MATH-283 (Calculus III) Curriculum

Session 07 - Midterm Review 1

Handout Worksheet

Fall Semester 2026

Session Objectives & Overview

By the end of this session, you will be able to:

Decompose 3D force and displacement vectors into parallel and perpendicular components using scalar and vector projections, apply dot and cross products to compute work, torque, angles, and geometric areas/volumes, and distinguish scalar vs vector physical quantities.
Formulate vector, parametric, and symmetric equations of lines in 3D space, calculate perpendicular point-to-line clearance envelopes, and systematically classify spatial line pairs as parallel, intersecting, skew, or coincident using direction vector ratios and two-parameter linear systems.
Differentiate and integrate vector-valued space curves $\vec{r}(t)$ , evaluate arc length and construct intrinsic unit-speed parametrizations $\vec{r}(s)$ , calculate curvature $\kappa(t)$ and osculating radii $\rho(t)$ , and decompose kinematic acceleration into tangential and normal components ( $\vec{a} = a_T \hat{T} + a_N \hat{N}$ ) to audit structural and biomechanical dynamic limits.

Micro-Lecture

1. Midterm 1 Strategy & The Physicist-Mathematician-Engineer Triad

Mathematical Foundation: Unit 1 of Calculus III transitions our mathematical universe from flat 2D calculus to 3D spatial modeling $\mathbb{R}^3$ . Success on Midterm Exam 1 requires mastery of the three distinct operational tiers:
1. Spatial Representation: Mapping geometry using position vectors $\vec{r} = \langle x, y, z \rangle = x\hat{i} + y\hat{j} + z\hat{k}$ (see Vectors) and direction vectors $\vec{v} = \langle a, b, c \rangle$ .
2. Vector Multiplication Algebra: Distinguishing scalar output operations (the Dot Product $\vec{u}\cdot\vec{v}$ and Vector Projections) from orthogonal vector output operations (the Cross Product $\vec{u}\times\vec{v}$ and Triple Scalar Product and Spatial Volume).
3. Space Curve Differential Geometry: Analyzing particle kinematics ( $\vec{v}(t), \vec{a}(t)$ ), intrinsic distance through Arc Length of a Space Curve, turning sharpness via Curvature of a Space Curve, and dynamic force partitioning via Tangential and Normal Components of Acceleration.
Engineering Context: On exams, engineers do not perform random algebraic substitutions. We classify the physical system, select the governing vector formula, and state the symbolic equation before executing numerical calculations.

2. Formula Hierarchy & Problem Classification

Master the formula hierarchy: identify whether a prompt asks for a scalar (speed, arc length, distance, curvature, work, volume) or a vector (position, velocity, acceleration, unit tangent, normal vector, torque, projection).

Skill Block 1

Worked Example A: Vector Projections, Decomposition, and Cross Product

Given vectors $\vec{u} = \langle 2, -1, 2 \rangle$ and $\vec{v} = \langle 1, 2, 2 \rangle$ in $\mathbb{R}^3$ :

Compute the dot product $\vec{u} \cdot \vec{v}$ and evaluate the angle $\theta$ between them.
Find the scalar projection $\text{scal}_{\vec{u}}\vec{v}$ and vector projection $\text{proj}_{\vec{u}}\vec{v}$ .
Compute the orthogonal vector component $\text{orth}_{\vec{u}}\vec{v}$ and verify that it is perpendicular to $\vec{u}$ .
Compute the cross product $\vec{u} \times \vec{v}$ and find the exact geometric area of the parallelogram spanned by $\vec{u}$ and $\vec{v}$ .

Worked Solution:

Compute Vector Magnitudes and Dot Product: State the Vector Magnitude Formula:
$\|\vec{u}\| = \sqrt{u_1^2 + u_2^2 + u_3^2} = \sqrt{2^2 + (-1)^2 + 2^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3$ $\|\vec{v}\| = \sqrt{v_1^2 + v_2^2 + v_3^2} = \sqrt{1^2 + 2^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3$

State the Dot Product Formula:
$\vec{u} \cdot \vec{v} = u_1 v_1 + u_2 v_2 + u_3 v_3 = (2)(1) + (-1)(2) + (2)(2) = 2 - 2 + 4 = 4$ (Defined in Dot Product.)

Evaluate angle $\theta$ :
$\cos\theta = \frac{\vec{u} \cdot \vec{v}}{\|\vec{u}\|\|\vec{v}\|} = \frac{4}{(3)(3)} = \frac{4}{9} \implies \theta = \arccos\left(\frac{4}{9}\right) \approx 63.61^\circ$
Calculate Scalar and Vector Projections: State the Scalar Projection Formula:
$\text{scal}_{\vec{u}}\vec{v} = \frac{\vec{u} \cdot \vec{v}}{\|\vec{u}\|} = \frac{4}{3}$

State the Vector Projection Formula:
$\text{proj}_{\vec{u}}\vec{v} = \left( \frac{\vec{u} \cdot \vec{v}}{\|\vec{u}\|^2} \right) \vec{u} = \frac{4}{9}\langle 2, -1, 2 \rangle = \left\langle \frac{8}{9}, -\frac{4}{9}, \frac{8}{9} \right\rangle$ (Defined in Vector Projections.)
Compute and Verify the Orthogonal Component: State the Orthogonal Component Formula:
$\text{orth}_{\vec{u}}\vec{v} = \vec{v} - \text{proj}_{\vec{u}}\vec{v} = \langle 1, 2, 2 \rangle - \left\langle \frac{8}{9}, -\frac{4}{9}, \frac{8}{9} \right\rangle = \left\langle \frac{9 - 8}{9}, \frac{18 - (-4)}{9}, \frac{18 - 8}{9} \right\rangle = \left\langle \frac{1}{9}, \frac{22}{9}, \frac{10}{9} \right\rangle$

Verify orthogonality:
$\text{orth}_{\vec{u}}\vec{v} \cdot \vec{u} = \left(\frac{1}{9}\right)(2) + \left(\frac{22}{9}\right)(-1) + \left(\frac{10}{9}\right)(2) = \frac{2 - 22 + 20}{9} = \frac{0}{9} = 0\quad \checkmark$
Compute Cross Product and Parallelogram Area: State the Cross Product Determinant Formula:
$\vec{u} \times \vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -1 & 2 \\ 1 & 2 & 2 \end{vmatrix}$ (Defined in Cross Product.)

