NDA Maths · Vectors

Cross Product and Triple Product

The vector product whose magnitude is the area of a parallelogram, its direction the right-hand-rule perpendicular — plus the scalar and vector triple products built from it.

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Why this matters

The cross product takes two vectors and produces a THIRD vector — perpendicular to both, with magnitude equal to the area of the parallelogram they span. That single idea opens up a family of geometric tools: computing areas of triangles and parallelograms, building unit vectors perpendicular to a plane, expressing torque/moment of a force about a point, and detecting when three vectors lie in one plane (via the scalar triple product). The seven concepts below take you from the basic algebra (anti-commutative, NOT associative) through the Lagrange identity, the triple products, and the vector triple product (BAC-CAB rule). 37 PYQs across 2017–2026, with 27% rated HARD — the densest and toughest Vectors subtopic. Master these seven and the chapter's HARD tail collapses.

Concept 1 of 7

Cross product — algebra and properties

Intuition

The cross product of two 3-D vectors produces a third vector perpendicular to both, with magnitude equal to the area of the parallelogram they span. Algebraically it is anti-commutative (swap and the sign flips) and distributive over addition, but — unlike multiplication of numbers — NOT associative. A null cross product means the two vectors are parallel (or one of them is the zero vector).

Definition

For vectors $\vec{a}, \vec{b}, \vec{c}$ and scalar $k$ : $\vec{a}\times\vec{b} = -\vec{b}\times\vec{a}$ (anti-commutative); $\vec{a}\times(\vec{b}+\vec{c}) = \vec{a}\times\vec{b} + \vec{a}\times\vec{c}$ (distributive); $(k\vec{a})\times\vec{b} = k(\vec{a}\times\vec{b})$ (scalar associative); $\vec{a}\times\vec{a} = \vec{0}$ ; $\vec{a}\times\vec{b} = \vec{0} \iff \vec{a}\,\|\,\vec{b}$ (or one is zero); $(\vec{a}\times\vec{b})\times\vec{c} \neq \vec{a}\times(\vec{b}\times\vec{c})$ in general.

Difference-of-squares-style identity

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

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

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

Express

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

in terms of

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

Distribute: $3(\vec{a}\times\vec{a}) - 6(\vec{a}\times\vec{b}) + (\vec{b}\times\vec{a}) - 2(\vec{b}\times\vec{b})$ .
Drop the zero diagonal terms $\vec{a}\times\vec{a} = \vec{b}\times\vec{b} = \vec{0}$ .
Apply anti-commutativity $\vec{b}\times\vec{a} = -\vec{a}\times\vec{b}$ : $-6(\vec{a}\times\vec{b}) - (\vec{a}\times\vec{b})$ .
Combine: $-7\,\vec{a}\times\vec{b}$ .

Answer:

(3\vec{a} + \vec{b}) \times (\vec{a} - 2\vec{b}) = -7\,\vec{a}\times\vec{b}

Practice this conceptself-check · 4 quick reps

Try it yourself

Express

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

in terms of

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

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
$\vec{a}\times\vec{a} = ?$
2.
$\vec{a}\times\vec{b} = ?$ in terms of $\vec{b}\times\vec{a}$ .
3.
If $\vec{a}\times\vec{b} = \vec{0}$ (both non-zero), the vectors are?
4.
Is the cross product associative?

From the bank · past-year question

Example 1VectorsEASY

What is

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

equal to?

[Q68 · Sep · 2018]

Cross product is NOT associative

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

and

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

are generally different vectors — both are linear combinations of

\vec{a}

and

\vec{b}

(or

\vec{b}

and

\vec{c}

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

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

It means

\vec{a}

and

\vec{b}

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

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

and

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

together imply

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

for some scalar

\lambda

Drill 4 more on cross product — algebra and properties

Concept 2 of 7

Cross-product magnitude, area, and the Lagrange identity

Intuition

The magnitude of

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

is the area of the parallelogram on

\vec{a}

and

\vec{b}

, measured by

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

. Half of that is the area of the triangle. Combining the cross-product magnitude with the dot-product magnitude gives the Lagrange identity — a one-line bridge from one to the other.

