JEE Mains Maths · Vector Algebra

Cross Product: Areas and Perpendicular Vectors

The cross product as area — of triangles, parallelograms and quadrilaterals — and as the direction perpendicular to two vectors, together with the identity that links it to the dot product.

All Vector Algebra notes JEE Mains Maths strategy

Why this matters

Thirty-seven PYQs. Nearly half are areas, from a triangle's sides, a parallelogram's diagonals or a quadrilateral's vertices. The rest need a direction perpendicular to two vectors, or the identity |a × b|² + (a · b)² = |a|²|b|². Three ideas cover the page.

Concept 1 of 3: Areas of triangles, parallelograms and quadrilaterals

|\vec a\times\vec b|

is the area of the parallelogram with sides

\vec a

and

\vec b

, so a triangle with those sides has half of it. When a parallelogram is given by its diagonals

\vec d_1,\vec d_2

, its area is

\frac12|\vec d_1\times\vec d_2|

, and the same formula gives any quadrilateral from its two diagonals. Scaling passes straight through:

(2\vec a)\times(3\vec b)=6(\vec a\times\vec b)

Definition

Parallelogram with sides $\vec a,\vec b$ : $|\vec a\times\vec b|$ . Triangle: $\frac12|\vec a\times\vec b|$ .
Parallelogram or quadrilateral with diagonals $\vec d_1,\vec d_2$ : $\frac12|\vec d_1\times\vec d_2|$ .
$\vec a\times\vec b$ in components: the determinant with rows $\hat i,\hat j,\hat k$ ; $\vec a$ ; $\vec b$ .

Triangle area

\text{Area}=\tfrac12\,|\overrightarrow{AB}\times\overrightarrow{AC}|

Worked example

Find the area of the triangle with vertices

(0,0,0)

(1,2,0)

(0,1,3)

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2024 · 30 Jan 2024 · Q78Moderate

Example 1 · Vector Algebra · Cross Product: Areas and Perpendicular Vectors

Let

A(2,3,5)

and

C( - 3,4, - 2)

be opposite vertices of a parallelogram

ABCD

if the diagonal

\overrightarrow{BD}=\widehat{i}+ 2\widehat{j}+ 3\widehat{k}

then the area of the parallelogram is equal to

Diagonals give half

With sides the parallelogram's area is

|\vec a\times\vec b|

; with diagonals it is half of

|\vec d_1\times\vec d_2|

. Using the wrong one doubles or halves the answer.

Drill 15 more on areas of triangles, parallelograms and quadrilaterals

Concept 2 of 3: A vector perpendicular to two others

\vec a\times\vec b

is perpendicular to both

\vec a

and

\vec b

, so any vector perpendicular to both is

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

, and one more condition fixes

\lambda

. A line lying in two planes is perpendicular to both normals, so its direction is

\vec n_1\times\vec n_2

Definition

Perpendicular to $\vec a$ and $\vec b$ : $\lambda(\vec a\times\vec b)$ .
Unit vectors perpendicular to both: $\pm\frac{\vec a\times\vec b}{|\vec a\times\vec b|}$ .
$\hat i\times\hat j=\hat k$ , $\hat j\times\hat k=\hat i$ , $\hat k\times\hat i=\hat j$ ; reversing the order changes the sign.

Common perpendicular

\vec c\perp\vec a,\ \vec c\perp\vec b\ \Rightarrow\ \vec c=\lambda(\vec a\times\vec b)

Worked example

\vec c

is perpendicular to

(1,1,0)

and

(0,1,1)

, and

\vec c\cdot\hat i=2

. Find

\vec c

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2021 · Paper 6 · Q89Moderate

Example 2 · Vector Algebra · Cross Product: Areas and Perpendicular Vectors

Let

\overrightarrow{c}

be a vector perpendicular to the vectors

\overrightarrow{a}=\widehat{i}+\widehat{j}-\widehat{k}

and

\overrightarrow{b}=\widehat{i}+ 2\widehat{j}+\widehat{k}

. If

\overrightarrow{c}\cdot (\widehat{i}+\widehat{j}+ 3\widehat{k}) = 8

then the value of

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

is equal to

Two unit vectors, not one

Both

\vec a\times\vec b

and

\vec b\times\vec a

are perpendicular to the pair. A condition in the question (a sign, a positive component) decides which.

Drill 10 more on a vector perpendicular to two others

Concept 3 of 3: |a × b|² + (a · b)² = |a|²|b|²

Since

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

and

\vec a\cdot\vec b=|\vec a||\vec b|\cos\theta

, their squares add to

|\vec a|^2|\vec b|^2

. So any two of the three quantities give the third, without ever finding the angle. It also shows

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

combines neatly with a dot term.

Definition

$|\vec a\times\vec b|^2+(\vec a\cdot\vec b)^2=|\vec a|^2|\vec b|^2$ .
$(\vec a-\vec b)\times(\vec a+\vec b)=2(\vec a\times\vec b)$ .
$|(\vec a\times\vec b)\times\vec c|=|\vec a\times\vec b||\vec c|\sin\phi$ , $\phi$ the angle between $\vec a\times\vec b$ and $\vec c$ .

Lagrange's identity

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

Worked example

|\vec a|=3

|\vec b|=5

\vec a\cdot\vec b=9

. Find

|\vec a\times\vec b|

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2023 · 31 January 2023 · Q87Moderate

Example 3 · Vector Algebra · Cross Product: Areas and Perpendicular Vectors

Let

\overrightarrow{a}

and

\overrightarrow{b}

be two vectors such that

|\overrightarrow{a}| =\sqrt{14},|\overrightarrow{b}| =\sqrt{6}

and

|\overrightarrow{a}\times\overrightarrow{b}| =\sqrt{48}

. Then

(\overrightarrow{a}\cdot\overrightarrow{b})^{2}

is equal to

The identity is in squares

Subtract the squares and take the root at the end.

|\vec a\times\vec b|

is not

|\vec a||\vec b|-\vec a\cdot\vec b

Drill 9 more on |a × b|² + (a · b)² = |a|²|b|²

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 (3)

Areas of triangles, parallelograms and quadrilaterals
Triangle area
$\text{Area}=\tfrac12\,|\overrightarrow{AB}\times\overrightarrow{AC}|$
A vector perpendicular to two others
Common perpendicular
$\vec c\perp\vec a,\ \vec c\perp\vec b\ \Rightarrow\ \vec c=\lambda(\vec a\times\vec b)$
|a × b|² + (a · b)² = |a|²|b|²
Lagrange's identity
$|\vec a\times\vec b|^2+(\vec a\cdot\vec b)^2=|\vec a|^2|\vec b|^2$

Watch out for (3)

Diagonals give half
→ Areas of triangles, parallelograms and quadrilaterals
Two unit vectors, not one
→ A vector perpendicular to two others
The identity is in squares
→ |a × b|² + (a · b)² = |a|²|b|²

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