The Hook: Thor’s Hammer and Work Done
In Avengers: Age of Ultron, when Thor tries to pull his hammer Mjolnir up but it doesn’t move, he’s applying force vertically upward.
Question: Is Thor doing any work?
Answer: NO! Because work = Force × Displacement × cos(angle)
If displacement is zero (hammer doesn’t move), work = 0.
But if displacement is perpendicular to force (like pushing a wall while walking sideways), work is also zero because cos(90°) = 0!
This “multiplication considering angle” is the scalar product - the dot product.
JEE Weightage: Scalar product appears in 3-4 questions in JEE Main, 4-5 in JEE Advanced. Essential for work-energy, projections!
The Core Concept
What is Scalar Product?
The scalar product (or dot product) of two vectors $\vec{a}$ and $\vec{b}$ is defined as:
$$\boxed{\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos\theta}$$where θ is the angle between the two vectors.
Result: A scalar (just a number, no direction!)
In simple terms: Dot product measures “how much” one vector goes in the direction of another.
Interactive Demo: Vector Operations Visualizer
Explore vector operations interactively. Switch to “A . B” tab to see dot product, projection, and perpendicularity in action.
Why “Dot” Product?
Written with a dot: $\vec{a} \cdot \vec{b}$ (read as “a dot b”)
Alternative Notation
$$\vec{a} \cdot \vec{b} = ab\cos\theta$$where $a = |\vec{a}|$ and $b = |\vec{b}|$
Component Form (Most Useful for JEE!)
If $\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}$ and $\vec{b} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k}$, then:
$$\boxed{\vec{a} \cdot \vec{b} = a_1b_1 + a_2b_2 + a_3b_3}$$Memory Trick: “Multiply Corresponding Components, then Add” → MCCA
For 2D vectors:
$$\vec{a} \cdot \vec{b} = a_1b_1 + a_2b_2$$Find $\vec{a} \cdot \vec{b}$ where $\vec{a} = 2\hat{i} + 3\hat{j} + \hat{k}$ and $\vec{b} = \hat{i} - 2\hat{j} + 3\hat{k}$.
Solution:
$$\vec{a} \cdot \vec{b} = (2)(1) + (3)(-2) + (1)(3)$$ $$= 2 - 6 + 3 = -1$$Answer: -1 (a scalar!)
Properties of Scalar Product
1. Commutative Law
$$\boxed{\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a}}$$Meaning: Order doesn’t matter.
2. Distributive Law
$$\boxed{\vec{a} \cdot (\vec{b} + \vec{c}) = \vec{a} \cdot \vec{b} + \vec{a} \cdot \vec{c}}$$Meaning: Can distribute dot product over addition.
3. Scalar Multiplication
$$\boxed{(m\vec{a}) \cdot \vec{b} = m(\vec{a} \cdot \vec{b}) = \vec{a} \cdot (m\vec{b})}$$where m is a scalar.
4. Not Associative
$$(\vec{a} \cdot \vec{b}) \cdot \vec{c} \text{ is meaningless!}$$Why? $\vec{a} \cdot \vec{b}$ gives a scalar, and you can’t dot a scalar with a vector!
5. Dot Product with Itself
$$\boxed{\vec{a} \cdot \vec{a} = |\vec{a}|^2 = a^2}$$Super useful! This gives magnitude squared.
Dot Product of Unit Vectors
Dot product of same unit vectors:
$$\hat{i} \cdot \hat{i} = \hat{j} \cdot \hat{j} = \hat{k} \cdot \hat{k} = 1$$Dot product of different unit vectors:
$$\hat{i} \cdot \hat{j} = \hat{j} \cdot \hat{k} = \hat{k} \cdot \hat{i} = 0$$Memory Trick: “Same = 1, Different = 0” (because perpendicular unit vectors!)
