Chapter 3: Vectors and Their Components

3-1 VECTORS AND THEIR COMPONENTS

• Learning objectives
  • 3.01 Add vectors by drawing them in head-to-tail arrangements, applying the commutative and associative laws.
  • 3.02 Subtract a vector from a second one.
  • 3.03 Calculate the components of a vector on a given coordinate system, showing them in a drawing.
  • 3.04 Given the components of a vector, draw the vector and determine its magnitude and orientation.
  • 3.05 Convert angle measures between degrees and radians.
• Key ideas
  • Scalars vs vectors
  • Scalars have magnitude only and are described by a number with a unit (e.g., 10°C).
  • Vectors have magnitude and direction (e.g., 5 m, north) and follow vector algebra.
  • Geometric addition of vectors
  • Two vectors a and b can be added by drawing them to a common scale and placing them head to tail.
  • The resultant vector is the vector from the tail of the first to the head of the second (the vector sum). This sum is not the same as ordinary algebraic addition in one dimension.
  • Subtracting b from a is equivalent to adding a with −b (reverse the direction of b).
  • Vector addition is commutative and obeys the associative law: a + b = b + a and (a + b) + c = a + (b + c).
  • Displacement vectors
  • The displacement vector tells the overall change of position, not the actual path taken.
  • Different paths from A to B can represent the same displacement vector.
  • Visual procedure for adding two-dimensional vectors
    1) Draw vector a to a convenient scale and proper angle.
    2) Draw vector b with tail at the head of a, same scale and proper angle.
    3) The vector sum is the vector from the tail of a to the head of b.
  • Notation and interpretation
  • A symbol with an overhead arrow (e.g., ) denotes a vector (magnitude + direction).
  • A plain italic symbol (e.g., a) denotes a magnitude only (scalar).
  • Important properties
  • The order of addition does not matter (commutative law).
  • When adding more than two vectors, grouping in any order yields the same result (associative law).
  • Subtleties with subtraction
  • The vector −b has the same magnitude as b but opposite direction; a − b = a + (−b).
  • Scope of vector arithmetic
  • Rules apply to all vector quantities (displacements, velocities, accelerations, etc.).
  • You can add vectors only if they represent the same kind of quantity (e.g., displacements with displacements).
• Checkpoint 1 (conceptual)
  • Given the magnitudes of displacements |a| = 3 m and |b| = 4 m and that a and b may have various orientations, what are (a) the maximum possible magnitude of a − b and (b) the minimum possible magnitude?
• Components of vectors (2D)
  • Two-dimensional components arise from projections onto the coordinate axes.
  • If a vector
    \vec{a}) makes an angle θ with the positive x-axis, then its components are
    $a<em>x = a \cos\theta, \quad a</em>y = a \sin\theta,$
    where the sign of a component indicates direction along the corresponding axis.
  • The magnitude and orientation of the vector can be recovered from its components via
    $a = \sqrt{a<em>x^2 + a</em>y^2}, \quad \tan \phi = \frac{a<em>y}{a</em>x},$
    where φ is the angle the vector makes with the positive x-axis.
  • In 3D, a vector requires three components (ax, ay, a_z) or magnitude with two angles (α, β).
• Resolving vectors geometrically vs. algebraically
  • Resolving into components is equivalent to projecting onto axes; resolving is preserved under rigid shifts of the vector if the magnitude and direction do not change.
  • Once resolved, components can be manipulated like scalars to perform addition or subtraction component-wise.
• 3-1 VECTORS AND THEIR COMPONENTS (continuation)
  • The diagrammatic relationships show that the components and the vector form a right triangle (the vector as the hypotenuse).
  • Reconstructing a vector from its components: arrange the components head-to-tail to form the vector.
• 3-1 Sample problems (highlights)
  • Sample Problem 3.01 (Adding vectors in a drawing, orienteering): Using three displacement vectors in a head-to-tail arrangement, the greatest final distance from base camp occurs for the arrangement a, b, and −c (order does not affect the resultant due to commutativity). The resultant magnitude is read from the diagram.
  • Sample Problem 3.02 (Finding components, airplane flight): Given a displacement of magnitude 215 km at 22° east of due north, convert to x (east) and y (north) components:
  • The angle relative to the x-axis is 90° − 22° = 68°.
  • Components: $d<em>x = 215 \cos 68^{\circ} \approx 81\ \text{km (east)}, \quad d</em>y = 215 \sin 68^{\circ} \approx 199\ \text{km (north)}.$
  • Sample Problem 3.03 (Searching through a hedge maze): Compute the net displacement by summing three displacement vectors both in x and y components, then construct the resultant and determine its magnitude and direction via $|\vec{d}| = \sqrt{d<em>x^2 + d</em>y^2}$ and $\tan\theta = d<em>y/d</em>x.$ The example yields a net displacement with magnitude $|\vec{d}| \approx 13.9\ \text{m}$ and direction $\theta \approx -12.7^{\circ}$ (clockwise from the +x axis).
  • Sample Problem 3.04 (Adding vectors, unit-vector components): Given vectors in component form, assemble the sum and express it in unit-vector notation and also in magnitude-angle form.
• Tactics for vectors and trigonometry (p.6)
  • Tactic 1: Angles — Degrees and radians
  • Angles measured from the +x axis are positive counterclockwise and negative clockwise.
  • Convert between degrees and radians using $360^{\circ} = 2\pi\ \text{rad}$; e.g., $\text{radians} = \text{degrees} \times \frac{\pi}{180}$ and $\text{degrees} = \text{radians} \times \frac{180}{\pi}$.
  • Tactic 2: Trig functions
  • Know sine, cosine, and tangent definitions and their sign behavior in quadrants; be able to judge reasonableness of calculator results.
  • Tactic 3: Inverse trig functions
  • Inverse trig functions have restricted ranges; multiple angles can satisfy the same sine, cosine, or tangent value. Use physical context to pick the most reasonable answer.
  • Tactic 4: Measuring vector angles
  • The formulas for cos and sin in the x-axis frame are valid when the angle is measured from the positive x-axis.
  • If measured from another direction, you may need to swap cos/sin or invert the ratio; often safer to convert the angle to the +x axis frame before using trigonometric relations.
• 3-2 UNIT VECTORS, ADDING VECTORS BY COMPONENTS

