Exhaustive Applied Physics Study Guide: Vectors and Vector Calculus

Foundations of Physics and Applied Physics

• Etymology and Definition:

  • Derived from the Greek word physis, which translates to "nature".

  • Physics is the branch of science dedicated to studying matter, energy, and the fundamental interactions between them.

  • Core phenomena investigated within physical interactions include:

  • Motion

  • Force

  • Heat

  • Light

  • Sound

  • Electricity

  • Magnetism

• Core Objective and Philosophy:

  • Aims to describe and explain the mechanisms of nature using a small number of universal basic laws.

• Applied Physics Applications:

  • Applies fundamental physical principles to develop, design, and improve engineering solutions and technologies across multiple sectors:

  • Electronics

  • Mechanical systems and machinery

  • Advanced materials science

  • Telecommunications and signal processing

  • Medical imaging technologies

  • Energy conversion and power systems

• Quantitative Nature of Physics:

  • Physics is inherently quantitative; its laws are formulated mathematically.

  • Every physical concept must be precisely measured and expressed as a numerical value combined with an appropriate physical unit.

Physical Quantities, Scalars, and Vectors

• Definition of Physical Quantity:

  • Any measurable property or attribute of a physical system.

  • Standard mathematical formulation:     $\text{Physical Quantity} = \text{number} \times \text{unit}$

  • Standard examples:

  • Mass: $5\,kg$

  • Velocity: $20\,m/s$

  • Acceleration due to gravity: $9.8\,m/s^2$

  • Energy / Work: $100\,J$

• Classification Test (The Direction Diagnostic):

  • To classify a physical quantity, ask: Does direction matter?

  • If direction is irrelevant to its complete description: It is a scalar.

  • If direction is essential to its complete description: It is a vector.

Scalar Quantities

• Definition:

  • Physical quantities that are completely specified by magnitude alone (a numerical value along with an appropriate unit).

• Arithmetic Addition Rules:

  • Scalars combine according to standard rules of ordinary arithmetic.

  • Example: $2\,kg + 3\,kg = 5\,kg$ under all circumstances.

• Comprehensive Table of Scalar Quantities:

  • Mass: Example value $50\,kg$ (Direction required: No)

  • Time: Example value $10\,s$ (Direction required: No)

  • Temperature: Example value $37\,^\circ\text{C}$ (Direction required: No)

  • Distance: Example value $5\,km$ (Direction required: No)

  • Speed: Example value $60\,km/h$ (Direction required: No)

  • Energy / Work: Example value $100\,J$ (Direction required: No)

Vector Quantities

• Definition:

  • Physical quantities that require both a magnitude and a spatial direction to be completely specified.

• Arithmetic Addition Rules:

  • Vectors do not combine via ordinary arithmetic.

  • The resultant value depends directly on the relative directions of the combining vectors.

  • Illustrative Example: A force of $3\,N$ and a force of $4\,N$ can combine to yield any net force magnitude between $1\,N$ (when antiparallel at $180^\circ$) and $7\,N$ (when parallel at $0^\circ$). When acting at right angles ($90^\circ$), they yield a resultant force of exactly 5\,N$.\n- **Comprehensive Table of Vector Quantities**:\n - Displacement: Example 5\,km towards north\n - Velocity: Example 60\,km/h towards east\n - Acceleration: Example 9.8\,m/s^2 vertically downward\n - Force: Example 10\,N$at$30^\circ above the horizontal\n - Weight: Example 500\,N vertically downward\n\n# Vector Representation, Components, and Magnitudes\n\n- **Graphical Representation**:\n - Represented visually as a directed line segment (an arrow).\n - **Tail**: The initial starting point of the vector.\n - **Head (Arrowhead)**: Indicates the spatial direction of the vector.\n - **Length**: Drawn to scale to represent the magnitude of the vector.\n - **Direction Angle (\theta)**: Specifies the direction relative to a coordinate reference axis.\n\n\n\n- **Notational Conventions**:\n - **Printed Text**: Bold capital letter, e.g., \mathbf{A}.\n - **Handwritten Text**: Letter with an overhead arrow, e.g., \vec{A}.\n - **Magnitude Notation**: Written as |\mathbf{A}|$or simply as a non-bold letter$A$.