$\vec{u} \times \vec{v} = \langle (-1)(2) - (2)(2), \; (2)(1) - (2)(2), \; (2)(2) - (-1)(1) \rangle = \langle -2 - 4, \; 2 - 4, \; 4 + 1 \rangle = \langle -6, -2, 5 \rangle$

Area of Parallelogram:
$\text{Area} = \|\vec{u} \times \vec{v}\| = \sqrt{(-6)^2 + (-2)^2 + 5^2} = \sqrt{36 + 4 + 25} = \sqrt{65}\text{ units}^2$

Practice Problem 1: Concept & Reading Check

Open this Skill Block on the Memex to locate the information needed to answer the following:

(a) Explain the geometric and dimensional difference between the scalar projection $\text{scal}_{\vec{u}}\vec{v}$ and the vector projection $\text{proj}_{\vec{u}}\vec{v}$ .
(b) In Worked Example A step 3, state the mathematical definition of orthogonality and show how the dot product proves perpendicularity.
(c) In Suspension Bridge Guy-Wire Force Decomposition (Civil & Mechanical Engineering), why is the overturning moment torque about the tower base zero?

Solution Key:

(a) Scalar vs Vector Projection: The scalar projection $\text{scal}_{\vec{u}}\vec{v} = \frac{\vec{u}\cdot\vec{v}}{\|\vec{u}\|}$ represents the signed scalar length of the shadow cast by $\vec{v}$ onto the axis spanned by $\vec{u}$ . The vector projection $\text{proj}_{\vec{u}}\vec{v} = (\text{scal}_{\vec{u}}\vec{v})\hat{u} = \left(\frac{\vec{u}\cdot\vec{v}}{\|\vec{u}\|^2}\right)\vec{u}$ scales the unit direction vector $\hat{u}$ by this scalar length, producing a true vector in $\mathbb{R}^3$ .
(b) Orthogonality Condition: Two non-zero vectors $\vec{a}$ and $\vec{b}$ are perpendicular if the angle between them is $\theta = 90^\circ$ . Since $\cos(90^\circ) = 0$ , the governing equation is $\vec{a} \cdot \vec{b} = \|\vec{a}\|\|\vec{b}\|\cos(90^\circ) = 0$ . In step 3, computing $(\text{orth}_{\vec{u}}\vec{v}) \cdot \vec{u} = 0$ confirms that the residual vector is strictly perpendicular to $\vec{u}$ .
(c) Zero Overturning Moment: Torque is defined by the cross product $\vec{\tau} = \vec{r} \times \vec{F} = \|\vec{r}\|\|\vec{F}\|\sin\theta \hat{n}$ . When the force vector acts directly along the line of the position vector, the angle is $\theta = 0^\circ$ (or $180^\circ$ ). Because $\sin(0^\circ) = 0$ , the cross product vanishes ( $\vec{r} \times \vec{F} = \vec{0}$ ), producing zero overturning moment about the footing.

Show your steps and calculations here.

Practice Problem 2

Given points $A(1, 1, 0)$ , $B(3, 2, 2)$ , and $C(1, 4, 0)$ in $\mathbb{R}^3$ :

Determine the displacement vectors $\vec{AB}$ and $\vec{AC}$ in component form.
Decompose $\vec{AB}$ into a vector component parallel to $\vec{AC}$ ( $\text{proj}_{\vec{AC}}\vec{AB}$ ) and a component perpendicular to $\vec{AC}$ ( $\text{orth}_{\vec{AC}}\vec{AB}$ ).
Calculate the exact area of triangle $\triangle ABC$ .

Solution Key:

Find Displacement Vectors: State the Displacement Vector Formula:
$\vec{AB} = \langle x_B - x_A, \; y_B - y_A, \; z_B - z_A \rangle = \langle 3 - 1, \; 2 - 1, \; 2 - 0 \rangle = \langle 2, 1, 2 \rangle$ $\vec{AC} = \langle x_C - x_A, \; y_C - y_A, \; z_C - z_A \rangle = \langle 1 - 1, \; 4 - 1, \; 0 - 0 \rangle = \langle 0, 3, 0 \rangle$
Compute Projections: State the Dot Product and Magnitude:
$\vec{AB} \cdot \vec{AC} = (2)(0) + (1)(3) + (2)(0) = 3$ $\|\vec{AC}\|^2 = 0^2 + 3^2 + 0^2 = 9$

State the Vector Projection Formula:
$\text{proj}_{\vec{AC}}\vec{AB} = \left( \frac{\vec{AB} \cdot \vec{AC}}{\|\vec{AC}\|^2} \right) \vec{AC} = \frac{3}{9}\langle 0, 3, 0 \rangle = \frac{1}{3}\langle 0, 3, 0 \rangle = \langle 0, 1, 0 \rangle$

State the Orthogonal Component Formula:
$\text{orth}_{\vec{AC}}\vec{AB} = \vec{AB} - \text{proj}_{\vec{AC}}\vec{AB} = \langle 2, 1, 2 \rangle - \langle 0, 1, 0 \rangle = \langle 2, 0, 2 \rangle$ (Notice that $\langle 2, 0, 2 \rangle \cdot \langle 0, 3, 0 \rangle = 0$ , confirming orthogonality).
Compute Triangle Area: State the Cross Product Formula:
$\vec{AB} \times \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & 2 \\ 0 & 3 & 0 \\ \end{vmatrix} = \langle (1)(0) - (2)(3), \; (2)(0) - (2)(0), \; (2)(3) - (1)(0) \rangle = \langle -6, 0, 6 \rangle$

State the Triangle Area Formula:
$\text{Area}(\triangle ABC) = \frac{1}{2}\|\vec{AB} \times \vec{AC}\| = \frac{1}{2}\sqrt{(-6)^2 + 0^2 + 6^2} = \frac{1}{2}\sqrt{36 + 36} = \frac{1}{2}\sqrt{72} = \frac{1}{2}(6\sqrt{2}) = 3\sqrt{2}\text{ units}^2$

Show your steps and calculations here.

