JEE Mains Physics · Motion in a Plane

Vectors: Resultant, Components and Products

Two vectors at an angle θ add to a resultant of size √(A² + B² + 2AB cos θ); any vector can be split into x and y components and added component by component; the dot product A·B = AB cos θ tests perpendicularity and the cross product gives a vector at right angles to both.

All Motion in a Plane notes JEE Mains Physics strategy

Why this matters

Thirty PYQs, nineteen of them multiple choice, and none from 2026. Eleven find the size or angle of a resultant, eight resolve vectors into components, and eleven use the dot or cross product. Eleven of the thirty ask for a number, more than on any other page of the chapter, so the algebra has to be exact. Six come with a figure: read every angle off it before you resolve.

Concept 1 of 3: Resultant of two vectors at an angle

Place the tail of B at the head of A; the resultant runs from the tail of A to the head of B. Its size depends on the angle between them: largest (A + B) when they point the same way, smallest (|A − B|) when they are opposite, and √(A² + B²) when they are at right angles. Squaring the resultant turns every question about it into one line of algebra in cos θ.

Definition

Resultant: $R^{2} = A^{2} + B^{2} + 2AB\cos\theta$ ; its angle with A: $\tan\alpha = \dfrac{B\sin\theta}{A + B\cos\theta}$ .
Difference: $|\vec{A} - \vec{B}|^{2} = A^{2} + B^{2} - 2AB\cos\theta$ .
Equal magnitudes A: $|\vec{A} + \vec{B}| = 2A\cos\dfrac{\theta}{2}$ , $|\vec{A} - \vec{B}| = 2A\sin\dfrac{\theta}{2}$ , so $|\vec{A} - \vec{B}| = |\vec{A} + \vec{B}|\tan\dfrac{\theta}{2}$ .
Resultant perpendicular to A: the component of B along A cancels A, so $A + B\cos\theta = 0$ .
Equal magnitudes with $|\vec{A} + \vec{B}| = n|\vec{A} - \vec{B}|$ : $\dfrac{1 + \cos\theta}{1 - \cos\theta} = n^{2}$ .
Range of the resultant: $|A - B| \le R \le A + B$ .

Resultant of two vectors

R = \sqrt{A^{2} + B^{2} + 2AB\cos\theta} \qquad \tan\alpha = \frac{B\sin\theta}{A + B\cos\theta} \qquad |\vec{A} + \vec{B}| = 2A\cos\frac{\theta}{2}\ (A = B)

Worked example

Forces of 5 N and 8 N act at a point with an angle of

60^{\circ}

between them. Find the size of the resultant and the angle it makes with the 5 N force.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2024 · 4 April 2024 · Q29Moderate

Example 1 · Motion in a Plane · Vectors: Resultant, Components and Products

Two forces

{\overrightarrow{F}}_{1}

and

{\overrightarrow{F}}_{2}

are acting on a body. One force has magnitude thrice that of the other force and the resultant of the two forces is equal to the force of larger magnitude. The angle between

{\overrightarrow{F}}_{1}

and

{\overrightarrow{F}}_{2}

\cos^{- 1}\left( \frac{1}{n} \right)

. The value of

|n|

is___

Adding magnitudes

A 3 N and a 4 N force give 7 N only when they point the same way. At right angles the resultant is 5 N; at any other angle use R² = A² + B² + 2AB cos θ.

Perpendicular to A, not to B

If the resultant is perpendicular to A, the condition is A + B cos θ = 0: the part of B along A cancels A. Writing B + A cos θ = 0 makes the resultant perpendicular to B instead.

Drill 10 more on resultant of two vectors at an angle

Concept 2 of 3: Resolving vectors into components

Any vector in a plane is the sum of a piece along x and a piece along y. Resolve every vector this way, add all the x pieces and all the y pieces separately, and rebuild the resultant from the two totals. The only care needed is with the angle: the side adjacent to the given angle takes the cosine.

Definition

At angle θ from the x-axis: $A_x = A\cos\theta$ , $A_y = A\sin\theta$ . At angle θ from the y-axis the two swap: $A_y = A\cos\theta$ , $A_x = A\sin\theta$ .
Magnitude $A = \sqrt{A_x^{2} + A_y^{2} + A_z^{2}}$ ; direction $\tan\alpha = A_y/A_x$ , with the signs fixing the quadrant.
Several vectors: $R_x = \sum A_x$ , $R_y = \sum A_y$ , then $R = \sqrt{R_x^{2} + R_y^{2}}$ .
Unit vector $\hat{B} = \vec{B}/|\vec{B}|$ ; a vector of size A along B is $A\hat{B}$ .
Regular n-sided figure with centre O: the vectors from O to all n vertices add to zero, so from one vertex A to the other n − 1 vertices they add to $n\,\overrightarrow{AO}$ .