Why? Angle between same vectors = 0°, cos(0°) = 1 Angle between perpendicular vectors = 90°, cos(90°) = 0
Finding Angle Between Vectors
From $\vec{a} \cdot \vec{b} = ab\cos\theta$:
$$\boxed{\cos\theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}}$$This appears in 50% of JEE dot product problems!
Steps:
- Calculate $\vec{a} \cdot \vec{b}$ using component form
- Find $|\vec{a}|$ and $|\vec{b}|$
- Use $\cos\theta = \frac{\vec{a} \cdot \vec{b}}{ab}$
- Find θ = $\cos^{-1}(...)$
Find angle between $\vec{a} = \hat{i} + \hat{j} + \hat{k}$ and $\vec{b} = \hat{i} + \hat{j}$.
Solution:
- $\vec{a} \cdot \vec{b} = (1)(1) + (1)(1) + (1)(0) = 2$
- $|\vec{a}| = \sqrt{1+1+1} = \sqrt{3}$
- $|\vec{b}| = \sqrt{1+1} = \sqrt{2}$
- $\cos\theta = \frac{2}{\sqrt{3} \cdot \sqrt{2}} = \frac{2}{\sqrt{6}} = \frac{\sqrt{6}}{3}$
- $\theta = \cos^{-1}\left(\frac{\sqrt{6}}{3}\right) \approx 35.26°$
Perpendicular and Parallel Vectors
Perpendicular Vectors (Orthogonal)
$$\boxed{\vec{a} \perp \vec{b} \iff \vec{a} \cdot \vec{b} = 0}$$Why? θ = 90° → cos(90°) = 0
To check if vectors perpendicular: Just calculate dot product!
- If $\vec{a} \cdot \vec{b} = 0$ → Perpendicular ✓
- If $\vec{a} \cdot \vec{b} \neq 0$ → Not perpendicular ✗
No need to find angle! Saves 30 seconds.
Parallel Vectors (Collinear)
$$\boxed{\vec{a} \parallel \vec{b} \iff \vec{a} \cdot \vec{b} = \pm ab}$$Why? θ = 0° or 180° → cos θ = ±1
- Same direction (0°): $\vec{a} \cdot \vec{b} = ab$ (positive)
- Opposite direction (180°): $\vec{a} \cdot \vec{b} = -ab$ (negative)
For what value of k are vectors $\vec{a} = k\hat{i} + 2\hat{j} + \hat{k}$ and $\vec{b} = 2\hat{i} - \hat{j} + 3\hat{k}$ perpendicular?
Solution: For perpendicular vectors: $\vec{a} \cdot \vec{b} = 0$
$$k(2) + (2)(-1) + (1)(3) = 0$$ $$2k - 2 + 3 = 0$$ $$2k + 1 = 0$$ $$k = -\frac{1}{2}$$Projection of Vectors
Projection of $\vec{a}$ on $\vec{b}$:
$$\boxed{\text{proj}_{\vec{b}}\vec{a} = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|} = |\vec{a}|\cos\theta}$$Vector projection:
$$\boxed{\text{proj}_{\vec{b}}\vec{a} \cdot \hat{b} = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|^2}\vec{b}}$$Memory Trick: “Dot divided by bottom” → $\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}$
Scalar projection = just the number (can be positive/negative) Vector projection = projection in direction of $\vec{b}$
Find projection of $\vec{a} = 2\hat{i} + 3\hat{j}$ on $\vec{b} = \hat{i} + \hat{j}$.
Solution:
$$\text{proj}_{\vec{b}}\vec{a} = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}$$ $$= \frac{(2)(1) + (3)(1)}{\sqrt{1+1}} = \frac{5}{\sqrt{2}} = \frac{5\sqrt{2}}{2}$$Work Done (Physics Application)
Work done by force $\vec{F}$ causing displacement $\vec{s}$:
$$\boxed{W = \vec{F} \cdot \vec{s} = Fs\cos\theta}$$where θ is angle between force and displacement.
See: Work Energy Theorem
Special Cases:
- θ = 0° (force along displacement): W = Fs (maximum work)
- θ = 90° (force perpendicular): W = 0 (no work!)