  • Properties of Magnitude:

  • Magnitude is never negative ($|\mathbf{A}| \ge 0$).

  • Magnitude retains the physical unit of the quantity (e.g., $m$, $N$, $m/s$).

• Decomposition into Vector Components:

  • Decomposing a vector resolves it into perpendicular projections along coordinate axes.

• Two-Dimensional Component Formulas:

  • Horizontal x-component: $A_1 = A\cos(\theta)$

  • Vertical y-component: $A_2 = A\sin(\theta)$

  • Component representation: $\mathbf{A} = A_1\hat{i} + A_2\hat{j}$

• Three-Dimensional Expansion:

  • Adds a third orthogonal component along the z-axis: $\mathbf{A} = A_1\hat{i} + A_2\hat{j} + A_3\hat{k}$

  • The unit vectors $\hat{i}$, $\hat{j}$, and $\hat{k}$ point along the x, y, and z axes respectively.

• Worked Component Calculation Example:

  • Problem: Resolve a force vector of $10\,N$ acting at an angle of $30^\circ$ to the x-axis.

  • x-component calculation: $F_1 = 10\cos(30^\circ) = 10 \times 0.866 = 8.66\,N$

  • y-component calculation: $F_2 = 10\sin(30^\circ) = 10 \times 0.5 = 5.00\,N$

  • Complete vector notation: $\mathbf{F} = 8.66\hat{i} + 5.00\hat{j}\,N$

  • Mathematical Calculation of Magnitude and Direction:

• Vector components form a right-angled triangle, allowing the application of Pythagoras' theorem.

• Two-Dimensional Magnitude Formula:     $|\mathbf{A}| = \sqrt{A_1^2 + A_2^2}$

• Two-Dimensional Direction Angle Formula:     $\theta = \tan^{-1}\left(\frac{A_2}{A_1}\right)$     (where $\theta$ is measured counterclockwise relative to the positive x-axis).

• Three-Dimensional Magnitude Formula:     $|\mathbf{A}| = \sqrt{A_1^2 + A_2^2 + A_3^2}$

  • Step-by-Step Magnitude Worked Examples:

• Example 1 (2D Vector): $\mathbf{A} = 3\hat{i} + 4\hat{j}$

  • Step 1 (Read components): $A_1 = 3$, $A_2 = 4$

  • Step 2 (Square components): $3^2 = 9$, $4^2 = 16$

  • Step 3 (Add squared components): $9 + 16 = 25$

  • Step 4 (Square root sum): $|\mathbf{A}| = \sqrt{25} = 5$

  • Step 5 (Determine direction angle): $\theta = \tan^{-1}\left(\frac{4}{3}\right) = 53.13^\circ$

• Example 2 (3D Vector): $\mathbf{B} = 2\hat{i} - 3\hat{j} + 6\hat{k}$

  • Step 1 (Read components): $B_1 = 2$, $B_2 = -3$, $B_3 = 6$

  • Step 2 (Square components): $2^2 = 4$, $(-3)^2 = 9$, $6^2 = 36$ (squaring a component always yields a positive number)

  • Step 3 (Add squared components): $4 + 9 + 36 = 49$

  • Step 4 (Square root sum): $|\mathbf{B}| = \sqrt{49} = 7$

Unit Vectors

• Definition and Key Properties:

  • A unit vector is a vector with a magnitude equal to exactly 1.

  • It isolatedly captures directional information while scaling magnitude to unity.

  • Dimensionless: Unit vectors carry no units because physical units cancel during division.

  • Symbolized with a circumflex ("hat") over the capital letter: $\hat{\mathbf{A}}$ (read as "A hat").

• Mathematical Formulas:

  • Formula for Unit Vector:     $\hat{\mathbf{A}} = \frac{\mathbf{A}}{|\mathbf{A}|}$

  • Reconstituting a Vector:     $\mathbf{A} = |\mathbf{A}|\hat{\mathbf{A}}$     $\text{Any Vector} = \text{magnitude} \times \text{unit vector} = \text{size} \times \text{direction}$

• Standard Unit Vectors:

  • $\hat{i}$: Unit vector pointing along the positive x-axis ($|\hat{i}| = 1$).

