MHT-CET Maths · Formula sheet

Vectors formulas

59 formulas and 123 common traps for MHT-CET Maths Vectors, grouped by subtopic.

Magnitude, Components, and Unit Vectors

Vectors and component form

Component form

\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}

$a_1, a_2, a_3$ components along $\hat{i}, \hat{j}, \hat{k}$
$\hat{i}, \hat{j}, \hat{k}$ standard perpendicular unit vectors

Magnitude of a vector and distance between two points

Magnitude and distance

|\vec{a}| = \sqrt{a_1^2 + a_2^2 + a_3^2} \qquad AB = |\vec{b} - \vec{a}|

$a_1, a_2, a_3$ components of $\vec{a}$
$\vec{a}, \vec{b}$ position vectors of $A$ and $B$

Unit vector along a given direction

Unit vector

\hat{a} = \dfrac{\vec{a}}{|\vec{a}|} \qquad r\,\hat{a} = \dfrac{r}{|\vec{a}|}\,\vec{a}

$\hat{a}$ unit vector along $\vec{a}$
$|\vec{a}|$ magnitude of $\vec{a}$
$r$ desired magnitude of the scaled vector

Magnitude of a sum from angles or perpendicularity

Magnitude of a sum (squared)

|\vec{a} + \vec{b} + \vec{c}|^2 = \sum|\vec{a}|^2 + 2\sum(\vec{a}\cdot\vec{b}), \qquad \vec{a}\cdot\vec{b} = |\vec{a}||\vec{b}|\cos\theta

$\sum|\vec{a}|^2$ $|\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2$
$\sum(\vec{a}\cdot\vec{b})$ all three pairwise dot products
$\theta$ angle between the pair of vectors in a dot product

Magnitude of a sum with projection-equality and a perpendicular pair

Grouped expansion with a perpendicular pair

|\vec{a} + \vec{b} - \vec{c}|^2 = |\vec{a}|^2 + |\vec{b} - \vec{c}|^2 + 2\,\vec{a}\cdot(\vec{b} - \vec{c})

$\vec{a}\cdot(\vec{b}-\vec{c}) = 0$ from equal projections of $\vec{b}, \vec{c}$ along $\vec{a}$
$|\vec{b}-\vec{c}|^2 = |\vec{b}|^2 + |\vec{c}|^2$ from $\vec{b}\perp\vec{c}$

Unit vector parallel to a parallelogram's diagonal

Unit vector along a diagonal

\hat{d} = \dfrac{\vec{a} + \vec{b}}{|\vec{a} + \vec{b}|}

$\vec{a}, \vec{b}$ adjacent sides from the shared vertex
$\vec{a} + \vec{b}$ the diagonal through that vertex

Common traps

A missing axis means a zero component, not a 2-D vector

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

lives in 3-D with

a_2 = 0

. When you square components for a magnitude, the missing term contributes

0^2 = 0

— don't drop the slot or miscount the axes.

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

Reversing it gives

\overrightarrow{BA}

. The distance (magnitude) is the same either way, but the direction flips — and direction matters the moment the result feeds a dot product or an angle.

Magnitude needs every component squared

For

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

, the answer is

\sqrt{9 + 16} = 5

, not

3 + 4 = 7

. Add the SQUARES, then take ONE square root at the end.

Divide by the magnitude, don't subtract it

The unit vector is

\vec{a}/|\vec{a}|

— scale every component by the same

1/|\vec{a}|

. A common slip is normalising only one component or dividing by the wrong length.

Add the cross terms with the factor of 2

The expansion is

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

. Forgetting the

2

halves every cross term — a frequent wrong answer when the angle version has non-zero dot products.

Perpendicularity conditions cancel ALL cross terms at once

The three conditions

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

etc. don't say each individual dot product is zero — they say their SUM is zero. That's all you need: the whole

2\sum(\vec{a}\cdot\vec{b})

term drops, leaving just

\sqrt{\sum|\cdot|^2}

Take the square root at the very end

You compute

|\vec{a}+\vec{b}+\vec{c}|^2

first. For magnitudes

1, 8, 4

with zero cross terms it's

81

, and the answer is

\sqrt{81} = 9

— NOT

81

. The distractor that leaves the squared value un-rooted is the classic trap.

"Equal projections along $\vec{a}$ " means $(\vec{b}-\vec{c})\cdot\vec{a} = 0$ , not $\vec{b} = \vec{c}$

The projections being equal only forces

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

—

\vec{b}

and

\vec{c}

can still differ wildly. Use it to kill exactly the

\vec{a}

-cross term, nothing more.

Watch the sign on the vector you subtract

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

, the

\vec{c}

term is subtracted, but

|\vec{b}-\vec{c}|^2

still ADDS the squared lengths when

\vec{b}\perp\vec{c}

(the cross term

-2\vec{b}\cdot\vec{c}

is zero). The minus sign only matters through the dot product, which vanishes here.

Diagonal $\vec{a}+\vec{b}$ , not $\vec{a}-\vec{b}$

A parallelogram has two diagonals:

\vec{a} + \vec{b}

(through the shared vertex) and

\vec{a} - \vec{b}

(the other one). The question usually wants the SUM; reading it as the difference gives a different unit vector entirely.

Normalise the diagonal's own magnitude, not a stray $\sqrt{77}$

Compute

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

AFTER adding — for

3\hat{i}-6\hat{j}+2\hat{k}

that is

\sqrt{9+36+4} = 7

. Distractors often divide by an unrelated magnitude (like

\sqrt{77}

from

\sqrt{49 + 28}

); always re-square the summed components.

Vector Geometry — Section Formula, Triangle, and Parallelogram

Learn this subtopic in the notes

Section formula — internal, external, and midpoint

Section formula (internal / external / midpoint)

\vec{r} = \frac{m\vec{b} + n\vec{a}}{m + n} \qquad \vec{r} = \frac{m\vec{b} - n\vec{a}}{m - n} \qquad \vec{r} = \frac{\vec{a} + \vec{b}}{2}

$\vec{a}, \vec{b}$ position vectors of the endpoints $A, B$
$m : n$ ratio $AR : RB$ in which $R$ divides the segment
$\vec{r}$ position vector of the dividing point $R$

Centroid and median identities

Centroid and median vector

\vec{g} = \frac{\vec{a} + \vec{b} + \vec{c}}{3} \qquad \vec{AD} = \frac{\vec{AB} + \vec{AC}}{2} \qquad \vec{g}_{\text{tetra}} = \frac{\vec{a} + \vec{b} + \vec{c} + \vec{d}}{4}

$\vec{a}, \vec{b}, \vec{c}$ position vectors of the vertices
$\vec{g}$ position vector of the centroid
$D$ midpoint of $BC$ ; $\vec{AD}$ is the median from $A$

Finding the ratio, collinearity, and cevian intersection

Ratio recovery and external division

\vec{r} = \frac{m\vec{b} + n\vec{a}}{m + n} \;\Rightarrow\; \text{ratio } m:n \qquad \vec{r}_{\text{ext}} = \frac{m\vec{b} - n\vec{a}}{m - n}

$m : n$ the ratio recovered by comparing coefficients
$\vec{r}_{\text{ext}}$ external-division point — used for one branch of perpendicular-cevian problems

Incentre, orthocentre, and the angle bisector

Incentre and angle bisector

\vec{I} = \frac{a\vec{a} + b\vec{b} + c\vec{c}}{a + b + c} \qquad \text{bisector of }\angle AOB \parallel \frac{\vec{a}}{|\vec{a}|} + \frac{\vec{b}}{|\vec{b}|}

$a, b, c$ lengths of sides opposite $A, B, C$ : $a = |BC|$ , etc.
$\vec{I}$ position vector of the incentre
$\hat{a} + \hat{b}$ sum of unit vectors — the internal-bisector direction

Triangle and parallelogram applications

Right-angle test and parallelogram diagonals

\angle A = 90^\circ \iff \vec{AB}\cdot\vec{AC} = 0 \qquad \vec{PR} = \vec{PQ} + \vec{QR}, \;\; \vec{QS} = \vec{QP} + \vec{PS}

$\vec{AB}, \vec{AC}$ the two side-vectors leaving the right-angle vertex
$\vec{PR}, \vec{QS}$ diagonals of quadrilateral $PQRS$

Common traps

Internal adds, external subtracts

The only difference between the two formulas is the sign in the denominator (and numerator): internal uses

m + n

, external uses

m - n

. An external-division question with the internal formula (or vice versa) is the single most common slip — read whether

R

lies between the points or beyond them.