Definition

For non-zero $\vec{a}, \vec{b}$ at angle $\theta$ : $|\vec{a}\times\vec{b}| = |\vec{a}|\,|\vec{b}|\sin\theta$ . Area of parallelogram with sides $\vec{a}, \vec{b}$ is $|\vec{a}\times\vec{b}|$ ; area of triangle with the same sides is $\tfrac{1}{2}|\vec{a}\times\vec{b}|$ . Lagrange identity: $|\vec{a}\times\vec{b}|^2 + (\vec{a}\cdot\vec{b})^2 = |\vec{a}|^2\,|\vec{b}|^2$ .

Magnitude, area, and Lagrange

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

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

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 whose two adjacent sides are

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

and

\vec{b} = \hat{i} + 3\hat{j}

Cross product: $\vec{a}\times\vec{b} = (2\cdot 3 - 1\cdot 1)\hat{k} = 5\hat{k}$ , so $|\vec{a}\times\vec{b}| = 5$ .
Triangle area is half the parallelogram area: $\tfrac{1}{2}|\vec{a}\times\vec{b}| = \tfrac{1}{2}\cdot 5 = 2.5$ .

Answer:Area

= 2.5

square units

Practice this conceptself-check · 4 quick reps

Try it yourself

\vec{a}

and

\vec{b}

have magnitudes 3 and 5 with angle

30^\circ

between them. Find the area of the parallelogram they span.

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
Area of the parallelogram on $\vec{a}, \vec{b}$ ?
2.
$|\vec{a}\times\vec{b}|$ for $|\vec{a}| = 2$ , $|\vec{b}| = 3$ , angle $30^\circ$ ?
3.
Triangle area with sides $\vec{a}, \vec{b}$ ?
4.
Lagrange: $|\vec{a}\times\vec{b}|^2 + (\vec{a}\cdot\vec{b})^2 = ?$

From the bank · past-year question

Example 2VectorsEASY

|\vec{a}\times\vec{b}|^2+|\vec{a}\cdot\vec{b}|^2=144

and

|\vec{a}|=4

, then what is

|\vec{b}|

equal to?

[Q69 · Apr · 2020]

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

A frequently-tested statement: \"

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

is the area of a triangle with sides

\vec{a}

and

\vec{b}

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

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

Unlike

\cos\theta

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

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

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

\sin\theta

Drill 8 more on cross-product magnitude, area, and the lagrange identity

Concept 3 of 7

Unit vector perpendicular to two given vectors

Intuition

Any vector perpendicular to both

\vec{a}

and

\vec{b}

lies along the cross product direction (or its opposite). To get the unit perpendicular, take

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

and divide by its magnitude. There are exactly two such unit vectors, pointing in opposite directions.

Definition

If $\vec{a}, \vec{b}$ are not parallel, a unit vector perpendicular to both is $\hat{n} = \pm\dfrac{\vec{a}\times\vec{b}}{|\vec{a}\times\vec{b}|}$ . Both signs are valid answers unless the question specifies a direction (right-hand rule, towards a third vector, etc.).

Unit perpendicular

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

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

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 unit vector perpendicular to both

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

and

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

Compute $\vec{a}\times\vec{b}$ as a determinant: $\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 1 & -1 & 1 \end{vmatrix}$ .
Expand: $\hat{i}\big((1)(1) - (1)(-1)\big) - \hat{j}\big((1)(1) - (1)(1)\big) + \hat{k}\big((1)(-1) - (1)(1)\big) = \hat{i}(2) - \hat{j}(0) + \hat{k}(-2)$ .
Simplify: $\vec{a}\times\vec{b} = 2\hat{i} - 2\hat{k}$ .
Magnitude: $|\vec{a}\times\vec{b}| = \sqrt{4+0+4} = 2\sqrt{2}$ .
Normalise: $\hat{n} = \pm\dfrac{2\hat{i} - 2\hat{k}}{2\sqrt{2}} = \pm\dfrac{1}{\sqrt{2}}(\hat{i}-\hat{k})$ .

