MHT-CET Maths · Vectors

Cross Product, Angle, and Area

The vector product whose magnitude is the area of a parallelogram and whose direction is the right-hand-rule perpendicular — the engine behind areas, unit normals, angles, and a whole family of vector-equation problems.

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Why this matters

At 83 PYQs this is the chapter's biggest subtopic after the scalar triple product, and just as punishing — roughly 64% of these are rated HARD. Three themes dominate: AREA (triangle, parallelogram, from diagonals, or from a side-plus-diagonal), the PERPENDICULAR DIRECTION (unit normal, vector of a given magnitude perpendicular to two), and VECTOR EQUATIONS that mix a cross and a dot condition (solve for the unknown vector, find an unknown component, or expand a vector triple product with the BAC-CAB rule). Master the determinant computation and the |a×b| = |a||b|sin θ relation first — every concept below is built on them.

Concept 1 of 12: The cross product — definition and determinant form

The cross product of two 3-D vectors produces a THIRD vector that is perpendicular to both, with magnitude equal to the area of the parallelogram they span and direction fixed by the right-hand rule. In practice you almost never use the sin θ form to compute it — you expand a 3x3 determinant whose top row is the unit vectors and whose lower two rows are the components of the two vectors.

Definition

For $\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}$ and $\vec{b} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k}$ :

Geometric form: $\vec{a}\times\vec{b} = |\vec{a}||\vec{b}|\sin\theta\,\hat{n}$ , where $\hat{n}$ is the unit perpendicular by the right-hand rule
Determinant form: $\vec{a}\times\vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix}$
Anti-commutative: $\vec{a}\times\vec{b} = -\,\vec{b}\times\vec{a}$
Self / parallel: $\vec{a}\times\vec{a} = \vec{0}$ ; and $\vec{a}\times\vec{b} = \vec{0} \iff \vec{a}\,\|\,\vec{b}$ (or one is zero)
Standard products: $\hat{i}\times\hat{j} = \hat{k},\; \hat{j}\times\hat{k} = \hat{i},\; \hat{k}\times\hat{i} = \hat{j}$ (cyclic); reverse any pair and the sign flips

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

Diagram · drag to rotate a × b

Drag to orbit the scene. However you turn it, a × b stays perpendicular to the plane of a and b, on the side your right-hand fingers (curling a → b) point your thumb. Length is schematic; magnitude is |a||b| sin θ.

Worked example

Compute

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

for

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

and

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

Practise this concept5 quick reps

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

Concept 2 of 12: Magnitude of the cross product, angle, and the Lagrange identity

The magnitude of the cross product is

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

— a single number you can read off when you know the two magnitudes and the angle between them. Rearranged, it isolates

\sin\theta

; paired with the dot product it isolates

\cos\theta

. The Lagrange identity bundles both into one equation, letting you cross between cross and dot without ever finding the angle.

Definition

For non-zero $\vec{a}, \vec{b}$ at angle $\theta$ :

Magnitude: $|\vec{a}\times\vec{b}| = |\vec{a}||\vec{b}|\sin\theta$
Angle: $\sin\theta = \dfrac{|\vec{a}\times\vec{b}|}{|\vec{a}||\vec{b}|}$ and $\cos\theta = \dfrac{\vec{a}\cdot\vec{b}}{|\vec{a}||\vec{b}|}$
Lagrange identity: $|\vec{a}\times\vec{b}|^2 + (\vec{a}\cdot\vec{b})^2 = |\vec{a}|^2|\vec{b}|^2$

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 $|\vec{a}|^2|\vec{b}|^2$

Worked example

|\vec{a}| = 4

|\vec{b}| = 5

and the angle between them is

120^\circ

, find

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

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2024 · 12th May Shift 2 · Q133Moderate

Example 2 · Vectors · Cross Product, Angle, and Area

|\vec{u}| = 8

|\vec{v}| = 12

and the angle between them is

150^\circ

, then

|\vec{u} \times \vec{v}|

$\sin\theta$ 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.

