparison rule needs linear independence.
Treating $∣ a + b ∣ = ∣ a ∣ + ∣ b ∣$ as an identity instead of the special parallel case.
Using the internal section formula when the point lies outside the segment. External division needs $\frac{m b - n a}{m - n}$ .
Claiming the zero vector has no direction and therefore cannot be added. It adds perfectly well; only its direction is indeterminate.

Quick recap

$\overset{a}{^} = \frac{a}{∣ a ∣}$ , $A B = b - a$ , $∣ a ∣ = a_{1}^{2} + a_{2}^{2} + a_{3}^{2}$
Section formula: internal $\frac{m b + n a}{m + n}$ , external $\frac{m b - n a}{m - n}$ , midpoint $\frac{a + b}{2}$
Direction cosines $l, m, n$ satisfy $l^{2} + m^{2} + n^{2} = 1$ ; components are direction ratios
Angle bisector direction is $\overset{a}{^} + \hat{b}$ , never $a + b$ unless the magnitudes are equal
Non-coplanar triples form a basis, so coefficients can be compared; four vectors in space are always dependent
Centroid $\frac{a + b + c}{3}$ , incentre $\frac{a a + b b + c c}{a + b + c}$

Frequently asked questions

What is the difference between direction cosines and direction ratios?

Direction cosines are the cosines of the angles a vector makes with the positive coordinate axes, and they always satisfy $l^{2} + m^{2} + n^{2} = 1$ . Direction ratios are any three numbers proportional to them, so a vector has exactly one set of direction cosines up to sign but infinitely many sets of direction ratios.

Why is the direction of the zero vector called indeterminate?

Its initial and terminal points coincide, so no direction can be assigned to the segment. By convention it is treated as parallel to every vector, which keeps statements such as "collinear if one is a scalar multiple of the other" consistent.

Can two vectors ever be non-coplanar?

No. Any two vectors can be brought to a common initial point and a plane can always be passed through the two segments, so two vectors are always coplanar. Non-coplanarity is a property that first appears with three vectors.

When can I compare coefficients on both sides of a vector equation?

Only when the vectors whose coefficients you are comparing are linearly independent, that is, non-collinear for two vectors and non-coplanar for three. If they are dependent the expression is not unique and comparing coefficients gives wrong answers.

How do I decide quickly whether three given points are collinear?

Form two vectors from the three points, for example $A B$ and $A C$ , and check whether one is a scalar multiple of the other. Equivalently, look for scalars $x, y, z$ not all zero with $x a + y b + z c = 0$ and $x + y + z = 0$ .

Previous year questions on Basic Concepts of Vector Algebra

1 question from past papers, each with a step-by-step solution.

JEE Main 2026 Apr 5 Shift 2, Mathematics Q16

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