Answer:

\hat{n} = \pm\dfrac{1}{\sqrt{2}}(\hat{i}-\hat{k})

Practice this conceptself-check · 4 quick reps

Try it yourself

Find a unit vector perpendicular to both

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

and

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

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
Unit vector perpendicular to both $\vec{a}, \vec{b}$ ?
2.
How many unit vectors are perpendicular to two non-parallel vectors?
3.
$\hat{i}\times\hat{j} = ?$
4.
Unit vector perpendicular to both $\hat{i}$ and $\hat{j}$ ?

From the bank · past-year question

Example 3VectorsMODERATE

A unit vector perpendicular to each of the vectors

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

and

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

[Q65 · Sep · 2018]

Both $\pm$ signs give valid answers

If an MCQ offers

+\hat{n}

and the question doesn't pin down direction,

-\hat{n}

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

z

-component\" specifically to break this ambiguity.

Scalar multiples of a unit perpendicular are not unit

A vector like

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

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

|\hat{n}| = 1

before selecting an option.

Drill 2 more on unit vector perpendicular to two given vectors

Concept 4 of 7

Moment of a force (torque)

Intuition

The moment of a force about a point

O

measures how strongly the force tends to rotate the body about

O

. It is the cross product of the position vector (from

O

to the point of application) with the force itself. Direction follows the right-hand rule; magnitude is the perpendicular distance times the force.

Definition

If a force $\vec{F}$ acts at a point $P$ and we measure its moment about a point $O$ , then $\vec{M} = \overrightarrow{OP} \times \vec{F}$ . Moment magnitude is $|\vec{M}| = |\overrightarrow{OP}||\vec{F}|\sin\theta = d\,|\vec{F}|$ where $d$ is the perpendicular distance from $O$ to the line of action.

Moment of a force

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

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

Diagram · torque τ = r × F (drag to rotate)

Torque about the pivot is τ = r × F, where r reaches the point where the force F acts. It points perpendicular to the plane of r and F (right-hand rule, here out of that plane), with magnitude |τ| = r·F·sinθ — largest when the force is perpendicular to the arm, zero when it's along it.

Worked example

A force

\vec{F} = \hat{i} + 3\hat{j} - 2\hat{k}

acts at the point

A(3, -1, 2)

. Find its moment about the point

B(1, 0, 1)

Position vector from the pivot: $\overrightarrow{BA} = A - B = 2\hat{i} - \hat{j} + \hat{k}$ .
Set up the cross product $\overrightarrow{BA} \times \vec{F}$ as a determinant: $\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -1 & 1 \\ 1 & 3 & -2 \end{vmatrix}$ .
Expand: $\hat{i}\big((-1)(-2) - (1)(3)\big) - \hat{j}\big((2)(-2) - (1)(1)\big) + \hat{k}\big((2)(3) - (-1)(1)\big) = -\hat{i} + 5\hat{j} + 7\hat{k}$ .

Answer:

\vec{M} = -\hat{i} + 5\hat{j} + 7\hat{k}

Practice this conceptself-check · 4 quick reps

Try it yourself

A force

\vec{F} = \hat{i} + 2\hat{j} + 3\hat{k}

acts at

P(1, 1, 1)

. Find its moment about the origin.

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
Moment of force $\vec{F}$ acting at $P$ about $O$ ?
2.
Is the moment $\vec{r}\times\vec{F}$ or $\vec{F}\times\vec{r}$ ?
3.
$\overrightarrow{OP} = \hat{i}$ , $\vec{F} = \hat{j}$ . Moment about $O$ ?
4.
Moment of a force about a point — vector or scalar?

From the bank · past-year question

Example 4VectorsMODERATE

A force

\vec{F}=2\hat{i}-\lambda\hat{j}+5\hat{k}

is applied at the point

A(1,2,5)

. If its moment about the point

B(-1,-2,3)

16\hat{i}-6\hat{j}+2\lambda\hat{k}

, then what is the value of

\lambda

[Q70 · Sep · 2025]

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

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

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

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

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

onto that line — a scalar, not a vector.

