Author: Rishabh Kumar (IIT + ISI | Global Math Mentor)**
Published: October 2025
Category: Vectors | Geometry in Space
🔹 Introduction
Vectors describe direction and magnitude — but to analyze how two vectors relate, we need tools like the scalar (dot) product and vector projection.
These concepts reveal:
- How much one vector aligns with another,
- The angle between them, and
- How to break a vector into meaningful geometric components.
Scalar product measures “alignment”; projection measures “influence.”
⚡️ 1️⃣ The Scalar (Dot) Product
The scalar product (or dot product) of two vectors gives a single number — a scalar — that quantifies their directional relationship.
For vectors a\mathbf{a}a and b\mathbf{b}b: a⋅b=∣a∣ ∣b∣ cosθ\mathbf{a} \cdot \mathbf{b} = |\mathbf{a}|\,|\mathbf{b}|\,\cos\thetaa⋅b=∣a∣∣b∣cosθ
Where θ\thetaθ = angle between a\mathbf{a}a and b\mathbf{b}b.
🔹 Algebraic Form
If a=a1i+a2j+a3k,b=b1i+b2j+b3k,\mathbf{a} = a_1\mathbf{i} + a_2\mathbf{j} + a_3\mathbf{k}, \quad \mathbf{b} = b_1\mathbf{i} + b_2\mathbf{j} + b_3\mathbf{k},a=a1i+a2j+a3k,b=b1i+b2j+b3k,
then a⋅b=a1b1+a2b2+a3b3\mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3a⋅b=a1b1+a2b2+a3b3
✅ Produces a scalar value (number), not a vector.
🔹 Geometric Meaning
a⋅b=∣a∣∣b∣cosθ\mathbf{a} \cdot \mathbf{b} = |\mathbf{a}||\mathbf{b}|\cos\thetaa⋅b=∣a∣∣b∣cosθ
- If θ=0∘\theta = 0^\circθ=0∘ → vectors are parallel, a⋅b=∣a∣∣b∣\mathbf{a}\cdot\mathbf{b} = |\mathbf{a}||\mathbf{b}|a⋅b=∣a∣∣b∣
- If θ=90∘\theta = 90^\circθ=90∘ → vectors are perpendicular, a⋅b=0\mathbf{a}\cdot\mathbf{b} = 0a⋅b=0
- If θ=180∘\theta = 180^\circθ=180∘ → vectors are opposite, a⋅b=−∣a∣∣b∣\mathbf{a}\cdot\mathbf{b} = -|\mathbf{a}||\mathbf{b}|a⋅b=−∣a∣∣b∣
The sign of the dot product shows whether vectors point in roughly the same or opposite direction.
🔹 Example 1 — Finding the Angle Between Vectors
a=2i+3j,b=4i+j\mathbf{a} = 2\mathbf{i} + 3\mathbf{j}, \quad \mathbf{b} = 4\mathbf{i} + \mathbf{j}a=2i+3j,b=4i+j a⋅b=(2)(4)+(3)(1)=11\mathbf{a}\cdot\mathbf{b} = (2)(4) + (3)(1) = 11a⋅b=(2)(4)+(3)(1)=11 ∣a∣=22+32=13,∣b∣=42+12=17|\mathbf{a}| = \sqrt{2^2 + 3^2} = \sqrt{13}, \quad |\mathbf{b}| = \sqrt{4^2 + 1^2} = \sqrt{17}∣a∣=22+32=13,∣b∣=42+12=17 cosθ=111317=11221\cos\theta = \frac{11}{\sqrt{13}\sqrt{17}} = \frac{11}{\sqrt{221}}cosθ=131711=22111 θ=cos−1(11221)\theta = \cos^{-1}\left(\frac{11}{\sqrt{221}}\right)θ=cos−1(22111)
✅ Angle ≈ 42.3°
🔹 Example 2 — Testing Perpendicularity
a=3i−2j,b=2i+3j\mathbf{a} = 3\mathbf{i} – 2\mathbf{j}, \quad \mathbf{b} = 2\mathbf{i} + 3\mathbf{j}a=3i−2j,b=2i+3j a⋅b=(3)(2)+(−2)(3)=0\mathbf{a}\cdot\mathbf{b} = (3)(2) + (-2)(3) = 0a⋅b=(3)(2)+(−2)(3)=0
✅ Vectors are perpendicular.
