NDA Mathematics · Formula sheet

Vectors formulas

27 formulas and 55 common traps for NDA Mathematics Vectors, grouped by subtopic.

Foundations: Vectors, Operations, and Position

Position vectors and displacement vectors

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

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})

$\vec{a} + \vec{b}$ 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

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

Component form: the i, j, k unit vectors

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

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

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

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

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

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}

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

Common traps

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.

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.

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.

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}

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).

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.

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.

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.

Magnitude, Components, Projection, Direction Cosines

Learn this subtopic in the notes

Magnitude of a vector and distance between two points

Magnitude and distance

|\vec{v}| = \sqrt{v_1^2 + v_2^2 + v_3^2} \qquad AB = |\vec{b} - \vec{a}|

$v_1, v_2, v_3$ components of $\vec{v}$ along $\hat{i}, \hat{j}, \hat{k}$
$\vec{a}, \vec{b}$ position vectors of the endpoints

Direction Cosines

Direction-cosine identities

\cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1 \qquad \sin^2\alpha + \sin^2\beta + \sin^2\gamma = 2

$\alpha, \beta, \gamma$ angles between $\vec{v}$ and the positive $x, y, z$ axes
$l, m, n$ direction cosines (the unit vector's components)

Scalar projection of one vector on another

Scalar and vector projection

\text{proj}_{\vec{b}}\vec{a} = \dfrac{\vec{a}\cdot\vec{b}}{|\vec{b}|} \qquad \overrightarrow{\text{proj}_{\vec{b}}\vec{a}} = \dfrac{\vec{a}\cdot\vec{b}}{|\vec{b}|^2}\,\vec{b}

$\vec{a}$ vector being projected
$\vec{b}$ vector providing the direction
$|\vec{b}|$ magnitude of $\vec{b}$ (NOT $|\vec{b}|^2$ for scalar version)

Unit vectors and direction-given construction

Unit vector and direction construction

\hat{v} = \dfrac{\vec{v}}{|\vec{v}|} \qquad \vec{v} = r(\cos\alpha\,\hat{i} + \cos\beta\,\hat{j} + \cos\gamma\,\hat{k})

$\hat{v}$ unit vector along $\vec{v}$
$r$ desired magnitude of the constructed vector
$\alpha, \beta, \gamma$ angles with the positive coordinate axes

Common traps

$\overrightarrow{AB}$ is $\vec{b} - \vec{a}$ — head minus tail

Reverse the subtraction and you get

\overrightarrow{BA}

— the magnitude is the same but the displacement points the other way. Direction matters whenever the result feeds into a dot product or angle.

Lagrange identity gives you the missing magnitude

Whenever a question gives

|\vec{a}\times\vec{b}|^2 + (\vec{a}\cdot\vec{b})^2 = k

together with one magnitude, use the identity

|\vec{a}|^2|\vec{b}|^2 = |\vec{a}\times\vec{b}|^2 + (\vec{a}\cdot\vec{b})^2

to read the other magnitude off directly. (The same identity is the central formula of

\vec{a}\times\vec{b}

magnitude work — see the cross-product note.)

Factor-of-2 trap: $\sin^2\alpha + \sin^2\beta + \sin^2\gamma = 2$ , not 1

From

\sum\cos^2 = 1

and

\sin^2 = 1 - \cos^2

, summing three times:

\sum\sin^2 = 3 - \sum\cos^2 = 3 - 1 = 2

. The distractor

= 1

(copying the cosine identity) is the single most common wrong answer in this concept.

Direction cosines can be negative

An obtuse angle with an axis gives a negative cosine — totally fine. Some students try to force

l, m, n \geq 0

; don't. The identity

l^2 + m^2 + n^2 = 1

holds with signs.

Divide by $|\vec{b}|$ , not $|\vec{b}|^2$ , for the scalar projection

Vector projection has

|\vec{b}|^2

in the denominator because it carries the direction

\vec{b}

back into the answer; scalar projection drops the direction and divides only once. Mixing the two is a frequent factor-of-

|\vec{b}|

bug.

Sign of the scalar projection encodes obtuse/acute

If the projection comes out negative, the angle between

\vec{a}

and

\vec{b}

is obtuse. Don't reach for absolute value automatically — the sign is the information.

