Comprehensive Vector Mathematics: Dot Product, Cross Product, and 3D Vector Operations

Scalar Product (Dot Product) Fundamentals

  • Definition and Alternative Terminology:

    • The dot product is also formally known as the scalar product.

    • The result of a dot product between two vectors is a scalar quantity (a magnitude with no direction).

    • When representing the scalar result RR, no vector arrow is written above the letter RR.

  • General Formula:

    • For any two vectors aˉ\bar{a} and bˉ\bar{b}, the scalar product is given by:         aˉ⋅bˉ=∣aˉ∣∣bˉ∣cos⁡(θ)\bar{a} \cdot \bar{b} = |\bar{a}| |\bar{b}| \cos(\theta)

    • ∣aˉ∣|\bar{a}| represents the magnitude of vector aˉ\bar{a}.

    • ∣bˉ∣|\bar{b}| represents the magnitude of vector bˉ\bar{b}.

    • θ\theta represents the angle between the directions of vectors aˉ\bar{a} and bˉ\bar{b}.

  • Three-Dimensional Cartesian Coordinate Representation:

    • Unit vectors define directions along the orthogonal axes: i^\hat{i} for the x-direction, j^\hat{j} for the y-direction, and k^\hat{k} for the z-direction.

    • Vector aˉ\bar{a} in unit vector component form:         aˉ=axi^+ayj^+azk^\bar{a} = a_x \hat{i} + a_y \hat{j} + a_z \hat{k}

    • Vector bˉ\bar{b} in unit vector component form:         bˉ=bxi^+byj^+bzk^\bar{b} = b_x \hat{i} + b_y \hat{j} + b_z \hat{k}

Derivation of Vector Dot Product Formula in 3D Coordinate Systems

  • Distributive Property Application:

    • Dotting two vectors in component form involves multiplying two algebraic expressions using the distributive method:         aˉ⋅bˉ=(axi^+ayj^+azk^)⋅(bxi^+byj^+bzk^)\bar{a} \cdot \bar{b} = (a_x \hat{i} + a_y \hat{j} + a_z \hat{k}) \cdot (b_x \hat{i} + b_y \hat{j} + b_z \hat{k})

  • Behavior of Parallel and Antiparallel Unit Vectors:

    • For unit vectors in the same direction (parallel vectors, such as i^⋅i^\hat{i} \cdot \hat{i}, j^⋅j^\hat{j} \cdot \hat{j}, k^⋅k^\hat{k} \cdot \hat{k}), the angle between them is θ=0∘\theta = 0^\circ.

    • Evaluating cosine for parallel unit vectors:         cos⁡(0∘)=1\cos(0^\circ) = 1

    • Therefore, the dot product of identical unit vectors equals positive one:         i^⋅i^=1\hat{i} \cdot \hat{i} = 1         j^⋅j^=1\hat{j} \cdot \hat{j} = 1         k^⋅k^=1\hat{k} \cdot \hat{k} = 1

    • If two directions are antiparallel, the angle between them is θ=180∘\theta = 180^\circ, yielding cos⁡(180∘)=−1\cos(180^\circ) = -1.

  • Behavior of Perpendicular (Orthogonal) Unit Vectors:

    • In a 3D Cartesian coordinate system, the x, y, and z axes cross each other at right angles (θ=90∘\theta = 90^\circ).

    • Evaluating cosine for orthogonal unit vectors:         cos⁡(90∘)=0\cos(90^\circ) = 0

    • Therefore, the dot product of mutually perpendicular unit vectors equals zero:         i^⋅j^=0\hat{i} \cdot \hat{j} = 0         i^⋅k^=0\hat{i} \cdot \hat{k} = 0         j^⋅i^=0\hat{j} \cdot \hat{i} = 0         j^⋅k^=0\hat{j} \cdot \hat{k} = 0         k^⋅i^=0\hat{k} \cdot \hat{i} = 0         k^⋅j^=0\hat{k} \cdot \hat{j} = 0

  • Final Standard Dot Product Component Formula:

    • Eliminating all orthogonal terms (00) and preserving parallel terms (11) simplifies the expansion to:         aˉ⋅bˉ=axbx+ayby+azbz\bar{a} \cdot \bar{b} = a_x b_x + a_y b_y + a_z b_z

Worked Examples: Calculating Dot Products

  • Example 1: Dot Product of Two 2D Vectors Using Magnitudes and Angle

    • Problem Statement: Vector aˉ\bar{a} has a magnitude of 15 m15\,\text{m} directed at 30∘30^\circ North of East. Vector bˉ\bar{b} has a magnitude of 20 m20\,\text{m} directed at 15∘15^\circ West of North. Calculate the scalar resultant R=aˉ⋅bˉR = \bar{a} \cdot \bar{b}.

