The Hook: Spider-Man’s Web Swing
When Spider-Man swings through Manhattan, he shoots two webs simultaneously at different buildings. Each web pulls him with a force:
- Web 1: 500 N towards Empire State Building
- Web 2: 300 N towards Chrysler Building
The Question: What’s the total (resultant) force on Spider-Man? Can we just add 500 + 300 = 800 N?
NO! Forces have direction. We need vector addition, not simple arithmetic.
This is why you can’t lift a car by pulling from 10 different directions with 100 N each!
JEE Weightage: Vector addition appears in 3-4 questions in JEE Main, 4-5 in JEE Advanced. Essential for all mechanics!
The Core Concept
What is Vector Addition?
Vector addition combines two or more vectors to find a single resultant vector that has the same effect.
Key Insight: You CANNOT add magnitudes directly. Direction matters!
In simple terms: If two forces pull you, you don’t move in either direction alone - you move along the resultant direction.
Interactive Demo: Vector Addition Visualization
Experiment with adding different vectors visually! See how the parallelogram law and triangle law work in real-time. Try different magnitudes and angles to build intuition.
Method 1: Triangle Law of Vector Addition
If two vectors $\vec{a}$ and $\vec{b}$ are represented by two sides of a triangle taken in order, then their resultant is the third side taken in opposite order.
$$\boxed{\vec{R} = \vec{a} + \vec{b}}$$How to Apply Triangle Law
Steps:
- Draw vector $\vec{a}$
- From the head of $\vec{a}$, draw vector $\vec{b}$
- The resultant goes from tail of $\vec{a}$ to head of $\vec{b}$
Memory Trick: “Tail to Head, Then Head” → TH-TH
Visual Representation
→b
-----→
↗
→a Resultant R = a + b (closing the triangle)
A person walks 3 km East, then 4 km North. Find resultant displacement.
Solution:
- $\vec{a} = 3\hat{i}$ (East)
- $\vec{b} = 4\hat{j}$ (North)
- Resultant: $\vec{R} = 3\hat{i} + 4\hat{j}$
Magnitude: $|\vec{R}| = \sqrt{9 + 16} = 5$ km
Direction: $\tan\theta = \frac{4}{3} \Rightarrow \theta = 53.13°$ North of East
Method 2: Parallelogram Law of Vector Addition
If two vectors $\vec{a}$ and $\vec{b}$ are represented by two adjacent sides of a parallelogram, then their resultant is the diagonal passing through their common point.
$$\boxed{\vec{R} = \vec{a} + \vec{b}}$$How to Apply Parallelogram Law
Steps:
- Draw both vectors $\vec{a}$ and $\vec{b}$ from the same point
- Complete the parallelogram
- Draw diagonal from common point - this is resultant
Magnitude of Resultant
If angle between $\vec{a}$ and $\vec{b}$ is $\theta$:
$$\boxed{|\vec{R}| = \sqrt{a^2 + b^2 + 2ab\cos\theta}}$$Memory Trick: “Resultant Resembles Cosine Law” → Like $c^2 = a^2 + b^2 - 2ab\cos C$ but with +2ab cos θ
Direction of Resultant
If $\vec{R}$ makes angle $\alpha$ with $\vec{a}$:
$$\boxed{\tan\alpha = \frac{b\sin\theta}{a + b\cos\theta}}$$Special Cases (High-Yield for JEE!)
| Angle θ | Condition | Resultant Magnitude | Direction |
|---|---|---|---|
| 0° | Parallel, same direction | $R = a + b$ | Along both vectors |
| 90° | Perpendicular | $R = \sqrt{a^2 + b^2}$ | $\tan\alpha = \frac{b}{a}$ |
| 180° | Opposite directions | $R = | a - b |
| 60° | Common in JEE | $R = \sqrt{a^2 + b^2 + ab}$ | - |
| 120° | Common in JEE | $R = \sqrt{a^2 + b^2 - ab}$ | - |
When $|\vec{a}| = |\vec{b}| = a$ (equal magnitudes):
$$\boxed{|\vec{R}| = 2a\cos\frac{\theta}{2}}$$Special values:
- θ = 60° → $R = a\sqrt{3}$
- θ = 90° → $R = a\sqrt{2}$
- θ = 120° → $R = a$
Appears in 50% of JEE vector problems!
Two forces 5 N and 10 N act at 60° to each other. Find resultant.