Drill 3 more on moment of a force (torque)

Concept 5 of 7

Scalar triple product and coplanarity

Intuition

The scalar triple product

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

is the signed volume of the parallelepiped on the three vectors. Three vectors are coplanar precisely when that volume is zero — equivalently, the determinant of their component matrix vanishes. This is the single most-used test in HARD PYQs.

Definition

For $\vec{a}, \vec{b}, \vec{c}$ in $\mathbb{R}^3$ , the scalar triple product is $[\vec{a}\,\vec{b}\,\vec{c}] = \vec{a}\cdot(\vec{b}\times\vec{c}) = \begin{vmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{vmatrix}$ . Coplanarity criterion: $\vec{a}, \vec{b}, \vec{c}$ are coplanar iff $[\vec{a}\,\vec{b}\,\vec{c}] = 0$ . Volume of the parallelepiped on the three vectors is $|[\vec{a}\,\vec{b}\,\vec{c}]|$ .

STP as determinant + coplanarity test

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

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

Diagram · triple product = box volume (SVG, drag to rotate)

[a b c] = a · (b × c) = 13.22 (signed volume of the box)

The box spanned by a, b, c has volume |[a b c]|. Painter's-ordered faces fake the solidity — edges don't truly hide behind nearer faces, which is the SVG limit this comparison is testing.

Worked example

Find

\lambda

such that

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

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

and

\vec{c} = \hat{i} + \lambda\hat{j} + 3\hat{k}

are coplanar.

Coplanar $\Rightarrow$ the determinant $[\vec{a}\,\vec{b}\,\vec{c}]$ is zero.
Set up: $\begin{vmatrix} 1 & 1 & -1 \\ 2 & -1 & 1 \\ 1 & \lambda & 3 \end{vmatrix} = 0$ .
Expand along the first row: $1\cdot\big[(-1)(3) - (1)(\lambda)\big] - 1\cdot\big[(2)(3) - (1)(1)\big] + (-1)\cdot\big[(2)(\lambda) - (-1)(1)\big]$ .
Simplify: $(-3 - \lambda) - (6 - 1) - (2\lambda + 1) = -9 - 3\lambda = 0$ .
Solve: $\lambda = -3$ .

Answer:

\lambda = -3

Practice this conceptself-check · 4 quick reps

Try it yourself

Find the volume of the parallelepiped with edge vectors

\vec{a} = \hat{i}

\vec{b} = \hat{j}

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

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
$[\vec{a}\,\vec{b}\,\vec{c}] = 0$ means the vectors are?
2.
Volume of the parallelepiped on $\vec{a}, \vec{b}, \vec{c}$ ?
3.
$[\hat{i}\,\hat{j}\,\hat{k}] = ?$
4.
Write the STP as a dot-cross: $[\vec{a}\,\vec{b}\,\vec{c}] = ?$

From the bank · past-year question

Example 5VectorsMODERATE

If the vectors

\vec{a}=2\hat{i}-3\hat{j}+\hat{k}

\vec{b}=\hat{i}+2\hat{j}-3\hat{k}

and

\vec{c}=\hat{j}+p\hat{k}

are coplanar, then what is the value of

p

[Q70 · Apr · 2020]

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

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

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

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

2\times 2

minor.

Drill 2 more on scalar triple product and coplanarity

Concept 6 of 7

STP cyclic property and derived linear-combo identities

Intuition

The scalar triple product is symmetric under cyclic rotation of its three vectors —

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

— and flips sign under any single swap. Combined with linearity, this generates a family of derived identities that PYQs love to test, including the famous

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

. When given a constraint like

\vec{a}+2\vec{b}+3\vec{c}=\vec{0}

, cross-multiply and apply linearity to extract a

\lambda

value.