Practice Problem 3

Given vectors $\vec{u} = \langle 2, 1, 0 \rangle$ , $\vec{v} = \langle 1, 3, -1 \rangle$ , and $\vec{w} = \langle 0, 2, 4 \rangle$ in $\mathbb{R}^3$ :

Compute the cross product $\vec{v} \times \vec{w}$ .
Compute the triple scalar product $\vec{u} \cdot (\vec{v} \times \vec{w})$ .
Determine the volume of the parallelepiped spanned by $\vec{u}, \vec{v}, \vec{w}$ , and state whether the three vectors are coplanar.

Solution Key:

Compute Cross Product $\vec{v} \times \vec{w}$ : State the Cross Product Determinant Formula:
$\vec{v} \times \vec{w} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 3 & -1 \\ 0 & 2 & 4 \end{vmatrix} = \langle (3)(4) - (-1)(2), \; (-1)(0) - (1)(4), \; (1)(2) - (3)(0) \rangle = \langle 14, -4, 2 \rangle$
Compute Triple Scalar Product: State the Triple Scalar Product Formula:
$\vec{u} \cdot (\vec{v} \times \vec{w}) = \langle 2, 1, 0 \rangle \cdot \langle 14, -4, 2 \rangle = (2)(14) + (1)(-4) + (0)(2) = 28 - 4 + 0 = 24$ (Defined in Triple Scalar Product and Spatial Volume.)
Determine Volume and Test Coplanarity: State the Parallelepiped Volume Formula:
$\text{Volume} = |\vec{u} \cdot (\vec{v} \times \vec{w})| = |24| = 24\text{ units}^3$ Because the triple scalar product evaluates to a non-zero value ( $24 \ne 0$ ), the three vectors span a three-dimensional volume and are not coplanar.

Show your steps and calculations here.

Regroup 1

Projection Direction Trap: $\text{proj}_{\vec{u}}\vec{v}$ points strictly along vector $\vec{u}$ , NOT $\vec{v}$ . Always divide by $\|\vec{u}\|^2$ and multiply by vector $\vec{u}$ .
Cross Product Non-Commutativity: $\vec{u} \times \vec{v} = -(\vec{v} \times \vec{u})$ . Order matters for direction, though magnitude (area) is preserved.

Skill Block 2

Worked Example A: Classifying 3D Lines and Computing Shortest Clearances

Consider line $L_1$ passing through point $P(1, 0, 2)$ with direction vector $\vec{v}_1 = \langle 2, 1, -1 \rangle$ , and line $L_2$ passing through point $Q(0, 3, 1)$ with direction vector $\vec{v}_2 = \langle 1, -1, 2 \rangle$ .

Write the vector and parametric equations for $L_1$ (parameter $t$ ) and $L_2$ (parameter $s$ ).
Classify the geometric relationship between $L_1$ and $L_2$ (Parallel, Intersecting, Skew, or Coincident).
Compute the shortest perpendicular distance from point $P(1, 0, 2)$ on $L_1$ to line $L_2$ .

Worked Solution:

Formulate Equations of Lines: State the Vector Equation of a Line Formula:
$\vec{r}(t) = \vec{r}_0 + t\vec{v}$
(Defined in Equations of Lines in 3D.)
- Line $L_1$ : $\vec{r}_1(t) = \langle 1 + 2t, \; t, \; 2 - t \rangle$ $\text{Parametric: } x_1 = 1 + 2t, \quad y_1 = t, \quad z_1 = 2 - t$
- Line $L_2$ : $\vec{r}_2(s) = \langle s, \; 3 - s, \; 1 + 2s \rangle$ $\text{Parametric: } x_2 = s, \quad y_2 = 3 - s, \quad z_2 = 1 + 2s$
Classify the Spatial Relationship: State the Line Classification Protocol: (Defined in Relationships Between Lines in Space.)
- Parallel Check: Compare direction vectors: $\vec{v}_1 = \langle 2, 1, -1 \rangle, \quad \vec{v}_2 = \langle 1, -1, 2 \rangle$ $\frac{2}{1} \ne \frac{1}{-1} \implies \vec{v}_1 \ne k\vec{v}_2 \quad (\text{Not Parallel})$
- Intersection Check: Set $\vec{r}_1(t) = \vec{r}_2(s)$ : $\begin{cases} 1 + 2t = s & (1) \\ t = 3 - s & (2) \\ 2 - t = 1 + 2s & (3) \end{cases}$ Substitute (1) into (2): $t = 3 - (1 + 2t) = 2 - 2t \implies 3t = 2 \implies t = \frac{2}{3}$ Then from (1): $s = 1 + 2\left(\frac{2}{3}\right) = \frac{7}{3}$ . Test $\left(t = \frac{2}{3}, s = \frac{7}{3}\right)$ in equation (3): $\text{LHS} = 2 - \frac{2}{3} = \frac{4}{3}$ $\text{RHS} = 1 + 2\left(\frac{7}{3}\right) = 1 + \frac{14}{3} = \frac{17}{3}$ $\frac{4}{3} \ne \frac{17}{3} \quad (\text{Inconsistent})$
- Classification: Because the direction vectors are non-parallel and the linear system has no solution, lines $L_1$ and $L_2$ are Skew.
Compute Perpendicular Distance from $P(1, 0, 2)$ to Line $L_2$ : State the Point-to-Line Distance Formula:
$d = \frac{\|\vec{QP} \times \vec{v}_2\|}{\|\vec{v}_2\|}$
(Defined in Distance Between a Point and a Line in Space.)
1. Find displacement vector $\vec{QP}$ from $Q(0, 3, 1)$ to $P(1, 0, 2)$ : $\vec{QP} = \langle 1 - 0, \; 0 - 3, \; 2 - 1 \rangle = \langle 1, -3, 1 \rangle$
2. Compute cross product $\vec{QP} \times \vec{v}_2$ : $\vec{QP} \times \vec{v}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -3 & 1 \\ 1 & -1 & 2 \end{vmatrix} = \langle (-3)(2) - (1)(-1), \; (1)(1) - (1)(2), \; (1)(-1) - (-3)(1) \rangle$ $\vec{QP} \times \vec{v}_2 = \langle -6 + 1, \; 1 - 2, \; -1 + 3 \rangle = \langle -5, -1, 2 \rangle$
3. Compute magnitudes: $\|\vec{QP} \times \vec{v}_2\| = \sqrt{(-5)^2 + (-1)^2 + 2^2} = \sqrt{25 + 1 + 4} = \sqrt{30}$ $\|\vec{v}_2\| = \sqrt{1^2 + (-1)^2 + 2^2} = \sqrt{1 + 1 + 4} = \sqrt{6}$
4. Calculate exact distance: $d = \frac{\sqrt{30}}{\sqrt{6}} = \sqrt{\frac{30}{6}} = \sqrt{5}\text{ units}$

