Part 1 · Chapter 03

Vectors

The mathematical language for quantities that carry a direction

Fundamentals of Physics Prof. Mithun Mondal Reading time ≈ 28 min
i What you'll learn
  • The difference between a scalar and a vector.
  • How to add and subtract vectors geometrically (tip-to-tail).
  • How to resolve a vector into components and rebuild it.
  • Unit-vector notation and adding vectors component by component.
  • The two products: the dot product (a scalar) and the cross product (a vector).
Section 3-1

What Is Physics?

Many physical quantities carry both a size and a direction — and to handle them we need a special language: the language of vectors. You already use it informally: "go five blocks down this street, then turn left" is a vector instruction. Physics and engineering lean on vectors constantly, especially once we reach rotation and magnetic forces.

Section 3-2

Vectors vs Scalars

Vector

Has magnitude and direction and follows special rules of combination. Examples: displacement, velocity, acceleration, force. Drawn as an arrow.

# Scalar

Has magnitude only and obeys ordinary algebra. Examples: temperature, mass, time, energy. A signed number with a unit is enough.

The simplest vector is displacement — a change of position drawn as an arrow from start to finish. A displacement vector says nothing about the path taken; only the overall change from beginning to end. And a vector can be slid around the page freely: as long as its length and direction don't change, it's the same vector.

Section 3-3

Adding Vectors Geometrically

To add two vectors by hand, use the tip-to-tail rule: draw the first to scale, draw the second starting at the head of the first, and the sum (or resultant) runs from the tail of the first to the head of the second.

Vector sum
\[ s = a + b \]
The "+" here means vector addition — it combines magnitudes and directions, not just numbers.

Vector addition obeys two familiar laws:

Two laws of vector addition
\[\begin{gathered} a + b = b + a \qquad \text{(commutative)} \\ (a + b) + c = a + (b + c) \qquad \text{(associative)} \end{gathered}\]
Order and grouping don't matter — the resultant is the same.

Subtraction is just adding the reverse. The vector \(-b\) has the same length as \(b\) but points the opposite way, so:

Vector subtraction
\[ d = a - b = a + (-b) \]
! Add only vectors of the same kind

You can add two displacements or two velocities, but adding a displacement to a velocity is meaningless — like trying to add 21 s to 12 m.

Quick check. If |a| = 3 m and |b| = 4 m, the resultant \(c = a + b\) can range from a maximum of 7 m (same direction) down to a minimum of 1 m (opposite directions).
Section 3-4

Components of a Vector

Drawing arrows is tidy but tedious. The powerful method is algebraic: drop a vector onto a set of axes and work with its components — its projections on the x and y axes. Finding them is called resolving the vector.

Resolving a vector (angle θ from +x axis)
\[ a_{x} = a \cos \theta \qquad a_{y} = a \sin \theta \]
A component's sign shows its direction along that axis. The vector and its two components form a right triangle.

Going the other way — from components back to magnitude and direction:

Components → magnitude and angle
\[ a = \sqrt{a_{x}^{2} + a_{y}^{2}} \qquad \tan \theta = a_{y} / a_{x} \]
! Watch the inverse-tangent quadrant

A calculator's \(\tan ^{-1}\) only returns angles in two quadrants. Always sketch the vector: if the components place it in a different quadrant than the calculator's answer, add 180° to fix it. The right answer is the one that matches your drawing.

