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JEE Mains Maths · Formula sheet

Vector Algebra formulas

15 formulas and 15 common traps for JEE Mains Maths Vector Algebra, grouped by subtopic.

Full notes with worked examples

Dot Product: Angles and Projections

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Angle between two vectors; perpendicular, acute and obtuse

Angle from components

cos⁡θ=a1b1+a2b2+a3b3∣a⃗∣ ∣b⃗∣\cos\theta=\frac{a_1b_1+a_2b_2+a_3b_3}{|\vec a|\,|\vec b|}

Projections and components along and across a vector

Scalar projection

projb⃗a⃗=a⃗⋅b⃗∣b⃗∣\text{proj}_{\vec b}\vec a=\frac{\vec a\cdot\vec b}{|\vec b|}

A vector in the plane of two others

In the plane of a and b

v⃗=x a⃗+y b⃗\vec v=x\,\vec a+y\,\vec b

Common traps

A negative dot product includes π

a⃗⋅b⃗<0\vec a\cdot\vec b<0 also holds when the vectors point in opposite directions. If the question wants a strictly obtuse angle, exclude the antiparallel case.

|b| for the length, |b|² for the vector

The scalar projection divides by ∣b⃗∣|\vec b|; the projection vector divides by ∣b⃗∣2|\vec b|^2 and then multiplies by b⃗\vec b. Mixing them scales the answer by ∣b⃗∣|\vec b|.

The direction is not the vector

(a⃗×b⃗)×c⃗(\vec a\times\vec b)\times\vec c fixes only the direction. Its length and sign still come from the remaining condition.

Magnitudes and Unit-Vector Identities

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Expanding the square of a sum

Square of a sum

∣a⃗+b⃗∣2=∣a⃗∣2+2 a⃗⋅b⃗+∣b⃗∣2|\vec a+\vec b|^2=|\vec a|^2+2\,\vec a\cdot\vec b+|\vec b|^2

Combinations with a cross-product term

Length with a cross term (unit vectors)

∣λa⃗+μb⃗+ν(a⃗×b⃗)∣2=λ2+μ2+2λμcos⁡θ+ν2sin⁡2θ|\lambda\vec a+\mu\vec b+\nu(\vec a\times\vec b)|^2=\lambda^2+\mu^2+2\lambda\mu\cos\theta+\nu^2\sin^2\theta

Common traps

Square before you add

∣a⃗+b⃗∣|\vec a+\vec b| is ∣a⃗∣+∣b⃗∣|\vec a|+|\vec b| only when the vectors point the same way. Always square, expand, and take the root at the end.

Only for unit vectors

∣a⃗×b⃗∣=sin⁡θ|\vec a\times\vec b|=\sin\theta needs ∣a⃗∣=∣b⃗∣=1|\vec a|=|\vec b|=1. In general it is ∣a⃗∣∣b⃗∣sin⁡θ|\vec a||\vec b|\sin\theta.

Cross Product: Areas and Perpendicular Vectors

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Areas of triangles, parallelograms and quadrilaterals

Triangle area

Area=12 ∣AB→×AC→∣\text{Area}=\tfrac12\,|\overrightarrow{AB}\times\overrightarrow{AC}|

A vector perpendicular to two others

Common perpendicular

c⃗⊥a⃗, c⃗⊥b⃗ ⇒ c⃗=λ(a⃗×b⃗)\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

∣a⃗×b⃗∣2+(a⃗⋅b⃗)2=∣a⃗∣2∣b⃗∣2|\vec a\times\vec b|^2+(\vec a\cdot\vec b)^2=|\vec a|^2|\vec b|^2

Common traps

Diagonals give half

With sides the parallelogram's area is ∣a⃗×b⃗∣|\vec a\times\vec b|; with diagonals it is half of ∣d⃗1×d⃗2∣|\vec d_1\times\vec d_2|. Using the wrong one doubles or halves the answer.

Two unit vectors, not one

Both a⃗×b⃗\vec a\times\vec b and b⃗×a⃗\vec b\times\vec a are perpendicular to the pair. A condition in the question (a sign, a positive component) decides which.

