Physics
Chapter 2: Vectors
Vector Quantities
Physical quantities specified by both magnitude (number of units) and direction.
Important
Vectors cannot be divided.
Examples:
- Displacement
- Velocity
- Position
- Force
- Torque
In mathematics, vectors are represented by arrows.
Figure 2.2: A vector is drawn from its origin (tail) to its end (head). Magnitude is its length.
Representing Vectors
Notation
- Vector: Bold letter with an arrow, e.g., \(\vec{d}\) (displacement).
- Scalar: Italic letter, e.g., \(d\) (distance).
Graphical Representation
Vectors are drawn to scale.
- An arrow’s length represents its magnitude.
- An arrow’s direction represents the vector’s direction.
Tip
Example: “Walk 6 km northeast from my tent.”
- Magnitude: 6 km
- Direction: Northeast
Displacement Vector
Describes a change in position.
- From point A (initial) to point B (final).
- Vector \(\vec{D}\) originates at A and ends at B.
- Magnitude \(D \equiv |\vec{D}|\) (always positive).
Figure 2.3: Displacement \(\vec{D}\) from A to B is the straight arrow, regardless of the path taken.
Vector Relations
Parallel, Antiparallel, and Orthogonal
- Parallel Vectors: Identical directions. If \(\vec{A}\) and \(\vec{B}\) are parallel and \(|\vec{A}| = |\vec{B}|\), then \(\vec{A} = \vec{B}\).
- Antiparallel Vectors: Opposite directions (180\(^\circ\) apart).
Figure 2.5: (d) Equal vectors, (b) Antiparallel vectors.
- Orthogonal Vectors: Perpendicular directions (90\(^\circ\) apart).
- See Figure 2.5 (e) for orthogonal vectors.
Algebra of Vectors in One Dimension
Vectors can be:
- Multiplied by scalars
- Added to other vectors
- Subtracted from other vectors
Let’s use the fishing trip example (Figure 2.6).
Figure 2.6: Displacement vectors for a fishing trip.
Vector-Scalar Multiplication
If a vector \(\vec{A}\) is multiplied by a scalar \(\alpha\): \(\vec{B} = \alpha \vec{A}\)
Properties:
- Direction:
- If \(\alpha\) is positive, \(\vec{B}\) is parallel to \(\vec{A}\).
- If \(\alpha\) is negative, \(\vec{B}\) is antiparallel to \(\vec{A}\).
- Magnitude: \(|\vec{B}| = |\alpha| |\vec{A}|\)
Note
This is a scalar equation because magnitudes are scalar quantities.
Example:
Displacement \(\vec{D}_{AC} = 0.75 \vec{D}_{AB}\) (from A to C).
- \(\vec{D}_{AC}\) is parallel to \(\vec{D}_{AB}\).
- \(|\vec{D}_{AC}| = 0.75 \times 6 \text{ km} = 4.5 \text{ km}\).
Figure 2.7 (a): Vector-scalar multiplication. \(\vec{B} = 2\vec{A}\) (parallel, double length), \(\vec{C} = -2\vec{A}\) (antiparallel, double length).
Vector Addition and Subtraction (1D)
Addition
The sum of two (or more) vectors is called the resultant vector.
\(\vec{D}_{AD} = \vec{D}_{AC} + \vec{D}_{CD}\)
If \(\vec{D}_{AC} = 0.75 \vec{D}_{AB}\) and \(\vec{D}_{CD} = -0.2 \vec{D}_{AB}\):
\(\vec{D}_{AD} = (0.75 - 0.2) \vec{D}_{AB} = 0.55 \vec{D}_{AB}\)
Figure 2.7 (b): Addition of two parallel vectors in 1D. Resultant \(\vec{R} = \vec{A} + \vec{B}\).
Subtraction
Vector subtraction is equivalent to adding an antiparallel vector.