Drill 1 more on magnitude of the cross product, angle, and the lagrange identity

Concept 3 of 12: Area of a triangle from two side vectors

Build two edge vectors from a common vertex, cross them, and halve the magnitude — that is the area of the triangle. The half is the whole point: the cross product gives the parallelogram, and a triangle is exactly half of it. When the area is GIVEN and a coordinate is unknown, set up the same equation and solve.

Definition

For a triangle with vertices $A, B, C$ , form $\overrightarrow{AB} = B - A$ and $\overrightarrow{AC} = C - A$ . Then Area $= \dfrac{1}{2}\,|\overrightarrow{AB} \times \overrightarrow{AC}|$ . Equivalently, for a triangle whose two adjacent SIDES are the vectors $\vec{a}$ and $\vec{b}$ , the area is $\dfrac{1}{2}|\vec{a}\times\vec{b}|$ .

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

Visualization · the parallelogram area is |a × b|

a₁: 5a₂: 1b₁: 1b₂: 4

|a × b| = 19 (parallelogram area)triangle = 9.5direction: out of the page

|a × b| = |a₁b₂ − a₂b₁| is exactly the parallelogram area; the triangle on a and b is half of it. Make a and b parallel and the area — and the cross product — collapse to zero. The fill colour flips with the right-hand-rule direction (out of vs into the page).

Worked example

Find the area of the triangle with vertices

A(1,1,1)

B(2,3,1)

and

C(1,2,3)

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 2nd May Shift 2 · Q128Moderate

Example 3 · Vectors · Cross Product, Angle, and Area

The area of the triangle whose vertices are

A(1,-1,2)

B(2,1,-1)

and

C(3,-1,2)

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.

Drill 2 more on area of a triangle from two side vectors

Concept 4 of 12: Area of a parallelogram — from sides, diagonals, or a side and a diagonal

A parallelogram on two adjacent side vectors has area

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

— no half this time. If instead you are handed the two DIAGONALS, the area is HALF the magnitude of their cross product. And if you are given one side plus one diagonal, recover the other side by subtraction, then cross the two sides.

Definition

From two adjacent sides $\vec{a}, \vec{b}$ : Area $= |\vec{a}\times\vec{b}|$ .
From the two diagonals $\vec{d_1}, \vec{d_2}$ : Area $= \dfrac{1}{2}|\vec{d_1}\times\vec{d_2}|$ .
From a side and a diagonal: if $\vec{a}$ is a side and $\vec{c}$ the diagonal from the same vertex, the adjacent side is $\vec{b} = \vec{c} - \vec{a}$ ; then Area $= |\vec{a}\times\vec{b}|$ .

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 $\tfrac{1}{2}$

Diagram · parallelogram diagonals = a + b and a − b

From a shared corner, sides a and b span the parallelogram. The diagonal from that corner is a + b; the diagonal between the side tips is a − b. They bisect each other, and |a + b|² + |a − b|² = 2(|a|² + |b|²).

Worked example

Find the area of the parallelogram whose diagonals are

\vec{d_1} = 2\hat{i} + \hat{j} - 2\hat{k}

and

\vec{d_2} = \hat{i} + 2\hat{j} + 2\hat{k}

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 3rd May Shift 1 · Q112Moderate

Example 4 · Vectors · Cross Product, Angle, and Area

The area (in sq. units) of the parallelogram whose diagonals are along the vectors

8\hat{i}-6\hat{j}

and

3\hat{i}+4\hat{j}-12\hat{k}

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.

Drill 2 more on area of a parallelogram — from sides, diagonals, or a side and a diagonal

Concept 5 of 12: Bilinear expansion and area-scaling identities

When both sides of a cross product are linear combinations of two base vectors, expand using distributivity and anti-commutativity — the

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

and

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

terms vanish, leaving a clean multiple of

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

. The signature special case is

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

. This turns 'find the new area' problems into a single coefficient times the old area.

