Vector Algebra Formula Sheet — JEE Main Mathematics

Every key Vector Algebra formula, definition and theorem for JEE Main Mathematics in one place — with common examiner traps and worked examples. Free to read; blurt from memory, then check your gaps.

Syllabus — topics coveredNTA · 5 sub-topics

Vectors and scalars
Addition of vectors
Components of a vector in two dimensions and three dimensional space
Scalar and vector products
Scalar and vector triple product

Vectors — Types, Components & Algebra

Vector: A quantity with both magnitude and direction, written

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

. Its magnitude is

∣ a ∣ = a_{1}^{2} + a_{2}^{2} + a_{3}^{2}

and its unit vector is

\overset{a}{^} = \frac{a}{∣ a ∣}

(same direction, length

1

Types of vectors

▸ $Zero/null$ $0$ (magnitude $0$ ); $unit$ (magnitude $1$ ).
▸ $Equal$ : same magnitude AND direction.
▸ $Collinear$ (parallel): $b = λ a$ ; $coplanar$ : lie in one plane.
▸ $Position vector$ $OP$ from a fixed origin O.

★ Remember · Addition: triangle & parallelogram laws

Place vectors head-to-tail and join start to finish (triangle law), or as adjacent sides of a parallelogram. Addition is

commutative

(

a + b = b + a

) and

associative

;

A B = b - a

🎯 Exam · Magnitude of a sum

∣ a + b ∣^{2} = ∣ a ∣^{2} + ∣ b ∣^{2} + 2∣ a ∣∣ b ∣ cos θ

. So

∣ a + b ∣ = ∣ a - b ∣

exactly when

a ⊥ b

(the diagonals of a rectangle are equal).

⚠️ Watch out · Magnitude isn't additive

∣ a + b ∣ \neq = ∣ a ∣ + ∣ b ∣

in general — equality holds only when

a, b

point the

same way

. By the triangle inequality,

∣ a ∣ - ∣ b ∣ \leq ∣ a + b ∣ \leq ∣ a ∣ + ∣ b ∣

🚫 Examiner Trap · Examiner traps

(1) Magnitude of

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

a_{1}^{2} + a_{2}^{2} + a_{3}^{2}

— components are SIGNED, square them. (2) A unit vector is

\overset{a}{^} = \frac{a}{∣ a ∣}

(only for

a \neq = 0

). (3) The zero vector has magnitude

0

and NO definite direction. (4) Collinear

\Leftrightarrow b = λ a

; coplanar means lying in one plane — don't confuse the two.

Position Vectors, Section Formula & Collinearity

★ Remember · Section formula

The point dividing AB in ratio m:n internally has position vector

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

(midpoint

\frac{a + b}{2}

); externally

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

. The centroid of a triangle is

\frac{a + b + c}{3}

Collinearity & coplanarity

▸Two vectors collinear $\Leftrightarrow$ $b = λ a$ .
▸Points A,B,C collinear $\Leftrightarrow$ $A B = λ A C$ (i.e. $A B \times A C = 0$ ).
▸Three vectors coplanar $\Leftrightarrow$ one is a linear combination of the others ( $[a b c] = 0$ ; the triple product is not in the JEE Main 2026 syllabus).
▸ $r = x a + y b + z c$ with $x + y + z = 1$ $\Rightarrow$ the four points are coplanar.

🎯 Exam · Vectors give clean geometry proofs

Express each point as a position vector, then 'collinear', 'midpoint', 'parallel' and 'perpendicular' all become simple vector equations. This converts a geometry problem into algebra you can just compute.

⚠️ Watch out · Watch the order in the section formula

The point dividing AB in ratio m:n takes

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

— the

far

endpoint

b

pairs with m. Swapping the weights gives the point that divides it in ratio n:m instead.

🚫 Examiner Trap · Examiner traps

(1)

A B = b - a

(terminal MINUS initial) — not

a - b

. (2) Internal section is

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

; EXTERNAL division uses

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

(sign change). (3) Points A,B,C collinear

\Leftrightarrow A B = λ A C

i.e.