The identity is in squares

Subtract the squares and take the root at the end. ∣a⃗×b⃗∣|\vec a\times\vec b| is not ∣a⃗∣∣b⃗∣−a⃗⋅b⃗|\vec a||\vec b|-\vec a\cdot\vec b.

Solving Vector Equations

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r × a = b × a: the difference is parallel to a

The parallel form

r⃗×a⃗=b⃗×a⃗ ⇒ r⃗=b⃗+λa⃗\vec r\times\vec a=\vec b\times\vec a\ \Rightarrow\ \vec r=\vec b+\lambda\vec a

a × c = b with a · c given: cross again with a

Solving a × c = b

c⃗=(a⃗⋅c⃗) a⃗−a⃗×b⃗∣a⃗∣2\vec c=\frac{(\vec a\cdot\vec c)\,\vec a-\vec a\times\vec b}{|\vec a|^2}

Common traps

c × b is minus b × c

a⃗×c⃗=c⃗×b⃗\vec a\times\vec c=\vec c\times\vec b gives (a⃗+b⃗)×c⃗=0⃗(\vec a+\vec b)\times\vec c=\vec0, not (a⃗−b⃗)(\vec a-\vec b): moving c⃗×b⃗\vec c\times\vec b across flips it to b⃗×c⃗\vec b\times\vec c.

Check solvability first

If a⃗⋅b⃗≠0\vec a\cdot\vec b\ne0, no c⃗\vec c satisfies a⃗×c⃗=b⃗\vec a\times\vec c=\vec b. A question asking how many such vectors exist can have the answer 0.

Triple Products and Coplanarity

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Scalar triple product: volume and coplanarity

Coplanarity

[a⃗ b⃗ c⃗]=a⃗⋅(b⃗×c⃗)=0[\vec a\ \vec b\ \vec c]=\vec a\cdot(\vec b\times\vec c)=0

Vector triple product: a × (b × c)

BAC − CAB

a⃗×(b⃗×c⃗)=(a⃗⋅c⃗) b⃗−(a⃗⋅b⃗) c⃗\vec a\times(\vec b\times\vec c)=(\vec a\cdot\vec c)\,\vec b-(\vec a\cdot\vec b)\,\vec c

Common traps

Points need edges, not positions

Four points are coplanar when the three edges from one of them are coplanar. Testing the four position vectors directly tests something else.

Not associative

a⃗×(b⃗×c⃗)\vec a\times(\vec b\times\vec c) and (a⃗×b⃗)×c⃗(\vec a\times\vec b)\times\vec c are different vectors. Check which pair is inside the bracket before expanding.

Vectors in Geometry and Rotation

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Section points, centroid and collinearity

Section formula

p⃗=n a⃗+m b⃗m+n\vec p=\frac{n\,\vec a+m\,\vec b}{m+n}

Triangle centres and angle bisectors

Internal bisector

direction=a⃗∣a⃗∣+b⃗∣b⃗∣\text{direction}=\frac{\vec a}{|\vec a|}+\frac{\vec b}{|\vec b|}

Rotating a vector, and rotating the axes

Plane rotation

(x,y)↦(xcos⁡θ−ysin⁡θ, xsin⁡θ+ycos⁡θ)(x,y)\mapsto(x\cos\theta-y\sin\theta,\ x\sin\theta+y\cos\theta)

Common traps

Cross the weights

For AP:PB=m:nAP:PB=m:n, a⃗\vec a carries weight nn and b⃗\vec b weight mm. Putting mm on a⃗\vec a gives the point dividing in n:mn:m.

Add unit vectors

a⃗+b⃗\vec a+\vec b bisects the angle only when ∣a⃗∣=∣b⃗∣|\vec a|=|\vec b|. Normalise first: a^+b^\hat a+\hat b.

Which way it turns

A right-angle rotation has two possible results, ±\pm a perpendicular vector. The question's words — counterclockwise, 'passing through the y-axis' — pick one; check it against them.

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