Comprehensive Study Guide on Vector Analysis and Physics Applications

Introduction to Vector Quantities

  • Vector Representation:
    • Vector quantities are characterized by both magnitude and direction.
    • If the magnitude of a vector is known, it must be written in a specific format to be usable in mathematical formulas and physical equations.

Unit Vectors in Cartesian Systems

  • Definition and Properties:

    • A unit vector has a magnitude equal to 11 (using the unit of the specific quantity of interest).
    • In the Cartesian coordinate system, the unit vectors are denoted as i^\hat{i} and j^\hat{j}.
    • These unit vectors are treated as constants.
    • Directionality: i^\hat{i} is always parallel to the x-axis, and j^\hat{j} is always parallel to the y-axis.
  • Vectors Parallel to an Axis:

    • Example: A car moves along the x-axis toward the right.
    • Given magnitude of velocity: v=5 m/s−1v = 5\,m/s^{-1}.
    • Vector notation for this velocity: v=5i^ m/s−1\mathbf{v} = 5\hat{i}\,m/s^{-1}.

Drawing and Scaling Vectors

  • Vector Components:

    • A force vector F\mathbf{F} can be represented as the sum of its horizontal and vertical components.
    • Example: For a vector with an x-component of 33 and a y-component of 22, the force is written as: F=(3i^+2j^) N\mathbf{F} = (3\hat{i} + 2\hat{j})\,N.
  • Proportionality and Magnitude:

    • The drawn length of a vector is directly proportional to its magnitude.
    • Example comparison:
      • If F1=2 N\mathbf{F}_1 = 2\,N and F2=1 N\mathbf{F}_2 = 1\,N, then the length of the green vector representing F2\mathbf{F}_2 is half the length of the red vector representing F1\mathbf{F}_1.

Vector Addition in Two Dimensions

  • Example Calculation:
    • Given forces:
      • F1=(2i^+j^) N\mathbf{F}_1 = (2\hat{i} + \hat{j})\,N
      • F2=(2i^) N\mathbf{F}_2 = (2\hat{i})\,N
    • To find the total force (Sum): Ftot=F1+F2\mathbf{F}_{tot} = \mathbf{F}_1 + \mathbf{F}_2
    • Calculation: (2i^+j^+2i^) N=(4i^+j^) N(2\hat{i} + \hat{j} + 2\hat{i})\,N = (4\hat{i} + \hat{j})\,N
    • Correction based on slide 7 illustration notes: If F1\mathbf{F}_1 is drawn as i^+j^\hat{i} + \hat{j} and F2\mathbf{F}_2 as 2i^2\hat{i}, then the result is (3i^+j^) N(3\hat{i} + \hat{j})\,N.

Analytical Properties of 2D Vectors

  • Magnitude (Length):

    • For a vector a=axi^+ayj^\mathbf{a} = a_x \hat{i} + a_y \hat{j}, the magnitude is calculated as:
    • a=∣a∣=ax2+ay2a = |\mathbf{a}| = \sqrt{a_x^2 + a_y^2}
  • Scalar (Dot) Product:

    • For two vectors a=axi^+ayj^\mathbf{a} = a_x \hat{i} + a_y \hat{j} and b=bxi^+byj^\mathbf{b} = b_x \hat{i} + b_y \hat{j}:
    • a⋅b=axbx+ayby\mathbf{a} \cdot \mathbf{b} = a_x b_x + a_y b_y

Exercise 1: Drafting Vectors from Geometry

  • Scenario: An equilateral triangle with side length a=2 ma = 2\,m. The side BΓB\Gamma is parallel to the x-axis.
  • Part A: Writing vector BABA in terms of i^\hat{i} and j^\hat{j}:
    • The angle in an equilateral triangle is 60∘60^{\circ}.
    • X-component: a×cos⁡(60∘)=2×0.5=1 ma \times \cos(60^{\circ}) = 2 \times 0.5 = 1\,m.
    • Y-component: a×sin⁡(60∘)=2×32≈1.73 ma \times \sin(60^{\circ}) = 2 \times \frac{\sqrt{3}}{2} \approx 1.73\,m.
    • Result: BA=(1i^+1.73j^) m\mathbf{BA} = (1\hat{i} + 1.73\hat{j})\,m.
  • Part B: Finding the unit vector parallel to BABA:
    • A unit vector BA^\hat{BA} has a magnitude of 11 and no units.
    • It is calculated by dividing the vector by its magnitude:
    • BA^=BA∣BA∣=(1i^+1.73j^) m2 m\hat{BA} = \frac{\mathbf{BA}}{|\mathbf{BA}|} = \frac{(1\hat{i} + 1.73\hat{j})\,m}{2\,m}
    • Result: BA^=(0.5i^+0.87j^)\hat{BA} = (0.5\hat{i} + 0.87\hat{j}).

