Session 04 - Lines and Planes in Space | Tony's Intertwingled Memex

📅 Micro-Lecture

Engineering Context: Designing in 3D

In this session, we investigate the mathematical principles of Lines and Planes in Space and explore how engineers apply these concepts to analyze real-world physical and structural systems.

Segment 1 of 8

Skill Block 1

📚 1. Core Theory 💡 2. Worked Example ✍️ 3. Active Practice

Explanations and derivations of the core concepts.

Worked Example

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

Find the vector equation of the line. Evaluate at a few values of $t$ .
Find the parametric equation of the line.
Convert the parametric equation into a symmetric equation.

💡 Reveal Worked Solution

Here is the step-by-step solution to the problem presented in the image.

Given Data

Point $P$ : $(-4, -6, -2)$
Point $Q$ : $(10, 2, 1)$

1. Vector Equation of the Line

To find the vector equation, we first need a position vector $\vec{r}_0$ for a point on the line and a direction vector $\vec{v}$ that is parallel to the line.

Find the direction vector $\vec{v}$ by finding the vector from $P$ to $Q$ :

\vec{v} = \vec{PQ} = \langle 10 - (-4), 2 - (-6), 1 - (-2) \rangle

\vec{v} = \langle 14, 8, 3 \rangle

Use point $P$ as the initial position vector $\vec{r}_0$ :

\vec{r}_0 = \langle -4, -6, -2 \rangle

Write the vector equation $\vec{r}(t) = \vec{r}_0 + t\vec{v}$ :

\vec{r}(t) = \langle -4, -6, -2 \rangle + t\langle 14, 8, 3 \rangle

Evaluating at a few values of $t$ :

For $t = 0$ :

\vec{r}(0) = \langle -4, -6, -2 \rangle \quad (\text{This is point } P)

For $t = 1$ :

\vec{r}(1) = \langle -4+14, -6+8, -2+3 \rangle = \langle 10, 2, 1 \rangle \quad (\text{This is point } Q)

For $t = -1$ :

\vec{r}(-1) = \langle -4-14, -6-8, -2-3 \rangle = \langle -18, -14, -5 \rangle

2. Parametric Equations of the Line

The parametric equations break the vector equation down into separate components for $x$ , $y$ , and $z$ using the format:

$x = x_0 + at$
$y = y_0 + bt$
$z = z_0 + ct$

Substituting our point $P(-4, -6, -2)$ and direction vector components $\langle 14, 8, 3 \rangle$ :

$x = -4 + 14t$
$y = -6 + 8t$
$z = -2 + 3t$

3. Symmetric Equation of the Line

To convert the parametric equations into a symmetric equation, we solve each individual equation for $t$ and set them equal to each other:

Solve for $t$ in the $x$ equation:

x = -4 + 14t \implies x + 4 = 14t \implies t = \frac{x + 4}{14}

Solve for $t$ in the $y$ equation:

y = -6 + 8t \implies y + 6 = 8t \implies t = \frac{y + 6}{8}

Solve for $t$ in the $z$ equation:

z = -2 + 3t \implies z + 2 = 3t \implies t = \frac{z + 2}{3}

Equating all three expressions for $t$ gives the symmetric equation:

\frac{x + 4}{14} = \frac{y + 6}{8} = \frac{z + 2}{3}

Active Practice Problem

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

Find the vector equation of the line. Evaluate at a few values of $t$ .
Find the Parametric equation of the line.
Convert the parametric equation into a symmetric equation.

Practice Space Solve this problem on paper or tablet.

Refer to the 'Core Theory' tab or review the Worked Example demonstration for this skill.

Answer:

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

Step-by-Step Execution

Here is the complete solution for Problem S1.1 from the activity:

Given Data

Point $A$ : $(1, 4, -2)$
Point $B$ : $(-3, 5, 0)$

1. Vector Equation of the Line

To construct the vector equation, we need an initial position vector $\vec{r}_0$ and a direction vector $\vec{v}$ .

Find the direction vector $\vec{v}$ by finding the displacement vector from $A$ to $B$ :

\vec{v} = \vec{AB} = \langle -3 - 1, \ 5 - 4, \ 0 - (-2) \rangle

\vec{v} = \langle -4, 1, 2 \rangle

Set point $A$ as your initial position vector $\vec{r}_0$ :

\vec{r}_0 = \langle 1, 4, -2 \rangle

Assemble the vector equation $\vec{r}(t) = \vec{r}_0 + t\vec{v}$ :

\vec{r}(t) = \langle 1, 4, -2 \rangle + t\langle -4, 1, 2 \rangle

Evaluating at a few values of $t$ :

For $t = 0$ :

\vec{r}(0) = \langle 1, 4, -2 \rangle \quad (\text{Point } A)