Section 3-5

Unit Vectors

A unit vector has magnitude exactly 1 and no units — its only job is to point. The unit vectors along the x, y, and z axes are written \(\hat{\imath}\), \(\hat{\jmath}\), and \(\hat{k}\) (the "hat" marks them as unit vectors). With them, any vector becomes a clean sum:

Unit-vector notation
\[ a = a_{x} \hat{\imath} + a_{y} \hat{\jmath} + a_{z} \hat{k} \]
Here ax, ay, az are the scalar components, and axî etc. are the vector components.
Section 3-6

Adding Vectors by Components

This is the everyday workhorse. To add vectors, just add their components axis by axis:

Component addition (r = a + b)
\[ r_{x} = a_{x} + b_{x} \qquad r_{y} = a_{y} + b_{y} \qquad r_{z} = a_{z} + b_{z} \]
Two vectors are equal only if all their corresponding components are equal.
  1. Resolve every vector into its x, y (and z) components.
  2. Add the components separately, axis by axis, to get the components of the sum.
  3. Recombine — either leave the answer in unit-vector form, or convert to magnitude-and-angle.
Subtraction is the same trick: for \(d = a - b\), just subtract components — \(d_{x} = a_{x} - b_{x}\), and so on.
Section 3-7

Vectors & the Laws of Physics

We're free to orient the axes however we like. Rotate them, and a vector's components change — but its magnitude and direction do not. Every choice of axes describes the same vector.

🧭
Why physics speaks in vectors
The relations among vectors don't depend on the coordinate system.

That coordinate-independence is exactly why the laws of physics are written as vector equations. One compact equation like \(r = a + b\) stands in for three component equations at once.

Section 3-8

Multiplying Vectors

There are three distinct "multiplications," and none works quite like ordinary arithmetic.

1 · Vector times a scalar

Multiplying a vector by a scalar \(s\) scales its length by \(|s|\). The direction is unchanged if \(s \gt 0\) and reversed if \(s \lt 0\).

2 · The scalar (dot) product → gives a number

Dot product
\[ a \cdot b = a b \cos \varphi = a_{x}b_{x} + a_{y}b_{y} + a_{z}b_{z} \]
φ is the angle between the vectors. The result is a scalar. It is largest when the vectors are parallel (φ = 0) and zero when they are perpendicular (φ = 90°). It is commutative: a · b = b · a.

Think of the dot product as "how much of one vector lies along the other." You'll meet it first as work (force along displacement) in Chapter 7.

3 · The vector (cross) product → gives a new vector

Cross product magnitude
\[ |a \times b| = a b \sin \varphi \]
Here φ is the smaller angle between the vectors. The result is zero when they are parallel and largest when perpendicular.
The right-hand rule
a × b points perpendicular to the plane of a and b.

Point your right-hand fingers along \(a\) and curl them toward \(b\) through the smaller angle; your thumb gives the direction of \(a \times b\). Order matters: \(b \times a = -(a \times b)\). You'll meet the cross product as torque in Chapter 11.

Cross product in components
\[ a \times b = (a_{y}b_{z} - b_{y}a_{z}) \hat{\imath} + (a_{z}b_{x} - b_{z}a_{x}) \hat{\jmath} + (a_{x}b_{y} - b_{x}a_{y}) \hat{k} \]
Worked Examples

Putting It to Work

1 Finding components — airplane flight

Problem. A plane is sighted 215 km away, in a direction 22° east of due north. How far east and how far north is it?

Set up. Take +x east and +y north. Measured from the +x axis, the angle is \(\theta = 90^{\circ} - 22^{\circ} = 68^{\circ}\).

Components
\[\begin{gathered} d_{x} = 215 \cos 68^{\circ} \approx 81 \mathrm{km} \text{(east)} \\ d_{y} = 215 \sin 68^{\circ} \approx 199 \mathrm{km} \approx 2.0 \times 10^{2} \mathrm{km} \text{(north)} \end{gathered}\]

So the plane is about 81 km east and 2.0 × 10² km north of the airport.

2 Adding by components

Problem. Add \(a = 4.2\hat{\imath} - 1.5\hat{\jmath}\), \(b = -1.6\hat{\imath} + 2.9\hat{\jmath}\), and \(c = -3.7\hat{\jmath}\) (metres).