Check that the given angles are consistent with $\sum\cos^2 = 1$

A vector cannot make

\alpha = 60^\circ

and

\beta = 45^\circ

with the

x

and

y

axes AND have

\gamma

acute unless the third cosine fits. From

\cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1

\cos^2\gamma = 1 - \tfrac{1}{4} - \tfrac{1}{2} = \tfrac{1}{4}

, so

\gamma = 60^\circ

(acute) or

120^\circ

Equally inclined to two axes only fixes one component pair

\"Vector inclined equally to

x

and

y

axes\" means

a_1 = a_2

, which combined with the magnitude pins down both. Don't read it as

a_1 = a_2 = a_3

— that's three axes, a different constraint.

Dot Product and Angle

Learn this subtopic in the notes

Dot product — components form and work done

Dot product (components form)

\vec{a}\cdot\vec{b} = a_1 b_1 + a_2 b_2 + a_3 b_3 \qquad W = \vec{F}\cdot\vec{d}

$a_i, b_i$ components of $\vec{a}, \vec{b}$ along $\hat{i}, \hat{j}, \hat{k}$
$W$ work done by a constant force $\vec{F}$ through displacement $\vec{d}$

Perpendicularity Test

Equivalent perpendicularity statements

\vec{a}\perp\vec{b} \;\Longleftrightarrow\; \vec{a}\cdot\vec{b} = 0 \;\Longleftrightarrow\; |\vec{a}+\vec{b}| = |\vec{a}-\vec{b}|

$\vec{a}\cdot\vec{b}$ scalar dot product
$|\vec{a}\pm\vec{b}|$ magnitudes of the diagonals of the parallelogram on $\vec{a}, \vec{b}$

Angle between two vectors via the dot-product formula

Angle from dot product

\cos\theta = \dfrac{\vec{a}\cdot\vec{b}}{|\vec{a}|\,|\vec{b}|}

$\theta$ angle between $\vec{a}$ and $\vec{b}$ , measured in $[0, \pi]$
$\vec{a}\cdot\vec{b}$ dot product (scalar)
$|\vec{a}|, |\vec{b}|$ magnitudes (always positive)

Solving for an angle from a perpendicularity / magnitude constraint

Expansion template

(\alpha\vec{a}+\beta\vec{b})\cdot(\gamma\vec{a}+\delta\vec{b}) = \alpha\gamma|\vec{a}|^2 + (\alpha\delta+\beta\gamma)\,\vec{a}\cdot\vec{b} + \beta\delta|\vec{b}|^2

$\alpha, \beta, \gamma, \delta$ given scalar coefficients
$|\vec{a}|, |\vec{b}|$ given (often $=1$ for unit vectors)
$\vec{a}\cdot\vec{b}$ unknown — solve for it, then read off $\theta$

Unit vectors, orthogonal triples, and decomposition

Orthonormal-triple identities

\vec{a}\cdot\vec{a} = 1, \quad \vec{a}\cdot\vec{b} = 0 \;\;(\text{if } \vec{a}\neq\vec{b}) \quad \text{for an orthonormal triple}

$\vec{a}, \vec{b}, \vec{c}$ three mutually-perpendicular unit vectors
$p, q, r$ decomposition coefficients along $\vec{a}, \vec{b}, \vec{a}\times\vec{b}$

Common traps

Dot product gives a scalar; cross product gives a vector

An MCQ option that returns a vector for

\vec{a}\cdot\vec{b}

, or a scalar for

\vec{a}\times\vec{b}

, can be eliminated on type grounds alone. This dimension-check rejects ~25% of trap options.

Work done is signed — negative work is fine

If the force has any component opposite to the displacement, the dot product (and the work) can come out negative. Don't reach for absolute value automatically; the sign tells you whether the force is helping or hindering.

$|\vec{a}+\vec{b}| = |\vec{a}-\vec{b}|$ means $\vec{a}\perp\vec{b}$ , not $\vec{a}=\vec{b}$

Square both sides:

|\vec{a}|^2 + 2\vec{a}\cdot\vec{b} + |\vec{b}|^2 = |\vec{a}|^2 - 2\vec{a}\cdot\vec{b} + |\vec{b}|^2

collapses to

4\vec{a}\cdot\vec{b} = 0

. Geometrically: the two diagonals of a parallelogram have equal length iff the parallelogram is a rectangle.