Solution:
$$R = \sqrt{5^2 + 10^2 + 2(5)(10)\cos 60°}$$ $$= \sqrt{25 + 100 + 2(50)\left(\frac{1}{2}\right)}$$ $$= \sqrt{25 + 100 + 50} = \sqrt{175} = 5\sqrt{7} \text{ N}$$Direction: $\tan\alpha = \frac{10\sin 60°}{5 + 10\cos 60°} = \frac{10 \cdot \frac{\sqrt{3}}{2}}{5 + 10 \cdot \frac{1}{2}} = \frac{5\sqrt{3}}{10} = \frac{\sqrt{3}}{2}$
Method 3: Component Method (Most Powerful!)
Most efficient for JEE! Break vectors into components along perpendicular axes, add separately, then combine.
For vectors in 2D:
$$\vec{a} = a_x\hat{i} + a_y\hat{j}$$ $$\vec{b} = b_x\hat{i} + b_y\hat{j}$$Resultant:
$$\boxed{\vec{R} = (a_x + b_x)\hat{i} + (a_y + b_y)\hat{j}}$$For 3D Vectors
$$\boxed{\vec{R} = (a_x + b_x)\hat{i} + (a_y + b_y)\hat{j} + (a_z + b_z)\hat{k}}$$Simply: Add corresponding components!
Steps for Component Method
- Resolve each vector into components
- Add all x-components → $R_x$
- Add all y-components → $R_y$
- Add all z-components → $R_z$ (if 3D)
- Resultant: $\vec{R} = R_x\hat{i} + R_y\hat{j} + R_z\hat{k}$
Magnitude: $|\vec{R}| = \sqrt{R_x^2 + R_y^2 + R_z^2}$
Find resultant of:
- $\vec{a} = 2\hat{i} + 3\hat{j} - \hat{k}$
- $\vec{b} = \hat{i} - 2\hat{j} + 3\hat{k}$
- $\vec{c} = -\hat{i} + \hat{j} + 2\hat{k}$
Solution:
$$\vec{R} = (2+1-1)\hat{i} + (3-2+1)\hat{j} + (-1+3+2)\hat{k}$$ $$= 2\hat{i} + 2\hat{j} + 4\hat{k}$$Magnitude: $|\vec{R}| = \sqrt{4 + 4 + 16} = \sqrt{24} = 2\sqrt{6}$
Vector Subtraction
Simply: Reverse direction of $\vec{b}$, then add!
Component Method for Subtraction
$$\vec{a} - \vec{b} = (a_x - b_x)\hat{i} + (a_y - b_y)\hat{j} + (a_z - b_z)\hat{k}$$$\vec{a} = 5\hat{i} + 2\hat{j} + 3\hat{k}$, $\vec{b} = 2\hat{i} + \hat{j} + \hat{k}$. Find $\vec{a} - \vec{b}$.
Solution:
$$\vec{a} - \vec{b} = (5-2)\hat{i} + (2-1)\hat{j} + (3-1)\hat{k} = 3\hat{i} + \hat{j} + 2\hat{k}$$Properties of Vector Addition
1. Commutative Law
$$\boxed{\vec{a} + \vec{b} = \vec{b} + \vec{a}}$$Meaning: Order doesn’t matter in addition.
2. Associative Law
$$\boxed{(\vec{a} + \vec{b}) + \vec{c} = \vec{a} + (\vec{b} + \vec{c})}$$Meaning: Grouping doesn’t matter.
3. Additive Identity
$$\boxed{\vec{a} + \vec{0} = \vec{a}}$$Meaning: Adding zero vector doesn’t change anything.
4. Additive Inverse
$$\boxed{\vec{a} + (-\vec{a}) = \vec{0}}$$Meaning: Vector plus its negative gives zero vector.
Memory Tricks & Patterns
Mnemonic for Resultant Formula
Memory Trick: “Always Bring Cosine Twice” → $a^2 + b^2 + 2ab\cos\theta$
Compare with distance formula: In triangles, we subtract (law of cosines). In vector addition, we add!
Quick Angle Recognition
See these in problem? Use shortcuts:
| Given | Think | Formula |
|---|---|---|
| “Two equal forces” | $a = b$ | $R = 2a\cos\frac{\theta}{2}$ |
| “Perpendicular forces” | θ = 90° | $R = \sqrt{a^2 + b^2}$ |
| “Opposite directions” | θ = 180° | $R = |
| “60° or 120° angle” | Common JEE | Substitute cos 60° = 1/2 |
Saves 1 minute per problem!
Component Method Decision Tree
Triangle Law:
- Only 2 vectors
- Geometric/graphical representation needed
Parallelogram Law:
- 2 vectors, both from same point
- Need to find magnitude/direction formula
Component Method: ← USE THIS 90% OF THE TIME
- 3 or more vectors
- Vectors given in $\hat{i}, \hat{j}, \hat{k}$ form
- Any complex calculation
- Fastest for JEE!