For $t = 1$ :

\vec{r}(1) = \langle 1 - 4, \ 4 + 1, \ -2 + 2 \rangle = \langle -3, 5, 0 \rangle \quad (\text{Point } B)

For $t = 2$ :

\vec{r}(2) = \langle 1 - 8, \ 4 + 2, \ -2 + 4 \rangle = \langle -7, 6, 2 \rangle

2. Parametric Equations of the Line

By breaking the vector equation down into its distinct scalar components ( $x$ , $y$ , and $z$ ) using the format $x = x_0 + at$ , $y = y_0 + bt$ , and $z = z_0 + ct$ , we get:

$x = 1 - 4t$
$y = 4 + t$
$z = -2 + 2t$

3. Symmetric Equation of the Line

To find the symmetric equation, we isolate $t$ in each parametric component and set them equal to one another:

Solve for $t$ in $x$ :

x = 1 - 4t \implies x - 1 = -4t \implies t = \frac{x - 1}{-4}

Solve for $t$ in $y$ :

y = 4 + t \implies t = \frac{y - 4}{1}

Solve for $t$ in $z$ :

z = -2 + 2t \implies z + 2 = 2t \implies t = \frac{z + 2}{2}

Equating these gives the symmetric equation:

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

Segment 2 of 8

📅 Regroup 1

Review common misconceptions and clarify key notations.

Segment 3 of 8

Skill Block 2

📚 1. Core Theory 💡 2. Worked Example ✍️ 3. Active Practice

Explanations and derivations of the core concepts.

Worked Example

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

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

💡 Reveal Worked Solution

Here is the step-by-step solution for Skill 2 Demo using the distance formula provided in the activity.

Given Data

Point $M$ (anywhere in space): $(1, 1, 3)$
Symmetric equation of the line: $\frac{x-3}{4} = \frac{y+1}{2} = z-3$

1. Extract Information from the Line

From the symmetric equation form $\frac{x-x_0}{a} = \frac{y-y_0}{b} = \frac{z-z_0}{c}$ , we can extract:

A point $P$ on the line: $(3, -1, 3)$
The direction vector $\vec{v}$ along the line: $\langle 4, 2, 1 \rangle$ (note that $z-3$ is equivalent to $\frac{z-3}{1}$ )

2. Find the Vector $\vec{PM}$

Now, find the vector pointing from the line point $P$ to the external point $M$ :

\vec{PM} = \langle x_M - x_P, \ y_M - y_P, \ z_M - z_P \rangle

\vec{PM} = \langle 1 - 3, \ 1 - (-1), \ 3 - 3 \rangle = \langle -2, 2, 0 \rangle

3. Compute the Cross Product $\vec{PM} \times \vec{v}$

The formula requires the cross product of $\vec{PM}$ and the direction vector $\vec{v}$ :

\vec{PM} \times \vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -2 & 2 & 0 \\ 4 & 2 & 1 \end{vmatrix}

Expand the determinant by the top row:

\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} = \hat{i}(2) - \hat{j}(-2) + \hat{k}(-4 - 8)

\vec{PM} \times \vec{v} = \langle 2, 2, -12 \rangle

4. Calculate the Magnitudes

Now find the magnitudes (lengths) of both the cross product vector and the direction vector $\vec{v}$ :

Magnitude of $\vec{PM} \times \vec{v}$ :

||\vec{PM} \times \vec{v}|| = \sqrt{2^2 + 2^2 + (-12)^2} = \sqrt{4 + 4 + 144} = \sqrt{152}

\sqrt{152} = \sqrt{4 \times 38} = 2\sqrt{38}

Magnitude of $\vec{v}$ :

||\vec{v}|| = \sqrt{4^2 + 2^2 + 1^2} = \sqrt{16 + 4 + 1} = \sqrt{21}

5. Calculate the Distance $d$

Substitute these magnitudes into the distance formula $d = \frac{||\vec{PM} \times \vec{v}||}{||\vec{v}||}$ :

d = \frac{2\sqrt{38}}{\sqrt{21}}

To write it under a single radical or rationalize it:

d = 2\sqrt{\frac{38}{21}} = \frac{2\sqrt{798}}{21} \approx 2.69

The exact distance between the point and the line is $\frac{2\sqrt{798}}{21}$ (or approximately $2.69$ units).

Active Practice Problem

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

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

Practice Space Solve this problem on paper or tablet.

Refer to the 'Core Theory' tab or review the Worked Example demonstration for this skill.

Answer: The exact distance between the point and the line is $\frac{\sqrt{70}}{7}$ (or approximately $1.20$ units).