  • $\hat{j}$: Unit vector pointing along the positive y-axis ($|\hat{j}| = 1$).

  • $\hat{k}$: Unit vector pointing along the positive z-axis ($|\hat{k}| = 1$).

• Step-by-Step Unit Vector Worked Examples:

  • Example 1: Given $\mathbf{A} = 3\hat{i} + 4\hat{j}$

  • Step 1 (Find magnitude): $|\mathbf{A}| = 5$

  • Step 2 (Divide components by magnitude): $\hat{\mathbf{A}} = \left(\frac{3}{5}\right)\hat{i} + \left(\frac{4}{5}\right)\hat{j}$

  • Step 3 (Express in decimal format): $\hat{\mathbf{A}} = 0.6\hat{i} + 0.8\hat{j}$

  • Step 4 (Verification check): $\sqrt{0.6^2 + 0.8^2} = \sqrt{0.36 + 0.64} = \sqrt{1} = 1$

  • Example 2: Given $\mathbf{B} = 2\hat{i} - 3\hat{j} + 6\hat{k}$

  • Step 1 (Find magnitude): $|\mathbf{B}| = 7$

  • Step 2 (Divide components by magnitude): $\hat{\mathbf{B}} = \left(\frac{2}{7}\right)\hat{i} - \left(\frac{3}{7}\right)\hat{j} + \left(\frac{6}{7}\right)\hat{k}$

  • Step 3 (Verification check): $\sqrt{\left(\frac{2}{7}\right)^2 + \left(-\frac{3}{7}\right)^2 + \left(\frac{6}{7}\right)^2} = \sqrt{\frac{4 + 9 + 36}{49}} = \sqrt{\frac{49}{49}} = 1$

Position Vector, Displacement Vector, and Kinematics

• Position Vector ($\mathbf{r}$):

  • Definition: An arrow drawn directly from the origin $O(0,0,0)$ to a target spatial point $P(x, y, z)$.

  • Expresses the location of a point relative to a coordinate origin.

• Vector Equation:     $\mathbf{r} = x\hat{i} + y\hat{j} + z\hat{k}$

• Distance from Origin:     $|\mathbf{r}| = \sqrt{x^2 + y^2 + z^2}$

• Worked Position Vector Example:

  • Point $P(2, 3, 6)\,m$

  • Position vector: $\mathbf{r} = 2\hat{i} + 3\hat{j} + 6\hat{k}\,m$

  • Distance from origin: $|\mathbf{r}| = \sqrt{2^2 + 3^2 + 6^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7\,m$

  • Displacement Vector ($\Delta\mathbf{r}$):

• Definition: Represents the change in position of a particle, drawn as a vector from an initial point $P_1$ to a final point P_2$.\n\n\n\n - **Vector Equation**:\n    \Delta\mathbf{r} = \mathbf{r}_2 - \mathbf{r}_1\n    *(Always final position minus initial position)*\n - **Worked Displacement Vector Example**:\n - Motion from P_1(1, 2)\,m$to$P_2(4, 6)\,m\n - Position vectors: \mathbf{r}_1 = 1\hat{i} + 2\hat{j}\,m$,$\mathbf{r}_2 = 4\hat{i} + 6\hat{j}\,m\n - Displacement: \Delta\mathbf{r} = (4 - 1)\hat{i} + (6 - 2)\hat{j} = 3\hat{i} + 4\hat{j}\,m\n - Magnitude of displacement: |\Delta\mathbf{r}| = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5\,m\n\n- **Distinction Between Distance and Displacement**:\n - Key differences between distance and displacement:\n - **Quantity Category**: Distance is a **scalar**; Displacement is a **vector**.\n - **Physical Definition**: Distance is the total path length traversed; Displacement is the direct straight-line distance from start to finish with direction.\n - **Path Dependency**: Distance depends directly on the path taken; Displacement is **path independent** (depends only on starting and ending points).\n - **Zero Value Condition**: Distance can **never** be zero after motion occurs; Displacement **can be zero** if motion terminates at the starting position.\n - **Worked Circular Track Example**:\n - Scenario: A student walks once around a circular track of radius r = 50\,m$.