Practice Problem 4: Concept & Reading Check

Open this Skill Block on the Memex to locate the information needed to answer the following:

(a) Why must you use two distinct parameters ( $t$ and $s$ ) when testing whether two lines in 3D space intersect?
(b) Deconstruct Worked Example A step 2: what algebraic outcome confirms that lines are skew rather than parallel or intersecting?
(c) In Autonomous Drone Flight Path Separation (Aerospace & Computer Engineering), what geometric threshold determines whether an emergency avoidance maneuver is required?

Solution Key:

(a) Parameter Decoupling: In physical kinematics, the parameter $t$ often represents time. If you use the same parameter $t$ for both lines, setting $\vec{r}_1(t) = \vec{r}_2(t)$ only tests whether the two objects will collide at the exact same instant. Two physical paths can cross at the same spatial coordinates $(x,y,z)$ at completely different times ( $t \ne s$ ). Using independent parameters $t$ and $s$ tests for true geometric intersection of the curves.
(b) Skew Line Criteria: Skew lines are defined as lines in $\mathbb{R}^3$ that are non-coplanar, which requires two distinct conditions: (1) their direction vectors are non-parallel ( $\vec{v}_1 \ne k\vec{v}_2$ ), and (2) their paths do not intersect, proven algebraically when the $(t, s)$ values obtained from two coordinate equations fail to satisfy the third coordinate equation.
(c) Clearance Threshold: In the autonomous drone scenario, the FAA clearance standard mandates that the minimum perpendicular distance between the drone's position and the neighboring flight path corridor must satisfy $d \ge 2.0\text{ m}$ . Computing $d = 2.89\text{ m} \ge 2.0\text{ m}$ confirms compliance without requiring an emergency trajectory override.

Show your steps and calculations here.

Practice Problem 5

Line $L_1$ is given by symmetric equations $\frac{x-2}{1} = \frac{y+1}{2} = \frac{z-3}{1}$ , and line $L_2$ is given by parametric equations $x = 1 + 2s$ , $y = 3 + 4s$ , $z = 2 + 2s$ .

Extract direction vectors $\vec{v}_1$ and $\vec{v}_2$ and determine whether the lines are parallel.
Determine whether $L_1$ and $L_2$ are coincident (the same line) or distinct parallel lines.
Compute the shortest perpendicular distance between the two parallel lines.

Solution Key:

Extract Direction Vectors and Test Parallelism:
- From symmetric equations of $L_1$ : reference point $P(2, -1, 3)$ and direction vector $\vec{v}_1 = \langle 1, 2, 1 \rangle$ .
- From parametric equations of $L_2$ : reference point $Q(1, 3, 2)$ and direction vector $\vec{v}_2 = \langle 2, 4, 2 \rangle$ .
- Proportionality check: $\vec{v}_2 = \langle 2, 4, 2 \rangle = 2\langle 1, 2, 1 \rangle = 2\vec{v}_1$ Because $\vec{v}_2 = 2\vec{v}_1$ , lines $L_1$ and $L_2$ are Parallel.
Test for Coincidence: Test whether point $P(2, -1, 3) \in$ line $L_1$ lies on line $L_2$ :
$x: 1 + 2s = 2 \implies 2s = 1 \implies s = \frac{1}{2}$ Test $s = 1/2$ in the $y$ -equation of $L_2$ : $y\left(\frac{1}{2}\right) = 3 + 4\left(\frac{1}{2}\right) = 3 + 2 = 5 \ne -1$ Since $P(2, -1, 3)$ does not satisfy $L_2$ , the lines are Distinct Parallel Lines.
Compute Perpendicular Distance between Parallel Lines: State the Point-to-Line Distance Formula:
$d = \frac{\|\vec{QP} \times \vec{v}_1\|}{\|\vec{v}_1\|}$
(Defined in Distance Between a Point and a Line in Space.)
1. Find displacement vector $\vec{QP}$ from $Q(1, 3, 2)$ to $P(2, -1, 3)$ : $\vec{QP} = \langle 2 - 1, \; -1 - 3, \; 3 - 2 \rangle = \langle 1, -4, 1 \rangle$
2. Compute cross product $\vec{QP} \times \vec{v}_1$ : $\vec{QP} \times \vec{v}_1 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -4 & 1 \\ 1 & 2 & 1 \end{vmatrix} = \langle (-4)(1) - (1)(2), \; (1)(1) - (1)(1), \; (1)(2) - (-4)(1) \rangle$ $\vec{QP} \times \vec{v}_1 = \langle -4 - 2, \; 1 - 1, \; 2 + 4 \rangle = \langle -6, 0, 6 \rangle$
3. Compute magnitudes: $\|\vec{QP} \times \vec{v}_1\| = \sqrt{(-6)^2 + 0^2 + 6^2} = \sqrt{36 + 0 + 36} = \sqrt{72} = 6\sqrt{2}$ $\|\vec{v}_1\| = \sqrt{1^2 + 2^2 + 1^2} = \sqrt{1 + 4 + 1} = \sqrt{6}$
4. Calculate distance $d$ : $d = \frac{6\sqrt{2}}{\sqrt{6}} = \frac{6}{\sqrt{3}} = \frac{6\sqrt{3}}{3} = 2\sqrt{3}\text{ units}$