    • Method 1: Direct Formula using Smaller Inter-Vector Angle

      • Angle Calculation: Vector aˉ\bar{a} is at 30∘30^\circ above the positive x-axis (East), leaving 90∘−30∘=60∘90^\circ - 30^\circ = 60^\circ in Quadrant I. Vector bˉ\bar{b} is 15∘15^\circ West of North in Quadrant II. The smaller total angle θ\theta between the two vectors is:             θ=60∘+15∘=75∘\theta = 60^\circ + 15^\circ = 75^\circ

      • Scalar Evaluation:             R=∣aˉ∣∣bˉ∣cos⁡(θ)R = |\bar{a}| |\bar{b}| \cos(\theta)             R=(15 m)(20 m)cos⁡(75∘)R = (15\,\text{m})(20\,\text{m}) \cos(75^\circ)             R=(300 m2)(0.258819)=77.65 m2R = (300\,\text{m}^2)(0.258819) = 77.65\,\text{m}^2

    • Method 2: Component Conversion and Algebraic Dot Product

      • Vector aˉ\bar{a} Components:             ax=15cos⁡(30∘)=12.99 m≈13 ma_x = 15 \cos(30^\circ) = 12.99\,\text{m} \approx 13\,\text{m}             ay=15sin⁡(30∘)=7.5 ma_y = 15 \sin(30^\circ) = 7.5\,\text{m}             aˉ=13 m i^+7.5 m j^\bar{a} = 13\,\text{m}\,\hat{i} + 7.5\,\text{m}\,\hat{j}

      • Vector bˉ\bar{b} Components: Standard reference angle from positive x-axis is 90∘+15∘=105∘90^\circ + 15^\circ = 105^\circ             bx=20cos⁡(105∘)=−5.18 mb_x = 20 \cos(105^\circ) = -5.18\,\text{m}             by=20sin⁡(105∘)=19.3 mb_y = 20 \sin(105^\circ) = 19.3\,\text{m}             bˉ=−5.18 m i^+19.3 m j^\bar{b} = -5.18\,\text{m}\,\hat{i} + 19.3\,\text{m}\,\hat{j}

      • Dot Product Calculation:             aˉ⋅bˉ=axbx+ayby\bar{a} \cdot \bar{b} = a_x b_x + a_y b_y             aˉ⋅bˉ=(13 m)(−5.18 m)+(7.5 m)(19.3 m)\bar{a} \cdot \bar{b} = (13\,\text{m})(-5.18\,\text{m}) + (7.5\,\text{m})(19.3\,\text{m})             aˉ⋅bˉ=−67.34 m2+144.75 m2=77.41 m2\bar{a} \cdot \bar{b} = -67.34\,\text{m}^2 + 144.75\,\text{m}^2 = 77.41\,\text{m}^2

      • Note on Rounding Differences: Small discrepancies between 77.65 m277.65\,\text{m}^2 and 77.41 m277.41\,\text{m}^2 (or 77.4 m277.4\,\text{m}^2 vs 77.6 m277.6\,\text{m}^2) stem from intermediate decimal rounding; both are standard valid calculation approaches.

  • Example 2: Finding the Angle Between Two 3D Vectors

    • Problem Statement: Given vector cˉ=2 m i^+2 m j^+2 m k^\bar{c} = 2\,\text{m}\,\hat{i} + 2\,\text{m}\,\hat{j} + 2\,\text{m}\,\hat{k} and vector dˉ=1 m i^+0 m j^+2 m k^\bar{d} = 1\,\text{m}\,\hat{i} + 0\,\text{m}\,\hat{j} + 2\,\text{m}\,\hat{k}, calculate the angle θ\theta between them.

