Class 11 Physics · NCERT notes for NEET

Motion in a Plane: NCERT Notes, Formulas and Tricks for NEET

Vectors, then their use in 2D motion: projectiles, uniform circular motion and relative velocity. NEET asks a question from here almost every year, mostly projectile ratios, vector resultants and centripetal acceleration.

1. Vectors: addition, resolution, products
2. Projectile motion 🔒
3. Uniform circular motion 🔒
4. Relative velocity in two dimensions 🔒

Topic 1 is open for everyone. The full chapter, quick revision and practice open with a plan, or start free with 5 full chapters.

Open the interactive version (simulations and tap-to-answer quick checks).

Topic 1

Vectors: addition, resolution, products

Two people pull a box with 4 N and 3 N at right angles. The box feels 5 N, not 7 N. Directions change the answer.

Scalars have only magnitude (mass, distance, time, work). Vectors have magnitude and direction (displacement, velocity, force) and add by the triangle or parallelogram law.
Two vectors are equal only if both magnitude and direction match. $λ A$ with $λ > 0$ only stretches $A$ ; $λ < 0$ also reverses it.
Addition is commutative and associative. Subtraction means adding the reverse: $A - B = A + (- B)$ .
Null vector $0$ : zero magnitude, no definite direction. $A + 0 = A$ , $A - A = 0$ , $0 A = 0$ .
Unit vector $\overset{n}{^} = A /∣ A ∣$ has magnitude 1 and no unit or dimension. $\hat{i}, \hat{j}, \hat{k}$ point along $x, y, z$ and are mutually perpendicular.
Resolution: a vector at angle $θ$ with the $x$ -axis has $A_{x} = A cos θ$ , $A_{y} = A sin θ$ . In space $A_{x} = A cos α$ , $A_{y} = A cos β$ , $A_{z} = A cos γ$ with $cos^{2} α + cos^{2} β + cos^{2} γ = 1$ .
Analytical addition: add component by component, $R_{x} = A_{x} + B_{x}$ , $R_{y} = A_{y} + B_{y}$ , $R_{z} = A_{z} + B_{z}$ .
The resultant of two vectors always lies between $∣ A - B ∣$ (opposite) and $A + B$ (parallel). Perpendicular vectors give $A^{2} + B^{2}$ .
Dot product $A \cdot B = A B cos θ$ is a scalar; zero when $A ⊥ B$ . $\hat{i} \cdot \hat{i} = 1$ , $\hat{i} \cdot \hat{j} = 0$ .
Cross product $∣ A \times B ∣ = A B sin θ$ , direction by the right-hand rule, zero for parallel vectors; $A \times B = - B \times A$ . Cyclic: $\hat{i} \times \hat{j} = \hat{k}$ , $\hat{j} \times \hat{k} = \hat{i}$ , $\hat{k} \times \hat{i} = \hat{j}$ .

Parallelogram law: put A and B tail to tail; the diagonal from O is R. theta is the angle between A and B, alpha is the angle of R from A. — Parallelogram law: put $A$ and $B$ tail to tail; the diagonal from O is $R$ . $θ$ is the angle between $A$ and $B$ , $α$ is the angle of $R$ from $A$ .

Resolution: drop perpendiculars from the head of A. Cos goes with the side next to theta, sin with the side opposite. — Resolution: drop perpendiculars from the head of $A$ . Cos goes with the side next to $θ$ , sin with the side opposite.

For A = 4, B = 3 the resultant slides from 7 (same direction) to 1 (opposite). At 90^circ it is 5. Any answer outside 1 to 7 is impossible. — For $A = 4$ , $B = 3$ the resultant slides from 7 (same direction) to 1 (opposite). At $9 0^{\circ}$ it is 5. Any answer outside 1 to 7 is impossible.

Simulation: Add two vectors: change the angleTry it

Formulas

$R = A^{2} + B^{2} + 2 A B cos θ$
Magnitude of $A + B$ ; $θ$ = angle between them
$tan α = \frac{B sin θ}{A + B cos θ}$
Direction of $R$ measured from $A$
$\frac{R}{sin θ} = \frac{A}{sin β} = \frac{B}{sin α}$
Law of sines in the vector triangle
$∣ A - B ∣ = A^{2} + B^{2} - 2 A B cos θ$
Magnitude of the difference (note the minus sign)
$A = A_{x}^{2} + A_{y}^{2} + A_{z}^{2}$ , $tan θ = A_{y} / A_{x}$
Magnitude and direction from components
$A \cdot B = A_{x} B_{x} + A_{y} B_{y} + A_{z} B_{z}$
Then $cos θ = A \cdot B / (A B)$
$∣ A \times B ∣ = A B sin θ$
Also the area of the parallelogram made by $A$ and $B$

Tips, tricks and hacks

Two equal vectorsFor equal magnitudes

A

at angle

θ

R = 2 A cos (θ /2)

. Memorise:

0^{\circ} \to 2 A

6 0^{\circ} \to 3 A

9 0^{\circ} \to 2 A

12 0^{\circ} \to A

18 0^{\circ} \to 0

Range check firstBefore using the formula, strike out options outside

∣ A - B ∣ \leq R \leq A + B

. Forces 5 N and 8 N can give 3 N to 13 N only, so 2 N or 14 N is wrong at once.

Dot zero testTo check perpendicularity just multiply matching components.

(2 \hat{i} + 3 \hat{j}) \cdot (3 \hat{i} - 2 \hat{j}) = 6 - 6 = 0

, so they are at

9 0^{\circ}

. No angle formula needed.

Cyclic i-j-kWrite i, j, k on a circle clockwise. Going with the arrow gives

+

(

\hat{i} \times \hat{j} = \hat{k}

), against it gives

-

(

\hat{j} \times \hat{i} = - \hat{k}

). Same letter twice gives zero.

Common mistakes

✗ Adding magnitudes: 3 N + 4 N = 7 N always.

✓ Only when they point the same way. In general use $A^{2} + B^{2} + 2 A B cos θ$ ; at $9 0^{\circ}$ it is 5 N.

✗ Using $+ 2 A B cos θ$ for $∣ A - B ∣$ .

✓ Reversing $B$ changes the angle to $18 0^{\circ} - θ$ , so the sign becomes minus.

✗ Taking $θ$ as the angle when vectors are drawn head to tail.

✓ $θ$ in the formula is the angle with tails joined. Shift the vectors tail to tail before reading the angle.

✗ Thinking $A \times B = B \times A$ like the dot product.

✓ Cross product is anti-commutative; swapping the order flips the direction.

NEET traps

Two vectors of unequal size can never add to zero; three coplanar vectors can (if they form a closed triangle).

Current and pressure have a direction in daily talk but are scalars; current does not add by the vector law.

A unit vector has no unit. " $\overset{n}{^}$ has unit of the original vector" is a wrong option.

Quick check: Two forces of equal magnitude

F

act on a body and the resultant also has magnitude

F

. The angle between the forces is

$6 0^{\circ}$
$9 0^{\circ}$
$12 0^{\circ}$
$15 0^{\circ}$

Answer: C. $2 F cos (θ /2) = F$ gives $cos (θ /2) = 1/2$ , so $θ /2 = 6 0^{\circ}$ and $θ = 12 0^{\circ}$ .

Topic 2 · 🔒

Projectile motion

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Topic 3 · 🔒

Uniform circular motion

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Topic 4 · 🔒

Relative velocity in two dimensions

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