Show your steps and calculations here.

Practice Problem 6

Line $L_1$ is given by $\vec{r}_1(t) = \langle 1 + 2t, \; -2 + 3t, \; 4 - t \rangle$ and line $L_2$ is given by $\vec{r}_2(s) = \langle 1 + s, \; 3 - s, \; -1 + 2s \rangle$ .

Extract direction vectors $\vec{v}_1$ and $\vec{v}_2$ and verify that the lines are not parallel.
Formulate the linear system $\vec{r}_1(t) = \vec{r}_2(s)$ with independent parameters $t$ and $s$ to classify the relationship (Intersecting or Skew). If the lines intersect, determine the exact point of intersection $P_{\text{int}}(x, y, z)$ .
Compute the shortest perpendicular distance from point $Q(1, 3, -1)$ on $L_2$ to line $L_1$ .

Solution Key:

Verify Non-Parallel Direction Vectors:
- Direction vector of $L_1$ : $\vec{v}_1 = \langle 2, 3, -1 \rangle$ .
- Direction vector of $L_2$ : $\vec{v}_2 = \langle 1, -1, 2 \rangle$ . Because $\langle 2, 3, -1 \rangle \ne k\langle 1, -1, 2 \rangle$ , the direction vectors are not proportional, so $L_1$ and $L_2$ are not parallel.
Formulate and Solve 2-Parameter Linear System: Set $\vec{r}_1(t) = \vec{r}_2(s)$ :
$\begin{cases} 1 + 2t = 1 + s & \text{(Equation 1)} \\ -2 + 3t = 3 - s & \text{(Equation 2)} \\ 4 - t = -1 + 2s & \text{(Equation 3)} \end{cases}$ From Equation 1: $s = 2t$ Substitute $s = 2t$ into Equation 2: $-2 + 3t = 3 - 2t \implies 5t = 5 \implies t = 1$ Then: $s = 2(1) = 2$ Test $(t=1, s=2)$ in Equation 3: $\text{LHS} = 4 - (1) = 3$ $\text{RHS} = -1 + 2(2) = -1 + 4 = 3$ Because $\text{LHS} = \text{RHS} = 3$ , the system is strictly consistent. Therefore, the lines Intersect.

Intersection Point:
$P_{\text{int}} = \vec{r}_1(1) = \langle 1 + 2(1), \; -2 + 3(1), \; 4 - 1 \rangle = (3, 1, 3)$
Compute Perpendicular Clearance from $Q(1, 3, -1)$ to $L_1$ : State the Point-to-Line Distance Formula:
$d = \frac{\|\vec{P_1 Q} \times \vec{v}_1\|}{\|\vec{v}_1\|}$ Where $P_1(1, -2, 4)$ is the reference anchor point on $L_1$ : $\vec{P_1 Q} = \langle 1 - 1, \; 3 - (-2), \; -1 - 4 \rangle = \langle 0, 5, -5 \rangle$ Compute cross product $\vec{P_1 Q} \times \vec{v}_1$ : $\vec{P_1 Q} \times \vec{v}_1 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 5 & -5 \\ 2 & 3 & -1 \end{vmatrix} = \langle (5)(-1) - (-5)(3), \; (-5)(2) - (0)(-1), \; (0)(3) - (5)(2) \rangle$ $\vec{P_1 Q} \times \vec{v}_1 = \langle -5 + 15, \; -10 - 0, \; 0 - 10 \rangle = \langle 10, -10, -10 \rangle$ Compute magnitudes: $\|\vec{P_1 Q} \times \vec{v}_1\| = \sqrt{10^2 + (-10)^2 + (-10)^2} = \sqrt{100 + 100 + 100} = \sqrt{300} = 10\sqrt{3}$ $\|\vec{v}_1\| = \sqrt{2^2 + 3^2 + (-1)^2} = \sqrt{4 + 9 + 1} = \sqrt{14}$ Calculate distance: $d = \frac{10\sqrt{3}}{\sqrt{14}} = \frac{10\sqrt{42}}{14} = \frac{5\sqrt{42}}{7}\text{ units}$

Show your steps and calculations here.

Regroup 2

The Skew Line Checklist: Remember that 3D lines are skew only when they are BOTH non-parallel AND non-intersecting. Always execute the direction vector proportionality test first before solving the linear system.
Symmetric Equation Conversion: If a symmetric component has a denominator of zero (e.g., $\frac{x-1}{2} = \frac{y-3}{4}, z = 5$ ), the direction vector is $\vec{v} = \langle 2, 4, 0 \rangle$ .