🎯 2️⃣ Vector Projection
While the dot product tells you how aligned two vectors are, the projection tells you how much of one vector lies along another.
Think of shining a light — the shadow of one vector on another is its projection.
🔹 Scalar Projection (Magnitude Only)
The scalar projection of a\mathbf{a}a on b\mathbf{b}b is: compb(a)=a⋅b∣b∣\text{comp}_{\mathbf{b}}(\mathbf{a}) = \frac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{b}|}compb(a)=∣b∣a⋅b
✅ This gives a number (positive or negative) depending on direction.
🔹 Vector Projection (With Direction)
The vector projection of a\mathbf{a}a on b\mathbf{b}b is: projb(a)=a⋅b∣b∣2 b\text{proj}_{\mathbf{b}}(\mathbf{a}) = \frac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{b}|^2}\,\mathbf{b}projb(a)=∣b∣2a⋅bb
✅ This gives an actual vector lying on b\mathbf{b}b.
🔹 Example — Projection of One Vector on Another
a=3i+4j,b=4i+0j\mathbf{a} = 3\mathbf{i} + 4\mathbf{j}, \quad \mathbf{b} = 4\mathbf{i} + 0\mathbf{j}a=3i+4j,b=4i+0j a⋅b=(3)(4)+(4)(0)=12\mathbf{a}\cdot\mathbf{b} = (3)(4) + (4)(0) = 12a⋅b=(3)(4)+(4)(0)=12 ∣b∣2=42+02=16|\mathbf{b}|^2 = 4^2 + 0^2 = 16∣b∣2=42+02=16 projb(a)=1216 b=34(4i)=3i\text{proj}_{\mathbf{b}}(\mathbf{a}) = \frac{12}{16}\,\mathbf{b} = \frac{3}{4}(4\mathbf{i}) = 3\mathbf{i}projb(a)=1612b=43(4i)=3i
✅ Vector projection = 3i3\mathbf{i}3i
(only the horizontal component of a\mathbf{a}a lies along b\mathbf{b}b).
🔹 Geometric Insight
If a\mathbf{a}a makes an angle θ\thetaθ with b\mathbf{b}b:
- Projection length = ∣a∣cosθ|\mathbf{a}|\cos\theta∣a∣cosθ
- Projection vector = ∣a∣cosθ b^|\mathbf{a}|\cos\theta \, \hat{\mathbf{b}}∣a∣cosθb^, where b^=b∣b∣\hat{\mathbf{b}} = \frac{\mathbf{b}}{|\mathbf{b}|}b^=∣b∣b
Scalar product = (Projection length) × (Magnitude of other vector).
🔹 Example — Work Done in Physics
Work WWW done by a constant force F\mathbf{F}F moving an object through displacement s\mathbf{s}s: W=F⋅s=∣F∣∣s∣cosθW = \mathbf{F}\cdot\mathbf{s} = |\mathbf{F}||\mathbf{s}|\cos\thetaW=F⋅s=∣F∣∣s∣cosθ
✅ The scalar product measures the useful component of force along motion — a real-world projection example.
🔹 Summary Table
| Concept | Formula | Result Type | Meaning |
|---|---|---|---|
| Dot Product | ( \mathbf{a}\cdot\mathbf{b} = | \mathbf{a} | |
| Scalar Projection | ( \frac{\mathbf{a}\cdot\mathbf{b}}{ | \mathbf{b} | } ) |
| Vector Projection | ( \frac{\mathbf{a}\cdot\mathbf{b}}{ | \mathbf{b} | ^2}\mathbf{b} ) |
🔹 Common Mistakes
- ❌ Forgetting the direction in vector projection.
- ❌ Mixing up ∣b∣|\mathbf{b}|∣b∣ and ∣b∣2|\mathbf{b}|^2∣b∣2 in denominator.
- ❌ Using degrees without converting to radians (in trig-based dot products).
- ❌ Thinking projection is perpendicular — it’s along the other vector.
🌟 Why It Matters
Scalar product and projections unite geometry and physics — used in:
- Calculating work, forces, and angles,
- Understanding orthogonality and perpendicularity,
- Analyzing motion and mechanics,
- Computing components in 3D vector spaces (IB HL / A Level P4).
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