A B \times A C = 0

. (4)

r = x a + y b + z c

with

x + y + z = 1

\Rightarrow

the points are coplanar.

Dot (Scalar) Product

Dot (scalar) product:

a \cdot b = ∣ a ∣∣ b ∣ cos θ = a_{1} b_{1} + a_{2} b_{2} + a_{3} b_{3}

— a

scalar

. It is commutative, and

a \cdot b = 0

exactly when the vectors are

perpendicular

(for non-zero vectors).

The dot product definition, projection of one vector onto another, its properties, and its uses (angle, projection, magnitude of sums, work) — The dot product, projection, and what it gives you.

★ Remember · Angle & projection

cos θ = \frac{a \cdot b}{∣ a ∣∣ b ∣}

— the sign of

a \cdot b

alone tells you acute (

+

) vs obtuse (

-

). Scalar projection of

a

b

\frac{a \cdot b}{∣ b ∣}

; the vector projection is

\frac{a \cdot b}{∣ b ∣ ^{2}} b

Key properties

▸ $a \cdot a = ∣ a ∣^{2}$ ; $\hat{i} \cdot \hat{i} = 1$ , $\hat{i} \cdot \hat{j} = 0$ .
▸Distributive: $a \cdot (b + c) = a \cdot b + a \cdot c$ .
▸ $∣ a \pm b ∣^{2} = ∣ a ∣^{2} + ∣ b ∣^{2} \pm 2 a \cdot b$ .
▸Work done by a force: $W = F \cdot d$ .

🎯 Exam · Perpendicularity is a dot product

To impose

a ⊥ b

, just set

a \cdot b = 0

— a single linear equation in the components. This is the standard way to find a vector orthogonal to a given one, or to locate a foot of perpendicular.

⚠️ Watch out · The dot product is a number

a \cdot b

is a

scalar

, so expressions like

(a \cdot b) \times c

are meaningless (you can't cross a scalar with a vector). Keep track of which products yield scalars and which yield vectors.

🚫 Examiner Trap · Examiner traps

(1) Dot product is a SCALAR;

a \cdot b = 0 \Leftrightarrow

perpendicular (for non-zero vectors). (2) Scalar projection of

a

b

\frac{a \cdot b}{∣ b ∣}

, but the VECTOR projection is

\frac{a \cdot b}{∣ b ∣ ^{2}} b

— don't mix them. (3) The sign of

a \cdot b

tells acute (

+

) vs obtuse (

-

). (4)

∣ a \pm b ∣^{2} = ∣ a ∣^{2} + ∣ b ∣^{2} \pm 2 a \cdot b

(cosine rule).

Cross (Vector) Product

Cross (vector) product:

a \times b = ∣ a ∣∣ b ∣ sin θ \overset{n}{^}

— a

vector

perpendicular to both

a

and

b

, with

\overset{n}{^}

given by the right-hand rule. It is

anticommutative

(

b \times a = - a \times b

) and is

0

exactly when the vectors are parallel.

In an oblique 3-D sketch, a points right and b points back into the page, with the parallelogram on them shaded (area equal to the magnitude of a cross b). The vector a cross b points straight up, perpendicular to both, by the right-hand rule. — $a \times b$ is perpendicular to the plane of $a, b$ ; its length is the parallelogram's area.

★ Remember · Area from the cross product

∣ a \times b ∣

is the area of the

parallelogram

a, b

, so the triangle area is

\frac{1}{2} ∣ A B \times A C ∣

. A unit vector perpendicular to both is

\overset{n}{^} = \frac{a \times b}{∣ a \times b ∣}

Key properties

▸Component form: the $\hat{i}, \hat{j}, \hat{k}$ determinant of the two component rows.
▸ $\hat{i} \times \hat{j} = \hat{k}$ , $\hat{j} \times \hat{k} = \hat{i}$ , $\hat{k} \times \hat{i} = \hat{j}$ (cyclic); $\hat{i} \times \hat{i} = 0$ .
▸Distributive but $not$ commutative or associative.
▸Moment (torque) of a force: $τ = r \times F$ .