Add axis by axis
\[\begin{gathered} r_{x} = 4.2 - 1.6 + 0 = 2.6 m \\ r_{y} = -1.5 + 2.9 - 3.7 = -2.3 m \\ r = 2.6\hat{\imath} - 2.3\hat{\jmath} \end{gathered}\]

As a magnitude and angle: \(r = \sqrt{2.6^{2} + 2.3^{2}} \approx 3.5 m\) at \(\tan ^{-1}(-2.3 / 2.6) \approx -41^{\circ}\) (i.e. 41° clockwise from +x).

3 Angle between vectors — using the dot product

Problem. Find the angle between \(a = 3.0\hat{\imath} - 4.0\hat{\jmath}\) and \(b = -2.0\hat{\imath} + 3.0\hat{k}\).

Solution. Magnitudes: \(a = \sqrt{3^{2} + 4^{2}} = 5.00\), \(b = \sqrt{2^{2} + 3^{2}} = 3.61\). Component dot product: \(a \cdot b = (3)(-2) + (-4)(0) + (0)(3) = -6.0\).

Solve a · b = ab cos φ for φ
\[ \varphi = \cos ^{-1}[ -6.0 / (5.00 \times 3.61) ] \approx 110^{\circ} \]

The negative dot product already told us the angle is obtuse (more than 90°).

Review

Chapter Summary

Scalars & vectors

Scalars have magnitude only; vectors have magnitude and direction and obey vector algebra.

Geometric addition

Tip-to-tail. Addition is commutative and associative; subtract by adding the reverse vector.

Components

\(a_{x} = a \cos \theta\), \(a_{y} = a \sin \theta\); rebuild with \(a = \sqrt{a_{x}^{2} + a_{y}^{2}}\), \(\tan \theta = a_{y}/a_{x}\).

Unit vectors

\(a = a_{x}\hat{\imath} + a_{y}\hat{\jmath} + a_{z}\hat{k}\). Add vectors component by component.

Dot product

\(a \cdot b = ab \cos \varphi\) → a scalar; zero when perpendicular.

Cross product

\(|a \times b| = ab \sin \varphi\) → a vector ⟂ to both, by the right-hand rule; zero when parallel.

Practice

Problems

Sketch each situation first — a quick diagram catches most sign and quadrant errors.

  1. A vector in the xy plane has magnitude 7.3 m and points 250° counterclockwise from the +x axis. Find its (a) x and (b) y components.
  2. Vector A has \(A_{x} = -25.0 m\) and \(A_{y} = 40.0 m\). Find (a) its magnitude and (b) the angle it makes with the +x axis.
  3. A ship meant to sail 120 km due north is blown to a point 100 km due east of its start. (a) How far and (b) in what direction must it now sail to reach its destination?
  4. A person walks 3.1 km north, 2.4 km west, then 5.2 km south. (a) How far and (b) in what direction is the final point from the start?
  5. Add \(a = 4.0\hat{\imath} + 3.0\hat{\jmath}\) and \(b = -13.0\hat{\imath} + 7.0\hat{\jmath}\) (metres). Give the sum's (a) unit-vector form, (b) magnitude, and (c) direction.
  6. A car drives 50 km east, 30 km north, then 25 km at 30° east of north. Find the (a) magnitude and (b) angle of the total displacement.
  7. For \(a = 3.0\hat{\imath} + 5.0\hat{\jmath}\) and \(b = 2.0\hat{\imath} + 4.0\hat{\jmath}\), compute (a) \(a \times b\) and (b) \(a \cdot b\).
  8. Vectors C and D have magnitudes 3 and 4 units. Find the angle between them if \(C \cdot D\) equals (a) 0, (b) +12, and (c) −12 units.
  9. Vector A has magnitude 6.00 and B has magnitude 7.00, with \(A \cdot B = 14.0\). What is the angle between them?
  10. Given \(a = 3.0\hat{\imath} - 4.0\hat{\jmath}\), find (a) the magnitude and (b) the direction of \(a\), then (c) the magnitude and (d) direction of \(-2.0a\).
Tip: a zero dot product means the vectors are perpendicular; a zero cross product means they are parallel. Those two facts shortcut a surprising number of problems.