MATH-283 (Calculus III) Curriculum

Session 04 - Vector-Valued Functions (Lines in Space)

Handout Worksheet

Fall Semester 2026

Session Objectives & Overview

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

Construct vector, parametric, and symmetric equations of lines in 3D space from points and direction vectors, and interpret vector-valued functions $\vec{r}(t) = \vec{r}_0 + t\vec{v}$ as spatial trajectories.
Calculate the shortest perpendicular distance between an external point and a 3D line using the geometric cross-product formula $d = \frac{\|\vec{PM} \times \vec{v}\|}{\|\vec{v}\|}$ .
Systematically classify the geometric relationship between two lines in 3D space (Equal, Parallel, Intersecting, or Skew) using direction vector ratios and independent two-parameter linear systems.

Micro-Lecture

1. Lines in 3D Space: The Failure of 2D Slope ( $y = mx + b$ )

Mathematical Foundation: In 2D algebra, a single scalar slope $m = \frac{\Delta y}{\Delta x}$ uniquely defines the inclination of a line. In 3D space ( $\mathbb{R}^3$ ), a line can point in infinitely many spatial directions, making a single slope value mathematically meaningless. Instead, we define a line using a known reference point $P_0(x_0, y_0, z_0)$ (position vector $\vec{r}_0 = \langle x_0, y_0, z_0 \rangle$ ) and a non-zero direction vector $\vec{v} = \langle a, b, c \rangle = a\hat{i} + b\hat{j} + c\hat{k}$ . The line is generated by the vector-valued function $\vec{r}(t) = \vec{r}_0 + t\vec{v}$ .
Engineering Context: In aerospace guidance and robotics, the scalar parameter $t$ represents physical time or normalized flight progress, allowing engineers to track 3D velocity vectors and exact waypoint coordinates along straight flight corridors.

2. Spatial Clearances and Point-to-Line Distance

Mathematical Foundation: Finding the perpendicular distance from an external point $M$ to a line in space uses the Cross Product property where the magnitude $\|\vec{PM} \times \vec{v}\|$ represents the geometric area of the parallelogram spanned by displacement vector $\vec{PM}$ and direction vector $\vec{v}$ . Because $\text{Area} = \text{Base} \times \text{Height} = \|\vec{v}\| d$ , the shortest distance is calculated via the Point-to-Line Distance Formula $d = \frac{\|\vec{PM} \times \vec{v}\|}{\|\vec{v}\|}$ .
Engineering Context: Used to enforce automated safety clearance envelopes between robotic arm links and high-voltage factory fixtures or structural conduits.

3. Spatial Line Relationships: The 3D Skew Phenomenon

Mathematical Foundation: In 2D, non-parallel lines must intersect. In 3D, two lines can be non-parallel and non-intersecting—a state called Skew. Evaluating line relationships requires a two-step algorithm: checking direction vector proportionality ( $\vec{u} = k\vec{v}$ ) and testing system consistency with independent parameters ( $t$ and $s$ ) using matrix techniques from Solving Systems of Linear Equations with Matrix Form.
Engineering Context: Fundamental for collision detection in game engines (ray tracing) and air traffic control separation algorithms.

Skill Block 1

Worked Example A: Three Standard Forms from Two Points

Given a line passing through the points $P = (-4, -6, -2)$ and $Q = (10, 2, 1)$ :

Find the direction vector $\vec{v} = \langle a, b, c \rangle$ and state the vector equation of the line. Evaluate it at $t = 0, 1, -1$ .
Find the parametric equations of the line.
Convert the parametric equations into symmetric equations.

Worked Solution:

Find Direction Vector $\vec{v} = \langle a, b, c \rangle$ via the Two-Point Displacement Formula: State the Two-Point Displacement Formula:
$\vec{v} = \vec{PQ} = \langle x_Q - x_P, y_Q - y_P, z_Q - z_P \rangle$ (This formula is defined in Equations of Lines in 3D.)

