Notes: Restricted Vectors and Vector Algebra

Vector Basics

• A vector is a quantity that has both magnitude and direction.
• A vector can be represented by a directed arrow in space, denoted as \vec{A}.
• Magnitude:
  $|\vec{A}| = A$
• Direction is given by the unit vector in the direction of \vec{A}, denoted as \hat{A} = \vec{A}/|\vec{A}|.
• Therefore, the vector can be written as
  $\vec{A} = A \, \hat{A}.$
• Two vectors are identical iff they have the same magnitude and the same direction:
  $\vec{A} = \vec{B} \iff A = B \text{ and } \, \hat{A} = \, \hat{B}.$

Coordinate Systems and Vectors

• In Cartesian coordinates, the positive directions are described by unit vectors:
  • x-axis: \hat{i},
  • y-axis: \hat{j},
  • z-axis: \hat{k}.
• A vector in Cartesian form is
  $\vec{A} = A<em>x \hat{i} + A</em>y \hat{j} + A_z \hat{k}.$
• If the vector lies in the x-y plane, then
  $\vec{A} = A<em>x \hat{i} + A</em>y \hat{j},$
  with
  $A<em>x = A \cos \theta, \quad A</em>y = A \sin \theta, \quad A = \sqrt{A<em>x^2 + A</em>y^2}.$
• Example (along a unit vector): if the magnitude is 5 and it points in the direction of \hat{i}, then
  $\vec{A} = 5 \, \hat{i}.$

Examples and Practice Questions (Conceptual)

• Concept Question 1 (vector quantity from right graph):

  • A. \vec{A} = 5 \cos 30^{\circ} \, \hat{i} + 5 \sin 30^{\circ} \, \hat{j}.
  • B. \vec{A} = \cos 30^{\circ} \, \hat{i} + \sin 30^{\circ} \, \hat{j}.
  • C. \vec{A} = 5 \sin 30^{\circ} \, \hat{i} + 5 \cos 30^{\circ} \, \hat{j}.
  • D. \vec{A} = \sin 30^{\circ} \, \hat{i} + \cos 30^{\circ} \, \hat{j}.
  • Answer (from slide): A.

• Concept Question 1 - Solution:

  • Correct form for a vector with magnitude 5 and direction given by 30° above the x-axis is
    $\vec{A} = 5 \cos 30^{\circ} \, \hat{i} + 5 \sin 30^{\circ} \, \hat{j}.$

• Concept Question 2 (different angle for same magnitude):

  • A. \vec{A} = 5 \cos 30^{\circ} \, \hat{i} + 5 \sin 30^{\circ} \, \hat{j}.
  • B. \vec{A} = 5 \cos 60^{\circ} \, \hat{i} + 5 \sin 60^{\circ} \, \hat{j}.
  • Answer (from slide): depends on given angle; use the angle from the x-axis.

• Concept Question 3 (vector components):

  • A. \vec{A} = 5 \, \hat{i} + 2 \, \hat{j}.
  • B. \vec{A} = 2 \, \hat{i} + 5 \, \hat{j}.

• Concept Question 3 - Solution: (depends on the pictured vector; options shown were A and B).

• Concept Question 4 (magnitude):

  • A. |\vec{A}| = 5.
  • B. |\vec{A}| = 2.
  • C. |\vec{A}| = \sqrt{5^2 + 2^2}.
  • Answer: |\vec{A}| = \sqrt{5^2 + 2^2}.

Case Problems: Coordinate Expressions of Vectors

• Case Problem 1: Given a vector of magnitude 5 graphically represented with angle 60° to the negative x-axis (i.e., pointing left and up).
  • Solution:
    $\vec{A} = 5 \cos 60^{\circ} \, (-\hat{i}) + 5 \sin 60^{\circ} \, \hat{j}.$
  • Numeric form: $\vec{A} = -\frac{5}{2} \hat{i} + \frac{5\sqrt{3}}{2} \, \hat{j} \approx (-2.50, 4.33).$
• Case Problem 2: Magnitude 3, angle 45° in Quadrant III (both x and y negative).
  • Solution:
    $\vec{A} = 3 \cos 45^{\circ} \, (-\hat{i}) + 3 \sin 45^{\circ} \, (-\hat{j}).$
  • Numeric form:
    $\vec{A} = -\frac{3}{\sqrt{2}} \, \hat{i} - \frac{3}{\sqrt{2}} \, \hat{j}.$
• Case Problem 3: Magnitude 3, angle 45° in Quadrant IV (x positive, y negative).
  • Solution:
    $\vec{A} = 3 \cos 45^{\circ} \hat{i} + 3 \sin 45^{\circ} (-\hat{j}).$
  • Numeric form:
    $\vec{A} = \frac{3}{\sqrt{2}} \, \hat{i} - \frac{3}{\sqrt{2}} \, \hat{j}.$
• Case Problem 4: General force vector with magnitude F and angle θ in the third quadrant:
  • Solution:
    $\vec{F} = F \sin \theta \, (-\hat{i}) + F \cos \theta \, (-\hat{j}).$
• Case Problem 5: Vector with magnitude T and angle θ (components relative to +x and +y axes):
  • Solution:
    $\vec{T} = T \cos \theta \, \hat{i} + T \sin \theta \, (−\hat{j}).$
• Case Problem 6: Given \vec{A} = -3\hat{i} + 4\hat{j}.
  • Magnitude:
    $|\vec{A}| = \sqrt{(-3)^2 + 4^2} = 5.$
  • Angle: since x is negative and y is positive (Quadrant II),
    $\theta = \pi - \tan^{-1}\left(\frac{|A<em>y|}{|A</em>x|}\right) = \pi - \tan^{-1}\left(\frac{4}{3}\right) \approx 126.87^{\circ}.$