  • Distance calculated: $\text{Distance} = 2\pi r = 2\pi \times 50 = 314\,m$

  • Displacement calculated: $\text{Displacement} = 0\,m$ (since the starting and ending points are identical).

Basic Vector Operations

• Overview of Primary Operations:

  • Addition: $\mathbf{A} + \mathbf{B}$ (combines two vectors).

  • Subtraction: $\mathbf{A} - \mathbf{B}$ (adds the negative of $\mathbf{B}$ to $\mathbf{A}$).

  • Scalar Multiplication: $k\mathbf{A}$ (stretches, shrinks, or reverses a vector).

  • Scalar Division: $\frac{\mathbf{A}}{k}$ (equivalent to multiplying by $\frac{1}{k}$).

  • Important Constraint: Division by a vector is undefined.

Vector Addition

• Graphical Method (Head-to-Tail Rule):

  • Place the tail of vector $\mathbf{B}$ at the head of vector $\mathbf{A}$.

  • The resultant vector $\mathbf{R} = \mathbf{A} + \mathbf{B}$ is drawn from the tail of $\mathbf{A}$ to the head of $\mathbf{B}$.

• Algebraic Component Rule:

  • Add corresponding components together:     $\mathbf{A} + \mathbf{B} = (A_1 + B_1)\hat{i} + (A_2 + B_2)\hat{j} + (A_3 + B_3)\hat{k}$

• Algebraic Properties:

  • Commutative Property: $\mathbf{A} + \mathbf{B} = \mathbf{B} + \mathbf{A}$

  • Associative Property: $(\mathbf{A} + \mathbf{B}) + \mathbf{C} = \mathbf{A} + (\mathbf{B} + \mathbf{C})$

• Worked Step-by-Step Addition Example:

  • Given: $\mathbf{A} = 2\hat{i} + 3\hat{j} - \hat{k}$, $\mathbf{B} = 4\hat{i} - \hat{j} + 2\hat{k}$

  • Step 1 (Add x-components): $2 + 4 = 6$

  • Step 2 (Add y-components): $3 + (-1) = 2$

  • Step 3 (Add z-components): $(-1) + 2 = 1$

  • Step 4 (Write result): $\mathbf{A} + \mathbf{B} = 6\hat{i} + 2\hat{j} + \hat{k}$

  • Step 5 (Calculate magnitude): $|\mathbf{A} + \mathbf{B}| = \sqrt{6^2 + 2^2 + 1^2} = \sqrt{36 + 4 + 1} = \sqrt{41} \approx 6.40$

Vector Subtraction

• The Negative Vector:

  • The vector $-\mathbf{B}$ has the same magnitude as $\mathbf{B}$ but points in the opposite direction ($180^\circ$ opposite).

  • Formed by reversing the sign of every individual component of $\mathbf{B}$.

• Subtraction Rule:

  • Subtraction is performed by adding the negative vector: $\mathbf{A} - \mathbf{B} = \mathbf{A} + (-\mathbf{B})$.

  • Subtract corresponding components:     $\mathbf{A} - \mathbf{B} = (A_1 - B_1)\hat{i} + (A_2 - B_2)\hat{j} + (A_3 - B_3)\hat{k}$

• Non-Commutative Behavior:

  • $\mathbf{A} - \mathbf{B} \neq \mathbf{B} - \mathbf{A}$

  • Specifically, $\mathbf{B} - \mathbf{A} = -(\mathbf{A} - \mathbf{B})$.

• Worked Step-by-Step Subtraction Example:

  • Given same vectors: $\mathbf{A} = 2\hat{i} + 3\hat{j} - \hat{k}$, $\mathbf{B} = 4\hat{i} - \hat{j} + 2\hat{k}$

  • Step 1 (Write negative vector $-\mathbf{B}$): $-\mathbf{B} = -4\hat{i} + \hat{j} - 2\hat{k}$

  • Step 2 (Subtract x-components): $2 - 4 = -2$

  • Step 3 (Subtract y-components): $3 - (-1) = 4$

  • Step 4 (Subtract z-components): $(-1) - 2 = -3$

  • Step 5 (Write result): $\mathbf{A} - \mathbf{B} = -2\hat{i} + 4\hat{j} - 3\hat{k}$

Scalar Multiplication and Division

• Effect of Scalar Multiplier $k$ on Vector $\mathbf{A}$:

  • $k > 1$: Vector becomes longer (stretched); direction unchanged.