Zero dot product needs both vectors non-zero

Technically

\vec{0}\cdot\vec{a} = 0

for any

\vec{a}

, but

\vec{0}

has no direction, so we don't call it perpendicular. PYQs assume non-zero vectors implicitly; double-check the hypothesis if a question opens with \"if non-zero...\".

Obtuse angle iff $\vec{a}\cdot\vec{b} < 0$

Quadratic-in-

x

PYQs frequently ask for values of a parameter that make the angle obtuse. Set up the inequality

\vec{a}\cdot\vec{b} < 0

, solve as a quadratic, then exclude the boundary case

\vec{a}\cdot\vec{b} = 0

(which is perpendicular, not obtuse) AND values that make the vectors antiparallel (then

\theta = \pi

, the obtuse extreme).

Direction matters when comparing two angles

The angle between

\vec{a}

and

-\vec{a}

\pi

, not 0. If a problem asks for the angle between

\vec{a}

and

\vec{a} - \vec{b}

, don't carelessly subtract magnitudes — use the formula end-to-end.

Don't forget the cross terms when expanding

(\vec{a}+2\vec{b})\cdot(5\vec{a}-4\vec{b})

has four products, not two — there are two

\vec{a}\cdot\vec{b}

terms that combine into the coefficient

\alpha\delta + \beta\gamma

. A factor-of-2 distractor often results from dropping one of them.

Unit vectors mean $|\vec{a}|^2 = 1$ , not $\vec{a} = 1$

When the magnitudes are stated as 1, the

|\vec{a}|^2

and

|\vec{b}|^2

terms simplify to 1, NOT zero. Some students drop them by analogy with

\vec{a}\cdot\vec{a}

when

\vec{a}

is the zero vector — wrong.

Three unit vectors at equal pairwise angles need not be orthonormal

\vec{a}\cdot\vec{b} = \vec{b}\cdot\vec{c} = \vec{c}\cdot\vec{a} = k

, the triple is symmetric but only orthonormal when

k = 0

. For other

k

values (e.g.

k = -1/2

— three coplanar vectors at

120^\circ

) the magnitudes of linear combinations look quite different.

$\{\vec{a}, \vec{b}, \vec{a}\times\vec{b}\}$ is orthonormal iff $\vec{a}\perp\vec{b}$ and both unit

\vec{a}, \vec{b}

are unit and perpendicular, then

|\vec{a}\times\vec{b}| = \sin 90^\circ = 1

— so the triple is orthonormal. If

\vec{a}\cdot\vec{b} \neq 0

, the cross product still produces a perpendicular vector, but it's not a unit vector and the basis isn't orthonormal.

Cross Product and Triple Product

Learn this subtopic in the notes

Cross product — algebra and properties

Difference-of-squares-style identity

(\vec{a} - \vec{b}) \times (\vec{a} + \vec{b}) = 2\,\vec{a}\times\vec{b}

$\vec{a}\times\vec{a}, \vec{b}\times\vec{b}$ both equal $\vec{0}$
$\vec{a}\times\vec{b}, \vec{b}\times\vec{a}$ differ in sign — they survive in the expansion

Cross-product magnitude, area, and the Lagrange identity

Magnitude, area, and Lagrange

|\vec{a}\times\vec{b}| = |\vec{a}||\vec{b}|\sin\theta \qquad |\vec{a}\times\vec{b}|^2 + (\vec{a}\cdot\vec{b})^2 = |\vec{a}|^2 |\vec{b}|^2

$\theta$ angle between $\vec{a}$ and $\vec{b}$
$|\vec{a}\times\vec{b}|$ parallelogram area; triangle area is half of this
Lagrange identityfrom $\sin^2\theta + \cos^2\theta = 1$ multiplied by $|\vec{a}|^2|\vec{b}|^2$

Unit vector perpendicular to two given vectors

Unit perpendicular

\hat{n} = \pm\dfrac{\vec{a}\times\vec{b}}{|\vec{a}\times\vec{b}|}

$\vec{a}\times\vec{b}$ vector perpendicular to both $\vec{a}$ and $\vec{b}$
$|\vec{a}\times\vec{b}|$ magnitude — divide to normalise
$\pm$ two unit perpendiculars exist, in opposite directions

Moment of a force (torque)