Substitute coordinates $P(-4, -6, -2)$ and $Q(10, 2, 1)$ :
$\vec{v} = \langle 10 - (-4), 2 - (-6), 1 - (-2) \rangle = \langle 14, 8, 3 \rangle = 14\hat{i} + 8\hat{j} + 3\hat{k}$ Here, the directional components are $a = 14$ , $b = 8$ , and $c = 3$ .
Construct Vector Equation via the Vector Line Formula: State the Vector Line Formula:
$\vec{r}(t) = \vec{r}_0 + t\vec{v}$
where $\vec{r}_0 = \langle x_0, y_0, z_0 \rangle$ is the initial position vector and $\vec{v} = \langle a, b, c \rangle$ is the direction vector.

Using $P(-4, -6, -2)$ as initial position vector $\vec{r}_0 = \langle -4, -6, -2 \rangle$ :
$\vec{r}(t) = \langle -4, -6, -2 \rangle + t\langle 14, 8, 3 \rangle = \langle -4 + 14t, -6 + 8t, -2 + 3t \rangle$

Evaluate at specific parameter values:
- For $t = 0$ : $\vec{r}(0) = \langle -4, -6, -2 \rangle$ (Point $P$ )
- For $t = 1$ : $\vec{r}(1) = \langle -4 + 14(1), -6 + 8(1), -2 + 3(1) \rangle = \langle 10, 2, 1 \rangle$ (Point $Q$ )
- For $t = -1$ : $\vec{r}(-1) = \langle -4 - 14, -6 - 8, -2 - 3 \rangle = \langle -18, -14, -5 \rangle$
Construct Parametric Equations: State the Parametric Line Equations:
$x(t) = x_0 + at, \quad y(t) = y_0 + bt, \quad z(t) = z_0 + ct$ Substituting $(x_0, y_0, z_0) = (-4, -6, -2)$ and $(a, b, c) = (14, 8, 3)$ : $x = -4 + 14t, \quad y = -6 + 8t, \quad z = -2 + 3t$
Convert to Symmetric Equations via Parameter Elimination: State the Symmetric Line Formula:
$\frac{x - x_0}{a} = \frac{y - y_0}{b} = \frac{z - z_0}{c}$ Isolate parameter $t$ in each parametric equation: $t = \frac{x + 4}{14}, \quad t = \frac{y + 6}{8}, \quad t = \frac{z + 2}{3}$ Equating expressions gives the symmetric equations: $\frac{x + 4}{14} = \frac{y + 6}{8} = \frac{z + 2}{3}$

Worked Example B: Direction Vector with a Zero Component

Find the parametric and symmetric equations of the line passing through point $P_0(2, -1, 4)$ with direction vector $\vec{v} = \langle 3, 0, -5 \rangle$ .

Worked Solution:

Identify Direction Components: From $\vec{v} = 3\hat{i} + 0\hat{j} - 5\hat{k} = \langle 3, 0, -5 \rangle$ , we have $a = 3$ , $b = 0$ , and $c = -5$ . The initial point coordinates are $(x_0, y_0, z_0) = (2, -1, 4)$ .
Apply Parametric Equations Formula: State the Parametric Equations Formula: $x = x_0 + at, \quad y = y_0 + bt, \quad z = z_0 + ct$ Substitute the components: $x = 2 + 3t, \quad y = -1 + (0)t = -1, \quad z = 4 - 5t$
Formulate Symmetric Equations with Zero Denominator Rule: Because $b = 0$ , dividing by $b$ is undefined. We hold $y$ fixed at constant value $y = -1$ and equate the remaining parameters: $t = \frac{x - 2}{3} = \frac{z - 4}{-5}$ Symmetric form: $y = -1, \quad \frac{x - 2}{3} = \frac{z - 4}{-5}$ (This zero-component convention is defined in Equations of Lines in 3D.)

Practice Problem 1: Concept & Reading Check

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

(a) Under the Conceptual Framework section, state the direction vector $\vec{v}$ in both unit vector form and component bracket form. Explain what the individual components $a, b,$ and $c$ represent physically.
(b) Tracing Worked Example A, explain the algebraic procedure used to eliminate parameter $t$ from the parametric equations to obtain symmetric equations.
(c) Review the linked engineering application Aerospace Engineering (Satellite Orbit Path Modeling). How does the parameter $t$ relate to the satellite's flight trajectory, and what coordinate position corresponds to $t = 0.4$ ?

