Mathematics for Physics · Chapter 20

Vectors and Scalars

Physics quantities fall into two categories: scalars, which are fully described by a single number with units, and vectors, which require both magnitude and direction. Every force, velocity, and field in this curriculum is a vector.

Learning Goals
  • Distinguish scalars from vectors and represent vectors in component form using unit vectors.
  • Add and subtract vectors algebraically by combining corresponding components.
  • Compute the dot product and use it to find the angle between two vectors or the work done by a force.
  • Compute the cross product using the determinant formula and apply the right-hand rule for direction.
  • Recognize vector fields and compute the gradient of a scalar field.

20.1 Scalars and Vectors

A scalar is a quantity with magnitude but no direction: mass, temperature, time, speed, energy. You can add and multiply scalars with ordinary arithmetic.

A vector is a quantity with both magnitude and direction: displacement, velocity, acceleration, force, momentum, electric field. We write vectors in bold (v) or with an arrow (v⃗). The magnitude of v is |v| or v.

Definition 20.1 — Vector Components
In 3D Cartesian coordinates, a vector A is written in terms of unit vectors î, ĵ, k̂ along the x, y, z axes:A=Axı^+Ayȷ^+Azk^A = A_{x} î + A_{y} ĵ + A_{z} k̂Themagnitudeis∣A∣=(Ax2+Ay2+Az2).In2Dfromangleθabovex−axis:Ax=∣A∣cThe magnitude is |A| = \sqrt(A_{x}^{2} + A_{y}^{2} + A_{z}^{2}). In 2D from angle \theta above x-axis: A_{x} = |A| cosθ,Ay=∣A∣sinθos \theta, A_{y} = |A| sin \theta.

Vector addition follows the parallelogram rule (graphically) or component-wise (algebraically): if C = A + B, then C_x = A_x + B_x, C_y = A_y + B_y, C_z = A_z + B_z. Subtraction reverses the direction of the second vector before adding.

Example 20.1 — Resultant Force from Two Applied Forces

Force F1F_{1} = 30 N at 40° above the +x axis. Force F2F_{2} = 50 N at 120° above the +x axis. Find the resultant.

Components of F₁:F1x=30cos40°=22.98NFγ=30sin40°=19.28NF_{1x} = 30 cos 40° = 22.98 N \qquad F_{1}ᵧ = 30 sin 40° = 19.28 N
Components of F₂:F2x=50cos120°=−25.00NFγ=50sin120°=43.30NF_{2x} = 50 cos 120° = -25.00 N \qquad F_{2}ᵧ = 50 sin 120° = 43.30 N
Resultant:Fx=22.98−25.00=−2.02NFγ=19.28+43.30=62.58NF_{x} = 22.98 - 25.00 = -2.02 N \qquad Fᵧ = 19.28 + 43.30 = 62.58 N
Magnitude:|F∣=(2.022+62.582)=3907.3=62.5NF| = \sqrt(2.02^{2} + 62.58^{2}) = \sqrt3907.3 = 62.5 N
Direction:θ=arctan(62.58/−2.02)=180°−88.1°=91.8°above+xaxis\theta = arctan(62.58/-2.02) = 180° - 88.1° = 91.8° above +x axis

20.2 The Dot Product

The dot product (scalar product) of two vectors gives a scalar measuring how much one vector projects onto another:

A⋅B=∣A∣∣B∣cosθ=AxBx+AyBy+AzBzA \cdot B = |A||B| cos \theta = A_{x} B_{x} + A_{y} B_{y} + A_{z} B_{z}(20.1)

Key properties: the dot product is zero when the vectors are perpendicular (cos 90° = 0), maximum when parallel, and negative when the angle exceeds 90°. It is commutative (A·B = B·A) and distributive.

Physical uses of the dot product: work W = F·Δr = FΔr cos θ (only the force component along displacement does work); electric flux Φ = E·A; magnetic flux Φ_B = B·A.

Definition 20.2 — Finding the Angle Between Vectors
Fromthedotproductformula,theangleθbetweentwovectorsFrom the dot product formula, the angle \theta between two vectors:cosθ=(A⋅B)/(∣A∣∣B∣cos \theta = (A \cdot B) / (|A| |B|Ifthedotproductispositive:θIf the dot product is positive: \theta < 90°.Zero:perpendicular.Negative:θ90°. Zero: perpendicular. Negative: \theta> 90°.
Example 20.2 — Work Done by a Force

A force F = (3ı^+4ȷ^)NactsonanobjectdisplacedbyΔ3î + 4ĵ) N acts on an object displaced by \Deltar = (5ı^−2ȷ^)m.Findtheworkdone5î - 2ĵ) m. Find the work done

Dot product:W=F⋅\Deltar=(3)(5)+(4)(−2)=15−8=7JW = F \cdot \Deltar = (3)(5) + (4)(-2) = 15 - 8 = 7 J
Alternatively:|F∣=5N,∣\Deltar∣=29m.cosθ=7/(529)→θ=74.9°.W=F\Deltarcosθ=529×7/(529)=7F| = 5 N, |\Deltar| = \sqrt29 m. cos \theta = 7/(5\sqrt29) \to \theta = 74.9°. W = F\Deltar cos \theta = 5\sqrt29 \times 7/(5\sqrt29) = 7J ✓

20.3 The Cross Product

The cross product (vector product) produces a vector perpendicular to both input vectors, with magnitude equal to the area of the parallelogram they span:

∣A×B∣=∣A∣∣B∣sinθ|A \times B| = |A||B| sin \theta(20.2)

Direction: right-hand rule — curl the fingers from A toward B, and the thumb points in the direction of A × B. Important: the cross product is anti-commutative: B × A = −(A × B).