Definition

Cyclic identity: $\vec{a}\cdot(\vec{b}\times\vec{c}) = \vec{b}\cdot(\vec{c}\times\vec{a}) = \vec{c}\cdot(\vec{a}\times\vec{b})$ . Anti-cyclic: swapping any two negates the value, e.g. $\vec{a}\cdot(\vec{c}\times\vec{b}) = -\vec{a}\cdot(\vec{b}\times\vec{c})$ . Linearity in each slot: $[(\alpha\vec{u}+\beta\vec{v})\,\vec{b}\,\vec{c}] = \alpha[\vec{u}\,\vec{b}\,\vec{c}] + \beta[\vec{v}\,\vec{b}\,\vec{c}]$ . If $\vec{a}, \vec{b}, \vec{c}$ are coplanar, $[\vec{a}\,\vec{b}\,\vec{c}] = 0$ and the derived cyclic sum collapses to zero too.

Cyclic + sum identity

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

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

Worked example

\vec{a} + 2\vec{b} + 3\vec{c} = \vec{0}

and

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

, find

\lambda

From $\vec{a} + 2\vec{b} + 3\vec{c} = \vec{0}$ , write $\vec{a} = -2\vec{b} - 3\vec{c}$ .
Substitute into each cross product. $\vec{a}\times\vec{b} = (-2\vec{b} - 3\vec{c})\times\vec{b} = -2(\vec{b}\times\vec{b}) - 3(\vec{c}\times\vec{b}) = 0 + 3(\vec{b}\times\vec{c}) = 3\vec{b}\times\vec{c}$ .
Similarly $\vec{c}\times\vec{a} = \vec{c}\times(-2\vec{b} - 3\vec{c}) = -2(\vec{c}\times\vec{b}) - 3(\vec{c}\times\vec{c}) = 2(\vec{b}\times\vec{c}) + 0 = 2\vec{b}\times\vec{c}$ .
Sum: $\vec{a}\times\vec{b} + \vec{b}\times\vec{c} + \vec{c}\times\vec{a} = (3 + 1 + 2)(\vec{b}\times\vec{c}) = 6\,\vec{b}\times\vec{c}$ .
Hence $\lambda = 6$ .

Answer:

\lambda = 6

Practice this conceptself-check · 4 quick reps

Try it yourself

Simplify

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

using the cyclic property of the scalar triple product.

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
Is $[\vec{a}\,\vec{b}\,\vec{c}] = [\vec{b}\,\vec{c}\,\vec{a}]$ ?
2.
$[\vec{a}\,\vec{c}\,\vec{b}] = ?$ in terms of $[\vec{a}\,\vec{b}\,\vec{c}]$ .
3.
$[\vec{a}\,\vec{a}\,\vec{b}] = ?$
4.
$[\vec{a}\,\vec{b}\,\vec{c}] + [\vec{b}\,\vec{c}\,\vec{a}] + [\vec{c}\,\vec{a}\,\vec{b}] = ?$

From the bank · past-year question

Example 6VectorsEASY

\vec{a}

\vec{b}

and

\vec{c}

are coplanar, then what is

(2\vec{a}\times3\vec{b})\cdot4\vec{c}+(5\vec{b}\times3\vec{c})\cdot6\vec{a}

equal to?

[Q74 · Apr · 2021]

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

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

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

-1

on every other term — turning

\lambda = 6

into

\lambda = -6

or worse.

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

Both are cyclic rotations of

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

. An MCQ giving

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

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

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

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

Drill 2 more on stp cyclic property and derived linear-combo identities

Concept 7 of 7

Vector triple product (BAC-CAB rule)

Intuition

Triple products of three vectors that return a VECTOR (not a scalar) expand 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\" — a mnemonic for the order of the vectors. The result lies in the plane of the two innermost vectors, which is a structural fact you can exploit before any algebra.

Definition

For any $\vec{a}, \vec{b}, \vec{c} \in \mathbb{R}^3$ : $\vec{a}\times(\vec{b}\times\vec{c}) = (\vec{a}\cdot\vec{c})\,\vec{b} - (\vec{a}\cdot\vec{b})\,\vec{c}$ . By anti-commutativity, $(\vec{a}\times\vec{b})\times\vec{c} = -\,\vec{c}\times(\vec{a}\times\vec{b}) = (\vec{a}\cdot\vec{c})\,\vec{b} - (\vec{b}\cdot\vec{c})\,\vec{a}$ . Either form lies in the plane of the two inner vectors and is perpendicular to the outer one.