Step-by-Step Execution

Given Data

Point $M$ (external point): $(0, 3, 6)$
Parametric equations of the line: $x = 1 - t$ , $y = 1 + 2t$ , $z = 5 + 3t$

1. Extract Information from the Line

From the parametric equations form $x = x_0 + at$ , $y = y_0 + bt$ , and $z = z_0 + ct$ , we can easily read off a point on the line and its direction vector:

A point $P$ on the line (by looking at the constant terms where $t = 0$ ):

P = (1, 1, 5)

The direction vector $\vec{v}$ along the line (the coefficients of $t$ ):

\vec{v} = \langle -1, 2, 3 \rangle

2. Find the Vector $\vec{PM}$

Now, find the vector pointing from the line's point $P$ to the external point $M$ :

\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

3. Compute the Cross Product $\vec{PM} \times \vec{v}$

Next, find the cross product required by the formula:

\vec{PM} \times \vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 2 & 1 \\ -1 & 2 & 3 \end{vmatrix}

Expand the determinant along the top row:

\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)

\vec{PM} \times \vec{v} = \hat{i}(4) - \hat{j}(-2) + \hat{k}(0)

\vec{PM} \times \vec{v} = \langle 4, 2, 0 \rangle

4. Calculate the Magnitudes

Now compute the magnitudes (lengths) of our cross product vector and the line's direction vector $\vec{v}$ :

Magnitude of $\vec{PM} \times \vec{v}$ :

||\vec{PM} \times \vec{v}|| = \sqrt{4^2 + 2^2 + 0^2} = \sqrt{16 + 4 + 0} = \sqrt{20} = 2\sqrt{5}

Magnitude of $\vec{v}$ :

||\vec{v}|| = \sqrt{(-1)^2 + 2^2 + 3^2} = \sqrt{1 + 4 + 9} = \sqrt{14}

5. Calculate the Distance $d$

Substitute these values into the distance formula $d = \frac{||\vec{PM} \times \vec{v}||}{||\vec{v}||}$ :

d = \frac{\sqrt{20}}{\sqrt{14}} = \sqrt{\frac{20}{14}} = \sqrt{\frac{10}{7}}

To rationalize the denominator:

d = \frac{\sqrt{10}}{\sqrt{7}} \cdot \frac{\sqrt{7}}{\sqrt{7}} = \frac{\sqrt{70}}{7} \approx 1.20

The exact distance between the point and the line is $\frac{\sqrt{70}}{7}$ (or approximately $1.20$ units).

Segment 4 of 8

📅 Regroup 2

Reflect on the physical modeling applications and mathematical setups.

Segment 5 of 8

📅 Regroup 3

Verify calculations and mathematical reasoning.

Segment 7 of 8

📅 Synthesis Wrap-up

Core takeaways from Session 04 and overview of homework homework assignment: Equations of Lines and Planes in Space.

Segment 8 of 8

Lines and Planes in Space

Draft Handout - Content Not Finalized

📅 Micro-Lecture

Engineering Context: Designing in 3D

Skill Block 1

Worked Example

Given Data

1. Vector Equation of the Line

2. Parametric Equations of the Line

3. Symmetric Equation of the Line

Active Practice Problem

Step-by-Step Execution

Given Data

1. Vector Equation of the Line

2. Parametric Equations of the Line

3. Symmetric Equation of the Line

📅 Regroup 1

Skill Block 2

Worked Example

Given Data

1. Extract Information from the Line

2. Find the Vector PM⃗\vec{PM}PM

3. Compute the Cross Product PM⃗×v⃗\vec{PM} \times \vec{v}PM×v

4. Calculate the Magnitudes

5. Calculate the Distance ddd

Active Practice Problem

Step-by-Step Execution

Given Data

1. Extract Information from the Line

2. Find the Vector PM⃗\vec{PM}PM

3. Compute the Cross Product PM⃗×v⃗\vec{PM} \times \vec{v}PM×v

4. Calculate the Magnitudes

5. Calculate the Distance ddd

📅 Regroup 2

Skill Block 3

Worked Example

Given Data

Step 1: Extract and Compare Direction Vectors

Step 2: Check for an Intersection Point

Solve the system using equations (1) and (3):

Test the values in the remaining equation (2):

Conclusion

Active Practice Problem

Step-by-Step Execution

Given Data

Step 1: Find and Compare Direction Vectors

Step 2: Write Parametric Equations for Both Lines

Step 3: Check for an Intersection Point

Solve the system using equations (1) and (3):

Notice the contradiction:

Conclusion

📅 Regroup 3

📅 Synthesis Wrap-up

2. Find the Vector $\vec{PM}$

3. Compute the Cross Product $\vec{PM} \times \vec{v}$

5. Calculate the Distance $d$

2. Find the Vector $\vec{PM}$

3. Compute the Cross Product $\vec{PM} \times \vec{v}$

5. Calculate the Distance $d$