\(\vec{D}_{DB} = \vec{D}_{AB} - \vec{D}_{AD} = \vec{D}_{AB} + (-\vec{D}_{AD})\)
Substituting \(\vec{D}_{AD} = 0.55 \vec{D}_{AB}\):
\(\vec{D}_{DB} = (1.0 - 0.55) \vec{D}_{AB} = 0.45 \vec{D}_{AB}\)
Figure 2.7 (c): Subtraction of two vectors in 1D. Difference \(\vec{D} = \vec{A} - \vec{B}\).
Example 2.1: A Ladybug Walker
A ladybug makes five displacements along a stick.
- \(\vec{D}_1 = (15 \text{ cm}) (+ \hat{u})\)
- \(\vec{D}_2 = (56 \text{ cm}) (- \hat{u})\)
- \(\vec{D}_3 = (3 \text{ cm}) (+ \hat{u})\)
- \(\vec{D}_4 = (25 \text{ cm}) (+ \hat{u})\)
- \(\vec{D}_5 = (19 \text{ cm}) (- \hat{u})\)
Find the total displacement and final resting position.
Figure 2.8: Ladybug displacements. (+\(\hat{u}\) toward floor, -\(\hat{u}\) toward wall).
Solution Strategy:
- Define a unit vector direction (e.g., \(+\hat{u}\) toward the floor).
- Express each displacement using this unit vector.
- Sum the displacement vectors using the distributive law.
Solution:
\(\vec{D} = \vec{D}_1 + \vec{D}_2 + \vec{D}_3 + \vec{D}_4 + \vec{D}_5\)
\(\vec{D} = (15 - 56 + 3 + 25 - 19) \text{ cm} \hat{u}\)
\(\vec{D} = -32 \text{ cm} \hat{u}\)
The total displacement is 32 cm toward the wall.
Initial position: 100-cm mark.
Final position: \(100 \text{ cm} - 32 \text{ cm} = 68 \text{ cm}\) mark.
Algebra of Vectors in Two Dimensions
When vectors are in a plane, geometric methods using rulers, protractors, and trigonometry are often used.
Parallelogram Rule for Vector Addition
To find \(\vec{R} = \vec{A} + \vec{B}\):
- Translate vectors \(\vec{A}\) and \(\vec{B}\) so their tails meet at a common origin.
- Draw parallel lines from the head of \(\vec{A}\) (parallel to \(\vec{B}\)) and from the head of \(\vec{B}\) (parallel to \(\vec{A}\)) to form a parallelogram.
- The resultant \(\vec{R}\) is the diagonal from the common origin to the opposite corner.
Figure 2.10 (a): Parallelogram rule for \(\vec{R} = \vec{A} + \vec{B}\).
Parallelogram Rule for Vector Subtraction
To find \(\vec{D} = \vec{A} - \vec{B}\):
- Translate vectors \(\vec{A}\) and \(\vec{B}\) so their tails meet at a common origin.
- The difference \(\vec{D}\) is the diagonal from the head of \(\vec{B}\) to the head of \(\vec{A}\).
Figure 2.10 (b): Parallelogram rule for \(\vec{D} = \vec{A} - \vec{B}\).
Caution
The magnitude of the resultant or difference vector is generally NOT the simple sum or difference of their magnitudes.
Tail-to-Head Method for Multiple Vectors
For adding three or more vectors (e.g., \(\vec{R} = \vec{A} + \vec{B} + \vec{C} + \vec{D}\)):
- Place the tail of the second vector at the head of the first.
- Place the tail of the third vector at the head of the second, and so on.
- The resultant vector \(\vec{R}\) is drawn from the tail of the first vector to the head of the last vector.
Figure 2.11: Vacation route with multiple displacement vectors, summed using the tail-to-head method.
Figure 2.12: General tail-to-head method.