Components and magnitude

A_x = A\cos\theta,\ A_y = A\sin\theta \qquad A = \sqrt{A_x^{2} + A_y^{2} + A_z^{2}} \qquad \vec{A}\ \text{along}\ \vec{B} = A\hat{B}

Worked example

Three forces act at a point: 10 N along +x, 6 N along +y, and

4\sqrt{2}

N at

135^{\circ}

to +x. Find the resultant and its angle with +x.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2023 · 15 Apr 2023 · Q16Moderate

Example 2 · Motion in a Plane · Vectors: Resultant, Components and Products

A vector in

x - y

plane makes an angle of

30^{\circ}

with

y

-axis The magnitude of

y

-component of vector is

2\sqrt{3}

. The magnitude of

x

-component of the vector will be:

Angle with the y-axis

When the angle θ is measured from the y-axis, the y component is A cos θ and the x component is A sin θ. Using A cos θ for x by habit swaps the two.

Losing the quadrant

tan α = A_y/A_x gives the same value for (3, 3) and (−3, −3). Look at the signs of the components to place the vector before you quote its angle.

Drill 7 more on resolving vectors into components

Concept 3 of 3: Dot product, cross product and projection

The dot product measures how much two vectors point the same way: it is AB cos θ, a plain number, and zero when they are perpendicular. The cross product measures how much they point across each other: its size is AB sin θ, it points at right angles to both, and it is zero when they are parallel. In component form the dot product is a sum of three products and the cross product is a 3 × 3 determinant.

Definition

$\vec{A}\cdot\vec{B} = AB\cos\theta = A_xB_x + A_yB_y + A_zB_z$ ; perpendicular when it is zero.
Projection (scalar component) of A on B: $\dfrac{\vec{A}\cdot\vec{B}}{|\vec{B}|}$ ; the component vector is $(\vec{A}\cdot\hat{B})\,\hat{B}$ .
$\vec{A}\times\vec{B} = \begin{vmatrix}\hat{i} & \hat{j} & \hat{k}\\ A_x & A_y & A_z\\ B_x & B_y & B_z\end{vmatrix}$ , of size $AB\sin\theta$ ; $\vec{A}\times\vec{A} = 0$ and $\vec{B}\times\vec{A} = -\vec{A}\times\vec{B}$ .
Unit vector perpendicular to both: $\dfrac{\vec{A}\times\vec{B}}{|\vec{A}\times\vec{B}|}$ .
Three vectors are coplanar when $\vec{A}\cdot(\vec{B}\times\vec{C}) = 0$ , the 3 × 3 determinant of their components.
$\vec{A}\cdot\vec{B} = |\vec{A}\times\vec{B}|$ means $\cos\theta = \sin\theta$ , so $\theta = 45^{\circ}$ .

Products and projection

\vec{A}\cdot\vec{B} = AB\cos\theta \qquad |\vec{A}\times\vec{B}| = AB\sin\theta \qquad \text{proj}_{B}\vec{A} = \frac{\vec{A}\cdot\vec{B}}{|\vec{B}|}

Worked example

For

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

and

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

, find the angle between them and a unit vector perpendicular to both.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2025 · 23 January 2025 · Q22Moderate

Example 3 · Motion in a Plane · Vectors: Resultant, Components and Products

Two particles are located at equal distance from origin. The position vectors of those are represented by

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

and

\overrightarrow{B}= 2\widehat{i}- 2\widehat{j}+ 4p\widehat{k}

, respectively. If both the vectors are at right angle to each other, the value of

n^{- 1}

\_\_\_\_

Dividing the projection by the wrong length

The projection of A on B is A·B divided by |B|, the vector you project ON. Dividing by |A| gives the projection of B on A.

Order matters in a cross product

B × A = −(A × B). The size is the same, but the direction flips, which changes the answer whenever a question asks for a direction or a unit vector.

Drill 10 more on dot product, cross product and projection

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)

Resultant of two vectors at an angle
Resultant of two vectors
$R = \sqrt{A^{2} + B^{2} + 2AB\cos\theta} \qquad \tan\alpha = \frac{B\sin\theta}{A + B\cos\theta} \qquad |\vec{A} + \vec{B}| = 2A\cos\frac{\theta}{2}\ (A = B)$
Resolving vectors into components
Components and magnitude
$A_x = A\cos\theta,\ A_y = A\sin\theta \qquad A = \sqrt{A_x^{2} + A_y^{2} + A_z^{2}} \qquad \vec{A}\ \text{along}\ \vec{B} = A\hat{B}$
Dot product, cross product and projection
Products and projection
$\vec{A}\cdot\vec{B} = AB\cos\theta \qquad |\vec{A}\times\vec{B}| = AB\sin\theta \qquad \text{proj}_{B}\vec{A} = \frac{\vec{A}\cdot\vec{B}}{|\vec{B}|}$

Watch out for (6)

Adding magnitudes
→ Resultant of two vectors at an angle
Perpendicular to A, not to B
→ Resultant of two vectors at an angle
Angle with the y-axis
→ Resolving vectors into components
Losing the quadrant
→ Resolving vectors into components
Dividing the projection by the wrong length
→ Dot product, cross product and projection
Order matters in a cross product
→ Dot product, cross product and projection

Test yourself on Motion in a Plane

20 past JEE Mains questions from this chapter, timed at 48 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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