3-2 UNIT VECTORS, ADDING VECTORS BY COMPONENTS

• Key ideas
  • Unit vectors i, j, k have magnitudes 1 and point in the positive directions of the x, y, z axes in a right-handed coordinate system.
  • A vector can be written in unit-vector notation as
    $\mathbf{a} = a<em>x \hat{\mathbf{i}} + a</em>y \hat{\mathbf{j}} + a_z \hat{\mathbf{k}}.$
  • The unit vectors point along the axes; the axes form a right-handed system.
• Unit vectors and components
  • The vector components (ax, ay, a_z) are the scalar components of a along each axis.
  • The vector components (ax, ay, a_z) can be used interchangeably with the vector in unit-vector form or in magnitude-angle form:
  • Unit-vector notation: $\mathbf{a} = a<em>x \hat{\mathbf{i}} + a</em>y \hat{\mathbf{j}} + a_z \hat{\mathbf{k}}.$
• Adding vectors by components
  • If \mathbf{a} and \mathbf{b} have components (ax, ay, az) and (bx, by, bz), then
    $\mathbf{a} + \mathbf{b} = (a<em>x + b</em>x) \hat{\mathbf{i}} + (a<em>y + b</em>y) \hat{\mathbf{j}} + (a<em>z + b</em>z) \hat{\mathbf{k}}.$
  • Subtraction can be handled similarly by subtracting components: $\mathbf{a} - \mathbf{b} = (a<em>x - b</em>x) \hat{\mathbf{i}} + (a<em>y - b</em>y) \hat{\mathbf{j}} + (a<em>z - b</em>z) \hat{\mathbf{k}}.$
• Rotation of the coordinate system
  • For a given vector, rotating the coordinate system around the origin changes the components but not the vector itself. All valid component representations describe the same vector (magnitude and direction).
  • The general expression remains: $\mathbf{a} = a<em>x \hat{\mathbf{i}} + a</em>y \hat{\mathbf{j}} + a_z \hat{\mathbf{k}}.$
• 3-2 Sample problems (highlights)
  • Sample Problem 3.03 (unit-vector components exercise): Compute vector sums by resolving into components, then recombine into unit-vector form or magnitude-angle form.
• 3-3 MULTIPLYING VECTORS