Example 2.2: Geometric Construction of the Resultant
Given:
\(\vec{A}\) (magnitude 10.0, angle 35\(^\circ\))
\(\vec{B}\) (magnitude 7.0, angle -110\(^\circ\))
\(\vec{C}\) (magnitude 8.0, angle 30\(^\circ\))
Find graphically using ruler and protractor:
- \(\vec{R} = \vec{A} + \vec{B}\)
- \(\vec{D} = \vec{A} - \vec{B}\)
- \(\vec{S} = \vec{A} - 3\vec{B} + \vec{C}\)
Figure 2.13: Vectors for Example 2.2.
Solution Strategy:
- Draw vectors to scale at their specified angles.
- Apply parallelogram rule for (a) and (b).
- Apply tail-to-head method for (c), including scalar multiplication for \(-3\vec{B}\).
- Measure resultant magnitudes and angles.
Example 2.2 Solution (Partial)
- \(\vec{R} = \vec{A} + \vec{B}\)
- \(\vec{D} = \vec{A} - \vec{B}\)
Figure 2.14: (a) Resultant \(\vec{R}\) (red), (b) Difference \(\vec{D}\) (blue).
- From measurements:
- \(R \approx 5.8 \text{ cm}\), \(\theta_R \approx 0^\circ\)
- \(D \approx 16.2 \text{ cm}\), \(\theta_D \approx 49.3^\circ\)
- \(\vec{S} = \vec{A} - 3\vec{B} + \vec{C}\)
Figure 2.15: Vector \(\vec{S}\) (green).
- From measurements:
- \(S \approx 36.9 \text{ cm}\), \(\theta_S \approx 52.9^\circ\)
2.2 Coordinate Systems and Components of a Vector
Vector Components in 2D
A vector \(\vec{A}\) in a plane can be described by its x- and y-vector components, \(\vec{A}_x\) and \(\vec{A}_y\).
- \(\vec{A} = \vec{A}_x + \vec{A}_y\)
- \(\vec{A}_x\) is the projection of \(\vec{A}\) onto the x-axis.
- \(\vec{A}_y\) is the projection of \(\vec{A}\) onto the y-axis.
Figure 2.16: Vector \(\vec{A}\) and its components \(\vec{A}_x\) and \(\vec{A}_y\) in the Cartesian system.
Example 2.3: Displacement of a Mouse Pointer
Initial position: \((6.0 \text{ cm}, 1.6 \text{ cm})\)
Final position: \((2.0 \text{ cm}, 4.5 \text{ cm})\)
Find the displacement vector \(\vec{D}\).
Solution Strategy:
- Identify initial \((x_b, y_b)\) and final \((x_e, y_e)\) coordinates.
- Calculate scalar components \(D_x\) and \(D_y\).
- Write the displacement vector in component form.
Solution:
\(D_x = x_e - x_b = (2.0 - 6.0) \text{ cm} = -4.0 \text{ cm}\)
\(D_y = y_e - y_b = (4.5 - 1.6) \text{ cm} = +2.9 \text{ cm}\)
\(\vec{D} = (-4.0 \text{ cm}) \hat{i} + (2.9 \text{ cm}) \hat{j} = (-4.0 \hat{i} + 2.9 \hat{j}) \text{ cm}\)
Figure 2.17: The graph of the displacement vector \(\vec{D}\).
Magnitude and Direction from Scalar Components
Magnitude
Using the Pythagorean theorem for a right triangle formed by \(A_x\), \(A_y\), and \(A\):
\(A^2 = A_x^2 + A_y^2 \implies A = \sqrt{A_x^2 + A_y^2}\)
Direction Angle
The direction angle \(\theta_A\) is measured counterclockwise from the \(+x\)-axis to the vector.
\(\tan \theta_A = \frac{A_y}{A_x}\)
Figure 2.18: Magnitude \(A\) and direction angle \(\theta_A\) from components.
Quadrants:
The value of \(\theta_A\) depends on the quadrant of the vector.