  • $0 < k < 1$: Vector becomes shorter (shrunk); direction unchanged.

  • $k < 0$: Direction is reversed ($180^\circ$); length is multiplied by factor |k|$.\n - k = 0$: Reduces vector to the null/zero vector$\mathbf{0}.\n- **Worked Multiplication & Division Examples** (using \mathbf{A} = 2\hat{i} + 3\hat{j} - \hat{k}):\n - Multiplication by 3$:$3\mathbf{A} = 6\hat{i} + 9\hat{j} - 3\hat{k}\n - Multiplication by -2$:$-2\mathbf{A} = -4\hat{i} - 6\hat{j} + 2\hat{k}\n - Division by 2$:$\frac{\mathbf{A}}{2} = 1\hat{i} + 1.5\hat{j} - 0.5\hat{k}\n\n# Products of Two Vectors\n\n- Vectors can be multiplied together in two distinct ways:\n - **Dot Product (Scalar Product)**: Result is a pure scalar quantity (a number).\n - **Cross Product (Vector Product)**: Result is a vector quantity.\n\n## The Dot Product (Scalar Product)\n\n- **Geometric Definition**:\n  \mathbf{A} \cdot \mathbf{B} = AB\cos(\theta)\n  where A = |\mathbf{A}|$and$B = |\mathbf{B}|$are magnitudes, and$\theta$is the angle between vectors placed tail-to-tail ($0^\circ \le \theta \le 180^\circ).\n- **Angular Behavior and Values**:\n - \theta = 0^\circ$(Parallel):$\cos(0^\circ) = 1 \implies \mathbf{A} \cdot \mathbf{B} = AB (Maximum positive value).\n - \theta = 90^\circ$(Perpendicular):$\cos(90^\circ) = 0 \implies \mathbf{A} \cdot \mathbf{B} = 0 (Orthogonal vectors).\n - \theta = 180^\circ$(Antiparallel):$\cos(180^\circ) = -1 \implies \mathbf{A} \cdot \mathbf{B} = -AB (Maximum negative value).\n- **Physical Application**:\n - Mechanical work done: W = \mathbf{F} \cdot \mathbf{d} = Fd\cos(\theta).\n - A force operating perpendicular (90^\circ) to displacement does zero work.\n- **Dot Product with Standard Unit Vectors**:\n - Same unit vectors: \hat{i} \cdot \hat{i} = \hat{j} \cdot \hat{j} = \hat{k} \cdot \hat{k} = 1\n - Different unit vectors: \hat{i} \cdot \hat{j} = \hat{j} \cdot \hat{k} = \hat{k} \cdot \hat{i} = 0\n- **Component Formula**:\n - Multiply corresponding components and add results (cross terms vanish):\n    \mathbf{A} \cdot \mathbf{B} = A_1 B_1 + A_2 B_2 + A_3 B_3\n- **Algebraic Properties**:\n - **Commutative**: \mathbf{A} \cdot \mathbf{B} = \mathbf{B} \cdot \mathbf{A}\n - **Self Dot Product**: \mathbf{A} \cdot \mathbf{A} = A_1^2 + A_2^2 + A_3^2 = |\mathbf{A}|^2 = A^2\n- **Angle Determination Formula**:\n  \cos(\theta) = \frac{\mathbf{A} \cdot \mathbf{B}}{|\mathbf{A}||\mathbf{B}|} = \frac{A_1 B_1 + A_2 B_2 + A_3 B_3}{\sqrt{A_1^2 + A_2^2 + A_3^2}\sqrt{B_1^2 + B_2^2 + B_3^2}}\n- **Worked Example for Dot Product and Angle**:\n - Given: \mathbf{A} = 2\hat{i} + 3\hat{j} - \hat{k}$,$\mathbf{B} = 4\hat{i} - \hat{j} + 2\hat{k}\n - Step 1 (Multiply like components): (2)(4) + (3)(-1) + (-1)(2)\n - Step 2 (Sum component products): \mathbf{A} \cdot \mathbf{B} = 8 - 3 - 2 = 3\n - Step 3 (Magnitude of \mathbf{A}$):$|\mathbf{A}| = \sqrt{2^2 + 3^2 + (-1)^2} = \sqrt{4 + 9 + 1} = \sqrt{14} \approx 3.742\n - Step 4 (Magnitude of \mathbf{B}$):$|\mathbf{B}| = \sqrt{4^2 + (-1)^2 + 2^2} = \sqrt{16 + 1 + 4} = \sqrt{21} \approx 4.583\n - Step 5 (Calculate \cos(\theta)$):$\cos(\theta) = \frac{3}{3.742 \times 4.583} = \frac{3}{17.146} \approx 0.175\n - Step 6 (Calculate angle \theta$):$\theta = \cos^{-1}(0.175) = 79.9^\circ\n\n## The Cross Product (Vector Product)\n\n- **Geometric Definition**:\n  \mathbf{A} \times \mathbf{B} = (AB\sin(\theta))\hat{\mathbf{n}}\n  where AB\sin(\theta)$is the scalar magnitude, and$\hat{\mathbf{n}}$is a unit vector perpendicular to both$\mathbf{A}$and$\mathbf{B}.\n- **Right-Hand Rule for Direction**:\n - Point fingers of your right hand along vector \mathbf{A}.\n - Curl fingers toward vector \mathbf{B}$through the smaller intervening angle$\theta$.