🎯 Exam · Parallel is a cross product

To impose

a ∥ b

, set

a \times b = 0

(equivalently, the components are proportional). This is the cross-product counterpart of using the dot product for perpendicularity.

⚠️ Watch out · Order matters — it's anticommutative

b \times a = - a \times b

, so swapping the factors reverses the direction. When you compute an area you take the magnitude (sign-free), but for a normal direction the order is essential.

🚫 Examiner Trap · Examiner traps

(1) Cross product is a VECTOR and is ANTI-commutative:

b \times a = - a \times b

. (2)

a \times b = 0 \Leftrightarrow

PARALLEL (note

a \times a = 0

); compare with dot

= 0

for perpendicular. (3)

∣ a \times b ∣

is the area of the PARALLELOGRAM; the triangle is HALF that. (4) Direction by right-hand rule:

\hat{i} \times \hat{j} = \hat{k}

(cyclic) — reverse the order and the sign flips.

Scalar Triple Product (Box Product) — not in JEE Main 2026 syllabus

Scalar triple product:

[a b c] = a \cdot (b \times c)

— a

scalar

equal to the

3 \times 3

determinant of the components. Its absolute value is the

volume of the parallelepiped

built on

a, b, c

A parallelepiped whose three edges from one corner are the vectors a, b and c; its volume is the absolute value of the scalar triple product [a b c]. — The parallelepiped on $a, b, c$ has volume $∣ [a b c] ∣$ (not in JEE Main 2026 syllabus).

★ Remember · Coplanarity test

Three vectors are

coplanar

\Leftrightarrow

[a b c] = 0

(zero volume). Four points A,B,C,D are coplanar

\Leftrightarrow

[A B A C A D] = 0

. They are linearly independent iff

[a b c] \neq = 0

Properties

▸Cyclic: $[a b c] = [b c a] = [c a b]$ .
▸Swapping any two vectors flips the sign; two equal vectors give $0$ .
▸Dot and cross can be interchanged: $a \cdot (b \times c) = (a \times b) \cdot c$ .
▸Tetrahedron volume $= \frac{1}{6} ∣ [a b c] ∣$ .

🎯 Exam · Just evaluate a determinant

In components,

[a b c]

is the determinant whose rows are the components of

a, b, c

. Volume, coplanarity and linear-independence questions all reduce to evaluating (or zeroing) this one determinant.

⚠️ Watch out · Order changes the sign, not the volume

A swap of two vectors negates

[a b c]

, but the volume is

∣ [a b c] ∣

— always take the modulus for a volume. A negative triple product just means a left-handed ordering.

🚫 Examiner Trap · Examiner traps

(1)

[a b c] = a \cdot (b \times c)

is a SCALAR (the

3 \times 3

determinant). (2) It is CYCLIC (

[a b c] = [b c a]

) but swapping two vectors FLIPS the sign; two equal vectors

\Rightarrow 0

. (3)

[a b c] = 0 \Leftrightarrow

the three are COPLANAR (zero volume). (4) Volume of parallelepiped

= ∣ [a b c] ∣

; tetrahedron

= \frac{1}{6} ∣ [a b c] ∣

(don't forget the

\frac{1}{6}

Vector Triple Product & Identities — not in JEE Main 2026 syllabus

Vector triple product:

a \times (b \times c) = (a \cdot c) b - (a \cdot b) c

— the

BAC-CAB

rule. The result is a vector lying in the plane of

b

and

c

, and the operation is

not associative

★ Remember · BAC − CAB (and mind the brackets)

a \times (b \times c) = (a \cdot c) b - (a \cdot b) c

, but

(a \times b) \times c = (a \cdot c) b - (b \cdot c) a

. They differ, so the brackets cannot be dropped — vector products are not associative.

Useful identities

▸Lagrange: $(a \times b) \cdot (c \times d) = (a \cdot c) (b \cdot d) - (a \cdot d) (b \cdot c)$ .
▸ $∣ a \times b ∣^{2} = ∣ a ∣^{2} ∣ b ∣^{2} - (a \cdot b)^{2}$ .
▸ $[a \times b b \times c c \times a] = [a b c]^{2}$ .
▸The middle vector of the bracket ( $b$ ) gets the 'outer-dot-far' coefficient.