- If \(A_x\) is positive (Quadrant I or IV), \(\theta_A = \arctan(\frac{A_y}{A_x})\).
- If \(A_x\) is negative (Quadrant II or III), \(\theta_A = \arctan(\frac{A_y}{A_x}) + 180^\circ\).
Figure 2.19: Scalar components can be positive or negative depending on the quadrant.
Polar Coordinates
Alternative way to describe a point \(P\) in a plane:
- Radial coordinate (\(r\)): Distance from the origin to point \(P\).
- Angular coordinate (\(\phi\)): Angle the radial vector makes with the positive x-direction (usually counterclockwise).
Figure 2.20: Polar coordinates (\(r, \phi\)). Unit radial vector \(\hat{r}\) and orthogonal unit vector \(\hat{t}\).
Conversion between Polar and Cartesian Coordinates
From polar \((r, \phi)\) to Cartesian \((x, y)\):
\(x = r \cos \phi\)
\(y = r \sin \phi\)
From Cartesian \((x, y)\) to polar \((r, \phi)\):
\(r = \sqrt{x^2 + y^2}\)
\(\tan \phi = \frac{y}{x}\) (adjust \(\phi\) for quadrant)
Vectors in Three Dimensions
To locate a point in space, we need three coordinates \((x, y, z)\).
Unit Vectors for 3D Axes
- \(\hat{i}\): Positive x-direction
- \(\hat{j}\): Positive y-direction
- \(\hat{k}\): Positive z-direction (usually vertical)
This forms a right-handed coordinate system.
Figure 2.21: The right-handed Cartesian coordinate system with \(\hat{i}\), \(\hat{j}\), \(\hat{k}\) unit vectors.
Three-Dimensional Vector Form
A vector \(\vec{A}\) in 3D space is the sum of its three vector components:
\(\vec{A} = \vec{A}_x + \vec{A}_y + \vec{A}_z = A_x \hat{i} + A_y \hat{j} + A_z \hat{k}\)
Figure 2.22: A vector in 3D space and its components.
Scalar Components from Points (3D)
If a vector starts at \((x_b, y_b, z_b)\) and ends at \((x_e, y_e, z_e)\):
\(A_x = x_e - x_b\)
\(A_y = y_e - y_b\)
\(A_z = z_e - z_b\)
Magnitude (3D)
\(A = \sqrt{A_x^2 + A_y^2 + A_z^2}\)
Example 2.9 Solution (Continued)
- \(\vec{R} = \vec{A} + \vec{B} + \vec{C}\)
\(R_x = A_x + B_x + C_x = 8.19 - 2.39 + 6.93 = 12.73 \text{ cm}\)
\(R_y = A_y + B_y + C_y = 5.73 - 6.58 + 4.00 = 3.15 \text{ cm}\)
\(\vec{R} = (12.7 \hat{i} + 3.1 \hat{j}) \text{ cm}\)
- \(\vec{D} = \vec{A} - \vec{B}\)
\(D_x = A_x - B_x = 8.19 - (-2.39) = 10.58 \text{ cm}\)
\(D_y = A_y - B_y = 5.73 - (-6.58) = 12.31 \text{ cm}\)
\(\vec{D} = (10.6 \hat{i} + 12.3 \hat{j}) \text{ cm}\)
- \(\vec{S} = \vec{A} - 3\vec{B} + \vec{C}\)
\(S_x = A_x - 3B_x + C_x = 8.19 - 3(-2.39) + 6.93 = 22.29 \text{ cm}\)
\(S_y = A_y - 3B_y + C_y = 5.73 - 3(-6.58) + 4.00 = 29.47 \text{ cm}\)
\(\vec{S} = (22.3 \hat{i} + 29.5 \hat{j}) \text{ cm}\)
Figure 2.24: Graphical illustration of analytical results.