Solution Key:

(a) As detailed in the Conceptual Framework, the direction vector is expressed as $\vec{v} = a\hat{i} + b\hat{j} + c\hat{k} = \langle a, b, c \rangle$ . The scalar components $a, b,$ and $c$ represent the differential spatial change rates $a = \frac{\Delta x}{\Delta t}, b = \frac{\Delta y}{\Delta t}, c = \frac{\Delta z}{\Delta t}$ , acting as directional slopes that specify how rapidly the line translates across $x, y,$ and $z$ per unit parameter $t$ .
(b) In Worked Example A, the transition from parametric to symmetric form is performed by algebraically isolating the common parameter $t$ in each coordinate equation ( $t = \frac{x+4}{14}$ , $t = \frac{y+6}{8}$ , $t = \frac{z+2}{3}$ ) and equating all three ratios ( $t = t = t$ ).
(c) In Aerospace Engineering (Satellite Orbit Path Modeling), the parameter $t$ represents normalized progress along the transfer orbit leg. Evaluating the parametric equations at $t = 0.4$ ( $40\%$ leg completion) yields the exact spatial waypoint coordinate $(4260, 1548, 6380)\text{ km}$ .

Show your steps and calculations here.

Practice Problem 2

Given a line passing through points $A = (1, 4, -2)$ and $B = (-3, 5, 0)$ :

Find the direction vector $\vec{v}$ and assemble the vector equation $\vec{r}(t)$ . Evaluate $\vec{r}(t)$ at $t = 0, 1, 2$ .
Find the parametric equations of the line.
Convert the parametric equations into symmetric equations.

Solution Key:

Compute Direction Vector $\vec{v}$ via the Two-Point Displacement Formula: State the Two-Point Displacement Formula:
$\vec{v} = \vec{AB} = \langle x_B - x_A, y_B - y_A, z_B - z_A \rangle$ Substitute coordinates $A(1, 4, -2)$ and $B(-3, 5, 0)$ : $\vec{v} = \langle -3 - 1, 5 - 4, 0 - (-2) \rangle = \langle -4, 1, 2 \rangle = -4\hat{i} + 1\hat{j} + 2\hat{k}$ Direction components are $a = -4$ , $b = 1$ , and $c = 2$ .
Assemble Vector Equation: State the Vector Line Formula:
$\vec{r}(t) = \vec{r}_0 + t\vec{v}$
Using $A(1, 4, -2)$ as $\vec{r}_0 = \langle 1, 4, -2 \rangle$ :
$\vec{r}(t) = \langle 1, 4, -2 \rangle + t\langle -4, 1, 2 \rangle = \langle 1 - 4t, 4 + t, -2 + 2t \rangle$
- At $t = 0$ : $\vec{r}(0) = \langle 1, 4, -2 \rangle$ (Point $A$ )
- At $t = 1$ : $\vec{r}(1) = \langle 1 - 4(1), 4 + 1, -2 + 2(1) \rangle = \langle -3, 5, 0 \rangle$ (Point $B$ )
- At $t = 2$ : $\vec{r}(2) = \langle 1 - 4(2), 4 + 2, -2 + 2(2) \rangle = \langle -7, 6, 2 \rangle$
Parametric Equations:
$x = 1 - 4t, \quad y = 4 + t, \quad z = -2 + 2t$
Symmetric Equations: Isolate $t$ :
$t = \frac{x - 1}{-4} = \frac{y - 4}{1} = \frac{z + 2}{2} \implies \frac{x - 1}{-4} = y - 4 = \frac{z + 2}{2}$

Show your steps and calculations here.

Practice Problem 3

Find the parametric and symmetric equations of the line passing through point $P(3, -2, 1)$ that is parallel to the line given by $\frac{x - 1}{2} = \frac{y + 3}{-4} = z - 5$ . Then write the vector-valued function representing the finite line segment between $t = 0$ and $t = 3$ .