Definition

For any scalars: $(p\vec{a} + q\vec{b}) \times (r\vec{a} + s\vec{b}) = (ps - qr)\,(\vec{a}\times\vec{b})$ , because $\vec{a}\times\vec{a} = \vec{b}\times\vec{b} = \vec{0}$ and $\vec{b}\times\vec{a} = -\vec{a}\times\vec{b}$ . Special case: $(\vec{a}-\vec{b})\times(\vec{a}+\vec{b}) = 2(\vec{a}\times\vec{b})$ . So if the original area is $|\vec{a}\times\vec{b}|$ , the new parallelogram on $p\vec{a}+q\vec{b}$ and $r\vec{a}+s\vec{b}$ has area $|ps-qr|\,|\vec{a}\times\vec{b}|$ .

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

Worked example

If the parallelogram on

\vec{a}, \vec{b}

has area

20

, find the area of the parallelogram on

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

and

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

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2024 · 11th May Shift 2 · Q118Moderate

Example 5 · Vectors · Cross Product, Angle, and Area

If the area of the parallelogram with

\vec{a}

and

\vec{b}

as two adjacent sides is 16 sq.units, then the area of the parallelogram having

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

and

\vec{a}+3\vec{b}

as two adjacent sides (in sq.units) is

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.

Drill 2 more on bilinear expansion and area-scaling identities

Concept 6 of 12: Unit (and given-magnitude) vector perpendicular to two vectors

Any vector perpendicular to both

\vec{a}

and

\vec{b}

lies along their cross product. Normalise it (divide by its magnitude) to get a UNIT perpendicular; scale it to any required length. Because two opposite directions both qualify, there are always exactly TWO unit perpendiculars.

Definition

If $\vec{a}, \vec{b}$ are not parallel, a unit vector perpendicular to both is $\hat{n} = \pm\dfrac{\vec{a}\times\vec{b}}{|\vec{a}\times\vec{b}|}$ . A vector of magnitude $m$ perpendicular to both is $\pm m\,\dfrac{\vec{a}\times\vec{b}}{|\vec{a}\times\vec{b}|}$ . Both signs are valid unless the question fixes a direction.

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

Diagram · unit normal n̂ = (a×b)/|a×b|

A plane has exactly two unit normals, ±n̂. The cross product a × b picks one by the right-hand rule; b × a gives the other. Dividing by |a × b| rescales it to length 1.

Worked example

Find a vector of magnitude

10

perpendicular to both

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

and

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

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 10th May Shift 2 · Q112Moderate

Example 6 · Vectors · Cross Product, Angle, and Area

The vector of magnitude 6 units and perpendicular to vectors

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

and

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

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.

Drill 2 more on unit (and given-magnitude) vector perpendicular to two vectors

Concept 7 of 12: Solving a vector equation: a cross condition plus a scalar condition

A single equation

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

does not pin down

\vec{r}

— any multiple of

\vec{a}

can be added without changing the cross product. Pair it with a scalar (dot) condition like

\vec{r}\cdot\vec{c} = k

and the system becomes determined. Write

\vec{r} = (x,y,z)

, turn the cross equation into three component equations, add the dot condition, and solve the linear system.

Definition

To solve $\vec{a}\times\vec{r} = \vec{b}$ together with a scalar condition such as $\vec{a}\cdot\vec{r} = k$ :

Set $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$ and expand $\vec{a}\times\vec{r}$ as a determinant.
Equate components with $\vec{b}$ to get three (dependent) linear equations.
Add the scalar condition to close the system, then solve.

Pattern $\vec{a}\times\vec{b} = \vec{a}\times\vec{c}$ rearranges to $\vec{a}\times(\vec{b}-\vec{c}) = \vec{0}$ , so $\vec{b}-\vec{c} = \lambda\vec{a}$ — substitute into the scalar condition to find $\lambda$ .