Which point gets the weight $m$ ?

\dfrac{m\vec{b} + n\vec{a}}{m+n}

, the FAR endpoint

\vec{b}

carries

m

and the NEAR endpoint

\vec{a}

carries

n

, where the ratio is

AR:RB = m:n

. Swapping the weights places

R

at the mirror point. When unsure, sanity-check: a

2:1

point should sit closer to

B

Median length $\neq$ half the side it bisects

The median through

A

\dfrac{\vec{AB} + \vec{AC}}{2}

, NOT

\dfrac{1}{2}\vec{BC}

. Take half of the SUM of the two adjacent side-vectors, then take its magnitude — don't halve the opposite side's length.

Centroid uses position vectors, not side vectors

\vec{g} = \dfrac{\vec{a} + \vec{b} + \vec{c}}{3}

needs the position vectors of the vertices. If a problem hands you only

\vec{AB}

and

\vec{AC}

, the centroid relative to

A

\dfrac{\vec{AB} + \vec{AC}}{3}

— a different (and frequently tested) form.

Internal and external points use the SAME magnitude of ratio

When

R

and

S

divide

PQ

internally and externally in the same ratio

2:3

, use

\vec{OR} = \dfrac{2\vec{q} + 3\vec{p}}{5}

and

\vec{OS} = \dfrac{2\vec{q} - 3\vec{p}}{-1} = 3\vec{p} - 2\vec{q}

. Forgetting the sign flip in the external denominator is the classic error in OR

\perp

OS problems.

Cevian-intersection ratio is asked along ONE cevian

AP

and

BQ

meet at

G

, the question wants the ratio

AG:GP

(along

AP

) — not

BG:GQ

. Pin which cevian you are reporting the ratio on, and don't invert it (a

5:7

answer reads as

7:5

if you measure from the wrong end).

Incentre weights are OPPOSITE side lengths

Vertex

A

is weighted by

a = |BC|

— the side facing

A

, not

|AB|

|AC|

. Mis-pairing the weight with an adjacent side is the standard incentre trap; label the sides

a, b, c

opposite

A, B, C

first.

Bisector uses unit vectors — sum, not difference

The INTERNAL bisector is along

\hat{a} + \hat{b}

(sum of unit vectors); the EXTERNAL bisector is along

\hat{a} - \hat{b}

. Forgetting to normalise (using

\vec{a} + \vec{b}

instead of

\hat{a} + \hat{b}

) gives the wrong direction unless

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

Orthocentre $\neq$ centroid $\neq$ circumcentre

A perpendicularity condition like

(\vec{a} - \vec{d})\cdot(\vec{b} - \vec{c}) = 0

says

DA \perp BC

— an ALTITUDE, so

D

is the orthocentre. Equal distances to the vertices would mean circumcentre; equal angle-bisector weighting means incentre. Read which condition is given.

Equal diagonals $\to$ rectangle, perpendicular diagonals $\to$ rhombus

Don't mix the two tests. A parallelogram whose diagonals are EQUAL in length is a rectangle; one whose diagonals are PERPENDICULAR is a rhombus. A figure that is a parallelogram but neither (diagonals unequal AND not perpendicular) is the 'neither rhombus nor rectangle' answer.

Right-angle test needs side-vectors FROM the vertex

For a right angle at

A

, dot

\vec{AB}

with

\vec{AC}

(both leaving

A

) — not

\vec{AB}

with

\vec{BC}

. Using the wrong pair tests perpendicularity at the wrong vertex and gives a spurious value.

Linear Combinations, Collinearity, and Coplanarity

Learn this subtopic in the notes

Linear combination of vectors

Linear combination

\vec{r} = m\vec{a} + n\vec{b} + p\vec{c} \qquad r_i = ma_i + nb_i + pc_i

$m, n, p$ real scalar coefficients (any sign, including zero)
$\vec{r}$ the combined vector — built componentwise

Collinear vectors and collinear points (one scalar)

Collinearity (one scalar)

\vec{a} = k\vec{b} \qquad \dfrac{a_1}{b_1} = \dfrac{a_2}{b_2} = \dfrac{a_3}{b_3}

$k$ the single scalar; $k > 0$ same direction, $k < 0$ opposite

Coplanar vectors (two scalars)

Coplanarity (two scalars)

\vec{a} = m\vec{b} + n\vec{c} \qquad [\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

$m, n$ the two scalars — one per spanning vector
$[\vec{a}\ \vec{b}\ \vec{c}]$ scalar triple product; zero ⟺ coplanar

Forming a combination, then normalising / measuring it

Scale a combination to a target magnitude

M\,\hat{r} = \dfrac{M}{|\vec{r}|}\,\vec{r} \qquad \hat{r} = \dfrac{\vec{r}}{|\vec{r}|}

$\vec{r}$ the linear combination, built first
$M$ the required magnitude of the parallel vector

Collinearity of three points (and who lies between)

Three-point collinearity

\vec{QR} = k\,\vec{PQ} \quad\text{where}\quad \vec{PQ} = \vec{q} - \vec{p},\ \ \vec{QR} = \vec{r} - \vec{q}

$k$ the scalar; its sign/size fixes the order of the points

Express a vector as a combination of two others

Two-equation linear system

\begin{cases} a_1 = mb_1 + nc_1 \\ a_2 = mb_2 + nc_2 \end{cases}\ \Rightarrow\ m,\ n

$m, n$ coefficients to solve for; the 3rd component is the check

Linear dependence vs independence (the determinant test)

Dependence ⟺ vanishing determinant

x\vec{a} + y\vec{b} + z\vec{c} = \vec{0}\ (\text{not all }0) \iff \begin{vmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{vmatrix} = 0

$x, y, z$ scalars; a non-trivial solution means dependent

Chained-collinearity systems (\"no two collinear\")

Chained collinearity

\vec{a} + 2\vec{b} = t\vec{c}, \quad \vec{b} + 3\vec{c} = \lambda\vec{a} \;\Rightarrow\; t = -6,\ \lambda = -\tfrac{1}{2}

$t, \lambda$ one scalar per collinearity fact; matched via coefficient comparison

A vector lying in the plane of two others

Coplanar form + a second condition

\vec{v} = m\vec{b} + n\vec{c} \quad\text{with}\quad \vec{v}\cdot\vec{q} = 0 \ \text{ or }\ \dfrac{\vec{v}\cdot\vec{c}}{|\vec{c}|} = k

$m, n$ two scalars from coplanarity
second conditionbisector / perpendicular / projection / magnitude — fixes the scalars

Components of a vector against a transformed basis

Match coefficients of a basis

\begin{cases} -x + y - z = 4 \\ \ \ x - y - z = 3 \\ \ \ x + y + z = 5 \end{cases}

rowscoefficients of $\vec{p}, \vec{q}, \vec{r}$ equated to the given components

Common traps

A linear combination is built componentwise — three sums, not one

When you form

m\vec{a} + n\vec{b} + p\vec{c}

in 3-D you are doing three independent scalar sums: one for

\hat{i}

, one for

\hat{j}

, one for

\hat{k}

. Combine the wrong components and the answer is silently wrong — keep the three columns separated.

Collinear needs ONE scalar; coplanar needs TWO — keep the count straight

Whenever a question says \"collinear / parallel,\" introduce ONE unknown scalar (

\vec{u} = t\vec{v}

). If it says \"coplanar / lies in the plane of,\" introduce TWO (

\vec{u} = m\vec{v} + n\vec{w}

). Mixing up the count is the single biggest source of wrong setups in this subtopic.