3-3 MULTIPLYING VECTORS

• Learning objectives
  • 3.09 Multiply vectors by scalars.
  • 3.10 Distinguish three products: scalar multiplication (vector by scalar), dot product (scalar), and cross product (vector, perpendicular to the plane of the two vectors).
  • 3.11 Find the dot product in magnitude-angle notation and in unit-vector notation.
  • 3.12 Find the angle between two vectors using their dot product.
• Key ideas
  • Scalar multiplication
  • Multiplying a vector by a scalar s yields a vector whose magnitude is |s| times the magnitude of the original vector and whose direction is the original direction if s > 0 and opposite if s < 0.
  • Dot product (scalar product)
  • $\mathbf{a} \cdot \mathbf{b} = |\mathbf{a}| |\mathbf{\;b}| \cos \phi,$ where φ is the angle between a and b.
  • In unit-vector notation: $\mathbf{a} \cdot \mathbf{b} = (a<em>x \hat{\mathbf{i}} + a</em>y \hat{\mathbf{j}} + a<em>z \hat{\mathbf{k}}) \cdot (b</em>x \hat{\mathbf{i}} + b<em>y \hat{\mathbf{j}} + b</em>z \hat{\mathbf{k}})$
    which expands to $a<em>x b</em>x + a<em>y b</em>y + a<em>z b</em>z.$
  • The dot product is commutative: $\mathbf{a} \cdot \mathbf{b} = \mathbf{b} \cdot \mathbf{a}.$
  • Cross product (vector product)
  • $\mathbf{a} \times \mathbf{b} = |\mathbf{a}| |\mathbf{b}| \sin \phi\, \hat{\mathbf{n}},$ where φ is the smaller angle between a and b and \hat{\mathbf{n}} is a unit vector perpendicular to the plane containing a and b (direction given by the right-hand rule).
  • The magnitude of the cross product is maximal when a and b are perpendicular; it is zero when a and b are parallel or antiparallel.
  • In unit-vector notation: $\mathbf{a} \times \mathbf{b} = (a<em>x \hat{\mathbf{i}} + a</em>y \hat{\mathbf{j}} + a<em>z \hat{\mathbf{k}}) \times (b</em>x \hat{\mathbf{i}} + b<em>y \hat{\mathbf{j}} + b</em>z \hat{\mathbf{k}}).$
  • The cross product is not commutative: $\mathbf{a} \times \mathbf{b} = - (\mathbf{b} \times \mathbf{a}).$
  • Right-hand rule
  • To determine the direction of the cross product, point the fingers of your right hand from a toward b; your thumb then points along a × b.
• 3-3 Sample problems (highlights)
  • Sample Problem 3.05: Angle between two vectors using dot products
  • The angle θ between two vectors is found from $\cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}| |\mathbf{b}|}.$
  • Sample Problem 3.06: Cross product, right-hand rule
  • Given specific vectors a and b, compute the magnitude of a × b via $|\mathbf{a} × \mathbf{b}| = |\mathbf{a}| |\mathbf{b}| \sin \phi.$
  • Use the right-hand rule to determine the direction of the resulting vector perpendicular to the plane of a and b.
  • Sample Problem 3.07: Cross product, unit-vector notation
  • Compute the cross product using unit vectors and distributive law, e.g., (ax i + ay j + az k) × (bx i + by j + bz k).
• Checkpoint and conceptual notes
  • Dot and cross products provide different kinds of information: dot gives a projection along the other vector; cross gives a perpendicular direction.
  • The cross product does not obey the commutative law; order matters.
  • For any vector products, you can use either magnitude-angle form or unit-vector notation, converting between representations as needed.
• Example applications and techniques
  • Right-hand rule is essential for cross products to determine direction.
  • Determinants provide another method to compute cross products (e.g., using i, j, k components or a determinant form).
  • When working in three dimensions, ensure you consistently use a right-handed coordinate system to interpret signs and directions.
• Summary of the main formulas
  • Scalar multiplication: $s\mathbf{a}$ has magnitude $|s|\,|\mathbf{a}|$ and direction of \mathbf{a} if s > 0, opposite if s < 0.
  • Dot product: $\mathbf{a} \cdot \mathbf{b} = |\mathbf{a}| |\mathbf{b}| \cos \phi = a<em>x b</em>x + a<em>y b</em>y + a<em>z b</em>z.$
  • Cross product: $\mathbf{a} \times \mathbf{b} = |\mathbf{a}| |\mathbf{b}| \sin \phi\, \hat{\mathbf{n}} = (a<em>y b</em>z - a<em>z b</em>y) \hat{\mathbf{i}} + (a<em>z b</em>x - a<em>x b</em>z) \hat{\mathbf{j}} + (a<em>x b</em>y - a<em>y b</em>x) \hat{\mathbf{k}}.$