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

Worked example

Find

\vec{r}

such that

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

and

\vec{a}\cdot\vec{r} = 4

, where

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

and

\vec{b} = \hat{i} - \hat{k}

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2024 · 9th May Shift 2 · Q119Hard

Example 7 · Vectors · Cross Product, Angle, and Area

\vec{a} = \hat{i}+\hat{j}+\hat{k},\;\vec{b} = \hat{j}-\hat{k}

, then vector

\vec{r}

satisfying

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

and

\vec{a}\cdot\vec{r} = 3

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

Drill 7 more on solving a vector equation: a cross condition plus a scalar condition

Concept 8 of 12: Finding unknown components from a given cross product

When a vector has unknown scalars in it and you are TOLD what a cross product (or a projection, or an area) equals, match components on both sides to extract the unknowns. A given

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

gives three component equations; a given projection or area gives one more — together they solve for the unknowns.

Definition

If $\vec{b}$ and $\vec{c}$ carry unknown scalars and $\vec{b}\times\vec{c}$ is given, expand $\vec{b}\times\vec{c}$ as a determinant and EQUATE it component-by-component to the given vector. Combine with any other scalar datum — a projection $\dfrac{\vec{a}\cdot\vec{c}}{|\vec{c}|}$ , an area, or a dot product — to pin down every unknown.

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

Worked example

Let

\vec{b} = 3\hat{i} - \beta\hat{j} + 4\hat{k}

and

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

. If

\vec{b}\times\vec{c} = -6\hat{i} + 10\hat{j} + 7\hat{k}

, find

\beta

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 4th May Shift 1 · Q112Hard

Example 8 · Vectors · Cross Product, Angle, and Area

Let

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

\vec{b}=3\hat{i}-\beta\hat{j}+4\hat{k}

and

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

, where

\alpha,\beta\in\mathbb{R}

, be three vectors. If the projection of

\vec{a}

\vec{c}

\frac{10}{3}

and

\vec{b}\times\vec{c}=-6\hat{i}+10\hat{j}+7\hat{k}

, then the value of

2\alpha+\beta

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.).

Drill 1 more on finding unknown components from a given cross product

Concept 9 of 12: Parallelism, collinearity, and a vector along a×b

A zero cross product is the cleanest test for parallel vectors. So

\vec{a}\times\vec{b} = 2(\vec{a}\times\vec{c})

rearranges to

\vec{a}\times(\vec{b} - 2\vec{c}) = \vec{0}

, telling you

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

is PARALLEL to

\vec{a}

. And to build a vector along

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

with a prescribed dot value, take

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

and solve for

\lambda

Definition

Parallel test: $\vec{u}\times\vec{v} = \vec{0} \iff \vec{u} \parallel \vec{v}$ (so $\vec{u} = \lambda\vec{v}$ ).
Linear combinations: $\vec{a}\times\vec{b} = k(\vec{a}\times\vec{c}) \Rightarrow \vec{a}\times(\vec{b} - k\vec{c}) = \vec{0} \Rightarrow \vec{b} - k\vec{c} = \lambda\vec{a}$ . Take magnitudes to find $\lambda$ .
Normals parallel $\Rightarrow$ planes parallel: if $(\vec{a}\times\vec{b})\times(\vec{c}\times\vec{d}) = \vec{0}$ then the two plane-normals are parallel, so the planes are parallel (angle $0$ ).
Vector along $\vec{a}\times\vec{b}$ with $\vec{c}\cdot\vec{d} = k$ : write $\vec{d} = \lambda(\vec{a}\times\vec{b})$ , solve $\lambda = \dfrac{k}{\vec{c}\cdot(\vec{a}\times\vec{b})}$ .

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

Worked example

\vec{a}\times\vec{b} = 3(\vec{a}\times\vec{c})

with

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

|\vec{b}| = 5

\vec{b}\cdot\vec{c} = 3

and

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

, find

|\lambda|

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 16th May Shift 2 · Q143Hard

Example 9 · Vectors · Cross Product, Angle, and Area

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

are three vectors such that

\vec{a} \neq 0

and

\vec{a} \times \vec{b} = 2\vec{a} \times \vec{c}

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

|\vec{b}|=4

and

|\vec{b}\times\vec{c}|=\sqrt{15}

. If

\vec{b}-2\vec{c} = \lambda\vec{a}

, then

\lambda

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.