  • Your extended thumb points along the direction of $\mathbf{A} \times \mathbf{B}$ ($\hat{\mathbf{n}}$).

• Angular Behavior and Special Cases:

  • $\theta = 0^\circ$ or $180^\circ$ (Parallel / Antiparallel): $\sin(\theta) = 0 \implies \mathbf{A} \times \mathbf{B} = \mathbf{0}$.

  • $\theta = 90^\circ$ (Perpendicular): $\sin(90^\circ) = 1 \implies |\mathbf{A} \times \mathbf{B}| = AB$ (Maximum magnitude).

• Physical Application:

  • Torque: $\boldsymbol{\tau} = \mathbf{r} \times \mathbf{F}$.

• Geometric Significance:

  • Magnitude $|\mathbf{A} \times \mathbf{B}|$ equals the area of the parallelogram formed by vectors $\mathbf{A}$ and $\mathbf{B}$.

• Cross Product with Standard Unit Vectors:

  • Cyclic relations among $\hat{i}$, $\hat{j}$, and $\hat{k}$:

• Cyclic Order (Positive):

  • $\hat{i} \times \hat{j} = \hat{k}$

  • $\hat{j} \times \hat{k} = \hat{i}$

  • $\hat{k} \times \hat{i} = \hat{j}$

• Anti-Cyclic Order (Negative):

  • $\hat{j} \times \hat{i} = -\hat{k}$

  • $\hat{k} \times \hat{j} = -\hat{i}$

  • $\hat{i} \times \hat{k} = -\hat{j}$

• Self Cross Products:

  • $\hat{i} \times \hat{i} = \hat{j} \times \hat{j} = \hat{k} \times \hat{k} = \mathbf{0}$

  • Anti-Commutative Property:

• $\mathbf{B} \times \mathbf{A} = -(\mathbf{A} \times \mathbf{B})$

  • Matrix Determinant Computation:

• Set up a $3 \times 3$ determinant matrix:     $\mathbf{A} \times \mathbf{B} = \begin{pmatrix} \hat{i} & \hat{j} & \hat{k} \\ A_1 & A_2 & A_3 \\ B_1 & B_2 & B_3 \end{pmatrix}$

• Expansion along the top row:     $\mathbf{A} \times \mathbf{B} = (A_2 B_3 - A_3 B_2)\hat{i} - (A_1 B_3 - A_3 B_1)\hat{j} + (A_1 B_2 - A_2 B_1)\hat{k}$

• Note the alternating signs: $+$ for $\hat{i}$, -$ for \hat{j}$,$+$for$\hat{k}$.