BAC-CAB rule

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

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

Worked example

For

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

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

and

\vec{c} = \hat{j} + \hat{k}

, find

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

using the BAC-CAB rule.

Apply BAC-CAB: $\vec{a}\times(\vec{b}\times\vec{c}) = (\vec{a}\cdot\vec{c})\vec{b} - (\vec{a}\cdot\vec{b})\vec{c}$ .
Compute $\vec{a}\cdot\vec{c} = 2\cdot 0 + 1\cdot 1 + 0\cdot 1 = 1$ and $\vec{a}\cdot\vec{b} = 2\cdot 1 + 1\cdot 0 + 0\cdot(-1) = 2$ .
Substitute: $1\cdot\vec{b} - 2\cdot\vec{c} = (\hat{i} - \hat{k}) - 2(\hat{j} + \hat{k}) = \hat{i} - 2\hat{j} - 3\hat{k}$ .

Answer:

\vec{a}\times(\vec{b}\times\vec{c}) = \hat{i} - 2\hat{j} - 3\hat{k}

Practice this conceptself-check · 4 quick reps

Try it yourself

Using the BAC-CAB rule, find

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

for

\vec{a} = \hat{i}

\vec{b} = \hat{j}

\vec{c} = \hat{i} + \hat{k}

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
$\vec{a}\times(\vec{b}\times\vec{c}) = ?$ (BAC-CAB)
2.
$\vec{a}\times(\vec{b}\times\vec{c})$ lies in the plane of?
3.
$\hat{i}\times(\hat{i}\times\hat{j}) = ?$
4.
Does BAC-CAB apply to $\vec{a}\cdot(\vec{b}\times\vec{c})$ ?

From the bank · past-year question

Example 7VectorsMODERATE

Let

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

and

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

. If

\vec{a}\times(\vec{b}\times\vec{a})=\alpha\hat{i}-\beta\hat{j}+\gamma\hat{k}

, then what is the value of

\alpha+\beta+\gamma

[Q66 · Apr · 2024]

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

If the expression is

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

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

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

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

lies in the plane of

\vec{a}, \vec{b}

(the innermost pair);

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

lies in the plane of

\vec{b}, \vec{c}

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

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

Take magnitudes of both equations and use

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

with

\sin\theta \leq 1

to force

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

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

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

classic and the 2026 S10 set.

Drill 9 more on vector triple product (bac-cab rule)

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 (7)

Cross product — algebra and properties
Difference-of-squares-style identity
$(\vec{a} - \vec{b}) \times (\vec{a} + \vec{b}) = 2\,\vec{a}\times\vec{b}$
Cross-product magnitude, area, and the Lagrange identity
Magnitude, area, and Lagrange
$|\vec{a}\times\vec{b}| = |\vec{a}||\vec{b}|\sin\theta \qquad |\vec{a}\times\vec{b}|^2 + (\vec{a}\cdot\vec{b})^2 = |\vec{a}|^2 |\vec{b}|^2$
Unit vector perpendicular to two given vectors
Unit perpendicular
$\hat{n} = \pm\dfrac{\vec{a}\times\vec{b}}{|\vec{a}\times\vec{b}|}$
Moment of a force (torque)
Moment of a force
$\vec{M} = \overrightarrow{OP} \times \vec{F}$
Scalar triple product and coplanarity
STP as determinant + coplanarity test
$[\vec{a}\,\vec{b}\,\vec{c}] = \begin{vmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{vmatrix} = 0 \iff \vec{a},\vec{b},\vec{c}\text{ coplanar}$
STP cyclic property and derived linear-combo identities
Cyclic + sum identity
$(\vec{a}\times\vec{b})\cdot\vec{c} + (\vec{b}\times\vec{c})\cdot\vec{a} + (\vec{c}\times\vec{a})\cdot\vec{b} = 3[\vec{a}\,\vec{b}\,\vec{c}]$
Vector triple product (BAC-CAB rule)
BAC-CAB rule
$\vec{a}\times(\vec{b}\times\vec{c}) = (\vec{a}\cdot\vec{c})\,\vec{b} - (\vec{a}\cdot\vec{b})\,\vec{c}$