Drill 2 more on parallelism, collinearity, and a vector along a×b

Concept 10 of 12: Vector triple product — the BAC-CAB rule

A triple product with TWO crosses returns a VECTOR, and it expands by the BAC-CAB identity:

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

. Read it as 'middle dot far times middle, minus middle dot near times far'. Many HARD PYQs hand you an expanded form and ask you to match coefficients — recognising the BAC-CAB shape is the whole game.

Definition

For any $\vec{a}, \vec{b}, \vec{c}$ :

$\vec{a}\times(\vec{b}\times\vec{c}) = (\vec{a}\cdot\vec{c})\vec{b} - (\vec{a}\cdot\vec{b})\vec{c}$
$(\vec{a}\times\vec{b})\times\vec{c} = (\vec{a}\cdot\vec{c})\vec{b} - (\vec{b}\cdot\vec{c})\vec{a}$
Self-nested: $\vec{a}\times(\vec{a}\times\vec{c}) = (\vec{a}\cdot\vec{c})\vec{a} - |\vec{a}|^2\vec{c}$

The result of $\vec{a}\times(\vec{b}\times\vec{c})$ lies in the plane of $\vec{b}$ and $\vec{c}$ (and is perpendicular to $\vec{a}$ ).

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 $\vec{c}$

Worked example

Vectors

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

have magnitudes

2, 1, 1

. If

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

, find the acute angle between

\vec{a}

and

\vec{c}

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2024 · 10th May Shift 2 · Q112Hard

Example 10 · Vectors · Cross Product, Angle, and Area

(\vec{a}\times\vec{b})\times\vec{c}=-5\vec{a}+4\vec{b}

and

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

, then the value of

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

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

Drill 6 more on vector triple product — the bac-cab rule

Concept 11 of 12: Magnitude of a vector triple product with a given angle

When you need

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

and you know the angle between

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

and

\vec{c}

, treat

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

as a single vector and apply the plain magnitude formula. First compute

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

, then multiply by

|\vec{c}|\sin(\text{angle})

. If

\vec{a}

is perpendicular to

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

, the angle is

90^\circ

and

\sin = 1

Definition

Treating $\vec{a}\times\vec{b}$ as one vector $\vec{w}$ : $|(\vec{a}\times\vec{b})\times\vec{c}| = |\vec{a}\times\vec{b}|\,|\vec{c}|\sin\phi$ , where $\phi$ is the angle between $\vec{a}\times\vec{b}$ and $\vec{c}$ . Special case: if $\vec{a}\perp(\vec{b}\times\vec{c})$ , then $|\vec{a}\times(\vec{b}\times\vec{c})| = |\vec{a}|\,|\vec{b}\times\vec{c}|$ (since $\sin 90^\circ = 1$ ).

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

Worked example

Let

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

and

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

. If

|\vec{c}| = 1

and the angle between

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

and

\vec{c}

30^\circ

, find

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

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 2nd May Shift 1 · Q104Hard

Example 11 · Vectors · Cross Product, Angle, and Area

Let

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

and

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

. Let

\vec{c}

be a vector such that

|\vec{c} - \vec{a}| = 3

and

|(\vec{a} \times \vec{b}) \times \vec{c}| = 3

and the angle between

\vec{c}

and

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

30^\circ

, then

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

is equal to

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

Drill 6 more on magnitude of a vector triple product with a given angle

Concept 12 of 12: Angle and cross-magnitude from a vector constraint

Some problems give a constraint (a linear relation, or a vector defined via

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

) and ask for an angle or a cross-product magnitude. The trick is to use

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

(a vector is perpendicular to its own cross product) and to square the constraint to bring in dot products. From the resulting

\cos\theta

you read off the angle or, via

\sin\theta

, the cross magnitude.