Common Mistakes to Avoid
Wrong: Two forces 3 N and 4 N → Resultant = 7 N
Right: Depends on angle!
- If parallel (0°): R = 7 N ✓
- If perpendicular (90°): R = 5 N
- If opposite (180°): R = 1 N
JEE loves this trap in MCQs!
Wrong: Drawing both vectors from same point when using triangle law
Right:
- First vector: anywhere
- Second vector: starts from head of first vector
- Resultant: tail of first to head of second
Common Error: Using exterior angle instead of interior angle
Question: Two vectors at 120° to each other
Check: Is this the angle between vectors? If they’re adjacent sides of parallelogram, use this angle directly!
Wrong: $\vec{a} = 3\hat{i} - 2\hat{j}$, $\vec{b} = -\hat{i} + 4\hat{j}$ Addition: $(3+1)\hat{i} + (-2+4)\hat{j}$ = $4\hat{i} + 2\hat{j}$ ✗
Right: $(3 + (-1))\hat{i} + (-2 + 4)\hat{j}$ = $2\hat{i} + 2\hat{j}$ ✓
Tip: Write negative signs in brackets to avoid mistakes!
Wrong: Using $\tan\alpha = \frac{b}{a}$ always
Right:
- Perpendicular vectors (90°): $\tan\alpha = \frac{b}{a}$ ✓
- General angle θ: $\tan\alpha = \frac{b\sin\theta}{a + b\cos\theta}$ ✓
Practice Problems
Level 1: Foundation (NCERT Style)
Add vectors $\vec{a} = 3\hat{i} + 4\hat{j}$ and $\vec{b} = \hat{i} - 2\hat{j}$.
Solution:
$$\vec{R} = (3+1)\hat{i} + (4-2)\hat{j} = 4\hat{i} + 2\hat{j}$$Magnitude: $|\vec{R}| = \sqrt{16 + 4} = 2\sqrt{5}$
Two forces 12 N and 5 N act perpendicular to each other. Find resultant.
Solution:
$$R = \sqrt{12^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13 \text{ N}$$Direction: $\tan\alpha = \frac{5}{12}$, so $\alpha = \tan^{-1}(5/12) \approx 22.6°$
Two forces 8 N and 3 N act in opposite directions. Find resultant.
Solution:
$$R = |8 - 3| = 5 \text{ N in direction of larger force (8 N)}$$Level 2: JEE Main Type
Two equal forces each of magnitude 10 N act at 120° to each other. Find resultant.
Solution: Method 1: Using equal magnitude formula
$$R = 2(10)\cos\frac{120°}{2} = 20\cos 60° = 20 \times \frac{1}{2} = 10 \text{ N}$$Method 2: General formula
$$R = \sqrt{10^2 + 10^2 + 2(10)(10)\cos 120°}$$ $$= \sqrt{100 + 100 + 200(-\frac{1}{2})} = \sqrt{100} = 10 \text{ N}$$Find $\vec{a} + \vec{b} + \vec{c}$ where:
- $\vec{a} = 2\hat{i} - 3\hat{j} + 5\hat{k}$
- $\vec{b} = -\hat{i} + 4\hat{j} - 2\hat{k}$
- $\vec{c} = 3\hat{i} - \hat{j} - 3\hat{k}$
Solution:
$$\vec{R} = (2-1+3)\hat{i} + (-3+4-1)\hat{j} + (5-2-3)\hat{k}$$ $$= 4\hat{i} + 0\hat{j} + 0\hat{k} = 4\hat{i}$$Magnitude: $|\vec{R}| = 4$
Adjacent sides of parallelogram are $\vec{a} = 3\hat{i} + 4\hat{j}$ and $\vec{b} = 2\hat{i} - 3\hat{j}$. Find both diagonals.
Solution: Diagonal 1: $\vec{d_1} = \vec{a} + \vec{b} = 5\hat{i} + \hat{j}$, magnitude = $\sqrt{26}$
Diagonal 2: $\vec{d_2} = \vec{a} - \vec{b} = \hat{i} + 7\hat{j}$, magnitude = $\sqrt{50} = 5\sqrt{2}$
Level 3: JEE Advanced Type
If $\vec{a} = 2\hat{i} + 3\hat{j}$, $\vec{b} = \hat{i} - 2\hat{j}$, find $\vec{c}$ such that $2\vec{a} + 3\vec{b} - \vec{c} = \vec{0}$.