Solution Key:

Extract Direction Vector $\vec{v}$ : From the given symmetric line $\frac{x-1}{2} = \frac{y+3}{-4} = \frac{z-5}{1}$ , read the denominator direction components: $\vec{v} = \langle a, b, c \rangle = \langle 2, -4, 1 \rangle = 2\hat{i} - 4\hat{j} + 1\hat{k}$
Construct Parametric Equations through $P(3, -2, 1)$ : State the Parametric Line Equations: $x = x_0 + at, \quad y = y_0 + bt, \quad z = z_0 + ct$ Substitute $(x_0, y_0, z_0) = (3, -2, 1)$ and $(a, b, c) = (2, -4, 1)$ : $x = 3 + 2t, \quad y = -2 - 4t, \quad z = 1 + t$
Construct Symmetric Equations: State the Symmetric Line Formula: $\frac{x - x_0}{a} = \frac{y - y_0}{b} = \frac{z - z_0}{c} \implies \frac{x - 3}{2} = \frac{y + 2}{-4} = z - 1$
Define Finite Line Segment: State the Bounded Vector Line Segment Formula: $\vec{r}(t) = \vec{r}_0 + t\vec{v}, \quad 0 \le t \le t_{\text{max}}$ $\vec{r}(t) = \langle 3 + 2t, -2 - 4t, 1 + t \rangle, \quad 0 \le t \le 3$

Show your steps and calculations here.

Regroup 1

Emphasize that direction vector $\vec{v} = \langle a, b, c \rangle = a\hat{i} + b\hat{j} + c\hat{k}$ gives the trajectory rates, while initial position vector $\vec{r}_0$ anchors the line in space.
Reiterate that zero components in $\vec{v}$ mean a coordinate is locked to a constant value ( $b = 0 \implies y = y_0$ ), representing a vertical planar constraint.
Reinforce that symmetric equations represent a line as the simultaneous intersection of two planes in space.

Skill Block 2

Worked Example A: Point-to-Line Distance from Symmetric Equations

Find the exact distance between the point $M = (1, 1, 3)$ and the line:

\frac{x - 3}{4} = \frac{y + 1}{2} = z - 3

Worked Solution:

Extract Line Reference Point $P$ and Direction Vector $\vec{v}$ : From the symmetric form $\frac{x-x_0}{a} = \frac{y-y_0}{b} = \frac{z-z_0}{c}$ :
- Point on the line: $P = (3, -1, 3)$ (with $\vec{r}_0 = \langle 3, -1, 3 \rangle$ )
- Direction vector: $\vec{v} = \langle 4, 2, 1 \rangle = 4\hat{i} + 2\hat{j} + 1\hat{k}$
Construct Displacement Vector $\vec{PM}$ : State the Displacement Vector Formula:
$\vec{PM} = \langle x_M - x_P, y_M - y_P, z_M - z_P \rangle$ Substitute external point $M(1, 1, 3)$ and line point $P(3, -1, 3)$ : $\vec{PM} = \langle 1 - 3, 1 - (-1), 3 - 3 \rangle = \langle -2, 2, 0 \rangle = -2\hat{i} + 2\hat{j} + 0\hat{k}$
Compute Cross Product $\vec{PM} \times \vec{v}$ : State the Cross Product Matrix Determinant Formula:
$\vec{PM} \times \vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ u_x & u_y & u_z \\ v_x & v_y & v_z \end{vmatrix}$ (This formula is defined in Cross Product.)

Evaluate the determinant:
$\vec{PM} \times \vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -2 & 2 & 0 \\ 4 & 2 & 1 \end{vmatrix}$ $\vec{PM} \times \vec{v} = \hat{i}(2(1) - 0(2)) - \hat{j}((-2)(1) - 0(4)) + \hat{k}((-2)(2) - 2(4))$ $\vec{PM} \times \vec{v} = 2\hat{i} - (-2)\hat{j} + (-4 - 8)\hat{k} = 2\hat{i} + 2\hat{j} - 12\hat{k} = \langle 2, 2, -12 \rangle$
Compute Vector Magnitudes: State the Vector Magnitude Formula $\|\vec{w}\| = \sqrt{w_x^2 + w_y^2 + w_z^2}$ :
$\|\vec{PM} \times \vec{v}\| = \sqrt{2^2 + 2^2 + (-12)^2} = \sqrt{4 + 4 + 144} = \sqrt{152} = 2\sqrt{38}$ $\|\vec{v}\| = \sqrt{4^2 + 2^2 + 1^2} = \sqrt{16 + 4 + 1} = \sqrt{21}$
Calculate Perpendicular Distance $d$ via the Point-to-Line Distance Formula: State the Point-to-Line Distance Formula:
$d = \frac{\|\vec{PM} \times \vec{v}\|}{\|\vec{v}\|}$ (This formula and its variables are defined in Distance Between a Point and a Line in Space.)