Coordinate Systems and Vector Addition/Subtraction

• Vector addition: \vec{A} + \vec{B} = \vec{C}
  • Steps:
    1) Parallelly move the vectors so their tails meet.
    2) Construct a parallelogram.
    3) The resultant vector \vec{C} originates from the tails along the diagonal.
• Vector subtraction: \vec{A} - \vec{B} = \vec{A} + (-\vec{B}).
  • Steps:
    1) Construct -\vec{B}.
    2) Move tails of \vec{A} and -\vec{B} to meet tails.
    3) Build the parallelogram; the diagonal from the tails is \vec{C} = \vec{A} - \vec{B}.
• Algebraic form (2D): $\vec{A} = A<em>x \, \hat{i} + A</em>y \, \hat{j}, \vec{B} = B<em>x \, \hat{i} + B</em>y \, \hat{j}$
  • Sum: $\vec{A} + \vec{B} = (A<em>x + B</em>x) \, \hat{i} + (A<em>y + B</em>y) \, \hat{j}.$
  • Difference: $\vec{A} - \vec{B} = (A<em>x - B</em>x) \, \hat{i} + (A<em>y - B</em>y) \, \hat{j}.$

Dot Product (Scalar Product)

• Definition:
  $\vec{A} \cdot \vec{B} = A B \cos \theta,$
  where \theta is the angle between the two vectors.
• Nature: The dot product is a scalar (magnitude only, no direction).
• Properties:
  • Commutative: \vec{A} \cdot \vec{B} = \vec{B} \cdot \vec{A}.
  • Distributive: \vec{A} \cdot (\vec{C} + \vec{D}) = \vec{A} \cdot \vec{C} + \vec{A} \cdot \vec{D};
    (\vec{A} + \vec{B}) \cdot \vec{C} = \vec{A} \cdot \vec{C} + \vec{B} \cdot \vec{C}.
• In Cartesian coordinates (2D):
  $\vec{A} = A<em>x \, \hat{i} + A</em>y \, \hat{j}, \\vec{B} = B<em>x \, \hat{i} + B</em>y \, \hat{j}$
  $\vec{A} \cdot \vec{B} = A<em>x B</em>x + A<em>y B</em>y.$
• Dot product in physics (example): Work done by a force \vec{F} over displacement \vec{S}:
  $W = \vec{F} \cdot \vec{S}.$
• Maxwell’s equations reference: https://en.wikipedia.org/wiki/Maxwell%27s_equations (listed as related reading in slides).

Concept Question 6: Dot Product Sign and Angle

• Three cases compare the dot product magnitude/sign for angles between 0 and \pi:
  • i) 0 < \theta < \tfrac{\pi}{2} (cos positive)
  • ii) \tfrac{\pi}{2} < \theta < \pi (cos negative but magnitude large)
  • iii) \theta = \tfrac{\pi}{2} (cos = 0)
• Answer: D. ii > iii > i
  • Rationale: cos \theta is positive for 0

Angular Formulas and Magnitude-Angle Conversion

• Given a vector \vec{A} with components \Ax and \Ay:
  • Component form: $A<em>x = A \cos \theta, \ A</em>y = A \sin \theta.$
  • Magnitude and angle: $A = \sqrt{A<em>x^2 + A</em>y^2}, \ \theta = \tan^{-1} \left( \frac{A<em>y}{A</em>x} \right).$
• Special considerations for quadrant:
  • If Ax < 0 and Ay > 0, angle lies in quadrant II: $\theta = \pi - \tan^{-1} \left( \frac{|A<em>y|}{|A</em>x|} \right).$

Case Problem 6 (Vector in Quadrant II)

• Given \vec{A} = -3 \,\hat{i} + 4 \,\hat{j}:
  • Magnitude: $|\vec{A}| = \sqrt{(-3)^2 + 4^2} = 5.$
  • Angle: $\theta = \pi - \tan^{-1} \left(\frac{4}{3}\right) \approx 126.87^{\circ}.$

Summary of Key Formulas

• Vector representation in 3D:
  $\vec{A} = A<em>x \, \hat{i} + A</em>y \, \hat{j} + A_z \, \hat{k}.$
• In the plane:
  $\vec{A} = A<em>x \, \hat{i} + A</em>y \, \hat{j}, ext{ with } A<em>x = A \cos \theta, A</em>y = A \sin \theta, A = \sqrt{A<em>x^2 + A</em>y^2}.$
• Magnitude-angle relation:
  $A = \sqrt{A<em>x^2 + A</em>y^2}, \quad \theta = \tan^{-1} \left( \frac{A<em>y}{A</em>x} \right).$
• Unit vectors: \hat{i}, \hat{j}, \hat{k}.
• Dot product:
  $\vec{A} \cdot \vec{B} = A B \cos \theta,$
  $\vec{A} \cdot \vec{B} = A<em>x B</em>x + A<em>y B</em>y ext{ (in 2D)}.$
• Vector addition/subtraction (component form):
  • Sum: \vec{A} + \vec{B} = (Ax + Bx) \, \hat{i} + (Ay + By) \, \hat{j}.
  • Difference: \vec{A} - \vec{B} = (Ax - Bx) \, \hat{i} + (Ay - By) \, \hat{j}.
• Work (dot product with displacement):
  $W = \vec{F} \cdot \vec{S}.$