Solution:
$$\vec{c} = 2\vec{a} + 3\vec{b}$$ $$= 2(2\hat{i} + 3\hat{j}) + 3(\hat{i} - 2\hat{j})$$ $$= 4\hat{i} + 6\hat{j} + 3\hat{i} - 6\hat{j}$$ $$= 7\hat{i}$$Answer: $\vec{c} = 7\hat{i}$
Two vectors $\vec{a}$ and $\vec{b}$ have magnitudes 3 and 4. If $|\vec{a} + \vec{b}| = 5$, find angle between them.
Solution:
$$|\vec{a} + \vec{b}|^2 = a^2 + b^2 + 2ab\cos\theta$$ $$25 = 9 + 16 + 2(3)(4)\cos\theta$$ $$25 = 25 + 24\cos\theta$$ $$0 = 24\cos\theta$$ $$\cos\theta = 0$$ $$\theta = 90°$$Answer: Vectors are perpendicular!
If $\vec{a} = 2\hat{i} + 3\hat{j}$, $\vec{b} = \hat{i} - \hat{j}$, find vector $\vec{r}$ that divides $\vec{a} + \vec{b}$ and $\vec{a} - \vec{b}$ in ratio 2:1.
Solution:
$$\vec{a} + \vec{b} = 3\hat{i} + 2\hat{j}$$ $$\vec{a} - \vec{b} = \hat{i} + 4\hat{j}$$Using section formula (internal division):
$$\vec{r} = \frac{2(\hat{i}+4\hat{j}) + 1(3\hat{i}+2\hat{j})}{2+1}$$ $$= \frac{2\hat{i}+8\hat{j}+3\hat{i}+2\hat{j}}{3} = \frac{5\hat{i}+10\hat{j}}{3}$$Answer: $\vec{r} = \frac{5}{3}\hat{i} + \frac{10}{3}\hat{j}$
Connection to Physics
Critical Applications:
Equilibrium: When $\sum \vec{F} = \vec{0}$
Inclined Planes: Resolving weight into components
Projectile Motion: Velocity = horizontal + vertical components
- See: Projectile Motion
Relative Velocity: $\vec{v}_{AB} = \vec{v}_A - \vec{v}_B$
- See: Relative Motion
Electric/Magnetic Fields: Force on charge in multiple fields
- See: Electric Field
Pro Tip: Every JEE mechanics problem with 2+ forces needs vector addition. Master this, save 3-5 minutes per problem!
Quick Revision Box
| Situation | Method | Formula |
|---|---|---|
| 2 vectors, any angle | Parallelogram Law | $R = \sqrt{a^2 + b^2 + 2ab\cos\theta}$ |
| Equal magnitudes | Special formula | $R = 2a\cos(\theta/2)$ |
| Perpendicular (90°) | Pythagoras | $R = \sqrt{a^2 + b^2}$ |
| Parallel (0°) | Direct add | $R = a + b$ |
| Opposite (180°) | Subtract | $R = |
| Components given | Component method | Add each component |
| 3+ vectors | Always component | $R_x, R_y, R_z$ separately |
Direction Formula: $\tan\alpha = \frac{b\sin\theta}{a+b\cos\theta}$ (general case)
Interactive Visualization
Try This: Change angles and see how resultant changes!
- At θ = 0°: Resultant is maximum (a + b)
- At θ = 180°: Resultant is minimum (|a - b|)
- At θ = 90°: Resultant = $\sqrt{a^2 + b^2}$
What’s Next?
Now that you can add vectors, learn how to multiply them:
Scalar Product (Dot Product) - For work, projections, angles between vectors
Vector Product (Cross Product) - For torque, area, perpendicular vectors
Also explore:
- Motion in a Plane - Uses vector addition heavily
- 3D Geometry - Equations of lines using vectors
Teacher’s Summary
- Component method is king - use it unless specifically asked for triangle/parallelogram
- Never add magnitudes directly - direction always matters!
- Special angles (60°, 90°, 120°) appear in 70% of JEE problems - memorize formulas
- For equal magnitudes: $R = 2a\cos(\theta/2)$ is your best friend
- Triangle law = Tail to Head → Resultant closes the triangle
“In JEE, see multiple forces? Think vector addition. See components? Add them directly. This saves you 2 minutes per problem!”
Time-saving tip: For perpendicular vectors, directly use Pythagoras - no need for full formula!
Pro Tip: When exam gives you “$|\vec{a}| = 3$, $|\vec{b}| = 4$, $|\vec{a}+\vec{b}| = 5$” - immediately recognize 3-4-5 Pythagorean triplet → vectors are perpendicular! Saves 30 seconds!