Substitute the evaluated magnitudes:
$d = \frac{2\sqrt{38}}{\sqrt{21}} = \frac{2\sqrt{798}}{21} \approx 2.69\text{ units}$

Practice Problem 4: Concept & Reading Check

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

(a) Under the Conceptual Framework section, state the Point-to-Line Distance Formula and explain how it is derived from the parallelogram area formula ( $\text{Area} = \text{Base} \times \text{Height}$ ).
(b) In the distance formula, what is the geometric difference between vector $\vec{PM}$ and vector $\vec{v}$ ?
(c) Review the linked engineering application Robotics & Obstacle Avoidance (Arm Clearance Analysis). How do control engineers use this distance calculation to verify safety clearance between a robotic arm and a static power terminal box?

Solution Key:

(a) As explained in the Conceptual Framework, displacement vector $\vec{PM}$ and direction vector $\vec{v}$ form adjacent sides of a parallelogram in $\mathbb{R}^3$ . The magnitude of their cross product $\|\vec{PM} \times \vec{v}\|$ gives the total area of this parallelogram. Since the area is also $\text{Area} = \text{Base} \times \text{Height} = \|\vec{v}\| \cdot d$ , dividing the area by the base length $\|\vec{v}\|$ yields the Point-to-Line Distance Formula: $d = \frac{\|\vec{PM} \times \vec{v}\|}{\|\vec{v}\|}$
(b) Vector $\vec{v}$ is the direction vector running parallel along the line, whereas $\vec{PM}$ is a displacement vector extending from a known reference point $P$ on the line to the external target point $M$ in space.
(c) In Robotics & Obstacle Avoidance (Arm Clearance Analysis), control engineers calculate the minimum perpendicular clearance ( $4.48\text{ dm} = 44.8\text{ cm}$ ) from the robot forearm centerline to the power terminal box $M(4, 3, 7)$ . This clearance is checked against the physical casing radius and safety threshold to guarantee the arm will not collide with the terminal during automated sweeps.

Show your steps and calculations here.

Practice Problem 5

Find the exact distance between the point $M(0, 3, 6)$ and the line with parametric equations:

x = 1 - t, \quad y = 1 + 2t, \quad z = 5 + 3t

Solution Key:

Extract Line Data: From parametric equations $x = 1 - t, y = 1 + 2t, z = 5 + 3t$ :
- Point on line: $P(1, 1, 5)$
- Direction vector: $\vec{v} = \langle -1, 2, 3 \rangle = -\hat{i} + 2\hat{j} + 3\hat{k}$
Form Displacement Vector $\vec{PM}$ : State the Displacement Vector Formula $\vec{PM} = \langle x_M - x_P, y_M - y_P, z_M - z_P \rangle$ : $\vec{PM} = \langle 0 - 1, 3 - 1, 6 - 5 \rangle = \langle -1, 2, 1 \rangle = -\hat{i} + 2\hat{j} + 1\hat{k}$
Compute Cross Product $\vec{PM} \times \vec{v}$ : $\vec{PM} \times \vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 2 & 1 \\ -1 & 2 & 3 \end{vmatrix}$ $\vec{PM} \times \vec{v} = \hat{i}(2(3) - 1(2)) - \hat{j}((-1)(3) - 1(-1)) + \hat{k}((-1)(2) - 2(-1))$ $\vec{PM} \times \vec{v} = \hat{i}(6 - 2) - \hat{j}(-3 + 1) + \hat{k}(-2 + 2) = 4\hat{i} + 2\hat{j} + 0\hat{k} = \langle 4, 2, 0 \rangle$
Compute Vector Magnitudes: $\|\vec{PM} \times \vec{v}\| = \sqrt{4^2 + 2^2 + 0^2} = \sqrt{16 + 4 + 0} = \sqrt{20} = 2\sqrt{5}$ $\|\vec{v}\| = \sqrt{(-1)^2 + 2^2 + 3^2} = \sqrt{1 + 4 + 9} = \sqrt{14}$
Calculate Perpendicular Distance via Point-to-Line Distance Formula: State the Point-to-Line Distance Formula: $d = \frac{\|\vec{PM} \times \vec{v}\|}{\|\vec{v}\|}$ Substitute magnitudes: $d = \frac{\sqrt{20}}{\sqrt{14}} = \sqrt{\frac{20}{14}} = \sqrt{\frac{10}{7}} = \frac{\sqrt{70}}{7} \approx 1.20\text{ units}$

Show your steps and calculations here.