Moment of a force

\vec{M} = \overrightarrow{OP} \times \vec{F}

$O$ pivot / reference point for the moment
$\overrightarrow{OP}$ position vector from $O$ to the point of application $P$
$\vec{F}$ applied force vector

Scalar triple product and coplanarity

STP as determinant + coplanarity test

[\vec{a}\,\vec{b}\,\vec{c}] = \begin{vmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{vmatrix} = 0 \iff \vec{a},\vec{b},\vec{c}\text{ coplanar}

$[\vec{a}\,\vec{b}\,\vec{c}]$ scalar triple product (a single number)
$\vec{a}\cdot(\vec{b}\times\vec{c})$ equivalent dot-cross form
$|[\vec{a}\,\vec{b}\,\vec{c}]|$ volume of the parallelepiped on the three vectors

STP cyclic property and derived linear-combo identities

Cyclic + sum identity

(\vec{a}\times\vec{b})\cdot\vec{c} + (\vec{b}\times\vec{c})\cdot\vec{a} + (\vec{c}\times\vec{a})\cdot\vec{b} = 3[\vec{a}\,\vec{b}\,\vec{c}]

Cyclic ordering $\vec{a} \to \vec{b} \to \vec{c} \to \vec{a}$ — all three terms are STPs of the same value
$3[\vec{a}\,\vec{b}\,\vec{c}]$ the cyclic sum is three times any one of them

Vector triple product (BAC-CAB rule)

BAC-CAB rule

\vec{a}\times(\vec{b}\times\vec{c}) = (\vec{a}\cdot\vec{c})\,\vec{b} - (\vec{a}\cdot\vec{b})\,\vec{c}

$\vec{a}\cdot\vec{c}, \vec{a}\cdot\vec{b}$ scalar coefficients
$\vec{b}, \vec{c}$ vector basis of the resulting plane
Result directionlies in the plane of $\vec{b}$ and $\vec{c}$ , perpendicular to $\vec{a}$

Common traps

Cross product is NOT associative

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

and

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

are generally different vectors — both are linear combinations of

\vec{a}

and

\vec{b}

(or

\vec{b}

and

\vec{c}

), but with different coefficients given by BAC-CAB. An MCQ statement \"cross product is associative\" is always wrong.

$\vec{a}\times\vec{b} = \vec{0}$ does NOT mean both vectors are zero

It means

\vec{a}

and

\vec{b}

are parallel — they could be non-zero scalar multiples of each other. The right reading:

\vec{a}\times\vec{b} = \vec{0}

and

\vec{a}, \vec{b} \neq \vec{0}

together imply

\vec{a} = \lambda\vec{b}

for some scalar

\lambda

Area of a triangle is $\tfrac{1}{2}|\vec{a}\times\vec{b}|$ , NOT $|\vec{a}\times\vec{b}|$

A frequently-tested statement: \"

|\vec{a}\times\vec{b}|

is the area of a triangle with sides

\vec{a}

and

\vec{b}

\" — this is FALSE; it's the parallelogram area. The factor-of-2 lives here. Halving gives the triangle.

$\sin\theta$ is always non-negative for $\theta\in[0,\pi]$

Unlike

\cos\theta

, the sine in the cross-product magnitude formula never goes negative — the magnitude is a length. If a question gives

\vec{a}\times\vec{b}

as a specific vector and asks for the acute angle, take magnitudes of BOTH sides before solving for

\sin\theta

Both $\pm$ signs give valid answers

If an MCQ offers

+\hat{n}

and the question doesn't pin down direction,

-\hat{n}

is equally correct — accept whichever is listed. Some PYQs add a constraint like \"with positive

z

-component\" specifically to break this ambiguity.

Scalar multiples of a unit perpendicular are not unit

A vector like

\dfrac{1}{50}(-4\hat{i}-5\hat{j}+3\hat{k})

may point in the right direction but if its magnitude isn't 1, it isn't a unit perpendicular. Always confirm

|\hat{n}| = 1

before selecting an option.

Order is $\vec{r} \times \vec{F}$ , not $\vec{F} \times \vec{r}$

Switching the order flips the sign of the moment by the anti-commutative rule. The pivot point comes FIRST: position vector from pivot to application point, THEN cross with force.