    • Derived Angle Formula:         θ=arccos⁡(cˉ⋅dˉ∣cˉ∣∣dˉ∣)\theta = \arccos\left(\frac{\bar{c} \cdot \bar{d}}{|\bar{c}| |\bar{d}|}\right)

    • Step 1: Dot Product Calculation:         cˉ⋅dˉ=(2)(1)+(2)(0)+(2)(2)=2+0+4=6 m2\bar{c} \cdot \bar{d} = (2)(1) + (2)(0) + (2)(2) = 2 + 0 + 4 = 6\,\text{m}^2

    • Step 2: Magnitude of Vector cˉ\bar{c}:         ∣cˉ∣=(2 m)2+(2 m)2+(2 m)2=4+4+4=12≈3.46 m|\bar{c}| = \sqrt{(2\,\text{m})^2 + (2\,\text{m})^2 + (2\,\text{m})^2} = \sqrt{4 + 4 + 4} = \sqrt{12} \approx 3.46\,\text{m}

    • Step 3: Magnitude of Vector dˉ\bar{d}:         ∣dˉ∣=(1 m)2+(0 m)2+(2 m)2=1+0+4=5≈2.24 m|\bar{d}| = \sqrt{(1\,\text{m})^2 + (0\,\text{m})^2 + (2\,\text{m})^2} = \sqrt{1 + 0 + 4} = \sqrt{5} \approx 2.24\,\text{m}

    • Step 4: Angle Computation:         θ=arccos⁡(6 m2(3.46 m)(2.24 m))=arccos⁡(67.7504)=39.27∘\theta = \arccos\left(\frac{6\,\text{m}^2}{(3.46\,\text{m})(2.24\,\text{m})}\right) = \arccos\left(\frac{6}{7.7504}\right) = 39.27^\circ

Vector Product (Cross Product) Fundamentals

  • Definition and Terminology:

    • The cross product is formally termed the vector product because crossing two vectors yields a vector quantity (having both magnitude and direction).

    • The resultant vector equation is written as:         rˉ=aˉ×bˉ\bar{r} = \bar{a} \times \bar{b}

  • Magnitude Equation:

    • The magnitude of a cross product is given by:         ∣rˉ∣=∣aˉ∣∣bˉ∣sin⁡(θ)|\bar{r}| = |\bar{a}| |\bar{b}| \sin(\theta)

    • Unlike the dot product, cross product calculations involve the sine of the inter-vector angle θ\theta.

  • Directional Cyclic Conventions for Unit Vectors:

    • Clockwise / Positive Directions:         i^×j^=k^\hat{i} \times \hat{j} = \hat{k}         j^×k^=i^\hat{j} \times \hat{k} = \hat{i}         k^×i^=j^\hat{k} \times \hat{i} = \hat{j}

    • Counterclockwise / Anti-commutative Negative Directions:         j^×i^=−k^\hat{j} \times \hat{i} = -\hat{k}         k^×j^=−i^\hat{k} \times \hat{j} = -\hat{i}         i^×k^=−j^\hat{i} \times \hat{k} = -\hat{j}

    • The cross product is order-sensitive (anti-commutative): swapping vector positions reverses the direction sign.

  • Cross Product of Parallel Unit Vectors:

    • For parallel unit vectors, θ=0∘\theta = 0^\circ. Since sin⁡(0∘)=0\sin(0^\circ) = 0:         i^×i^=0\hat{i} \times \hat{i} = 0         j^×j^=0\hat{j} \times \hat{j} = 0         k^×k^=0\hat{k} \times \hat{k} = 0

Derivation of Vector Cross Product Formula

  • Full Expansion via Distributive Property:     aˉ×bˉ=(axi^+ayj^+azk^)×(bxi^+byj^+bzk^)\bar{a} \times \bar{b} = (a_x \hat{i} + a_y \hat{j} + a_z \hat{k}) \times (b_x \hat{i} + b_y \hat{j} + b_z \hat{k})

    • Expanding term-by-term:

      1. axi^×bxi^=0a_x \hat{i} \times b_x \hat{i} = 0

      2. axi^×byj^=axbyk^a_x \hat{i} \times b_y \hat{j} = a_x b_y \hat{k}

      3. axi^×bzk^=−axbzj^a_x \hat{i} \times b_z \hat{k} = -a_x b_z \hat{j}

      4. ayj^×bxi^=−aybxk^a_y \hat{j} \times b_x \hat{i} = -a_y b_x \hat{k}

      5. ayj^×byj^=0a_y \hat{j} \times b_y \hat{j} = 0

      6. ayj^×bzk^=aybzi^a_y \hat{j} \times b_z \hat{k} = a_y b_z \hat{i}

      7. azk^×bxi^=azbxj^a_z \hat{k} \times b_x \hat{i} = a_z b_x \hat{j}

      8. azk^×byj^=−azbyi^a_z \hat{k} \times b_y \hat{j} = -a_z b_y \hat{i}

      9. azk^×bzk^=0a_z \hat{k} \times b_z \hat{k} = 0

  • Grouped Component Formula:

    • Collecting unit vector components yields:         aˉ×bˉ=(aybz−azby)i^+(azbx−axbz)j^+(axby−aybx)k^\bar{a} \times \bar{b} = (a_y b_z - a_z b_y)\hat{i} + (a_z b_x - a_x b_z)\hat{j} + (a_x b_y - a_y b_x)\hat{k}

  • Worked Algebraic Cross Product Example:

    • Given aˉ=2 m i^+2 m j^+2 m k^\bar{a} = 2\,\text{m}\,\hat{i} + 2\,\text{m}\,\hat{j} + 2\,\text{m}\,\hat{k} and bˉ=1 m i^+0 m j^+2 m k^\bar{b} = 1\,\text{m}\,\hat{i} + 0\,\text{m}\,\hat{j} + 2\,\text{m}\,\hat{k}:

      • i^-component:(2)(2)−(2)(0)=4\hat{i}\text{-component}: (2)(2) - (2)(0) = 4

      • j^-component:(2)(1)−(2)(2)=2−4=−2\hat{j}\text{-component}: (2)(1) - (2)(2) = 2 - 4 = -2

      • k^-component:(2)(0)−(2)(1)=0−2=−2\hat{k}\text{-component}: (2)(0) - (2)(1) = 0 - 2 = -2

      • Resulting vector:             aˉ×bˉ=4 m2 i^−2 m2 j^−2 m2 k^\bar{a} \times \bar{b} = 4\,\text{m}^2\,\hat{i} - 2\,\text{m}^2\,\hat{j} - 2\,\text{m}^2\,\hat{k}

Matrix Determinant Method for Vector Cross Products

  • 3x3 Matrix Setup:

    • To evaluate aˉ×bˉ\bar{a} \times \bar{b} without memorizing the component formula, set up a 3×33 \times 3 matrix:         (i^amp;j^amp;k^axamp;ayamp;azbxamp;byamp;bz)\begin{pmatrix} \hat{i} & \hat{j} & \hat{k} \\ a_x & a_y & a_z \\ b_x & b_y & b_z \end{pmatrix}

  • Sarrus' Rule Expansion Steps:

    • Copy the first two columns to the right side of the matrix:         i^amp;j^amp;k^amp;i^amp;j^axamp;ayamp;azamp;axamp;aybxamp;byamp;bzamp;bxamp;by\begin{matrix} \hat{i} & \hat{j} & \hat{k} & \hat{i} & \hat{j} \\ a_x & a_y & a_z & a_x & a_y \\ b_x & b_y & b_z & b_x & b_y \end{matrix}

    • Downward Diagonal Products (Positive Sign):

      1. i^⋅ay⋅bz\hat{i} \cdot a_y \cdot b_z

      2. j^⋅az⋅bx\hat{j} \cdot a_z \cdot b_x

      3. k^⋅ax⋅by\hat{k} \cdot a_x \cdot b_y

    • Upward Diagonal Products (Negative Sign):

      1. −(bx⋅ay⋅k^)- (b_x \cdot a_y \cdot \hat{k})

      2. −(by⋅az⋅i^)- (b_y \cdot a_z \cdot \hat{i})

      3. −(bz⋅ax⋅j^)- (b_z \cdot a_x \cdot \hat{j})

  • Example using Determinant Method:

    • For aˉ=2 i^+2 j^+2 k^\bar{a} = 2\,\hat{i} + 2\,\hat{j} + 2\,\hat{k} and bˉ=1 i^+0 j^+2 k^\bar{b} = 1\,\hat{i} + 0\,\hat{j} + 2\,\hat{k}:         i^amp;j^amp;k^amp;i^amp;j^2amp;2amp;2amp;2amp;21amp;0amp;2amp;1amp;0\begin{matrix} \hat{i} & \hat{j} & \hat{k} & \hat{i} & \hat{j} \\ 2 & 2 & 2 & 2 & 2 \\ 1 & 0 & 2 & 1 & 0 \end{matrix}

    • Downward: i^(2)(2)+j^(2)(1)+k^(2)(0)=4 i^+2 j^+0 k^\hat{i}(2)(2) + \hat{j}(2)(1) + \hat{k}(2)(0) = 4\,\hat{i} + 2\,\hat{j} + 0\,\hat{k}

    • Upward: −(1)(2)k^−(0)(2)i^−(2)(2)j^=−2 k^−0 i^−4 j^-(1)(2)\hat{k} - (0)(2)\hat{i} - (2)(2)\hat{j} = -2\,\hat{k} - 0\,\hat{i} - 4\,\hat{j}

    • Combined Result: 4 i^+(2−4) j^+(0−2) k^=4 i^−2 j^−2 k^ m24\,\hat{i} + (2 - 4)\,\hat{j} + (0 - 2)\,\hat{k} = 4\,\hat{i} - 2\,\hat{j} - 2\,\hat{k}\,\text{m}^2

Combined Vector Operations and Validity

  • Analysis of Triple Operations:

    • Triple Dot Product (aˉ⋅bˉ)⋅cˉ(\bar{a} \cdot \bar{b}) \cdot \bar{c}: Not Possible. The expression (aˉ⋅bˉ)(\bar{a} \cdot \bar{b}) produces a scalar quantity. Taking the dot product of a scalar and a vector is mathematically undefined, as dot products require two vectors.

    • Scalar Triple Product (aˉ×bˉ)⋅cˉ(\bar{a} \times \bar{b}) \cdot \bar{c}: Valid Operation. The expression (aˉ×bˉ)(\bar{a} \times \bar{b}) yields a vector. Dotting this resulting vector with vector cˉ\bar{c} yields a valid scalar result.

  • Worked Scalar Triple Product Example:

    • Given Vectors:         aˉ=2 m i^+1 m j^+0 m k^\bar{a} = 2\,\text{m}\,\hat{i} + 1\,\text{m}\,\hat{j} + 0\,\text{m}\,\hat{k}         bˉ=0 m i^−3 m j^+4 m k^\bar{b} = 0\,\text{m}\,\hat{i} - 3\,\text{m}\,\hat{j} + 4\,\text{m}\,\hat{k}         cˉ=−1 m i^+1 m j^+2 m k^\bar{c} = -1\,\text{m}\,\hat{i} + 1\,\text{m}\,\hat{j} + 2\,\text{m}\,\hat{k}

    • Step 1: Compute aˉ×bˉ\bar{a} \times \bar{b}:         i^amp;j^amp;k^amp;i^amp;j^2amp;1amp;0amp;2amp;10amp;−3amp;4amp;0amp;−3\begin{matrix} \hat{i} & \hat{j} & \hat{k} & \hat{i} & \hat{j} \\ 2 & 1 & 0 & 2 & 1 \\ 0 & -3 & 4 & 0 & -3 \end{matrix}

      • Downward: i^(1)(4)+j^(0)(0)+k^(2)(−3)=4 i^+0 j^−6 k^\hat{i}(1)(4) + \hat{j}(0)(0) + \hat{k}(2)(-3) = 4\,\hat{i} + 0\,\hat{j} - 6\,\hat{k}

      • Upward: −(0)(1)k^−(−3)(0)i^−(4)(2)j^=0 k^−0 i^−8 j^-(0)(1)\hat{k} - (-3)(0)\hat{i} - (4)(2)\hat{j} = 0\,\hat{k} - 0\,\hat{i} - 8\,\hat{j}

      • aˉ×bˉ=4 i^−8 j^−6 k^ m2\bar{a} \times \bar{b} = 4\,\hat{i} - 8\,\hat{j} - 6\,\hat{k}\,\text{m}^2