Definition

Two recurring levers:

Self-perpendicularity: $\vec{b}\cdot(\vec{a}\times\vec{b}) = 0$ and $\vec{a}\cdot(\vec{a}\times\vec{b}) = 0$ — the cross product is perpendicular to each factor.
Square a linear constraint: from $\vec{a} + p\vec{b} + q\vec{c} = \vec{0}$ , isolate one vector and dot with another to extract $\vec{a}\cdot\vec{c}$ , then $\cos\theta$ and $|\vec{a}\times\vec{c}| = |\vec{a}||\vec{c}|\sin\theta$ .
Rotation in a plane: if a side is rotated until perpendicular to another, the new angle satisfies $\cos\alpha = \sin\theta_{\text{initial}}$ .

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

Worked example

|\vec{a}| = 2

|\vec{b}| = 2

\vec{a}\cdot\vec{b} = 2

and

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

, find the angle between

\vec{b}

and

\vec{c}

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2024 · 9th May Shift 2 · Q125Hard

Example 12 · Vectors · Cross Product, Angle, and Area

Vectors

\vec{a}

and

\vec{b}

are such that

|\vec{a}|=1

|\vec{b}|=4

and

\vec{a}\cdot\vec{b}=2

. If

\vec{c} = 2\vec{a}\times\vec{b} - 3\vec{b}

, then the angle between

\vec{b}

and

\vec{c}

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

Drill 3 more on angle and cross-magnitude from a vector constraint

Summary — formulas & gotchas at a glance

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

Formulas (12)

The cross product — definition and determinant form
Cross product as a determinant
$\vec{a}\times\vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix} = |\vec{a}||\vec{b}|\sin\theta\,\hat{n}$
Magnitude of the cross product, angle, and the Lagrange identity
Magnitude and Lagrange
$|\vec{a}\times\vec{b}| = |\vec{a}||\vec{b}|\sin\theta \qquad |\vec{a}\times\vec{b}|^2 + (\vec{a}\cdot\vec{b})^2 = |\vec{a}|^2|\vec{b}|^2$
Area of a triangle from two side vectors
Triangle area
$\text{Area} = \tfrac{1}{2}\,|\overrightarrow{AB} \times \overrightarrow{AC}|$
Area of a parallelogram — from sides, diagonals, or a side and a diagonal
Parallelogram areas
$\text{Area} = |\vec{a}\times\vec{b}| \qquad \text{Area}_{\text{diagonals}} = \tfrac{1}{2}|\vec{d_1}\times\vec{d_2}|$
Bilinear expansion and area-scaling identities
Bilinear cross-expansion
$(p\vec{a} + q\vec{b}) \times (r\vec{a} + s\vec{b}) = (ps - qr)\,(\vec{a}\times\vec{b})$
Unit (and given-magnitude) vector perpendicular to two vectors
Unit / scaled perpendicular
$\hat{n} = \pm\dfrac{\vec{a}\times\vec{b}}{|\vec{a}\times\vec{b}|} \qquad \vec{v}_{|m|} = \pm\, m\,\dfrac{\vec{a}\times\vec{b}}{|\vec{a}\times\vec{b}|}$
Solving a vector equation: a cross condition plus a scalar condition
Cross plus scalar condition
$\vec{a}\times\vec{r} = \vec{b}, \;\; \vec{a}\cdot\vec{r} = k \;\;\Longrightarrow\;\; \vec{r}\text{ is uniquely determined}$
Finding unknown components from a given cross product
Component matching
$\vec{b}\times\vec{c} = (\text{given vector}) \;\Longrightarrow\; \text{equate each of }\hat{i},\hat{j},\hat{k}\text{ components}$
Parallelism, collinearity, and a vector along a×b
Parallel via zero cross product
$\vec{a}\times\vec{b} = k(\vec{a}\times\vec{c}) \;\Longrightarrow\; \vec{a}\times(\vec{b} - k\vec{c}) = \vec{0} \;\Longrightarrow\; \vec{b} - k\vec{c} = \lambda\vec{a}$
Vector triple product — the BAC-CAB rule
BAC-CAB rule
$\vec{a}\times(\vec{b}\times\vec{c}) = (\vec{a}\cdot\vec{c})\,\vec{b} - (\vec{a}\cdot\vec{b})\,\vec{c}$
Magnitude of a vector triple product with a given angle
Triple-product magnitude
$|(\vec{a}\times\vec{b})\times\vec{c}| = |\vec{a}\times\vec{b}|\,|\vec{c}|\sin\phi$
Angle and cross-magnitude from a vector constraint
Perpendicularity of a cross product
$\vec{a}\cdot(\vec{a}\times\vec{b}) = 0 \qquad \vec{b}\cdot(\vec{a}\times\vec{b}) = 0$