  • Worked Step-by-Step Cross Product Example:

• Given: $\mathbf{A} = 2\hat{i} + 3\hat{j} - \hat{k}$, $\mathbf{B} = 4\hat{i} - \hat{j} + 2\hat{k}$

• Matrix setup:     $\mathbf{A} \times \mathbf{B} = \begin{pmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & -1 \\ 4 & -1 & 2 \end{pmatrix}$

• Step 1 ($\hat{i}$ component): $(3)(2) - (-1)(-1) = 6 - 1 = 5$

• Step 2 ($\hat{j}$ component): -$[(2)(2) - (-1)(4)] = -[4 + 4] = -8

• Step 3 ($\hat{k}$ component): $(2)(-1) - (3)(4) = -2 - 12 = -14$

• Step 4 (Combine components): $\mathbf{A} \times \mathbf{B} = 5\hat{i} - 8\hat{j} - 14\hat{k}$

• Step 5 (Orthogonality check with $\mathbf{A}$): $\mathbf{A} \cdot (\mathbf{A} \times \mathbf{B}) = (2)(5) + (3)(-8) + (-1)(-14) = 10 - 24 + 14 = 0$ (Confirmed orthogonal)

• Step 6 (Orthogonality check with $\mathbf{B}$): $\mathbf{B} \cdot (\mathbf{A} \times \mathbf{B}) = (4)(5) + (-1)(-8) + (2)(-14) = 20 + 8 - 28 = 0$ (Confirmed orthogonal)

• Step 7 (Calculate cross product magnitude): $|\mathbf{A} \times \mathbf{B}| = \sqrt{5^2 + (-8)^2 + (-14)^2} = \sqrt{25 + 64 + 196} = \sqrt{285} \approx 16.88$

Comparison: Dot Product vs Cross Product

• Result Type: Dot product is a scalar; Cross product is a vector.

• Magnitude Expression: Dot product uses $AB\cos(\theta)$; Cross product uses $AB\sin(\theta)$.

• Calculation Method: Dot product sums component products; Cross product uses matrix determinant expansion.

• Order Dependence: Dot product is commutative ($\mathbf{A} \cdot \mathbf{B} = \mathbf{B} \cdot \mathbf{A}$); Cross product is anti-commutative ($\mathbf{B} \times \mathbf{A} = -(\mathbf{A} \times \mathbf{B})$).

• Zero Value Condition: Dot product is zero when vectors are perpendicular ($90^\circ$); Cross product is zero when vectors are parallel/antiparallel ($0^\circ$ or $180^\circ$).

• Physical Examples: Dot product calculates Work ($W = \mathbf{F} \cdot \mathbf{d}$); Cross product calculates Torque ($\boldsymbol{\tau} = \mathbf{r} \times \mathbf{F}$).

• Worked Example Values (for $\mathbf{A} = 2\hat{i} + 3\hat{j} - \hat{k}$, $\mathbf{B} = 4\hat{i} - \hat{j} + 2\hat{k}$): Dot product = $3$; Cross product = $5\hat{i} - 8\hat{j} - 14\hat{k}$.

Application Problem: Angle Inside a Cube

• Problem Description:

  • Find the angle $\theta$ formed between a face diagonal and a body diagonal of a cube, both drawn from the same corner vertex.

• Geometric Coordinate Setup:

  • Consider a cube of edge length a$.\n - Place origin corner O$at$(0, 0, 0), with edges directed along positive x, y, and z axes.\n - Target face diagonal corner P$in x-y plane:$P(a, a, 0).\n - Target body diagonal corner Q$:$Q(a, a, a).\n- **Step-by-Step Mathematical Derivation**:\n - **Step 1: Identify Corner Coordinates**:\n    O = (0, 0, 0)$,$P = (a, a, 0)$,$Q = (a, a, a)\n - **Step 2: Construct Diagonal Vectors**:\n - Face diagonal vector: \mathbf{OP} = \mathbf{P} - \mathbf{O} = a\hat{i} + a\hat{j} + 0\hat{k}\n - Body diagonal vector: \mathbf{OQ} = \mathbf{Q} - \mathbf{O} = a\hat{i} + a\hat{j} + a\hat{k}\n - **Step 3: Calculate Vector Magnitudes**:\n - Face diagonal magnitude: |\mathbf{OP}| = \sqrt{a^2 + a^2 + 0^2} = \sqrt{2a^2} = \sqrt{2}\,a\n - Body diagonal magnitude: |\mathbf{OQ}| = \sqrt{a^2 + a^2 + a^2} = \sqrt{3a^2} = \sqrt{3}\,a\n - **Step 4: Compute Dot Product**:\n    \mathbf{OP} \cdot \mathbf{OQ} = (a)(a) + (a)(a) + (0)(a) = a^2 + a^2 + 0 = 2a^2\n - **Step 5: Apply Vector Angle Equation**:\n    \cos(\theta) = \frac{\mathbf{OP} \cdot \mathbf{OQ}}{|\mathbf{OP}||\mathbf{OQ}|} = \frac{2a^2}{\sqrt{2}\,a \times \sqrt{3}\,a} = \frac{2a^2}{\sqrt{6}\,a^2} = \frac{2}{\sqrt{6}}\n - **Step 6: Simplify and Evaluate Angle**:\n    \cos(\theta) = \frac{2}{\sqrt{6}} \approx 0.8165\n    \theta = \cos^{-1}(0.8165) = 35.26^\circ\n - **Conclusion**:\n - The edge length a$cancels out completely. This demonstrates that the angle between a face diagonal and a body diagonal originating from the same vertex is universally$35.26^\circ for every cube regardless of size.\n\n# Summary Reference Sheet\n\n- **Component Form**: \mathbf{A} = A_1\hat{i} + A_2\hat{j} + A_3\hat{k}$,$A_1 = A\cos(\theta)$,$A_2 = A\sin(\theta)\n- **Magnitude Formula**: |\mathbf{A}| = \sqrt{A_1^2 + A_2^2 + A_3^2}\n- **Unit Vector Formula**: \hat{\mathbf{A}} = \frac{\mathbf{A}}{|\mathbf{A}|}\n- **Position Vector Formula**: \mathbf{r} = x\hat{i} + y\hat{j} + z\hat{k}\n- **Displacement Vector Formula**: \Delta\mathbf{r} = \mathbf{r}_2 - \mathbf{r}_1\n- **Addition / Subtraction**: Combine like components\n- **Dot Product Equation**: \mathbf{A} \cdot \mathbf{B} = AB\cos(\theta) = A_1 B_1 + A_2 B_2 + A_3 B_3\n- **Cross Product Formula**: |\mathbf{A} \times \mathbf{B}| = AB\sin(\theta); evaluated via matrix determinant expansion\n- **Angle Between Vectors**: \cos(\theta) = \frac{\mathbf{A} \cdot \mathbf{B}}{|\mathbf{A}||\mathbf{B}|}\n\n# Practice Problems\n\n1. **Magnitude and Unit Vector**:\n   Find the magnitude and the unit vector of \mathbf{A} = 6\hat{i} - 2\hat{j} + 3\hat{k}.\n2. **Displacement Vector Calculation**:\n   A particle moves from P_1(2, -1, 3)\,m$to$P_2(5, 3, 3)\,m$. Find its displacement vector$\Delta\mathbf{r}$and its magnitude$|\Delta\mathbf{r}|.\n3. **Orthogonality Proof**:\n   Show that \mathbf{A} = 2\hat{i} + 3\hat{j}$and$\mathbf{B} = 3\hat{i} - 2\hat{j} are perpendicular to each other.\n4. **Vector Product Operations**:\n   For \mathbf{A} = \hat{i} + 2\hat{j} - 2\hat{k}$and$\mathbf{B} = 2\hat{i} - \hat{j} + 2\hat{k}, determine:\n - Dot product \mathbf{A} \cdot \mathbf{B}\n - The angle \theta between them\n - Cross product \mathbf{A} \times \mathbf{B}$$

1. Cube Geometry - Edge to Body Diagonal:    Find the angle formed between an edge and a body diagonal of a cube.

2. Cube Geometry - Intersecting Body Diagonals:    Find the angle formed between two intersecting body diagonals of a cube.