Practice Problem 6

Calculate the exact shortest distance from the coordinate origin $(0, 0, 0)$ to the line passing through points $P(2, -1, 3)$ and $Q(4, 1, 2)$ .

Solution Key:

Find Direction Vector $\vec{v} = \vec{PQ}$ via Two-Point Displacement Formula: $\vec{v} = \langle 4 - 2, 1 - (-1), 2 - 3 \rangle = \langle 2, 2, -1 \rangle = 2\hat{i} + 2\hat{j} - 1\hat{k}$
Form Displacement Vector $\vec{PM}$ from $P(2, -1, 3)$ to Origin $M(0, 0, 0)$ : $\vec{PM} = \langle 0 - 2, 0 - (-1), 0 - 3 \rangle = \langle -2, 1, -3 \rangle = -2\hat{i} + 1\hat{j} - 3\hat{k}$
Compute Cross Product $\vec{PM} \times \vec{v}$ : $\vec{PM} \times \vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -2 & 1 & -3 \\ 2 & 2 & -1 \end{vmatrix}$ $\vec{PM} \times \vec{v} = \hat{i}(1(-1) - (-3)(2)) - \hat{j}((-2)(-1) - (-3)(2)) + \hat{k}((-2)(2) - 1(2))$ $\vec{PM} \times \vec{v} = \hat{i}(-1 + 6) - \hat{j}(2 + 6) + \hat{k}(-4 - 2) = 5\hat{i} - 8\hat{j} - 6\hat{k} = \langle 5, -8, -6 \rangle$
Compute Vector Magnitudes: $\|\vec{PM} \times \vec{v}\| = \sqrt{5^2 + (-8)^2 + (-6)^2} = \sqrt{25 + 64 + 36} = \sqrt{125} = 5\sqrt{5}$ $\|\vec{v}\| = \sqrt{2^2 + 2^2 + (-1)^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3$
Calculate Perpendicular Distance via Point-to-Line Distance Formula: State the Point-to-Line Distance Formula: $d = \frac{\|\vec{PM} \times \vec{v}\|}{\|\vec{v}\|}$ Substitute magnitudes: $d = \frac{5\sqrt{5}}{3} \approx 3.73\text{ units}$

Show your steps and calculations here.

Regroup 2

Review common algebraic slips in cross product determinant expansions (especially remembering the negative sign on the middle $\hat{j}$ term).
Highlight that any arbitrary point $P$ on the line can be chosen for $\vec{PM}$ ; the resulting distance $d$ will be identical because the base-height ratio remains invariant along the line.
Remind students that distance is always a non-negative scalar quantity.

Skill Block 3

Worked Example A: Intersecting vs. Skew Classification

Determine whether the lines $L_1$ and $L_2$ are equal, parallel, skew, or intersecting:

L_1: x = 5t, \quad y = -13 - 6t, \quad z = 19 - t

L_2: x = -1 - 6s, \quad y = -12 - 12s, \quad z = 19 - 18s

Worked Solution:

Compare Direction Vectors via the Parallelism Ratio Test: State the Direction Vector Parallelism Condition:
$\vec{u} = k\vec{v} \iff \frac{u_x}{v_x} = \frac{u_y}{v_y} = \frac{u_z}{v_z}$
- Direction vector of $L_1$ : $\vec{u} = \langle 5, -6, -1 \rangle$
- Direction vector of $L_2$ : $\vec{v} = \langle -6, -12, -18 \rangle$
- Check component ratios: $\frac{-6}{5} \neq \frac{-12}{-6} = 2$ .
- Since component ratios are not equal, direction vectors are not parallel ( $\vec{u} \neq k\vec{v}$ ). The lines must be either Intersecting or Skew.
Set Up System of Equations with Independent Parameters ( $t$ and $s$ ): Equate corresponding $x, y, z$ coordinates:
1. $5t = -1 - 6s \implies 5t + 6s = -1$
2. $-13 - 6t = -12 - 12s \implies -6t + 12s = 1$
3. $19 - t = 19 - 18s \implies -t + 18s = 0 \implies t = 18s$
Solve the Linear System: Using Equation (3) $t = 18s$ , substitute into Equation (1):
$5(18s) + 6s = -1 \implies 90s + 6s = -1 \implies 96s = -1 \implies s = -\frac{1}{96}$ Compute parameter $t$ : $t = 18\left(-\frac{1}{96}\right) = -\frac{3}{16}$ (Note: Alternatively, writing Equations 1 and 2 in matrix form $\begin{bmatrix} 5 & 6 \\ -6 & 12 \end{bmatrix}\begin{bmatrix} t \\ s \end{bmatrix} = \begin{bmatrix} -1 \\ 1 \end{bmatrix}$ and solving $\vec{x} = A^{-1}\vec{b}$ via a matrix calculator yields $\begin{bmatrix} t \\ s \end{bmatrix} = \begin{bmatrix} -3/16 \\ -1/96 \end{bmatrix}$ , as detailed in Solving Systems of Linear Equations with Matrix Form.)
Perform Consistency Check in Equation (2): Substitute $t = -\frac{3}{16}$ and $s = -\frac{1}{96}$ into Equation (2):
- $\text{Left Side (LS): } -13 - 6\left(-\frac{3}{16}\right) = -13 + \frac{9}{8} = -\frac{95}{8} = -11.875$
- $\text{Right Side (RS): } -12 - 12\left(-\frac{1}{96}\right) = -12 + \frac{1}{8} = -\frac{95}{8} = -11.875$
Because $\text{LS} = \text{RS}$ , the system is consistent. The lines share a unique common point and are Intersecting.
Find Intersection Point Coordinates: Substitute $t = -\frac{3}{16}$ into $L_1$ :
$x = 5\left(-\frac{3}{16}\right) = -\frac{15}{16}, \quad y = -13 - 6\left(-\frac{3}{16}\right) = -\frac{95}{8}, \quad z = 19 - \left(-\frac{3}{16}\right) = \frac{307}{16}$ The lines intersect at the point $\left( -\frac{15}{16}, -\frac{95}{8}, \frac{307}{16} \right)$ .

Practice Problem 7: Concept & Reading Check

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

(a) Under the Conceptual Framework section, complete the 4-way classification table: what are the conditions on direction vectors and common points for lines to be classified as Skew vs. Parallel?
(b) In Worked Example A, why is it essential to use different parameter letters ( $t$ and $s$ ) when setting the parametric equations equal? What physical error occurs if both lines use parameter $t$ ?
(c) Review the linked engineering application Aerospace Engineering & Air Traffic Control (Flight Path Conflict). Why does proving flight paths are skew guarantee the aircraft centerlines will not collide?

Solution Key:

(a) According to the Conceptual Framework, lines are Skew when their direction vectors are non-parallel ( $\vec{u} \neq k\vec{v}$ ) and they share zero common points (inconsistent system). Lines are Parallel (Distinct) when their direction vectors are parallel ( $\vec{u} = k\vec{v}$ ) and they share zero common points.
(b) As demonstrated in Worked Example A, two physical objects can travel along paths that cross in space without colliding if they pass through the intersection point at different times. Assigning separate parameters ( $t$ and $s$ ) tests whether the geometric lines cross at any combination of parameter values. Using the single parameter $t$ would test the much narrower condition of a simultaneous collision.
(c) In Aerospace Engineering & Air Traffic Control (Flight Path Conflict), classifying the flight paths as skew confirms that the linear trajectories exist in non-intersecting planes at different altitudes/headings, mathematically proving that the geometric centerlines of the two aircraft will never intersect in space.

Show your steps and calculations here.

Practice Problem 8

Given Line A running through points $A_1 = (2, 0, 3)$ and $A_2 = (4, 7, 8)$ and Line B running through points $B_1 = (-3, -1, 4)$ and $B_2 = (-1, 5, 9)$ , determine if Line A and Line B are equal, parallel, skew, or intersecting.