Watch out for (28)

The cross product is a VECTOR, not a scalar
→ The cross product — definition and determinant form
Watch the SIGN on the $\hat{j}$ -component
→ The cross product — definition and determinant form
$\vec{a}\times\vec{b} = \vec{0}$ does NOT mean both vectors are zero
→ The cross product — definition and determinant form
$\sin\theta$ is the same for $\theta$ and $180^\circ - \theta$
→ Magnitude of the cross product, angle, and the Lagrange identity
Use Lagrange to skip finding the angle
→ Magnitude of the cross product, angle, and the Lagrange identity
The HALF is on the triangle, not the parallelogram
→ Area of a triangle from two side vectors
When area is GIVEN, expect TWO values of the unknown
→ Area of a triangle from two side vectors
Cross edges from the SAME vertex
→ Area of a triangle from two side vectors
SIDES use no $\tfrac{1}{2}$ ; DIAGONALS do
→ Area of a parallelogram — from sides, diagonals, or a side and a diagonal
A diagonal is the SUM of the two sides, not one of them
→ Area of a parallelogram — from sides, diagonals, or a side and a diagonal
Keep the cross-terms in order
→ Bilinear expansion and area-scaling identities
Area takes the ABSOLUTE value of the coefficient
→ Bilinear expansion and area-scaling identities
Both $\pm$ signs are valid answers
→ Unit (and given-magnitude) vector perpendicular to two vectors
For perpendicular to $\vec{a}+\vec{b}$ and $\vec{a}-\vec{b}$ , use the shortcut
→ Unit (and given-magnitude) vector perpendicular to two vectors
Confirm the magnitude is actually $1$
→ Unit (and given-magnitude) vector perpendicular to two vectors
One cross equation is NOT enough on its own
→ Solving a vector equation: a cross condition plus a scalar condition
$\vec{a}\times\vec{b} = \vec{a}\times\vec{c}$ does NOT mean $\vec{b} = \vec{c}$
→ Solving a vector equation: a cross condition plus a scalar condition
Pick the component that isolates the unknown
→ Finding unknown components from a given cross product
Use the right datum for the right unknown
→ Finding unknown components from a given cross product
Track the sign of the dot product
→ Parallelism, collinearity, and a vector along a×b
Normal parallel means PLANES parallel, not perpendicular
→ Parallelism, collinearity, and a vector along a×b
Grouping matters — the two triple products differ
→ Vector triple product — the BAC-CAB rule
BAC-CAB is for VECTOR triple products only
→ Vector triple product — the BAC-CAB rule
When comparing a×(a×c) problems, isolate $\vec{a}\cdot\vec{c}$
→ Vector triple product — the BAC-CAB rule
Compute $|\vec{a}\times\vec{b}|$ FIRST, then treat it as one vector
→ Magnitude of a vector triple product with a given angle
Recover $|\vec{c}|$ from the side conditions before using $\sin\phi$
→ Magnitude of a vector triple product with a given angle
A cross product contributes ZERO to a dot with its own factor
→ Angle and cross-magnitude from a vector constraint
Square the constraint to get dot products
→ Angle and cross-magnitude from a vector constraint

Test yourself on Vectors

20 past MHT-CET questions from this chapter, timed at 36 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

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