Parallel VECTORS vs collinear POINTS

Vectors are parallel when they share a direction; points are collinear when they share a LINE. Test points by displacements:

A, B, C

collinear iff

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

(they share point

A

), not merely \"some pair of vectors is parallel.\"

Coplanar / dependent ⇒ TWO scalars and a vanishing determinant

For coplanarity you have two equivalent tests: express the vector as

m\vec{b} + n\vec{c}

(good when you need

m, n

), or set the

3\times 3

determinant of the components to zero (good for a yes/no or for finding an unknown component). Pick the test that matches what the question asks for.

Build the combination BEFORE you normalise — don't normalise the parts

To find a unit vector along

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

you must combine first, get one vector, THEN divide by its magnitude. Normalising

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

separately and combining the unit vectors gives a different (wrong) direction.

\"Parallel of magnitude $M$ \" has TWO answers — $\pm M\hat{r}$

A vector parallel to

\vec{r}

can point the same way OR the opposite way, so both

+M\hat{r}

and

-M\hat{r}

qualify. MCQs usually list only one; if your computed direction isn't an option, check its negative before assuming an error.

Use displacements that SHARE a point

To conclude collinearity from

\vec{QR} = k\vec{PQ}

the two displacements must share a common point (here

Q

). Two parallel displacements that don't share a point only say the segments are parallel — not that all the points lie on ONE line.

Solve from two equations, but ALWAYS verify with the third

Two component-equations pin down

m, n

— but in 3-D there's a third equation. If it doesn't hold, no valid

m, n

exist (the vectors aren't coplanar). Skipping the check can hand you scalars that don't actually reproduce

\vec{a}

Linearly dependent = coplanar = zero determinant — three names, one idea

In 3-D these are the SAME condition. A question may phrase it as \"linearly dependent,\" \"coplanar,\" or \"scalar triple product is zero\" — all three send you to the same

3\times 3

determinant set to zero.

A second condition (like $|\vec{c}| = \sqrt{3}$ ) is part of the same problem

Dependence often fixes only ONE unknown (e.g.

\beta = 1

). A magnitude condition supplies the OTHER (e.g.

1 + \alpha^2 + 1 = 3 \Rightarrow \alpha = \pm 1

). Use both givens — don't stop after the determinant.

Match coefficients only because the vectors are independent

\"No two collinear\" (in fact, all three independent) is the hypothesis that LETS you equate coefficients of

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

on both sides. Without independence you couldn't conclude that equal vector-expressions force equal coefficients.

Read the target carefully: $\vec{a}+2\vec{b}$ vs $\vec{a}+2\vec{b}+6\vec{c}$

The same setup answers two different MCQ targets. If

\vec{a}+2\vec{b} = -6\vec{c}

, then the value of

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

-6\vec{c}

, but the value of

\vec{a}+2\vec{b}+6\vec{c}

\vec{0}

. Answer the one actually asked.

Angle bisector uses UNIT vectors, not the raw vectors

The internal bisector of

\vec{b}

and

\vec{c}

is along

\hat{b} + \hat{c} = \dfrac{\vec{b}}{|\vec{b}|} + \dfrac{\vec{c}}{|\vec{c}|}

. Adding

\vec{b} + \vec{c}

directly only bisects when

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

— otherwise you get the wrong direction.

Coplanar form first, condition second

Always WRITE the coplanar form

m\vec{b} + n\vec{c}

before imposing the perpendicular / projection / magnitude condition. Trying to satisfy the condition without restricting to the plane gives a vector that doesn't actually lie where the question requires.

Equating components needs an INDEPENDENT basis

Matching coefficients of

\vec{p}, \vec{q}, \vec{r}

on both sides is valid ONLY because they are non-coplanar (independent), giving a unique representation. If they were coplanar the components wouldn't be unique and the method collapses.

Answer the asked combination, not the raw $x, y, z$

Many of these end by asking for a combination like

2x + y + z

, not the individual scalars. Solve the system fully, then plug into the requested expression — a fraction in

y

and

z

often cancels cleanly there.

Dot Product, Angle, and Perpendicularity

Learn this subtopic in the notes

Dot product — the two faces of a·b

Dot product — geometric and component forms

\vec a\cdot\vec b = |\vec a|\,|\vec b|\cos\theta = a_1 b_1 + a_2 b_2 + a_3 b_3

$\theta$ angle between the vectors, in $[0,\pi]$
$a_i, b_i$ components along $\hat i, \hat j, \hat k$
$\vec a\cdot\vec a$ equals $|\vec a|^2$

Magnitude of a combination via the dot product

Magnitude of a linear combination

|p\vec a + q\vec b|^2 = p^2|\vec a|^2 + 2pq\,(\vec a\cdot\vec b) + q^2|\vec b|^2

$p, q$ scalar coefficients
$\vec a\cdot\vec b$ the only cross term — vanishes if $\vec a\perp\vec b$

Perpendicularity test and solving for a parameter

Perpendicularity and the perp-parameter

\vec a\perp\vec b \iff \vec a\cdot\vec b = 0 \qquad (\vec a + \lambda\vec b)\perp\vec c \Rightarrow \lambda = -\frac{\vec a\cdot\vec c}{\vec b\cdot\vec c}

$\lambda$ the unknown scalar to solve for
$\vec a\cdot\vec c,\ \vec b\cdot\vec c$ two scalar dot products, computed once each

Disguised perpendicularity — equal-diagonal and Pythagoras forms

Equivalent perpendicularity statements

\vec a\perp\vec b \iff |\vec a + \vec b| = |\vec a - \vec b| \iff |\vec a + \vec b|^2 = |\vec a|^2 + |\vec b|^2

$|\vec a \pm \vec b|$ lengths of the two parallelogram diagonals

Angle between two vectors via cosθ

Angle from the dot product

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

$\theta$ angle between the vectors, in $[0,\pi]$
$|\vec a|, |\vec b|$ magnitudes (always positive)

Angle from a unit-vector perpendicularity constraint

Expansion of a perpendicular constraint (unit vectors)

(\alpha\vec a + \beta\vec b)\cdot(\gamma\vec a + \delta\vec b) = \alpha\gamma + (\alpha\delta + \beta\gamma)\cos\theta + \beta\delta = 0

$\alpha,\beta,\gamma,\delta$ given coefficients in the two combinations
$\cos\theta$ equals $\vec a\cdot\vec b$ — the unknown to isolate

Scalar and vector projection

Scalar and vector projection of a on b

\text{scalar} = \frac{\vec a\cdot\vec b}{|\vec b|}, \qquad \text{vector} = \frac{\vec a\cdot\vec b}{|\vec b|^2}\,\vec b

$\vec a\cdot\vec b$ dot product (carries the sign)
$|\vec b|$ divide once for scalar; squared for vector

Projection onto the normal of a plane

Projection on the plane normal

\vec n = \vec p \times \vec q, \qquad \text{proj} = \frac{|\vec v\cdot\vec n|}{|\vec n|}

$\vec n = \vec p\times\vec q$ normal to the plane of $\vec p, \vec q$
$\vec v\cdot\vec n$ dot of the target vector with the normal

Mutually orthogonal vectors — solving a small system

Mutual-orthogonality system

\vec a\cdot\vec c = 0 \;\text{ and }\; \vec b\cdot\vec c = 0 \;\Rightarrow\; \text{solve for the unknown components}

$\vec c$ the vector with unknown components
two equationslinear $2\times 2$ system from the two dot products

Unit vector along a combination, and scalar-product conditions

Unit vector of a combination

\hat v = \frac{\vec v}{|\vec v|}, \qquad \frac{\vec w\cdot\vec s}{|\vec s|} = k

$\hat v$ unit vector (magnitude 1) along $\vec v$
$k$ the given scalar-product value to solve against