🎯 Exam · Expand, don't try to compute the crosses

For any

a \times (b \times c)

, immediately apply BAC

-

CAB to turn it into dot products and the vectors

b, c

. Computing the inner cross then the outer cross is far more error-prone.

⚠️ Watch out · The identity depends on the bracketing

Which two vectors get dotted depends on

where the bracket is

. For

a \times (b \times c)

it's the

a

-dots; for

(a \times b) \times c

it's the

c

-dots. Read the brackets carefully before expanding.

🚫 Examiner Trap · Examiner traps

(1) BAC–CAB:

a \times (b \times c) = (a \cdot c) b - (a \cdot b) c

— the result lies in the plane of

b, c

. (2) The vector triple product is NOT associative:

a \times (b \times c) \neq = (a \times b) \times c

in general — mind the brackets. (3) Get the dot-products' grouping right (it's

(a \cdot c) b

first). (4) Lagrange:

(a \times b) \cdot (c \times d) = (a \cdot c) (b \cdot d) - (a \cdot d) (b \cdot c)

Applications of Vectors

Geometry & physics

▸ $Angle$ : $cos θ = \frac{a \cdot b}{∣ a ∣∣ b ∣}$ .
▸ $Area$ : triangle $\frac{1}{2} ∣ A B \times A C ∣$ ; parallelogram $∣ a \times b ∣$ .
▸ $Volume$ (not in JEE Main 2026 syllabus): parallelepiped $∣ [a b c] ∣$ ; tetrahedron $\frac{1}{6} ∣ [a b c] ∣$ .
▸ $Work$ $W = F \cdot d$ ; $torque$ $τ = r \times F$ .

★ Remember · Linear independence (not in JEE Main 2026 syllabus)

Three vectors

a, b, c

are

linearly independent

iff

[a b c] \neq = 0

— they then span all of 3-space. If the triple product is

0

they are coplanar (dependent), and one is a combination of the others.

🎯 Exam · Translate geometry into dot/cross

Perpendicular

\to a \cdot b = 0

; parallel

\to a \times b = 0

; collinear points

\to A B \times A C = 0

; coplanar

\to [\cdot] = 0

. Most 'prove that...' problems are one of these set equal to its target.

★ Remember · Scalar vs vector — keep them straight

Dot and scalar-triple products give

numbers

(angle, work, volume); cross and vector-triple products give

vectors

(normal, area direction, torque). Matching the right product to what the question wants is half the work.

🚫 Examiner Trap · Examiner traps

(1) Area: triangle

= \frac{1}{2} ∣ A B \times A C ∣

, parallelogram

= ∣ a \times b ∣

(or

\frac{1}{2} ∣ d_{1} \times d_{2} ∣

from diagonals). (2) Work

W = F \cdot d

is a SCALAR; torque

τ = r \times F

is a VECTOR — match the product to the quantity. (3)

a, b, c

are linearly independent

\Leftrightarrow [a b c] \neq = 0

. (4) For geometry proofs, translate perpendicularity/area conditions into dot/cross equations.

More JEE Main Mathematics formula sheets

Frequently Asked Questions

What are the most important Vector Algebra formulas for JEE Main?

This Vector Algebra formula sheet covers all the high-yield Mathematics formulas, definitions and theorems you need for JEE Main, across Vectors and scalars, Addition of vectors, Components of a vector in two dimensions and three dimensional space, Scalar and vector products, Scalar and vector triple product — each shown with the key result and, where useful, a worked example.

Is this Vector Algebra formula sheet free?

Yes — the full chapter formula sheet is free to read online, no login or payment required.

How should I revise Vector Algebra formulas?

Blurt the Vector Algebra formulas from memory, then check against this sheet to find your gaps — and practise a few previous-year questions on the chapter to make sure you can apply them under time pressure.

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