Skill Block 3

Worked Example A: Space Curve Kinematics, Arc Length, Curvature, and Acceleration Decomposition

A high-speed test vehicle moves in $\mathbb{R}^3$ along the trajectory:

\vec{r}(t) = \langle 3\cos t, \; 3\sin t, \; 4t \rangle \quad \text{for } t \ge 0

Determine the velocity $\vec{v}(t)$ , acceleration $\vec{a}(t)$ , and scalar speed $\|\vec{v}(t)\|$ .
Derive the accumulated arc length function $s(t)$ starting at $t=0$ , and construct the intrinsic arc length parametrization $\vec{r}(s)$ .
Compute the unit tangent vector $\hat{T}(t)$ and curvature $\kappa(t)$ .
Decompose the acceleration vector into tangential and normal components ( $a_T$ and $a_N$ ).

Worked Solution:

Compute Kinematics and Scalar Speed: State the Vector Derivative Formulas:
$\vec{v}(t) = \vec{r}'(t) = \langle -3\sin t, \; 3\cos t, \; 4 \rangle$ $\vec{a}(t) = \vec{r}''(t) = \langle -3\cos t, \; -3\sin t, \; 0 \rangle$ (Defined in Derivatives of Vector-Valued Functions.)

State the Scalar Speed Formula:
$\|\vec{v}(t)\| = \sqrt{(-3\sin t)^2 + (3\cos t)^2 + 4^2} = \sqrt{9(\sin^2 t + \cos^2 t) + 16} = \sqrt{9 + 16} = \sqrt{25} = 5$ (The vehicle travels at a strictly constant speed of $5\text{ units/s}$ .)
Derive $s(t)$ and Parametrize by Arc Length $\vec{r}(s)$ : State the Accumulated Arc Length Function Formula:
$s(t) = \int_{0}^{t} \|\vec{r}'(u)\|\,du = \int_{0}^{t} 5\,du = [5u]_{0}^{t} = 5t$ (Defined in Arc Length of a Space Curve.)

Invert $s(t)$ to solve for $t$ :
$s = 5t \implies t(s) = \frac{s}{5}$

Substitute into $\vec{r}(t)$ :
$\vec{r}(s) = \left\langle 3\cos\left(\frac{s}{5}\right), \; 3\sin\left(\frac{s}{5}\right), \; 4\left(\frac{s}{5}\right) \right\rangle = \left\langle 3\cos\left(\frac{s}{5}\right), \; 3\sin\left(\frac{s}{5}\right), \; \frac{4}{5}s \right\rangle$ (Notice that $\|\vec{r}'(s)\| = \sqrt{(-3/5\sin)^2 + (3/5\cos)^2 + (4/5)^2} = \sqrt{9/25 + 16/25} = 1$ , confirming unit speed traversal as established in Arc Length Parametrization.)
Compute Unit Tangent Vector $\hat{T}(t)$ and Curvature $\kappa(t)$ : State the Unit Tangent Vector Formula:
$\hat{T}(t) = \frac{\vec{v}(t)}{\|\vec{v}(t)\|} = \frac{1}{5}\langle -3\sin t, \; 3\cos t, \; 4 \rangle = \left\langle -\frac{3}{5}\sin t, \; \frac{3}{5}\cos t, \; \frac{4}{5} \right\rangle$ (Defined in Unit Tangent Vector.)

State the Cross-Product Curvature Formula:
$\kappa = \frac{\|\vec{v} \times \vec{a}\|}{\|\vec{v}\|^3}$ (Defined in Curvature of a Space Curve.)

Calculate $\vec{v} \times \vec{a}$ :
$\vec{v} \times \vec{a} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -3\sin t & 3\cos t & 4 \\ -3\cos t & -3\sin t & 0 \end{vmatrix}$ $\vec{v} \times \vec{a} = \langle (3\cos t)(0) - (4)(-3\sin t), \; (4)(-3\cos t) - (-3\sin t)(0), \; (-3\sin t)(-3\sin t) - (3\cos t)(-3\cos t) \rangle$ $\vec{v} \times \vec{a} = \langle 12\sin t, \; -12\cos t, \; 9(\sin^2 t + \cos^2 t) \rangle = \langle 12\sin t, \; -12\cos t, \; 9 \rangle$

Compute magnitude:
$\|\vec{v} \times \vec{a}\| = \sqrt{(12\sin t)^2 + (-12\cos t)^2 + 9^2} = \sqrt{144(\sin^2 t + \cos^2 t) + 81} = \sqrt{144 + 81} = \sqrt{225} = 15$

Evaluate curvature:
$\kappa = \frac{15}{5^3} = \frac{15}{125} = \frac{3}{25} = 0.12\text{ units}^{-1}$ (Radius of curvature is $\rho = \frac{1}{\kappa} = \frac{25}{3} \approx 8.33\text{ units}$ .)
Compute Tangential and Normal Acceleration Components: State the Tangential Acceleration Formula:
$a_T = \frac{\vec{v} \cdot \vec{a}}{\|\vec{v}\|} = \frac{(-3\sin t)(-3\cos t) + (3\cos t)(-3\sin t) + (4)(0)}{5} = \frac{9\sin t\cos t - 9\sin t\cos t}{5} = 0$ (Because speed is constant, rate of speed change is identically zero).

State the Normal Acceleration Formula:
$a_N = \frac{\|\vec{v} \times \vec{a}\|}{\|\vec{v}\|} = \frac{15}{5} = 3\text{ units/s}^2$ (Notice that $a_N = \kappa \|\vec{v}\|^2 = \left(\frac{3}{25}\right)(25) = 3$ , matching exactly).