Identities and bounds — sum of squared differences

Sum-of-squared-differences identity and bound

\sum |\vec a - \vec b|^2 = 3\!\left(|\vec a|^2 + |\vec b|^2 + |\vec c|^2\right) - |\vec a + \vec b + \vec c|^2 \;\le\; 3\!\left(|\vec a|^2 + |\vec b|^2 + |\vec c|^2\right)

$|\vec a + \vec b + \vec c|^2$ the non-negative term that is subtracted; zero at the maximum

Dot products entangled with cross-product constraints

Magnitude expansion to extract a dot product

|\vec c - \vec a|^2 = |\vec c|^2 + |\vec a|^2 - 2\,\vec a\cdot\vec c \;\Rightarrow\; \vec a\cdot\vec c = \frac{|\vec c|^2 + |\vec a|^2 - |\vec c - \vec a|^2}{2}

$|\vec c|$ extracted from the cross-product / angle condition first
$\vec a\cdot\vec c$ the unknown dot product, isolated from the expansion

Obtuse angle for all x — a quadratic-inequality parameter

Negative-for-all-x conditions

Ax^2 + Bx + C < 0 \;\forall x \iff A < 0 \;\text{ and }\; B^2 - 4AC < 0

$A < 0$ downward-opening parabola
$B^2 - 4AC < 0$ no real roots — stays below the axis everywhere

Moving point a·cos t + b·sin t — farthest from the origin

Farthest-point magnitude and direction

|\overrightarrow{OP}|^2 = 1 + (\hat a\cdot\hat b)\sin 2t, \quad M = \sqrt{1 + \hat a\cdot\hat b}, \quad \hat u = \frac{\hat a + \hat b}{|\hat a + \hat b|}

$\hat a\cdot\hat b$ cosine of the (acute) angle between the unit vectors
$M$ maximum distance, at $t = \pi/4$

Reading perpendicularity geometrically — the orthocentre

Dot-zero on differences = perpendicular segments

(\vec a - \vec d)\cdot(\vec b - \vec c) = 0 \iff \overrightarrow{DA}\perp\overrightarrow{BC}

$\overrightarrow{DA}$ the segment $\vec a - \vec d$
altitudestwo perpendicularity conditions → their intersection is the orthocentre

Common traps

Dot product is a scalar — never a vector

An MCQ option that returns

\hat i, \hat j, \hat k

components for

\vec a\cdot\vec b

is wrong on type grounds alone. The dot product is always a single number; the cross product is the one that gives a vector.

Match components in the SAME direction only

\vec a\cdot\vec b

multiplies

a_1b_1 + a_2b_2 + a_3b_3

— the

\hat i

of one with the

\hat i

of the other. Multiplying

a_1 b_2

(an

\hat i

with a

\hat j

) is the most common slip, because those mixed products are exactly the zero terms.

Scale FIRST, then subtract — watch the double minus

|\vec a - 2\vec b|

, the

\hat j

-component becomes

a_2 - 2b_2

. If

b_2

is itself negative,

-2b_2

is positive — e.g.

a_2 = -2, b_2 = -4

gives

-2 - 2(-4) = 6

, not

-10

. The sign trap lives in the doubled, negative component.

Magnitude is never negative

|\vec v| = \sqrt{\,\cdot\,}

takes the positive root. If a derivation yields a negative number under the root or a negative final magnitude, the arithmetic is wrong — recompute the components.

A missing component is ZERO, not absent

When

\vec c = 3\hat i + \hat j

, its

\hat k

-component is

0

. In the dot product that term contributes

0

— but you must still write the slot so the

\hat i

and

\hat j

terms line up with the right components of the other vector.

Solve for $\lambda$ cleanly: $\lambda = -(\vec a\cdot\vec c)/(\vec b\cdot\vec c)$

After expanding, the equation is always linear in

\lambda

(\vec a\cdot\vec c) + \lambda(\vec b\cdot\vec c) = 0

. Compute the two dot products as numbers first, then divide. Sign errors creep in when people expand all the components at once instead.

Order matters: $\vec a + \lambda\vec b$ vs $\vec b + \lambda\vec a$

If the question reads

\vec b + \lambda\vec a

perpendicular to

\vec c

, then

\lambda = -(\vec b\cdot\vec c)/(\vec a\cdot\vec c)

— the roles flip. Read which vector carries the

\lambda

before substituting.

$|\vec a + \vec b| = |\vec a - \vec b|$ is NOT $\vec a = \vec b$

It is tempting to read equal magnitudes as equal vectors. Squaring shows it collapses to

\vec a\cdot\vec b = 0

— a perpendicularity condition, not an equality. Geometrically, the diagonals of a parallelogram are equal iff it is a rectangle.

Pythagoras only works on the PLUS combination for a right angle

|\vec a + \vec b|^2 = |\vec a|^2 + |\vec b|^2

signals perpendicularity. Don't confuse it with

|\vec a + \vec b|^2 = |\vec a|^2 + 2\vec a\cdot\vec b + |\vec b|^2

, which is the general expansion that holds for any angle.

Build the combinations BEFORE taking the angle

For the angle between

3\vec a + 5\vec b

and

5\vec a + 3\vec b

, you must first compute each combination's components, then dot and divide. You cannot shortcut by 'mixing' the angle between

\vec a

and

\vec b

directly.

Leave the answer as $\cos^{-1}(\cdot)$ when it is not a standard angle

\cos\theta = \tfrac{13}{19}

, the angle is

\cos^{-1}\tfrac{13}{19}

— not a round number. The correct option is the inverse-cosine form; don't force it into

\pi/3

\pi/4

Coefficient swap flips the angle: check WHICH combination

(\vec a + 2\vec b)\cdot(5\vec a - 4\vec b) = 0

gives

\cos\theta = +\tfrac{1}{2}

(so

\tfrac{\pi}{3}

), but

(5\vec a + 4\vec b)\cdot(\vec a - 2\vec b) = 0

gives

\cos\theta = -\tfrac{1}{2}

(so

\tfrac{2\pi}{3}

). The sign of the cross-term coefficient decides acute vs obtuse — read the coefficients exactly.

Unit vectors mean $|\vec a|^2 = 1$ , not $0$

The self-dot terms

|\vec a|^2

and

|\vec b|^2

become

1

, not

0

. Dropping them (as if

\vec a

were the zero vector) loses the constant terms and gives a wrong

\cos\theta

Don't drop a cross term — there are TWO middle products

(\alpha\vec a + \beta\vec b)\cdot(\gamma\vec a + \delta\vec b)

has FOUR products; the two middle ones (

\alpha\delta

and

\beta\gamma

) both contribute to the

\vec a\cdot\vec b

coefficient. A factor-of-2 wrong answer usually means one was dropped.

Divide by $|\vec b|$ — projection is NOT just the dot product

The most common projection error is reporting

\vec a\cdot\vec b

and forgetting to divide by the magnitude of the vector you project ONTO. Scalar projection

= (\vec a\cdot\vec b)/|\vec b|

; the magnitude in the denominator is non-negotiable.

Scalar projection can be negative; magnitude of projection is its absolute value

If a question asks for the 'magnitude of the projection', take

|(\vec a\cdot\vec b)/|\vec b||

. A negative scalar projection (e.g.

-\tfrac{1}{7}

) just means the shadow points opposite to

\vec b

— its magnitude is

\tfrac{1}{7}

Vector projection divides by $|\vec b|^2$ , then multiplies by $\vec b$

Don't confuse the two: scalar uses

|\vec b|^1

and gives a number; vector uses

|\vec b|^2

and multiplies by

\vec b

to give a vector. Writing

\tfrac{\vec a\cdot\vec b}{|\vec b|}\,\vec b

is dimensionally wrong.

'Perpendicular to the plane' means CROSS product, not dot

The phrase 'vector perpendicular to the plane containing

\vec p

and

\vec q

' is asking for

\vec p \times \vec q

. Trying to project onto

\vec p

\vec q

directly answers a different question.