- θ = 180° (force opposite): W = -Fs (negative work)
Force $\vec{F} = 3\hat{i} + 4\hat{j}$ N causes displacement $\vec{s} = 2\hat{i} - \hat{j}$ m. Find work done.
Solution:
$$W = \vec{F} \cdot \vec{s} = (3)(2) + (4)(-1) = 6 - 4 = 2 \text{ J}$$Answer: 2 Joules
Memory Tricks & Patterns
Mnemonic for Dot Product
Memory Trick: “Dot gives Decimal” → Result is a scalar (number)
Memory Trick for Components: “Multiply Corresponding Components Add” → MCCA
Pattern Recognition
If problem says “perpendicular” or “orthogonal”:
- Immediately write: $\vec{a} \cdot \vec{b} = 0$
- Solve for unknown
Appears in: 60% of vector perpendicularity problems
See $\hat{i}, \hat{j}, \hat{k}$ in calculation?
- Same vectors: 1
- Different vectors: 0
Quick expansion:
$$(\hat{i}+2\hat{j}) \cdot (3\hat{i}-\hat{j}) = 3(1) + (-2)(1) = 1$$No need to write full components!
Need magnitude? Use self dot product:
$$|\vec{a}|^2 = \vec{a} \cdot \vec{a}$$Example: $\vec{a} = 3\hat{i} + 4\hat{j}$
$$|\vec{a}|^2 = \vec{a} \cdot \vec{a} = 9 + 16 = 25 \Rightarrow |\vec{a}| = 5$$When to Use Dot Product
Use scalar product when:
- Finding angle between vectors
- Checking if vectors are perpendicular
- Finding projection of one vector on another
- Calculating work done (physics)
- Finding component of vector along another direction
Don’t use when:
- Need result to be a vector → Use cross product
- Finding area, torque → Use cross product
Common Mistakes to Avoid
Wrong: $\cos\theta = \frac{|\vec{a}||\vec{b}|}{\vec{a} \cdot \vec{b}}$ (inverted!)
Right: $\cos\theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}$ (dot product in numerator)
Wrong: $(\vec{a} \cdot \vec{b}) + \vec{c}$ (can’t add scalar to vector!)
Right: Can only add scalar to scalar, vector to vector
Question: “Find component of $\vec{a}$ along $\vec{b}$”
Wrong: Using $\frac{\vec{a} \cdot \vec{b}}{|\vec{a}|}$
Right: Using $\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}$ (divide by denominator’s magnitude!)
Component of A along B → divide by |B|
Scenario: Force opposes displacement
Wrong: Work = Fs (always positive)
Right: Check angle!
- 0° ≤ θ < 90°: Positive work
- θ = 90°: Zero work
- 90° < θ ≤ 180°: Negative work
Use: $W = Fs\cos\theta$ to get correct sign!
Wrong: $\hat{i} \cdot \hat{j} = 1$ (assuming all are 1)
Right:
- Same unit vectors: 1
- Different unit vectors: 0 (perpendicular!)
Remember: Standard unit vectors are mutually perpendicular!
Practice Problems
Level 1: Foundation (NCERT Style)
Find $\vec{a} \cdot \vec{b}$ where $\vec{a} = 3\hat{i} + 2\hat{j}$ and $\vec{b} = \hat{i} + 4\hat{j}$.
Solution:
$$\vec{a} \cdot \vec{b} = (3)(1) + (2)(4) = 3 + 8 = 11$$Find $|\vec{a}|$ where $\vec{a} = 2\hat{i} - 3\hat{j} + 6\hat{k}$ using dot product.
Solution:
$$|\vec{a}|^2 = \vec{a} \cdot \vec{a} = (2)^2 + (-3)^2 + (6)^2 = 4 + 9 + 36 = 49$$ $$|\vec{a}| = 7$$Calculate $(\hat{i} + \hat{j}) \cdot (2\hat{i} - 3\hat{j} + \hat{k})$.