Watch out for (15)

Cross product is NOT associative
→ Cross product — algebra and properties
$\vec{a}\times\vec{b} = \vec{0}$ does NOT mean both vectors are zero
→ Cross product — algebra and properties
Area of a triangle is $\tfrac{1}{2}|\vec{a}\times\vec{b}|$ , NOT $|\vec{a}\times\vec{b}|$
→ Cross-product magnitude, area, and the Lagrange identity
$\sin\theta$ is always non-negative for $\theta\in[0,\pi]$
→ Cross-product magnitude, area, and the Lagrange identity
Both $\pm$ signs give valid answers
→ Unit vector perpendicular to two given vectors
Scalar multiples of a unit perpendicular are not unit
→ Unit vector perpendicular to two given vectors
Order is $\vec{r} \times \vec{F}$ , not $\vec{F} \times \vec{r}$
→ Moment of a force (torque)
Moment depends on the pivot — moment of a force about a POINT is unique, but about a LINE is also a vector
→ Moment of a force (torque)
STP $= 0$ means coplanar — NOT \" $\vec{a}$ parallel to $\vec{b}$ \"
→ Scalar triple product and coplanarity
Determinant row/column expansion: pick the row with most zeros
→ Scalar triple product and coplanarity
Anti-cyclic = sign flip — don't accidentally drop it
→ STP cyclic property and derived linear-combo identities
$(\vec{a}\times\vec{b})\cdot\vec{c} = (\vec{c}\times\vec{a})\cdot\vec{b}$
→ STP cyclic property and derived linear-combo identities
BAC-CAB only applies to vector triple products — not scalar
→ Vector triple product (BAC-CAB rule)
Cross-then-cross is NOT cross-then-dot-with-different-grouping
→ Vector triple product (BAC-CAB rule)
Special triples: if $\vec{a}\times\vec{b} = \vec{c}$ and $\vec{b}\times\vec{c} = \vec{a}$ , the three vectors are an orthonormal pairwise-perpendicular triple
→ Vector triple product (BAC-CAB rule)

Mastery check — 5 interleaved questions

Try each one before clicking. Questions are interleaved across the concepts above, not grouped — interleaving sharpens transfer.

Example 1VectorsMODERATE

What is

3\alpha+2\beta

equal to if

(2\hat{i}+6\hat{j}+27\hat{k})\times(\hat{i}+\alpha\hat{j}+\beta\hat{k})

is a null vector?

[Q51 · Sep · 2024]

Example 2VectorsMODERATE

Consider the following statements : 1. The magnitude of

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

is same as the area of a triangle with sides

\vec{a}

and

\vec{b}

. 2. If

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

where

\vec{a} \neq \vec{0}

\vec{b} \neq \vec{0}

, then

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

. Which of the above statements is/are correct ?

[Q49 · Sep · 2019]

Example 3VectorsHARD

Let

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

\vec{\beta} = 2\hat{i}-\hat{j}+3\hat{k}

and

\vec{\gamma} = 2\hat{i}+\hat{j}+6\hat{k}

be three vectors. If

\vec{\alpha}

and

\vec{\beta}

are both perpendicular to the vector

\vec{\delta}

and

\vec{\delta}\cdot\vec{\gamma} = 10

, then what is the magnitude of

\vec{\delta}

[Q45 · Sep · 2017]

Example 4VectorsEASY

Consider the following in respect of moment of a force: (A) The moment of force about a point is independent of point of application of force. (B) The moment of a force about a line is a vector quantity. Which of the statements given above is/are correct?

[Q68 · Apr · 2024]

Example 5VectorsHARD

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

\vec{b}=2\hat{i}+3\hat{j}+2\hat{k}

and

\vec{c}=\hat{i}+m\hat{j}+n\hat{k}

are three coplanar vectors and

|\vec{c}|=\sqrt{6}

, then which one of the following is correct?

[Q66 · Apr · 2017]

Drill every past-year question on this subtopic

37 questions from the bank — paginated, with cart and Word-export support.

Drill the 37 questions

Drill every past-year question on this subtopic

Related notes