Project onto the NORMAL, then divide by $|\vec n|$

After computing

\vec n

, it is still an ordinary scalar projection:

(\vec v\cdot\vec n)/|\vec n|

. Forgetting the

|\vec n|

denominator (because

\vec n

was 'just computed') is the usual slip.

Two perpendicularity conditions → two equations, not one

'Perpendicular to both' is two separate dot-product-zero equations. Using only

\vec a\cdot\vec c = 0

leaves the components under-determined; you need

\vec b\cdot\vec c = 0

as well to pin down both unknowns.

Watch sign and ordering in the answer pair

Options frequently include

(-3,2)

(2,-3)

(3,-2)

(-2,3)

— all permutations/sign-flips of the right values. Solve the system carefully and match the unknowns to the right positions (which is

m

p

, which is

n

q

Divide the WHOLE vector by its magnitude

A unit vector keeps the direction and rescales the length to 1: every component is divided by the same

|\vec v|

. Dividing only one component, or forgetting the

\hat j

-component when it is zero, breaks the unit-length property.

Compute the combination's components BEFORE the magnitude

For

3\vec a + \vec b - 2\vec c

, do the scalar multiplications and the add/subtract per component first, then take

\sqrt{\sum(\cdot)^2}

. Taking magnitudes of

\vec a, \vec b, \vec c

individually and combining them is wrong.

'Does not exceed' = maximum, not the typical value

The phrase 'does not exceed' asks for the UPPER BOUND. Expand the identity, then set the subtracted

|\vec a + \vec b + \vec c|^2 = 0

to reach the maximum

3(|\vec a|^2 + |\vec b|^2 + |\vec c|^2)

. The distractors are

83

(the magnitude sum) and

166

(

2\times

, what you get by dropping the dot products instead of bounding them).

Each magnitude-squared appears TWICE in the expanded sum

When you add the three squared differences,

|\vec a|^2

shows up in both

|\vec a - \vec b|^2

and

|\vec c - \vec a|^2

. Forgetting that doubling gives

1\times

instead of

2\times

the magnitude sum.

Extract $|\vec c|$ FIRST from the cross/angle condition

The magnitude expansion needs

|\vec c|

. Get it from the given

|(\vec a\times\vec b)\times\vec c| = |\vec a\times\vec b||\vec c|\sin\phi

before plugging into

|\vec c - \vec a|^2

. Skipping this leaves an unknown that blocks the solve.

$\vec b\times\vec c = \vec b\times\vec a$ is NOT $\vec c = \vec a$

Cross products being equal only forces

\vec c - \vec a

to be PARALLEL to

\vec b

(i.e.

\vec c = \vec a + \lambda\vec b

), not equal vectors. The free

\lambda

is then fixed by a separate dot condition.

Obtuse-for-ALL-x needs BOTH conditions

A negative dot product at one

x

is not enough. 'For all

x

' forces the quadratic below the axis everywhere: leading coefficient negative AND discriminant negative. Using only one gives too wide an interval.

Obtuse is strict: exclude the perpendicular boundary

\vec a\cdot\vec b = 0

is a right angle, not obtuse. The discriminant must be strictly

< 0

(not

\le 0

) so the dot product never reaches zero for any

x

Maximise $|\overrightarrow{OP}|^2$ , then the direction is $\hat a + \hat b$ (plus, not minus)

t = \pi/4

, both

\cos t

and

\sin t

are positive and equal, so

\overrightarrow{OP} \propto \hat a + \hat b

. The 'minus' direction

\hat a - \hat b

is the MINIMUM (nearest), a common distractor.

The cross term is $(\hat a\cdot\hat b)\sin 2t$ , coefficient 1 not 2

After using

2\sin t\cos t = \sin 2t

, the magnitude-squared is

1 + (\hat a\cdot\hat b)\sin 2t

, so

M = (1 + \hat a\cdot\hat b)^{1/2}

. A distractor uses

(1 + 2\,\hat a\cdot\hat b)^{1/2}

— that keeps the stray factor of 2.

Altitudes → orthocentre, not circumcentre

Perpendicularity of

\overrightarrow{DA}

\overrightarrow{BC}

defines an ALTITUDE, and altitudes meet at the orthocentre. The circumcentre comes from equal DISTANCES (

|\vec d - \vec a| = |\vec d - \vec b|

); don't swap the two centres.

Translate each dot-zero into the correct perpendicular pair

(\vec a - \vec d)

\overrightarrow{DA}

and

(\vec b - \vec c)

\overrightarrow{CB}

. Get the segment endpoints right before naming the altitude, or you may attach D to the wrong vertex's altitude.

Cross Product, Angle, and Area

Learn this subtopic in the notes

The cross product — definition and determinant form

Cross product as a determinant

\vec{a}\times\vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix} = |\vec{a}||\vec{b}|\sin\theta\,\hat{n}

Top rowthe unit vectors $\hat{i}, \hat{j}, \hat{k}$
$\theta$ angle between $\vec{a}$ and $\vec{b}$ , in $[0,\pi]$
$\hat{n}$ unit perpendicular to both, by the right-hand rule

Magnitude of the cross product, angle, and the Lagrange identity

Magnitude 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

$\sin\theta$ non-negative for $\theta \in [0,\pi]$ — the magnitude is a length
Lagrange identityfrom $\sin^2\theta + \cos^2\theta = 1$ times $|\vec{a}|^2|\vec{b}|^2$

Area of a triangle from two side vectors

Triangle area

\text{Area} = \tfrac{1}{2}\,|\overrightarrow{AB} \times \overrightarrow{AC}|

$\overrightarrow{AB}, \overrightarrow{AC}$ two edge vectors from the SAME vertex $A$
$\tfrac{1}{2}$ a triangle is half the parallelogram on the same two edges

Area of a parallelogram — from sides, diagonals, or a side and a diagonal

Parallelogram areas

\text{Area} = |\vec{a}\times\vec{b}| \qquad \text{Area}_{\text{diagonals}} = \tfrac{1}{2}|\vec{d_1}\times\vec{d_2}|

$\vec{a}, \vec{b}$ two adjacent SIDES
$\vec{d_1}, \vec{d_2}$ two DIAGONALS — note the extra $\tfrac{1}{2}$

Bilinear expansion and area-scaling identities

Bilinear cross-expansion

(p\vec{a} + q\vec{b}) \times (r\vec{a} + s\vec{b}) = (ps - qr)\,(\vec{a}\times\vec{b})

$ps - qr$ the determinant of the coefficient matrix $\begin{vmatrix} p & q \\ r & s \end{vmatrix}$
$\vec{a}\times\vec{a}, \vec{b}\times\vec{b}$ both $\vec{0}$ , so they drop out

Unit (and given-magnitude) vector perpendicular to two vectors

Unit / scaled perpendicular

\hat{n} = \pm\dfrac{\vec{a}\times\vec{b}}{|\vec{a}\times\vec{b}|} \qquad \vec{v}_{|m|} = \pm\, m\,\dfrac{\vec{a}\times\vec{b}}{|\vec{a}\times\vec{b}|}

$\vec{a}\times\vec{b}$ perpendicular to both $\vec{a}$ and $\vec{b}$
$\pm$ two opposite unit perpendiculars exist
$m$ required magnitude, scaling the unit perpendicular

Solving a vector equation: a cross condition plus a scalar condition

Cross plus scalar condition

\vec{a}\times\vec{r} = \vec{b}, \;\; \vec{a}\cdot\vec{r} = k \;\;\Longrightarrow\;\; \vec{r}\text{ is uniquely determined}

Cross conditionfixes $\vec{r}$ only up to a multiple of $\vec{a}$
Scalar conditionremoves the remaining freedom

Finding unknown components from a given cross product

Component matching

\vec{b}\times\vec{c} = (\text{given vector}) \;\Longrightarrow\; \text{equate each of }\hat{i},\hat{j},\hat{k}\text{ components}