Moment depends on the pivot — moment of a force about a POINT is unique, but about a LINE is also a vector

MCQ statements like \"moment of a force is independent of point of application\" or \"moment about a line is a scalar\" are common wrong options. Moment of a force depends on both the line of action AND the pivot; the moment about a line is the projection of

\vec{r}\times\vec{F}

onto that line — a scalar, not a vector.

STP $= 0$ means coplanar — NOT \" $\vec{a}$ parallel to $\vec{b}$ \"

Three vectors coplanar means they all fit inside some 2-D plane through the origin. They need not be parallel to each other. Parallel-pair is a stronger condition that ALSO makes STP zero, but not the only one.

Determinant row/column expansion: pick the row with most zeros

If one row has a zero (very common in coplanarity problems), expand along it — two of the three cofactors drop out immediately, saving an entire

2\times 2

minor.

Anti-cyclic = sign flip — don't accidentally drop it

\vec{c}\times\vec{b} = -\vec{b}\times\vec{c}

. If you absorb a cross-product without tracking the sign, you'll be off by a factor of

-1

on every other term — turning

\lambda = 6

into

\lambda = -6

or worse.

$(\vec{a}\times\vec{b})\cdot\vec{c} = (\vec{c}\times\vec{a})\cdot\vec{b}$

Both are cyclic rotations of

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

. An MCQ giving

2[\vec{a}\,\vec{b}\,\vec{c}]

as the cyclic sum is a factor-of-3 wrong option. (

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

is a different beast — that one is zero by orthogonality.)

BAC-CAB only applies to vector triple products — not scalar

If the expression is

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

(no second cross), it's a scalar triple product, not BAC-CAB. Identify the SHAPE first: how many crosses, how many dots — that fixes which identity to use.

Cross-then-cross is NOT cross-then-dot-with-different-grouping

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

lies in the plane of

\vec{a}, \vec{b}

(the innermost pair);

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

lies in the plane of

\vec{b}, \vec{c}

. The two are different vectors and PYQs use this asymmetry as the load-bearing distractor.

Special triples: if $\vec{a}\times\vec{b} = \vec{c}$ and $\vec{b}\times\vec{c} = \vec{a}$ , the three vectors are an orthonormal pairwise-perpendicular triple

Take magnitudes of both equations and use

|\vec{a}\times\vec{b}| = |\vec{a}||\vec{b}|\sin\theta

with

\sin\theta \leq 1

to force

|\vec{a}| = |\vec{b}| = |\vec{c}| = 1

AND pairwise perpendicularity. This is the lever behind both the 2017

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

classic and the 2026 S10 set.

Vector Geometry — Triangles, Parallelograms, Quadrilaterals

Learn this subtopic in the notes

Triangle closed-loop and centroid formula

Loop identity + centroid

\vec{AB} + \vec{BC} + \vec{CA} = \vec{0} \qquad \vec{g} = \dfrac{\vec{a} + \vec{b} + \vec{c}}{3}

$\vec{a}, \vec{b}, \vec{c}$ position vectors of vertices $A, B, C$
$\vec{AB}$ side vector from $A$ to $B$ , equal to $\vec{b} - \vec{a}$
$\vec{g}$ position vector of the centroid $G$

Parallelogram properties and diagonal relations

Sides from diagonals

\vec{AB} = \tfrac{1}{2}(\vec{AC} - \vec{BD}), \quad \vec{AD} = \tfrac{1}{2}(\vec{AC} + \vec{BD}), \quad \vec{OA}+\vec{OC} = \vec{OB}+\vec{OD}

$ABCD$ parallelogram with vertices labelled in order
$\vec{AC}, \vec{BD}$ diagonal vectors
$O$ arbitrary origin (often the centre or an external point)

Angles and vertices from position vectors

Angle at vertex from position vectors

\cos C = \dfrac{(\vec{a} - \vec{c})\cdot(\vec{b} - \vec{c})}{|\vec{a} - \vec{c}|\,|\vec{b} - \vec{c}|}

$\vec{a}, \vec{b}, \vec{c}$ position vectors of the vertices
$C$ angle of the triangle at vertex $C$

Distance and perpendicularity identities in quadrilaterals

Parallelogram law and distance expansion

|\vec{p}+\vec{q}|^2 + |\vec{p}-\vec{q}|^2 = 2(|\vec{p}|^2 + |\vec{q}|^2)

$\vec{p}, \vec{q}$ any two vectors (often diagonals or sides)
$|\vec{p}+\vec{q}|, |\vec{p}-\vec{q}|$ diagonal magnitudes when $\vec{p}, \vec{q}$ are sides

Common traps

Direction matters in the loop — $\vec{BA} = -\vec{AB}$

The identity

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

requires the sides to be traversed in one consistent direction around the triangle. If a statement reads

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

, it's wrong — that's saying

\vec{AC}

instead of

\vec{CA}

, which reverses one side.

