1.1 Analytic Geometry, Trigonometry, and Vector Operations

1.1 Analytic Geometry, Trigonometry, and Vector Operations

Analytic geometry, trigonometry, and vector operations form the mathematical foundation for statics, dynamics, fluid mechanics, and structural mechanics on the Fundamentals of Engineering (FE) exam. Engineering problems routinely require transforming spatial coordinates, determining angles between force vectors, evaluating moments about arbitrary points, and modeling curved geometries using conic sections.

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1. Analytic Geometry & Conic Sections

Straight Lines in Two Dimensions

A line in the Cartesian plane can be expressed in several standard forms:

• Slope-Intercept Form: $y = mx + b$, where $m$ is the slope and $b$ is the y-intercept.
• Point-Slope Form: $y - y_1 = m(x - x_1)$, given a point $(x_1, y_1)$ and slope $m$.
• General Form: $Ax + By + C = 0$, where $A, B, C$ are real constants.

Distance between Two Points

The Euclidean distance $d$ between points $P_1(x_1, y_1, z_1)$ and $P_2(x_2, y_2, z_2)$ in three dimensions is:

$$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}$$

Distance from a Point to a Line

The perpendicular distance $d$ from a point $P(x_1, y_1)$ to the line $Ax + By + C = 0$ in 2D space is:

$$d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$$

Conic Sections

Conic sections are geometric curves formed by the intersection of a plane with a double-napped circular cone. They are defined algebraically by second-degree equations in $x$ and $y$. The general second-degree equation is $Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0$.

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2. Trigonometric Identities & Triangle Solution Rules

Trigonometry is indispensable for resolving forces into components, analyzing periodic wave behavior, and solving non-right triangles.

Essential Trigonometric Identities

• Pythagorean Identities:

  $$\sin^2\theta + \cos^2\theta = 1$$ $$1 + \tan^2\theta = \sec^2\theta$$ $$1 + \cot^2\theta = \csc^2\theta$$

• Angle Addition & Subtraction Formulas:

  $$\sin(\alpha \pm \beta) = \sin\alpha\cos\beta \pm \cos\alpha\sin\beta$$ $$\cos(\alpha \pm \beta) = \cos\alpha\cos\beta \mp \sin\alpha\sin\beta$$ $$\tan(\alpha \pm \beta) = \frac{\tan\alpha \pm \tan\beta}{1 \mp \tan\alpha\tan\beta}$$

• Double-Angle Identities:

  $$\sin(2\theta) = 2\sin\theta\cos\theta$$ $$\cos(2\theta) = \cos^2\theta - \sin^2\theta = 2\cos^2\theta - 1 = 1 - 2\sin^2\theta$$

Oblique Triangle Laws

For any triangle with sides $a, b, c$ opposite angles $A, B, C$:

• Law of Sines:

  $$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$$

  (where $R$ is the circumradius of the triangle)

• Law of Cosines:

  $$c^2 = a^2 + b^2 - 2ab\cos C$$ $$a^2 = b^2 + c^2 - 2bc\cos A$$ $$b^2 = a^2 + c^2 - 2ac\cos B$$

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3. Vector Algebra in 2D and 3D

A vector quantity possesses both magnitude and direction. In Cartesian component notation, a 3D vector $\mathbf{A}$ is written as:

$$\mathbf{A} = A_x\mathbf{i} + A_y\mathbf{j} + A_z\mathbf{k} = \begin{bmatrix} A_x \\\ A_y \\\ A_z \end{bmatrix}$$

where $\mathbf{i}, \mathbf{j}, \mathbf{k}$ are unit basis vectors along the $x, y, z$ axes.