Practice Problem 7: Concept & Reading Check

Open this Skill Block on the Memex to locate the information needed to answer the following:

(a) Explain why a particle traveling at a strictly constant scalar speed ( $v(t) = c$ ) has $a_T = 0$ , but can still experience a non-zero acceleration vector $\vec{a}(t)$ .
(b) Deconstruct the denominator of the cross-product curvature formula $\kappa = \frac{\|\vec{v} \times \vec{a}\|}{\|\vec{v}\|^3}$ : why is speed cubed rather than squared?
(c) In Roller Coaster Loop Structural Dynamics (Entertainment & Mechanical Engineering), how does normal centripetal acceleration $a_N$ relate to passenger safety G-force limits?

Solution Key:

(a) Tangential vs Normal Acceleration: Kinematic acceleration $\vec{a}(t) = a_T \hat{T} + a_N \hat{N}$ decomposes into two independent physical mechanisms. The tangential component $a_T = \frac{d}{dt}\|\vec{v}\|$ quantifies scalar throttle or braking. If speed is constant, $a_T = 0$ . However, whenever a curve changes directional heading in space, the unit tangent vector rotates ( $d\hat{T}/dt \ne \vec{0}$ ). This directional rate of change generates a strictly perpendicular centripetal acceleration vector $a_N \hat{N}$ , meaning $\vec{a}(t) = a_N \hat{N} \ne \vec{0}$ .
(b) Curvature Denominator Derivation: By definition, curvature is turning per unit distance: $\kappa = \frac{\|\hat{T}'(t)\|}{\|\vec{r}'(t)\|} = \frac{a_N}{\|\vec{v}\|^2}$ . Because the normal acceleration formula evaluates to $a_N = \frac{\|\vec{v} \times \vec{a}\|}{\|\vec{v}\|}$ , substituting $a_N$ into the curvature expression produces: $\kappa = \frac{\frac{\|\vec{v} \times \vec{a}\|}{\|\vec{v}\|}}{\|\vec{v}\|^2} = \frac{\|\vec{v} \times \vec{a}\|}{\|\vec{v}\|^3}$
(c) Biomechanical G-Force Limits: In amusement park and aerospace engineering, passenger bodies experience apparent weight and inertial load proportional to net acceleration via Newton's Second Law ( $\vec{F}_{\text{seat}} - m\vec{g} = m\vec{a}$ ). The normal component $a_N = \kappa v^2 = \frac{v^2}{\rho}$ produces centripetal load. Dividing $a_N$ by standard gravitational acceleration ( $g = 9.81\text{ m/s}^2$ ) gives the added G-force loading. Keeping $a_N$ within bounded limits ( $\le 1.0\text{ g}$ lateral, $\le 4.5\text{ g}$ vertical) prevents loss of consciousness (blackout) and severe neck strain.

Show your steps and calculations here.

Practice Problem 8

A precision automated machining cutter travels along the 3D space curve:

\vec{r}(t) = \langle 2t, \; t^2, \; \frac{1}{3}t^3 \rangle \quad \text{for } t \ge 0

Find the velocity $\vec{v}(t)$ , acceleration $\vec{a}(t)$ , and scalar speed $\|\vec{v}(t)\|$ .
Compute the exact arc length $s$ traversed by the tool from $t = 0$ to $t = 3$ .
At the waypoint $t = 1$ , compute the tangential acceleration $a_T(1)$ , normal acceleration $a_N(1)$ , and the curvature $\kappa(1)$ .

Solution Key:

Differentiate and Compute Scalar Speed: State the Vector Derivative Formulas:
$\vec{v}(t) = \vec{r}'(t) = \langle 2, \; 2t, \; t^2 \rangle$ $\vec{a}(t) = \vec{r}''(t) = \langle 0, \; 2, \; 2t \rangle$ (Defined in Derivatives of Vector-Valued Functions.)

State the Scalar Speed Formula:
$\|\vec{v}(t)\| = \sqrt{2^2 + (2t)^2 + (t^2)^2} = \sqrt{4 + 4t^2 + t^4}$ Factor the perfect square trinomial under the radical: $\|\vec{v}(t)\| = \sqrt{(t^2 + 2)^2} = t^2 + 2$
Compute Arc Length ( $0 \le t \le 3$ ): State the Definite Arc Length Formula:
$s = \int_{a}^{b} \|\vec{v}(t)\|\,dt$ (Defined in Arc Length of a Space Curve.)

$s = \int_{0}^{3} (t^2 + 2)\,dt = \left[ \frac{1}{3}t^3 + 2t \right]_{0}^{3} = \left( \frac{1}{3}(27) + 2(3) \right) - 0 = 9 + 6 = 15\text{ units}$
Evaluate at $t = 1$ :
- Velocity at $t = 1$ : $\vec{v}(1) = \langle 2, 2, 1 \rangle \implies \|\vec{v}(1)\| = 1^2 + 2 = 3$
- Acceleration at $t = 1$ : $\vec{a}(1) = \langle 0, 2, 2 \rangle \implies \|\vec{a}(1)\| = \sqrt{0^2 + 2^2 + 2^2} = \sqrt{8} = 2\sqrt{2}$
- Dot product $\vec{v}(1) \cdot \vec{a}(1)$ : $\vec{v}(1) \cdot \vec{a}(1) = (2)(0) + (2)(2) + (1)(2) = 0 + 4 + 2 = 6$
State the Tangential Acceleration Formula:
$a_T(1) = \frac{\vec{v}(1) \cdot \vec{a}(1)}{\|\vec{v}(1)\|} = \frac{6}{3} = 2\text{ units/s}^2$
(Notice that differentiating speed directly: $a_T = \frac{d}{dt}[t^2 + 2] = 2t$ , which at $t=1$ gives $2$ ).