Each componentone equation per axis — three in all
Extra scalar datumprojection / area / dot — closes the system

Parallelism, collinearity, and a vector along a×b

Parallel via zero cross product

\vec{a}\times\vec{b} = k(\vec{a}\times\vec{c}) \;\Longrightarrow\; \vec{a}\times(\vec{b} - k\vec{c}) = \vec{0} \;\Longrightarrow\; \vec{b} - k\vec{c} = \lambda\vec{a}

$\vec{a}\times(\cdots) = \vec{0}$ the bracket is parallel to $\vec{a}$
$\lambda$ scalar found by taking magnitudes

Vector triple product — the 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 of $\vec{b}$ and $\vec{c}$
Result planespanned by $\vec{b}$ and $\vec{c}$

Magnitude of a vector triple product with a given angle

Triple-product magnitude

|(\vec{a}\times\vec{b})\times\vec{c}| = |\vec{a}\times\vec{b}|\,|\vec{c}|\sin\phi

$\phi$ angle between the vector $\vec{a}\times\vec{b}$ and $\vec{c}$
$|\vec{a}\times\vec{b}|$ compute this first, as a single magnitude

Angle and cross-magnitude from a vector constraint

Perpendicularity of a cross product

\vec{a}\cdot(\vec{a}\times\vec{b}) = 0 \qquad \vec{b}\cdot(\vec{a}\times\vec{b}) = 0

$\vec{a}\times\vec{b}$ perpendicular to BOTH $\vec{a}$ and $\vec{b}$
Squaring a constraintturns a vector relation into scalar (dot) equations

Common traps

The cross product is a VECTOR, not a scalar

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

has three components — it is a vector. The dot product

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

is the scalar. Mixing them up (e.g. expecting a single number from a cross product) is the most common slip.

Watch the SIGN on the $\hat{j}$ -component

Expanding the determinant, the middle (

\hat{j}

) cofactor carries a leading minus:

-\big[a_1 b_3 - a_3 b_1\big]

. Forgetting this minus is the single most frequent computational error in this whole subtopic.

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

It means

\vec{a}

and

\vec{b}

are parallel — one is a scalar multiple of the other. Combined with

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

, it gives

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

for some scalar

\lambda

$\sin\theta$ is the same for $\theta$ and $180^\circ - \theta$

A magnitude of

48

150^\circ

and at

30^\circ

are identical because

\sin 150^\circ = \sin 30^\circ = \tfrac{1}{2}

. So

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

alone never fixes the angle — it cannot tell acute from obtuse. The dot product (with its sign) does.

Use Lagrange to skip finding the angle

When a problem gives two of

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

and asks for a third,

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

gives it directly — no

\theta

needed.

The HALF is on the triangle, not the parallelogram

Triangle area is

\tfrac{1}{2}|\vec{a}\times\vec{b}|

; parallelogram area is the full

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

. Dropping the

\tfrac{1}{2}

doubles your answer — a classic distractor option.

When area is GIVEN, expect TWO values of the unknown

Setting

\tfrac{1}{2}|\overrightarrow{AB}\times\overrightarrow{AC}| =

(given) leads to a quadratic in the unknown coordinate, so it usually has two roots. Pick the one that appears in the options — both may be geometrically valid.

Cross edges from the SAME vertex

Use

\overrightarrow{AB}

and

\overrightarrow{AC}

(both start at

A

) — not

\overrightarrow{AB}

and

\overrightarrow{BC}

. Mixing base points gives a wrong vector and a wrong area.

SIDES use no $\tfrac{1}{2}$ ; DIAGONALS do

Read the problem carefully:

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

when

\vec{a}, \vec{b}

are SIDES, but

\tfrac{1}{2}|\vec{d_1}\times\vec{d_2}|

when they are DIAGONALS. Treating diagonals as sides doubles the area.

A diagonal is the SUM of the two sides, not one of them

For a parallelogram on sides

\vec{a}, \vec{b}

the diagonals are

\vec{a}+\vec{b}

and

\vec{a}-\vec{b}

. So given a side and a diagonal, the missing side is the DIFFERENCE

\vec{c} - \vec{a}

— subtract, don't add.

Keep the cross-terms in order

When expanding

(p\vec{a}+q\vec{b})\times(r\vec{a}+s\vec{b})

you get

ps(\vec{a}\times\vec{b}) + qr(\vec{b}\times\vec{a})

. The second term flips sign to

-qr(\vec{a}\times\vec{b})

, leaving

(ps-qr)

— NOT

(ps+qr)

Area takes the ABSOLUTE value of the coefficient

ps - qr

is negative, the area is still

|ps-qr|\cdot|\vec{a}\times\vec{b}|

. A negative scaling factor doesn't make a negative area.

Both $\pm$ signs are valid answers

If the question doesn't pin down a direction,

+\hat{n}

and

-\hat{n}

are equally correct — accept whichever option is listed. Some PYQs add a constraint (e.g. 'with positive

z

') precisely to break this tie.

For perpendicular to $\vec{a}+\vec{b}$ and $\vec{a}-\vec{b}$ , use the shortcut

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

, which is parallel to

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

. So the unit perpendicular to those combinations is the same as the unit perpendicular to

\vec{a}

and

\vec{b}

— compute

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

directly and skip the longer cross product.

Confirm the magnitude is actually $1$

A vector pointing in the right direction is only a UNIT vector if its magnitude equals

1

. Always divide by

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

— don't select an un-normalised option.

One cross equation is NOT enough on its own

The three component equations from

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

are dependent (they sum to a consistency condition), so they leave one degree of freedom. You MUST use the accompanying scalar condition to get a unique

\vec{r}

$\vec{a}\times\vec{b} = \vec{a}\times\vec{c}$ does NOT mean $\vec{b} = \vec{c}$

Cancelling the cross product is illegal. The correct deduction is

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

, i.e.

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

— then a second condition fixes

\lambda

Pick the component that isolates the unknown

A given

\vec{b}\times\vec{c}

gives three equations, but only one or two contain the unknown scalar cleanly. Match the component where the unknown appears alone, rather than expanding all three.

Use the right datum for the right unknown

Typically the projection condition isolates one unknown (

\alpha

) and the cross-product condition the other (

\beta

). Solve them separately, then combine into whatever the question finally asks (

2\alpha+\beta

\alpha^2+\beta^2-\alpha\beta

, etc.).

Track the sign of the dot product

From

|\vec{b}\times\vec{c}|^2 = |\vec{b}|^2|\vec{c}|^2 - (\vec{b}\cdot\vec{c})^2

you get

(\vec{b}\cdot\vec{c})^2

, so

\vec{b}\cdot\vec{c} = \pm 1

. The sign changes

\lambda^2

and hence the magnitude of

\lambda

— match the answer key's intended sign.

Normal parallel means PLANES parallel, not perpendicular

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

makes the normals parallel, so the angle between the planes is

0

, NOT

\tfrac{\pi}{2}

. Perpendicular planes would need perpendicular normals.

Grouping matters — the two triple products differ

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

lies in the plane of

\vec{b}, \vec{c}

, but

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

lies in the plane of

\vec{a}, \vec{b}

. They are generally DIFFERENT vectors — the missing brackets are not optional.

BAC-CAB is for VECTOR triple products only

If the expression is

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

(a dot, then a cross) it is the SCALAR triple product — a single number — and BAC-CAB does not apply. Count the crosses and dots first.

When comparing a×(a×c) problems, isolate $\vec{a}\cdot\vec{c}$

For self-nested forms, write

\vec{a}\cdot\vec{c} = |\vec{a}||\vec{c}|\cos\theta

, expand by BAC-CAB, and take magnitudes — the equation collapses to one in

\cos\theta

(or

\sec^2\theta

Compute $|\vec{a}\times\vec{b}|$ FIRST, then treat it as one vector

Don't try to expand

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

by BAC-CAB when the angle is given — it's faster to find the single magnitude

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

and apply

|\vec{w}||\vec{c}|\sin\phi

Recover $|\vec{c}|$ from the side conditions before using $\sin\phi$

These problems usually hide

|\vec{c}|

inside a condition like

|\vec{c} - \vec{a}| = k

\vec{a}\cdot\vec{c} = |\vec{c}|

. Solve for

|\vec{c}|

first; only then multiply by

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

A cross product contributes ZERO to a dot with its own factor

\vec{b}\cdot(2(\vec{a}\times\vec{b}) - 3\vec{b})

, the

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

term is

0

— don't try to compute it, it vanishes by perpendicularity.