    • Step 2: Dot Result with cˉ\bar{c}:         (aˉ×bˉ)⋅cˉ=(4 i^−8 j^−6 k^ m2)⋅(−1 i^+1 j^+2 k^ m)(\bar{a} \times \bar{b}) \cdot \bar{c} = (4\,\hat{i} - 8\,\hat{j} - 6\,\hat{k}\,\text{m}^2) \cdot (-1\,\hat{i} + 1\,\hat{j} + 2\,\hat{k}\,\text{m})         (aˉ×bˉ)⋅cˉ=(4)(−1)+(−8)(1)+(−6)(2)(\bar{a} \times \bar{b}) \cdot \bar{c} = (4)(-1) + (-8)(1) + (-6)(2)         (aˉ×bˉ)⋅cˉ=−4−8−12=−24 m3(\bar{a} \times \bar{b}) \cdot \bar{c} = -4 - 8 - 12 = -24\,\text{m}^3

    • Correction & Physical Note on Scalar Sign:

      • Re-assessing component signs for a corrected dot evaluation yields −16 m3-16\,\text{m}^3 (or −16 m-16\,\text{m} depending on power standardizations).

      • Physical scalar quantities represent magnitudes without inherent spatial direction.

Course Announcements, Quiz Information, and Laboratory Schedule

  • Quiz 1 Details:

    • Total Points: 50 points.

    • Structure:

      • 40%40\% Multiple Choice and Identification questions.

      • 60%60\% Problem Solving calculations.

    • Covered Topics:

      • Overview of Physics.

      • Vectors and Scalars definitions.

      • Vector Representation Forms: Standard form, Component form, Unit vector form.

      • Vector Conversions between forms.

      • Vector Operations: Vector Addition, Vector Subtraction, Negative of a Vector, Vector Dot Product, Vector Cross Product.

  • Schedule Breakdown:

    • August 24 (Laboratory Day - Face-to-Face):

      • Perform Laboratory Experiment 1 (Errors and Uncertainty of Measurements) and Experiment 2 (Vector Quantities).

      • Laboratory manuals must be acquired beforehand from the Business Affairs Office or the University Store (U-Store).

    • August 25 (Lecture Day - Face-to-Face):

      • First 1.5 hours: Conduct Quiz 1.

      • Remaining 1.5 hours: Lecture presentation.

    • August 27 (Thursday - Face-to-Face):

      • Laboratory sessions converted to face-to-face instruction.

    • August 31:

      • Scheduled Online Day per Academic Calendar.

Questions & Discussion

  • Discussion on Angle Between Identical Directions:

    • Question: What is the angle between two directions that are parallel or in the same direction (i^\hat{i} and i^\hat{i})?

    • Response: The angle is 0∘0^\circ. If directions are antiparallel, the angle is 180∘180^\circ.

  • Discussion on Dot Product of Orthogonal Directions:

    • Question: What is the angle between i^\hat{i} and j^\hat{j} directions, or i^\hat{i} and k^\hat{k} directions in a 3D coordinate system?

    • Response: The angle is 90∘90^\circ. Because cos⁡(90∘)=0\cos(90^\circ) = 0, any dot product between perpendicular unit vectors evaluates to 00.

  • Discussion on Precision and Rounding:

    • Question: How should rounding off be handled when calculating dot products using components vs exact magnitude formulas?

    • Response: Keep solutions rounded off to two decimal places. Answers derived via exact formulas (77.65 m277.65\,\text{m}^2) or converted components (77.41 m277.41\,\text{m}^2) are both acceptable standards.

  • Discussion on Matrix Determinant Method Applicability:

    • Question: Is the matrix determinant method applicable at all times for solving cross products?

    • Response: Yes, the determinant of a matrix can always be used when numerical values are given. However, expanded algebraic cross product solutions are necessary in higher-level physics and engineering courses when dealing with continuous variables or technical proofs.

  • Discussion on Administrative / Attendance Concerns:

    • Question (Constantino): How is attendance recorded for late enrollees who missed sessions during the first or second week?

    • Response: Missed sessions due to late enrollment are formally marked as excused in the system records.