State the Normal Acceleration Pythagorean Formula:
$a_N(1) = \sqrt{\|\vec{a}(1)\|^2 - a_T(1)^2} = \sqrt{(2\sqrt{2})^2 - (2)^2} = \sqrt{8 - 4} = \sqrt{4} = 2\text{ units/s}^2$

State the Curvature Formula:
$\kappa(1) = \frac{a_N(1)}{\|\vec{v}(1)\|^2} = \frac{2}{3^2} = \frac{2}{9}\text{ units}^{-1}$
(Radius of curvature at $t = 1$ is $\rho = \frac{1}{\kappa} = \frac{9}{2} = 4.5\text{ units}$ ).

Show your steps and calculations here.

Practice Problem 9

A weather tracking drone ascends along the circular helix space curve:

\vec{r}(t) = \langle 4\cos t, \; 4\sin t, \; 3t \rangle \quad \text{for } t \ge 0

Determine the velocity vector $\vec{v}(t)$ , acceleration vector $\vec{a}(t)$ , and scalar speed $\|\vec{v}(t)\|$ .
Compute the unit tangent vector $\hat{T}(t)$ and principal unit normal vector $\hat{N}(t) = \frac{\hat{T}'(t)}{\|\hat{T}'(t)\|}$ .
Calculate the curvature $\kappa(t)$ , the radius of the osculating circle $\rho$ , and decompose the acceleration into tangential and normal components ( $a_T$ and $a_N$ ).

Solution Key:

Differentiate and Compute Scalar Speed: State the Vector Derivative Formulas:
$\vec{v}(t) = \vec{r}'(t) = \langle -4\sin t, \; 4\cos t, \; 3 \rangle$ $\vec{a}(t) = \vec{r}''(t) = \langle -4\cos t, \; -4\sin t, \; 0 \rangle$

State the Scalar Speed Formula:
$\|\vec{v}(t)\| = \sqrt{(-4\sin t)^2 + (4\cos t)^2 + 3^2} = \sqrt{16(\sin^2 t + \cos^2 t) + 9} = \sqrt{16 + 9} = \sqrt{25} = 5\text{ units/s}$
Compute Unit Tangent Vector $\hat{T}(t)$ and Principal Unit Normal Vector $\hat{N}(t)$ : State the Unit Tangent Vector Formula:
$\hat{T}(t) = \frac{\vec{v}(t)}{\|\vec{v}(t)\|} = \frac{1}{5}\langle -4\sin t, \; 4\cos t, \; 3 \rangle = \left\langle -\frac{4}{5}\sin t, \; \frac{4}{5}\cos t, \; \frac{3}{5} \right\rangle$

Differentiate $\hat{T}(t)$ with respect to $t$ :
$\hat{T}'(t) = \left\langle -\frac{4}{5}\cos t, \; -\frac{4}{5}\sin t, \; 0 \right\rangle$ Compute its magnitude: $\|\hat{T}'(t)\| = \sqrt{\left(-\frac{4}{5}\cos t\right)^2 + \left(-\frac{4}{5}\sin t\right)^2 + 0^2} = \sqrt{\frac{16}{25}(\cos^2 t + \sin^2 t)} = \frac{4}{5}$

State the Principal Unit Normal Formula:
$\hat{N}(t) = \frac{\hat{T}'(t)}{\|\hat{T}'(t)\|} = \frac{5}{4}\left\langle -\frac{4}{5}\cos t, \; -\frac{4}{5}\sin t, \; 0 \right\rangle = \langle -\cos t, \; -\sin t, \; 0 \rangle$ (Defined in Principal Unit Normal Vector.)
Compute Curvature, Osculating Radius, and Acceleration Components: State the Curvature Formula:
$\kappa = \frac{\|\hat{T}'(t)\|}{\|\vec{r}'(t)\|} = \frac{4/5}{5} = \frac{4}{25} = 0.16\text{ units}^{-1}$

State the Radius of Curvature Formula:
$\rho = \frac{1}{\kappa} = \frac{25}{4} = 6.25\text{ units}$

State the Tangential Acceleration Formula:
$a_T = \frac{d}{dt}\|\vec{v}(t)\| = \frac{d}{dt}[5] = 0\text{ units/s}^2$

State the Normal Acceleration Formula:
$a_N = \kappa \|\vec{v}\|^2 = \left(\frac{4}{25}\right)(5^2) = \left(\frac{4}{25}\right)(25) = 4\text{ units/s}^2$ (Notice that $\vec{a}(t) = a_T \hat{T} + a_N \hat{N} = 0\hat{T} + 4\langle -\cos t, -\sin t, 0 \rangle = \langle -4\cos t, -4\sin t, 0 \rangle$ , matching $\vec{r}''(t)$ exactly).

Show your steps and calculations here.

Regroup 3

Algebraic Speed Check: Always check if the quadratic under the radical in $\|\vec{v}(t)\|$ factors as a perfect square $(at^2 + b)^2$ . This is a classic exam technique that avoids intractable radical integrals.
Normal Acceleration Positivity: $a_N$ is a geometric magnitude ( $a_N = \kappa v^2 \ge 0$ ). It can NEVER be negative! If your calculation yields a negative value, recheck your dot product and magnitude substitutions.

Synthesis Wrap-up

Exam Formula Strategy: State every governing symbolic equation before substituting numerical values to secure partial credit and prevent arithmetic slips.
Test Review Center Recovery Protocol: Remember that logging 4 active hours in the Test Review Center allows you to convert one pure 0% homework score into 100% (up to 2 assignments), restoring your retroactive midterm grade replacement eligibility.
Exam Readiness Checklist:
1. Can you set up and calculate $\text{proj}_{\vec{u}}\vec{v}$ and $\vec{u} \times \vec{v}$ without mixing up formulas?
2. Can you write symmetric line equations and compute point-to-line perpendicular distances?
3. Can you find $\vec{v}(t), \vec{a}(t)$ , arc length $s$ , curvature $\kappa$ , and tangential/normal acceleration components $a_T, a_N$ ?