Square the constraint to get dot products

A constraint like

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

is a vector equation; take magnitudes (square it) or dot it with one of the vectors to convert it into scalar equations you can solve for

\cos\theta

Scalar Triple Product, Coplanarity, and Volume

Learn this subtopic in the notes

The scalar triple product — dot-cross and determinant form

Scalar triple product as a determinant

[\vec{a}\ \vec{b}\ \vec{c}] = \vec{a}\cdot(\vec{b}\times\vec{c}) = \begin{vmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{vmatrix}

Rowsthe components of $\vec{a}, \vec{b}, \vec{c}$ in order
$(\vec{a}\times\vec{b})\cdot\vec{c}$ equal value — dot and cross interchange
$[\hat{i}\ \hat{j}\ \hat{k}]$ $= 1$ , the unit box

Cyclic and sign properties of the scalar triple product

Cyclic and swap rules

[\vec{a}\ \vec{b}\ \vec{c}] = [\vec{b}\ \vec{c}\ \vec{a}] = [\vec{c}\ \vec{a}\ \vec{b}] = -[\vec{b}\ \vec{a}\ \vec{c}]

Cyclic rotation $\vec{a}\to\vec{b}\to\vec{c}\to\vec{a}$ : value unchanged
One swapvalue negated
Repeated rowvalue $= 0$

Computing the value of a scalar triple product

STP when one vector is perpendicular to the other two

[\vec{a}\ \vec{b}\ \vec{c}] = |\vec{a}|\,|\vec{b}\times\vec{c}| = |\vec{a}||\vec{b}||\vec{c}|\sin\theta

$\vec{a}\perp\vec{b}, \vec{a}\perp\vec{c}$ so $\vec{a}\,\|\,\vec{b}\times\vec{c}$
$\theta$ angle between $\vec{b}$ and $\vec{c}$
$\sin\theta$ from $|\vec{b}\times\vec{c}| = |\vec{b}||\vec{c}|\sin\theta$

Coplanarity of three vectors (and solving for a parameter)

Coplanarity criterion

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

$[\vec{a}\ \vec{b}\ \vec{c}] = 0$ zero box volume ⇒ all three lie in one plane
Parametersolve the determinant-equals-zero equation for it

Scalar triple product of linear combinations

Linear-combination identities

[\vec{a}+\vec{b}\ \ \vec{b}+\vec{c}\ \ \vec{c}+\vec{a}] = 2[\vec{a}\ \vec{b}\ \vec{c}] \qquad [m\vec{a}+\vec{b}\ \ m\vec{b}+\vec{c}\ \ m\vec{c}+\vec{a}] = (m^3+1)[\vec{a}\ \vec{b}\ \vec{c}]

Repeated-vector termsall vanish on expansion
Coefficient $= \det$ of the combination-coefficient matrix

Volume of a parallelepiped (and min/max problems)

Parallelepiped volume

V = |[\vec{a}\ \vec{b}\ \vec{c}]| = \left|\begin{vmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{vmatrix}\right|

$|\cdot|$ modulus — volume is always non-negative
$V'(m)=0$ stationary volume; $V''$ sign decides min/max

Volume of a tetrahedron

Tetrahedron volume

V = \tfrac{1}{6}\,\bigl|[\overrightarrow{AB}\ \overrightarrow{AC}\ \overrightarrow{AD}]\bigr|

$\tfrac{1}{6}$ a tetrahedron is one-sixth of the parallelepiped
$\overrightarrow{AB}, \overrightarrow{AC}, \overrightarrow{AD}$ three edges from the SAME vertex $A$

Reciprocal-basis identities and the STP-squared rule

Reciprocal pairings and STP-squared

\vec{a}\cdot\vec{p} = \vec{b}\cdot\vec{q} = \vec{c}\cdot\vec{r} = 1, \quad \vec{a}\cdot\vec{q} = \vec{b}\cdot\vec{p} = \dots = 0 \qquad [\vec{a}\times\vec{b}\ \ \vec{b}\times\vec{c}\ \ \vec{c}\times\vec{a}] = [\vec{a}\ \vec{b}\ \vec{c}]^2

Matched $\vec{a}\cdot\vec{p} = 1$ etc.
Mismatched $\vec{a}\cdot\vec{q} = 0$ (perpendicularity)
STP-squaredcross-of-pairs box $= [\vec{a}\ \vec{b}\ \vec{c}]^2$

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
$\vec{b}, \vec{c}$ the plane the result lies in
Resultperpendicular to $\vec{a}$ , inside the $\vec{b}$ - $\vec{c}$ plane

Vector orthogonal to one vector and coplanar with two others

Orthogonal-and-coplanar vector

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

Coplanar with $\vec{a}, \vec{b}$ it is a combination $\lambda\vec{a} + \mu\vec{b}$
Perpendicular to $\vec{c}$ by construction of the outer cross
Normalisedivide by its magnitude for a UNIT answer

Solving a vector equation: a cross condition plus a magnitude/dot condition

Cross condition reduces to a parallel offset

\vec{r}\times\vec{b} = \vec{c}\times\vec{b} \;\Longrightarrow\; \vec{r} = \vec{c} + t\,\vec{b}, \quad\text{then use the scalar condition for } t

$(\vec{r} - \vec{c})\times\vec{b} = \vec{0}$ so $\vec{r} - \vec{c}\,\|\,\vec{b}$
$t$ the one free scalar — fixed by the dot/magnitude condition

Common traps

The scalar triple product is a SCALAR

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

is a single number. The vector triple product

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

is a VECTOR. Identify which one the question asks for before choosing an identity.

Dot and cross can swap, but keep the order of the three vectors

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

— moving the dot/cross is free. But swapping two of the three vectors themselves flips the sign (see the cyclic-and-sign concept next).

Cyclic keeps the value; ANY single swap negates it

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

(cyclic, same value), but

[\vec{a}\ \vec{c}\ \vec{b}] = -[\vec{a}\ \vec{b}\ \vec{c}]

(one swap, flipped). Lose track of a swap and your final

\lambda

comes out with the wrong sign.

A repeated vector kills the product — spot it early

Whenever expanding by linearity, any term that ends up with two identical vectors (like

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

) is zero. Most coplanarity-combo problems are solved entirely by deleting these zero terms.

Perpendicular to BOTH means parallel to the cross product

\vec{a}\perp\vec{b}

and

\vec{a}\perp\vec{c}

, then

\vec{a}

lies along

\vec{b}\times\vec{c}

, so the box is a right prism and

[\vec{a}\ \vec{b}\ \vec{c}] = |\vec{a}|\,|\vec{b}\times\vec{c}|

. Don't try to plug a single

\sin

of the angle between

\vec{a}

and

\vec{b}

— the relevant angle is between

\vec{b}

and

\vec{c}

"Depends on x and y" — expand the determinant first

When the vectors carry parameters, compute

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

as a determinant and simplify. It frequently collapses to a constant, meaning the answer is independent of every parameter — a deliberately surprising option.

Coplanar ⇒ STP = 0, NOT "two of them are parallel"

Three vectors coplanar means they fit in one plane; they need not be parallel to each other. A parallel pair also makes the STP zero, but it is a stronger, separate condition — don't confuse the two.

A squared parameter can give TWO coplanarity values

When the unknown appears as

\lambda^2

(e.g.

-\lambda^2\hat{i}+\dots

), the determinant-equals-zero equation can have two distinct real roots. Count carefully — "number of values of

\lambda

" questions hinge exactly on this.