Vector Magnitude & Unit Vector

• Magnitude: $|\mathbf{A}| = A = \sqrt{A_x^2 + A_y^2 + A_z^2}$
• Unit Vector: A vector of length 1 pointing in the direction of $\mathbf{A}$: $$\mathbf{u}_A = \frac{\mathbf{A}}{|\mathbf{A}|} = \frac{A_x}{|\mathbf{A}|}\mathbf{i} + \frac{A_y}{|\mathbf{A}|}\mathbf{j} + \frac{A_z}{|\mathbf{A}|}\mathbf{k}$$

Direction Cosines & Direction Angles

The angles $\alpha, \beta, \gamma$ between vector $\mathbf{A}$ and the positive $x, y, z$ axes are the direction angles. Their cosines satisfy:

$$\cos\alpha = \frac{A_x}{|\mathbf{A}|}, \quad \cos\beta = \frac{A_y}{|\mathbf{A}|}, \quad \cos\gamma = \frac{A_z}{|\mathbf{A}|}$$ $$\cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1$$

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4. Dot Product, Cross Product, and Triple Products

Dot Product (Scalar Product)

The dot product multiplies two vectors to yield a scalar:

$$\mathbf{A} \cdot \mathbf{B} = A_x B_x + A_y B_y + A_z B_z = |\mathbf{A}||\mathbf{B}|\cos\theta$$

Key Properties of the Dot Product

1. Angle between Vectors: $\theta = \arccos\left(\frac{\mathbf{A} \cdot \mathbf{B}}{|\mathbf{A}||\mathbf{B}|}\right)$
2. Orthogonality Condition: Two non-zero vectors $\mathbf{A}$ and $\mathbf{B}$ are orthogonal (perpendicular) if and only if $\mathbf{A} \cdot \mathbf{B} = 0$.
3. Scalar Component of Projection: The magnitude of the projection of $\mathbf{A}$ onto line/vector $\mathbf{B}$ is: $$A_B = \mathbf{A} \cdot \mathbf{u}_B = \frac{\mathbf{A} \cdot \mathbf{B}}{|\mathbf{B}|}$$
4. Vector Projection: The vector projection of $\mathbf{A}$ onto $\mathbf{B}$ is: $$\text{proj}_{\mathbf{B}}\mathbf{A} = \left(\frac{\mathbf{A} \cdot \mathbf{B}}{|\mathbf{B}|^2}\right)\mathbf{B}$$

Cross Product (Vector Product)

The cross product yields a vector perpendicular to the plane formed by $\mathbf{A}$ and $\mathbf{B}$ according to the right-hand rule:

$$\mathbf{A} \times \mathbf{B} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\\ A_x & A_y & A_z \\\ B_x & B_y & B_z \end{vmatrix} = (A_y B_z - A_z B_y)\mathbf{i} - (A_x B_z - A_z B_x)\mathbf{j} + (A_x B_y - A_y B_x)\mathbf{k}$$

Key Properties of the Cross Product

1. Magnitude: $|\mathbf{A} \times \mathbf{B}| = |\mathbf{A}||\mathbf{B}|\sin\theta$, which equals the area of the parallelogram spanned by $\mathbf{A}$ and $\mathbf{B}$.
2. Anti-commutativity: $\mathbf{A} \times \mathbf{B} = -(\mathbf{B} \times \mathbf{A})$.
3. Parallel Vectors: $\mathbf{A} \times \mathbf{B} = \mathbf{0}$ if $\mathbf{A}$ and $\mathbf{B}$ are parallel or anti-parallel ($\theta = 0^\circ$ or $180^\circ$).
4. Moment of a Force: The moment vector $\mathbf{M}_O$ created by force $\mathbf{F}$ applied at position vector $\mathbf{r}$ relative to pivot $O$ is: $$\mathbf{M}_O = \mathbf{r} \times \mathbf{F}$$

Triple Scalar Product (Box Product)

The triple scalar product calculates the volume of a parallelepiped defined by three vectors $\mathbf{A}, \mathbf{B}, \mathbf{C}$:

$$V = |\mathbf{A} \cdot (\mathbf{B} \times \mathbf{C})| = \left| \begin{vmatrix} A_x & A_y & A_z \\\ B_x & B_y & B_z \\\ C_x & C_y & C_z \end{vmatrix} \right|$$

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5. Worked Engineering Problems

Worked Example 1: 3D Vector Operations & Projections

Problem: A structural cable connects anchor point $A(1, -2, 3)$ meters to support node $B(4, 2, 15)$ meters. A tension force of magnitude $T = 390\text{ N}$ acts along line $AB$ from $A$ to $B$.