Solution:
$$= (1)(2) + (1)(-3) + (0)(1) = 2 - 3 + 0 = -1$$Level 2: JEE Main Type
Find angle between $\vec{a} = 2\hat{i} + 3\hat{j} + \hat{k}$ and $\vec{b} = \hat{i} + 2\hat{j} - \hat{k}$.
Solution:
- $\vec{a} \cdot \vec{b} = (2)(1) + (3)(2) + (1)(-1) = 2 + 6 - 1 = 7$
- $|\vec{a}| = \sqrt{4+9+1} = \sqrt{14}$
- $|\vec{b}| = \sqrt{1+4+1} = \sqrt{6}$
- $\cos\theta = \frac{7}{\sqrt{14} \cdot \sqrt{6}} = \frac{7}{\sqrt{84}} = \frac{7}{2\sqrt{21}}$
- $\theta = \cos^{-1}\left(\frac{7}{2\sqrt{21}}\right)$
Find λ such that $\vec{a} = 2\hat{i} + \lambda\hat{j} + \hat{k}$ and $\vec{b} = \hat{i} - 2\hat{j} + 3\hat{k}$ are perpendicular.
Solution: For perpendicular: $\vec{a} \cdot \vec{b} = 0$
$$2(1) + \lambda(-2) + 1(3) = 0$$ $$2 - 2\lambda + 3 = 0$$ $$5 = 2\lambda$$ $$\lambda = \frac{5}{2}$$Force $\vec{F} = 2\hat{i} + 3\hat{j} - \hat{k}$ N displaces object by $\vec{s} = 3\hat{i} - 2\hat{j} + \hat{k}$ m. Find work done.
Solution:
$$W = \vec{F} \cdot \vec{s} = (2)(3) + (3)(-2) + (-1)(1)$$ $$= 6 - 6 - 1 = -1 \text{ J}$$Negative work means force opposes displacement!
Find projection of $\vec{a} = 2\hat{i} + 3\hat{j} + \hat{k}$ on $\vec{b} = \hat{i} + \hat{j} + \hat{k}$.
Solution:
$$\text{proj}_{\vec{b}}\vec{a} = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}$$ $$= \frac{(2)(1) + (3)(1) + (1)(1)}{\sqrt{1+1+1}} = \frac{6}{\sqrt{3}} = 2\sqrt{3}$$Level 3: JEE Advanced Type
If $|\vec{a}| = |\vec{b}| = 1$ and angle between them is 60°, find $|\vec{a} + \vec{b}|$.
Solution:
$$|\vec{a} + \vec{b}|^2 = (\vec{a} + \vec{b}) \cdot (\vec{a} + \vec{b})$$ $$= \vec{a} \cdot \vec{a} + 2\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{b}$$ $$= |\vec{a}|^2 + 2|\vec{a}||\vec{b}|\cos 60° + |\vec{b}|^2$$ $$= 1 + 2(1)(1)\left(\frac{1}{2}\right) + 1 = 1 + 1 + 1 = 3$$ $$|\vec{a} + \vec{b}| = \sqrt{3}$$If $\vec{a} = \hat{i} + 2\hat{j} + 3\hat{k}$, $\vec{b} = 2\hat{i} - \hat{j} + \hat{k}$, find vector $\vec{c}$ perpendicular to both and $\vec{c} \cdot (\hat{i} + \hat{j} + \hat{k}) = 6$.