The Scalar Product (Dot Product)
The scalar product \(\vec{A} \cdot \vec{B}\) of two vectors \(\vec{A}\) and \(\vec{B}\) is defined as:
\(\vec{A} \cdot \vec{B} = AB \cos \phi\)
Where:
- \(A\) and \(B\) are the magnitudes of \(\vec{A}\) and \(\vec{B}\).
- \(\phi\) is the angle between the two vectors (\(0^\circ \le \phi \le 180^\circ\)).
Figure 2.27 (a): The angle \(\phi\) between vectors \(\vec{A}\) and \(\vec{B}\) for the scalar product.
Properties
- Parallel Vectors (\(\phi = 0^\circ\)): \(\vec{A} \cdot \vec{B} = AB \cos 0^\circ = AB\)
- Antiparallel Vectors (\(\phi = 180^\circ\)): \(\vec{A} \cdot \vec{B} = AB \cos 180^\circ = -AB\)
- Orthogonal Vectors (\(\phi = 90^\circ\)): \(\vec{A} \cdot \vec{B} = AB \cos 90^\circ = 0\)
- Vector with Itself: \(\vec{A} \cdot \vec{A} = A^2\) (square of its magnitude)
The Vector Product (Cross Product)
The vector product \(\vec{A} \times \vec{B}\) (or cross product) of two vectors \(\vec{A}\) and \(\vec{B}\) is a vector whose direction is perpendicular to the plane containing \(\vec{A}\) and \(\vec{B}\).
The magnitude of the vector product is:
\(|\vec{A} \times \vec{B}| = AB \sin \phi\)
Where:
- \(A\) and \(B\) are the magnitudes of \(\vec{A}\) and \(\vec{B}\).
- \(\phi\) is the angle between \(\vec{A}\) (first vector) and \(\vec{B}\) (second vector), \(0^\circ \le \phi \le 180^\circ\).
Figure 2.29: The cross product \(\vec{C} = \vec{A} \times \vec{B}\) is perpendicular to both \(\vec{A}\) and \(\vec{B}\).
Properties
- Parallel or Antiparallel Vectors (\(\phi = 0^\circ\) or \(\phi = 180^\circ\)): \(\vec{A} \times \vec{B} = 0\) (magnitude is zero because \(\sin 0^\circ = \sin 180^\circ = 0\)).
- Anticommutative: \(\vec{A} \times \vec{B} = -(\vec{B} \times \vec{A})\)
Direction of the Cross Product: The Right-Hand Rule (Corkscrew Rule)
- Place a corkscrew perpendicular to the plane containing \(\vec{A}\) and \(\vec{B}\).
- Turn the handle from the first vector (\(\vec{A}\)) towards the second vector (\(\vec{B}\)).
- The direction the corkscrew moves (in or out) is the direction of \(\vec{A} \times \vec{B}\).
Figure 2.30: Right-Hand Rule. (a) \(\vec{A} \times \vec{B}\) is upward. (b) \(\vec{B} \times \vec{A}\) is downward.
Distributive Property
\(\vec{A} \times (\vec{B} + \vec{C}) = \vec{A} \times \vec{B} + \vec{A} \times \vec{C}\)
Cross Product in Component Form
Unit Vector Cross Products
- \(\hat{i} \times \hat{i} = \vec{0}\), \(\hat{j} \times \hat{j} = \vec{0}\), \(\hat{k} \times \hat{k} = \vec{0}\)
- Cyclic order:
- \(\hat{i} \times \hat{j} = +\hat{k}\)
- \(\hat{j} \times \hat{k} = +\hat{i}\)
- \(\hat{k} \times \hat{i} = +\hat{j}\)
- Reverse cyclic order:
- \(\hat{j} \times \hat{i} = -\hat{k}\)
- \(\hat{k} \times \hat{j} = -\hat{i}\)
- \(\hat{i} \times \hat{k} = -\hat{j}\)
Figure 2.32: Cyclic order of unit vectors for cross products.