1. Express force vector $\mathbf{T}$ in Cartesian component form.
2. Calculate the scalar projection of $\mathbf{T}$ along a second structural member oriented along vector $\mathbf{v} = 2\mathbf{i} - 2\mathbf{j} + 1\mathbf{k}$.

Solution:

Step 1: Determine position vector $\mathbf{r}_{AB}$ and its magnitude.

$$\mathbf{r}_{AB} = (4 - 1)\mathbf{i} + (2 - (-2))\mathbf{j} + (15 - 3)\mathbf{k} = 3\mathbf{i} + 4\mathbf{j} + 12\mathbf{k}\text{ m}$$ $$|\mathbf{r}_{AB}| = \sqrt{3^2 + 4^2 + 12^2} = \sqrt{9 + 16 + 144} = \sqrt{169} = 13\text{ m}$$

Step 2: Form unit vector $\mathbf{u}_{AB}$ and force vector $\mathbf{T}$.

$$\mathbf{u}_{AB} = \frac{3}{13}\mathbf{i} + \frac{4}{13}\mathbf{j} + \frac{12}{13}\mathbf{k}$$ $$\mathbf{T} = T \mathbf{u}_{AB} = 390 \left(\frac{3}{13}\mathbf{i} + \frac{4}{13}\mathbf{j} + \frac{12}{13}\mathbf{k}\right) = 90\mathbf{i} + 120\mathbf{j} + 360\mathbf{k}\text{ N}$$

Step 3: Calculate unit vector $\mathbf{u}_v$ and scalar projection of $\mathbf{T}$ onto $\mathbf{v}$.

$$|\mathbf{v}| = \sqrt{2^2 + (-2)^2 + 1^2} = \sqrt{4 + 4 + 1} = 3$$ $$\mathbf{u}_v = \frac{2}{3}\mathbf{i} - \frac{2}{3}\mathbf{j} + \frac{1}{3}\mathbf{k}$$ $$T_v = \mathbf{T} \cdot \mathbf{u}_v = (90)\left(\frac{2}{3}\right) + (120)\left(-\frac{2}{3}\right) + (360)\left(\frac{1}{3}\right)$$ $$T_v = 60 - 80 + 120 = 100\text{ N}$$

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Worked Example 2: Moment Vector via Cross Product

Problem: A robotic manipulator exerts a force $\mathbf{F} = 20\mathbf{i} - 10\mathbf{j} + 30\mathbf{k}\text{ N}$ at point $P(2, 3, -1)\text{ m}$. Calculate the vector moment $\mathbf{M}_O$ and magnitude of moment exerted by this force about origin $O(0,0,0)$.

Solution:

Step 1: Write position vector $\mathbf{r} = \mathbf{r}_P - \mathbf{r}_O$.

$$\mathbf{r} = 2\mathbf{i} + 3\mathbf{j} - 1\mathbf{k}\text{ m}$$

Step 2: Evaluate cross product $\mathbf{M}_O = \mathbf{r} \times \mathbf{F}$.

$$\mathbf{M}_O = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\\ 2 & 3 & -1 \\\ 20 & -10 & 30 \end{vmatrix}$$ $$\mathbf{M}_O = \mathbf{i}[(3)(30) - (-1)(-10)] - \mathbf{j}[(2)(30) - (-1)(20)] + \mathbf{k}[(2)(-10) - (3)(20)]$$ $$\mathbf{M}_O = \mathbf{i}[90 - 10] - \mathbf{j}[60 + 20] + \mathbf{k}[-20 - 60]$$ $$\mathbf{M}_O = 80\mathbf{i} - 80\mathbf{j} - 80\mathbf{k}\text{ N}\cdot\text{m}$$

Step 3: Compute magnitude of moment $|\mathbf{M}_O|$.

$$|\mathbf{M}_O| = \sqrt{80^2 + (-80)^2 + (-80)^2} = 80\sqrt{3} \approx 138.56\text{ N}\cdot\text{m}$$