Exercise 2: Position and Displacement

  • Scenario: Triangle vertices A(0,0)A(0,0), B(1,0)B(1,0), and Γ(2,1)\Gamma(2,1), with coordinates in cmcm.
  • Finding the vector BΓB\Gamma:
    • Calculation method: BΓ=End Point−Starting PointB\Gamma = \text{End Point} - \text{Starting Point}.
    • Calculation: BΓ=Γ−B=(2,1)−(1,0)=(1,1)\mathbf{B\Gamma} = \Gamma - B = (2, 1) - (1, 0) = (1, 1).
    • Vector format: BΓ=(i^+j^) cm\mathbf{B\Gamma} = (\hat{i} + \hat{j})\,cm.

Vectors in Three Dimensions (3D)

  • The 3D Coordinate System:

    • Includes three axes: xx, yy, and zz.
    • Uses a right-handed system (described by the "Rechte-hand-regel").
    • Locating a point (x,y,z)(x, y, z) (e.g., A(1,2,1)A(1, 2, 1)) involves three steps moving parallel to each respective axis.
  • Orientation relative to a 2D plane:

    • Vectors can be perpendicular to the paper/screen.
    • Direction from the paper/screen toward the viewer (represented visually by the top of a screw).
    • Direction away from the viewer and into the paper/screen (represented by the back of a screw).
  • 3D Vector Notation:

    • A vector in 3D is written as: a=axi^+ayj^+azk^\mathbf{a} = a_x\hat{i} + a_y\hat{j} + a_z\hat{k}.

Practical 3D Application: Hiking Displacement

  • Scenario: Hiking from a University to the summit of Mitsikeli.
  • Data Points:
    • University Altitude: 530 m530\,m.
    • Mitsikeli Summit Altitude: 1810 m1810\,m.
    • Horizontal displacement (x): Δx=6 km\Delta x = 6\,km.
    • Horizontal displacement (y): Δy=10.5 km\Delta y = 10.5\,km.
    • Vertical displacement (z): Δz=1810 m−530 m=1280 m=1.28 km\Delta z = 1810\,m - 530\,m = 1280\,m = 1.28\,km.
  • Displacement Vector:
    • Δr=(6i^+10.5j^+1.28k^) km\Delta \mathbf{r} = (6\hat{i} + 10.5\hat{j} + 1.28\hat{k})\,km.

Mathematical Operations with Vectors

  • Addition and Subtraction:

    • Performed by adding or subtracting the corresponding components of the vectors.
    • (axi^+ayj^+azk^)±(bxi^+byj^+bzk^)=(ax±bx)i^+(ay±by)j^+(az±bz)k^(a_x\hat{i} + a_y\hat{j} + a_z\hat{k}) \pm (b_x\hat{i} + b_y\hat{j} + b_z\hat{k}) = (a_x \pm b_x)\hat{i} + (a_y \pm b_y)\hat{j} + (a_z \pm b_z)\hat{k}.
  • The Resultant Vector (Fol\mathbf{F}_{ol} or Ftotal\mathbf{F}_{total}):

    • The resultant is the algebraic sum of all component forces.
    • The sign of a component indicates its direction and is not always positive.
    • Example: F1=(i^+3j^)\mathbf{F}_1 = (\hat{i} + 3\hat{j}) and F2=(2i^+j^)\mathbf{F}_2 = (2\hat{i} + \hat{j}) results in Fol=(3i^+4j^)\mathbf{F}_{ol} = (3\hat{i} + 4\hat{j}).
  • Interior (Dot) Product:

    • Geometric Definition: A scalar value defined by a⋅b=∣a∣∣b∣cos⁡(ϕ)\mathbf{a} \cdot \mathbf{b} = |\mathbf{a}| |\mathbf{b}| \cos(\phi), where ϕ\phi is the angle between the vectors.
    • Algebraic Definition: Equivalent result derived from unit vector properties:
      • i^⋅i^=j^⋅j^=k^⋅k^=1\hat{i} \cdot \hat{i} = \hat{j} \cdot \hat{j} = \hat{k} \cdot \hat{k} = 1
      • i^⋅j^=j^⋅k^=i^⋅k^=0\hat{i} \cdot \hat{j} = \hat{j} \cdot \hat{k} = \hat{i} \cdot \hat{k} = 0
    • Calculation: a⋅b=axbx+ayby+azbz\mathbf{a} \cdot \mathbf{b} = a_x b_x + a_y b_y + a_z b_z.
    • Example with the forces above: F1⋅F2=(1×2)+(3×1)=2+3=5\mathbf{F}_1 \cdot \mathbf{F}_2 = (1 \times 2) + (3 \times 1) = 2 + 3 = 5.

Vector Magnitude and Normalization

  • Calculating Magnitude in 3D:

    • Derived from the dot product with itself: a2=a⋅aa^2 = \mathbf{a} \cdot \mathbf{a}.
    • a=ax2+ay2+az2a = \sqrt{a_x^2 + a_y^2 + a_z^2}.
    • Example for F1=i^+3j^\mathbf{F}_1 = \hat{i} + 3\hat{j}: ∣F1∣=12+32=10≈3.16|\mathbf{F}_1| = \sqrt{1^2 + 3^2} = \sqrt{10} \approx 3.16.
  • Constructing Unit Vectors:

    • To find a unit vector a^\hat{a} in the same direction as a\mathbf{a}, divide the vector components by the magnitude: a^=axai^+ayaj^+azak^\hat{a} = \frac{a_x}{a}\hat{i} + \frac{a_y}{a}\hat{j} + \frac{a_z}{a}\hat{k}.
    • If a unit vector is multiplied by a scalar xx, the magnitude of the new vector becomes xx.

Computational Exercise: Bond Angle Calculation

  • Target: Calculate the H-C-H bond angle in a molecule.
  • Given Coordinates (in Angstroms [\u00c5]):
    • Vector a=−0.7145i^−0.4125j^\mathbf{a} = -0.7145\hat{i} - 0.4125\hat{j}
    • Vector b=+0.7145i^−0.4125j^\mathbf{b} = +0.7145\hat{i} - 0.4125\hat{j}
  • Magnitudes:
    • ∣a∣=0.825|\mathbf{a}| = 0.825
    • ∣b∣=0.825|\mathbf{b}| = 0.825
  • Calculating the Angle (\u03c6):
    • cos⁡(ϕ)=a⋅b∣a∣∣b∣\cos(\phi) = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}||\mathbf{b}|}
    • Dot product: (−0.7145)×(0.7145)+(−0.4125)×(−0.4125)=−0.5105+0.1701=−0.34(-0.7145) \times (0.7145) + (-0.4125) \times (-0.4125) = -0.5105 + 0.1701 = -0.34
    • cos⁡(ϕ)=−0.340.825×0.825=−0.4995\cos(\phi) = \frac{-0.34}{0.825 \times 0.825} = -0.4995
    • Solving for ϕ\phi: The common tetrahedral-related value often results in θ≈120∘\theta \approx 120^{\circ} in planar models or 114∘114^{\circ} based on specific molecular constraints shown.

Applications in Physics and Geometry

  • Common Vector Quantities:

    • Position (Θέση)
    • Displacement (Μετατόπιση)
    • Velocity (Ταχύτητα)
    • Acceleration (Επιτάχυνση)
    • Force (F\mathbf{F}, Δύναμη)
  • Analytical vs. Euclidean Geometry:

    • Analytical geometry is significantly more practical for physics than Euclidean geometry.
    • It allows physical laws to be written in concise vector form.
    • Enables algebraic solutions to complex problems and integration with calculus (differential and integral analysis).