$\vec{AG}$ is NOT $\vec{g}/3$ — it is $\vec{g} - \vec{a}$

A common factor-of-3 distractor.

\vec{AG}

is the displacement from

A

G

, so it equals

\vec{g} - \vec{a}

. After algebra

\vec{AG} = \dfrac{(\vec{b}-\vec{a}) + (\vec{c}-\vec{a})}{3}

— two-thirds of the median from

A

Vertex order $ABCD$ matters

If the vertices are listed in a non-cyclic order, the figure is NOT a parallelogram in the standard sense.

\vec{AB} = \vec{DC}

(not

\vec{CD}

) — the equal sides are the ones going in the SAME direction around the figure.

The fourth vertex of a parallelogram: $D = A + C - B$

A, B, C

are three consecutive vertices,

\vec{AD} = \vec{BC}

forces

\vec{d} = \vec{a} + \vec{c} - \vec{b}

. A factor-of-2 distractor here often offers

\vec{a} + \vec{c} - 2\vec{b}

; reject it.

Direction of side vectors changes the angle

The angle at

C

is between

\vec{CA}

and

\vec{CB}

— NOT between

\vec{AC}

and

\vec{BC}

. Reversing both flips the dot-product sign and gives

\pi - C

instead of

C

. Always start from the named vertex outwards.

Fourth-vertex problems: $D = A + C - B$ , not the midpoint

If a parallelogram

ABCD

lists

A, B, C

as consecutive vertices, the fourth vertex

D

satisfies

\vec{AD} = \vec{BC}

, giving

\vec{d} = \vec{a} + \vec{c} - \vec{b}

. Mid-segment formulas applied here are the typical wrong-option trap.

Direction of comparison matters for parallelism

Two vectors are parallel if one is a scalar multiple of the other — the scalar can be negative.

\vec{PQ}

and

\vec{RS}

point in opposite directions yet are still parallel. But if a PYQ asks whether

\vec{PQ}

and

\vec{SR}

are equal (not just parallel), the sign matters.

Expand squared distances algebraically — don't reach for coordinates first

An identity like

PQ^2 + 2QS^2 - 2PR^2 = ?

is much faster to verify by expanding each

|\cdots|^2

as a dot product and collecting terms in

\vec{p}\cdot\vec{q}

\vec{p}\cdot\vec{r}

, etc., than by plugging in coordinates and computing each squared distance separately.

Vectors formulas

Foundations: Vectors, Operations, and Position

Position vectors and displacement vectors

Addition of vectors (triangle, parallelogram, polygon laws)

Component form: the i, j, k unit vectors

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

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

Section Formula — Internal and External Division

Common traps

Magnitude, Components, Projection, Direction Cosines

Magnitude of a vector and distance between two points

Direction Cosines

Scalar projection of one vector on another

Unit vectors and direction-given construction

Common traps

Dot Product and Angle

Dot product — components form and work done

Perpendicularity Test

Angle between two vectors via the dot-product formula

Solving for an angle from a perpendicularity / magnitude constraint

Unit vectors, orthogonal triples, and decomposition

Common traps

Cross Product and Triple Product

Cross product — algebra and properties

Cross-product magnitude, area, and the Lagrange identity

Unit vector perpendicular to two given vectors

Moment of a force (torque)

Scalar triple product and coplanarity

STP cyclic property and derived linear-combo identities

Vector triple product (BAC-CAB rule)

Common traps

Vector Geometry — Triangles, Parallelograms, Quadrilaterals

Triangle closed-loop and centroid formula

Parallelogram properties and diagonal relations

Angles and vertices from position vectors

Distance and perpendicularity identities in quadrilaterals

Common traps

More NDA Mathematics formula sheets