Solution: Let $\vec{c} = x\hat{i} + y\hat{j} + z\hat{k}$
Perpendicular to $\vec{a}$: $x + 2y + 3z = 0$ … (1) Perpendicular to $\vec{b}$: $2x - y + z = 0$ … (2) Given condition: $x + y + z = 6$ … (3)
From (1): $x = -2y - 3z$ Substitute in (2): $2(-2y-3z) - y + z = 0$
$$-4y - 6z - y + z = 0$$ $$-5y - 5z = 0 \Rightarrow y = -z$$Substitute in (3): $x - z + z = 6 \Rightarrow x = 6$ From (1): $6 + 2y + 3z = 0$
Using $y = -z$: $6 - 2z + 3z = 0 \Rightarrow z = -6$, $y = 6$
Answer: $\vec{c} = 6\hat{i} + 6\hat{j} - 6\hat{k}$
Find angle between diagonals of parallelogram with adjacent sides $\vec{a} = 3\hat{i} + 4\hat{j}$ and $\vec{b} = \hat{i} + 2\hat{j}$.
Solution: Diagonals: $\vec{d_1} = \vec{a} + \vec{b} = 4\hat{i} + 6\hat{j}$ $\vec{d_2} = \vec{a} - \vec{b} = 2\hat{i} + 2\hat{j}$
$$\vec{d_1} \cdot \vec{d_2} = (4)(2) + (6)(2) = 8 + 12 = 20$$ $$|\vec{d_1}| = \sqrt{16+36} = \sqrt{52} = 2\sqrt{13}$$ $$|\vec{d_2}| = \sqrt{4+4} = 2\sqrt{2}$$ $$\cos\theta = \frac{20}{2\sqrt{13} \cdot 2\sqrt{2}} = \frac{20}{4\sqrt{26}} = \frac{5}{\sqrt{26}}$$ $$\theta = \cos^{-1}\left(\frac{5}{\sqrt{26}}\right)$$Connection to Physics
Critical Applications:
Work Done: $W = \vec{F} \cdot \vec{s}$
- See: Work Energy Theorem
Power: $P = \vec{F} \cdot \vec{v}$
- See: Power
Magnetic Force: $W = 0$ because $\vec{F}_B \perp \vec{v}$ (dot product = 0)
- See: Magnetic Force
Electric Potential: $V = -\int \vec{E} \cdot d\vec{l}$
- See: Electric Potential
Projection in Mechanics: Component of velocity along incline
Master dot product → Understand work-energy problems instantly!
Quick Revision Box
| Concept | Formula |
|---|---|
| Definition | $\vec{a} \cdot \vec{b} = ab\cos\theta$ |
| Component form | $\vec{a} \cdot \vec{b} = a_1b_1 + a_2b_2 + a_3b_3$ |
| Angle between vectors | $\cos\theta = \frac{\vec{a} \cdot \vec{b}}{ |
| Perpendicular condition | $\vec{a} \cdot \vec{b} = 0$ |
| Parallel condition | $\vec{a} \cdot \vec{b} = \pm ab$ |
| Magnitude | $ |
| Projection of a on b | $\frac{\vec{a} \cdot \vec{b}}{ |
| Work done | $W = \vec{F} \cdot \vec{s}$ |
| Unit vectors | Same = 1, Different = 0 |
What’s Next?
Now learn the other type of vector multiplication:
Vector Product (Cross Product) - For area, torque, perpendicular vectors
Triple Products - Scalar triple product (volume), vector triple product
Also explore:
- Work Energy Theorem - Uses dot product extensively
- Angle Between Lines - Uses dot product for angles
- Direction Cosines - Uses unit vector dot products
- Vectors Basics - Review fundamentals
Teacher’s Summary
- Dot product gives a scalar - just a number, no direction
- Component formula is fastest - multiply corresponding components, add
- Perpendicular ⟺ dot product = 0 - instant perpendicularity check
- Angle formula is high-yield - appears in 50% of problems
- Work = F·s - most important physics application
“In JEE, see ‘angle between vectors’? Think dot product. See ‘perpendicular’? Set dot product = 0. See ‘work done’? Use F·s. Master these three, score full marks!”
Time-saving tip: For perpendicularity, don’t find angle - just check if dot product is zero!
Pro Tip: When problem gives magnitudes and angle, use $ab\cos\theta$. When given components, use $a_1b_1 + a_2b_2 + a_3b_3$. Choose based on what’s given - saves 1 minute!