Theorem 20.1 — Cross Product in Component Form
A×B=(AyBz−AzBy)ı^−(AxBz−AzBx)ȷ^+(AxBy−AyBx)k^A \times B = (A_{y} B_{z} - A_{z} B_{y})î - (A_{x} B_{z} - A_{z} B_{x})ĵ + (A_{x} B_{y} - A_{y} B_{x})k̂Theunitvectorrelations:ı^×ȷ^=k^,ȷ^×k^=ı^,k^×ı^=ȷ^(cyclic),andı^×ı^=0The unit vector relations: î \times ĵ = k̂, \qquad ĵ \times k̂ = î, \qquad k̂ \times î = ĵ (cyclic), and î \times î = 0 (any vector crossed with itself is zero).

Physical uses of the cross product: torque τ = r × F; angular momentum L = r × p; magnetic force F = qv × B; magnetic field from a current element dB = (μ₀I/4π)(dl × r̂/r²).

Example 20.3 — Torque as a Cross Product

A force F = (2ı^+3ȷ^−k^)Nisappliedatposition2î + 3ĵ - k̂) N is applied at positionr = (î −2ȷ^+3k^)mfromthepivot.Findthetorque- 2ĵ + 3k̂) m from the pivot. Find the torque

Cross product:τ=r×F\tau = r \times F
τ_x:= ryFz−rzFy=(−2)(−1)−(3)(3)=2−9=−7N\cdotmr_{y} F_{z} - r_{z} F_{y} = (-2)(-1) - (3)(3) = 2 - 9 = -7 N\cdotm
τ_y:= rzFx−rxFz=(3)(2)−(1)(−1)=6+1=7N\cdotmr_{z} F_{x} - r_{x} F_{z} = (3)(2) - (1)(-1) = 6 + 1 = 7 N\cdotm
τ_z:= rxFy−ryFx=(1)(3)−(−2)(2)=3+4=7N\cdotmr_{x} F_{y} - r_{y} F_{x} = (1)(3) - (-2)(2) = 3 + 4 = 7 N\cdotm
Result:τ = (−7ı^+7ȷ^+7k^)N\cdotm∣τ∣=73≈12.1N\cdotm-7î + 7ĵ + 7k̂) N\cdotm \qquad |\tau| = 7\sqrt3 \approx 12.1 N\cdotm

20.4 Fields as Vector Functions

In physics, a vector field assigns a vector to every point in space. The electric field E(r) and magnetic fieldB(r) are vector fields — at every point in space they have a direction and magnitude. Visualizing vector fields is the key to understanding Maxwell's equations and fluid mechanics.

The gradient of a scalar field φ is a vector field pointing in the direction of steepest increase: ∇φ = (∂φ/∂x)î + (∂φ/∂y)ĵ + (∂φ/∂z)k̂. The electric field is the negative gradient of the potential: E = −∇V.

Definition 20.3 — Common Traps
  • Vectors are not just magnitudes: direction and components matter.
  • Dot products produce scalars: cross products produce vectors perpendicular to both inputs.
  • Components depend on basis: the geometric vector does not.
  • Unit vectors carry direction only: they should have magnitude 1.
Exercises — 20.1–20.4 Vectors
1.
F1=30Nat40°andF2=50Nat120°(bothabove+xaxis).FindthemagnitudeoftheresF_{1} = 30 N at 40° and F_{2} = 50 N at 120° (both above +x axis). Find the magnitude of the resultant.
N
Straightforward
2.
Force F = (3ı^+4ȷ^)Nactsoverdisplacement(5ı^−2ȷ^)m.Findtheworkdone3î + 4ĵ) N acts over displacement (5î - 2ĵ) m. Find the work done
J
Straightforward
3.
Find the magnitude of torque when r = (î−2ȷ^+3k^)mand-2ĵ+3k̂) m andF = (2ı^+3ȷ^−k^)N2î+3ĵ-k̂) N
N·m
Intermediate
4.Find the angle between A = (2ı^−3ȷ^+k^)and2î - 3ĵ + k̂) andB = (î + 4ȷ^−2k^4ĵ - 2k̂
Intermediate
5.DefinethetriplescalarproductA⋅(B\timesC)andproveitequalsthevolumeoftheparallelepiDefine the triple scalar product A\cdot(B\timesC) and prove it equals the volume of the parallelepiped spanned by A, B, C. What does it mean if this product is zero?
Challenging
Key Takeaways
  • Scalars have magnitude only; vectors have magnitude and direction — add component-wise.
  • DotproductA\cdotB=∣A∣∣B∣cosθ=\sumAiBi—scalar;measuresprojection;zeroforperpendicularDot product A\cdotB = |A||B|cos\theta = \sumA_{i}B_{i} — scalar; measures projection; zero for perpendicular vectors.
  • CrossproductA\timesBhasmagnitude∣A∣∣B∣sinθ,directiongivenbyright−handrule—vector;zCross product A\timesB has magnitude |A||B|sin\theta, direction given by right-hand rule — vector; zero for parallel vectors.
  • Crossproductisanti−commutative:B\timesA=−A\timesB.Unlikethedotproduct,ordermattersCross product is anti-commutative: B\timesA = -A\timesB. Unlike the dot product, order matters.
  • Work=F⋅\Deltar(dotproduct),torque=r\timesF(crossproduct),magneticforceF=qv\timesB(crossprWork = F\cdot\Deltar (dot product), torque = r\timesF (cross product), magnetic force F = qv\timesB (cross product).
  • Vectorfieldsassignavectortoeverypointinspace;theelectricfieldE=−\nablaVisthenVector fields assign a vector to every point in space; the electric field E = -\nablaV is the negative gradient of potential.