$[m\vec{a}+\vec{b}\ \dots] = (m^3+1)[\vec{a}\ \vec{b}\ \vec{c}]$ , not $m^3$

The cyclic

m\,\cdot

pattern gives

m^3 + 1

— the extra

+1

comes from the

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

cross-term. So

m = 3

gives

28

, not

27

. Dropping the

+1

is a deliberate distractor.

Track every sign through the swaps

Expansions like

[\vec{a}+2\vec{b}+3\vec{c}\ \dots]

generate many cross-terms; a single mis-signed swap throws off the coefficient. Use cyclic to standardise every surviving term to

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

before summing.

Volume is the MODULUS — never a negative number

The scalar triple product can be negative (it is a SIGNED volume), but a physical volume is

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

. When a volume is given (e.g. 158 cu units), set the modulus equal to it, which may give two parameter values

\pm

Min vs max: check the second derivative

V'(m) = 0

m = \pm\frac{1}{\sqrt{3}}

, but only one is a minimum.

V''(m) = 6m > 0

m = +\frac{1}{\sqrt{3}}

(minimum) and

< 0

m = -\frac{1}{\sqrt{3}}

(maximum). The two questions share roots but want opposite answers.

The one-sixth is on the tetrahedron, not the parallelepiped

Tetrahedron volume

= \tfrac{1}{6}|[\vec{a}\ \vec{b}\ \vec{c}]|

; the parallelepiped is the full

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

. Forgetting the

\tfrac{1}{6}

over-counts the volume six-fold — a classic distractor when a volume is given.

Build all three edges from the SAME vertex

Use

\overrightarrow{AB}, \overrightarrow{AC}, \overrightarrow{AD}

— all starting at

A

. Mixing base points (e.g.

\overrightarrow{AB}, \overrightarrow{BC}, \dots

) gives the wrong determinant and a wrong unknown.

Matched pairs are 1, mismatched pairs are 0

\vec{a}\cdot\dfrac{\vec{b}\times\vec{c}}{[\vec{a}\ \vec{b}\ \vec{c}]} = 1

, but

\vec{a}\cdot\dfrac{\vec{c}\times\vec{a}}{[\vec{a}\ \vec{b}\ \vec{c}]} = 0

because

\vec{c}\times\vec{a}\perp\vec{a}

. Read each dot pair to decide whether it survives — most of the terms in these sums vanish.

Cross-of-pairs box is the SQUARE, not the cube

[\vec{a}\times\vec{b}\ \ \vec{b}\times\vec{c}\ \ \vec{c}\times\vec{a}] = [\vec{a}\ \vec{b}\ \vec{c}]^2

— so the coefficient

\lambda

\lambda[\vec{a}\ \vec{b}\ \vec{c}]^2

1

, not

3

2

Inner pair sets the plane: $\vec{a}\times(\vec{b}\times\vec{c})$ is in the $\vec{b}$ - $\vec{c}$ plane

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

lies in the plane of

\vec{b}, \vec{c}

;

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

lies in the plane of

\vec{a}, \vec{b}

. They are DIFFERENT vectors.

Match coefficients only when the basis vectors are independent

Reading off

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

by comparing coefficients of

\vec{b}

and

\vec{c}

is valid because "

\vec{b}

not parallel to

\vec{c}

" makes them independent. Watch the sign: the coefficient of

\vec{c}

-(\vec{a}\cdot\vec{b})

, so a positive RHS coefficient gives a NEGATIVE dot product (obtuse angle).

Two crosses ⇒ BAC-CAB; one cross + one dot ⇒ scalar triple product

If the shape is

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

(a number), it is NOT a BAC-CAB case. Count the crosses and dots first to pick the right identity.

Coplanar means a COMBINATION, not just "in the same plane"

Encode coplanarity with

\vec{a}, \vec{b}

by writing the unknown as

\lambda\vec{a} + \mu\vec{b}

from the start. Then the only freedom left is the ratio

\lambda:\mu

, which the perpendicularity or projection condition pins down.

Both signs of the UNIT answer are valid

The normalised result is

\pm\hat{v}

— both directions satisfy "perpendicular and coplanar" unless the question fixes orientation. Pick whichever sign appears in the options.

Confirm orthogonality AND coplanarity at the end

A tempting distractor satisfies one condition but not the other. Verify

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

(perpendicular) and that

\vec{v}

is a combination of

\vec{a}, \vec{b}

(coplanar) before selecting.

You cannot cancel the cross product

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

does NOT give

\vec{r} = \vec{c}

. The correct deduction is

(\vec{r} - \vec{c})\times\vec{b} = \vec{0}

, i.e.

\vec{r} = \vec{c} + t\vec{b}

; the scalar condition then fixes

t

The cross condition alone leaves one free scalar

A single cross equation can never determine

\vec{r}

uniquely — any multiple of

\vec{b}

can be added. Always use the accompanying scalar (dot or magnitude) condition to close the system.

Vectors formulas

Magnitude, Components, and Unit Vectors

Vectors and component form

Magnitude of a vector and distance between two points

Unit vector along a given direction

Magnitude of a sum from angles or perpendicularity

Magnitude of a sum with projection-equality and a perpendicular pair

Unit vector parallel to a parallelogram's diagonal

Common traps

Vector Geometry — Section Formula, Triangle, and Parallelogram

Section formula — internal, external, and midpoint

Centroid and median identities

Finding the ratio, collinearity, and cevian intersection

Incentre, orthocentre, and the angle bisector

Triangle and parallelogram applications

Common traps

Linear Combinations, Collinearity, and Coplanarity

Linear combination of vectors

Collinear vectors and collinear points (one scalar)

Coplanar vectors (two scalars)

Forming a combination, then normalising / measuring it

Collinearity of three points (and who lies between)

Express a vector as a combination of two others

Linear dependence vs independence (the determinant test)

Chained-collinearity systems (\"no two collinear\")

A vector lying in the plane of two others

Components of a vector against a transformed basis

Common traps

Dot Product, Angle, and Perpendicularity

Dot product — the two faces of a·b

Magnitude of a combination via the dot product

Perpendicularity test and solving for a parameter

Disguised perpendicularity — equal-diagonal and Pythagoras forms

Angle between two vectors via cosθ

Angle from a unit-vector perpendicularity constraint

Scalar and vector projection

Projection onto the normal of a plane

Mutually orthogonal vectors — solving a small system

Unit vector along a combination, and scalar-product conditions

Identities and bounds — sum of squared differences

Dot products entangled with cross-product constraints

Obtuse angle for all x — a quadratic-inequality parameter

Moving point a·cos t + b·sin t — farthest from the origin

Reading perpendicularity geometrically — the orthocentre

Common traps

Cross Product, Angle, and Area

The cross product — definition and determinant form

Magnitude of the cross product, angle, and the Lagrange identity

Area of a triangle from two side vectors

Area of a parallelogram — from sides, diagonals, or a side and a diagonal

Bilinear expansion and area-scaling identities

Unit (and given-magnitude) vector perpendicular to two vectors

Solving a vector equation: a cross condition plus a scalar condition

Finding unknown components from a given cross product

Parallelism, collinearity, and a vector along a×b

Vector triple product — the BAC-CAB rule

Magnitude of a vector triple product with a given angle

Angle and cross-magnitude from a vector constraint

Common traps

Scalar Triple Product, Coplanarity, and Volume

The scalar triple product — dot-cross and determinant form

Cyclic and sign properties of the scalar triple product

Computing the value of a scalar triple product

Coplanarity of three vectors (and solving for a parameter)

Scalar triple product of linear combinations

Volume of a parallelepiped (and min/max problems)

Volume of a tetrahedron

Reciprocal-basis identities and the STP-squared rule

Vector triple product (BAC-CAB rule)

Vector orthogonal to one vector and coplanar with two others

Solving a vector equation: a cross condition plus a magnitude/dot condition

Common traps

More MHT-CET Maths formula sheets