Solution Key:

Compute and Compare Direction Vectors: State the Two-Point Displacement Formula $\vec{v} = \langle x_2 - x_1, y_2 - y_1, z_2 - z_1 \rangle$ :
- $\vec{u} = \vec{A_1A_2} = \langle 4 - 2, 7 - 0, 8 - 3 \rangle = \langle 2, 7, 5 \rangle$
- $\vec{v} = \vec{B_1B_2} = \langle -1 - (-3), 5 - (-1), 9 - 4 \rangle = \langle 2, 6, 5 \rangle$
Compare component ratios:
$\frac{u_x}{v_x} = \frac{2}{2} = 1, \quad \frac{u_y}{v_y} = \frac{7}{6}, \quad \frac{u_z}{v_z} = \frac{5}{5} = 1$
Since $1 \neq \frac{7}{6}$ , the direction vectors are not parallel ( $\vec{u} \neq k\vec{v}$ ).
Write Parametric Equations with Distinct Parameters ( $t, s$ ):
- Line A: $x = 2 + 2t, \quad y = 7t, \quad z = 3 + 5t$
- Line B: $x = -3 + 2s, \quad y = -1 + 6s, \quad z = 4 + 5s$
Set Up System of Equations:
1. $2 + 2t = -3 + 2s \implies 2t - 2s = -5 \implies t - s = -2.5$
2. $7t = -1 + 6s \implies 7t - 6s = -1$
3. $3 + 5t = 4 + 5s \implies 5t - 5s = 1 \implies t - s = 0.2$
Identify Inconsistency: Equation (1) requires $t - s = -2.5$ , whereas Equation (3) requires $t - s = 0.2$ . Because $-2.5 \neq 0.2$ , this linear system contains an internal contradiction and has no simultaneous solution.
Conclusion: Because the direction vectors are not parallel and the lines share no common points, Line A and Line B are Skew.

Show your steps and calculations here.

Practice Problem 9

Determine whether the lines $L_1$ and $L_2$ are equal, parallel but not equal, skew, or intersecting:

L_1: \frac{x - 1}{2} = \frac{y + 2}{4} = \frac{z - 3}{1}

L_2: \frac{x - 5}{4} = \frac{y - 6}{8} = \frac{z - 5}{2}

Solution Key:

Extract and Compare Direction Vectors:
- Direction vector of $L_1$ : $\vec{u} = \langle 2, 4, 1 \rangle$
- Direction vector of $L_2$ : $\vec{v} = \langle 4, 8, 2 \rangle$
- Check component ratios: $\frac{4}{2} = \frac{8}{4} = \frac{2}{1} = 2$ .
- The direction vectors are parallel ( $\vec{v} = 2\vec{u}$ ).
Test for Shared Points via Symmetric Substitution: Extract reference point $P_1(1, -2, 3)$ from $L_1$ and substitute its coordinates into the symmetric equation of $L_2$ :
$\frac{1 - 5}{4} = \frac{-4}{4} = -1$ $\frac{-2 - 6}{8} = \frac{-8}{8} = -1$ $\frac{3 - 5}{2} = \frac{-2}{2} = -1$ Since all three ratios equal $-1$ , point $P_1(1, -2, 3)$ lies directly on line $L_2$ .
Conclusion: Because their direction vectors are parallel and they share a common point, $L_1$ and $L_2$ represent the exact same line: Equal (Coincident).

Show your steps and calculations here.

Regroup 3

Reinforce that in 3D geometry, non-parallel lines are most often skew rather than intersecting.
Highlight the distinction between Equal lines (parallel vectors + shared point) and Parallel lines (parallel vectors + no shared point).
Emphasize checking system consistency across all three spatial dimensions before declaring an intersection.

Synthesis Wrap-up

Core Summary:
- Vector Equation of a Line: $\vec{r}(t) = \vec{r}_0 + t\vec{v}$ maps 3D spatial paths using an initial point and direction vector $\vec{v} = \langle a, b, c \rangle = a\hat{i} + b\hat{j} + c\hat{k}$ .
- Distance to a Line: $d = \frac{\|\vec{PM} \times \vec{v}\|}{\|\vec{v}\|}$ derives from the cross-product parallelogram area divided by base length.
- Classifying Lines in Space: A 2-step decision process tests direction proportionality ( $\vec{u} = k\vec{v}$ ) and solves a 3-equation linear system with independent parameters ( $t, s$ ) to distinguish Equal, Parallel, Intersecting, and Skew configurations using matrix methods.
Homework Assignment (Knewton Alta):
- Equations of Lines and Planes in Space