Session 03 - Dot and Cross Product | Tony's Intertwingled Memex

📅 Micro-Lecture

Engineering Context: Quantifying Force

In this session, we investigate the mathematical principles of Dot and Cross Product 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

Find the dot product of $\vec{u} = \langle 6, 7 \rangle$ and $\vec{v} = \langle 7, -6 \rangle$ .

💡 Reveal Worked Solution

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

Problem Statement

Find the dot product of $\vec{u} = \langle 6, 7 \rangle$ and $\vec{v} = \langle 7, -6 \rangle$ .

Formula

For two 2D vectors $\vec{u} = \langle u_x, u_y \rangle$ and $\vec{v} = \langle v_x, v_y \rangle$ , the dot product is calculated as:

\vec{u} \cdot \vec{v} = (u_x \cdot v_x) + (u_y \cdot v_y)

Step-by-Step Calculation

Identify the components of each vector:

$u_x = 6$ , $u_y = 7$
$v_x = 7$ , $v_y = -6$

Substitute the values into the formula:

\vec{u} \cdot \vec{v} = (6 \cdot 7) + (7 \cdot -6)

Simplify the terms:

\vec{u} \cdot \vec{v} = 42 + (-42)

Combine the results:

\vec{u} \cdot \vec{v} = 0

Final Answer

The dot product is $0$ .

(Note: Since the dot product is exactly zero, these two vectors are orthogonal or perpendicular to each other).

Active Practice Problem

Find the dot product of $\vec{u} = 3\hat{i} - 2\hat{j}$ and $\vec{v} = 4\hat{i} - 5\hat{j}$ .

Workspace

Solution

Here is the step-by-step solution to the problem in the image:

Problem Statement

Find the dot product of $\vec{u} = 3\hat{i} - 2\hat{j}$ and $\vec{v} = 4\hat{i} - 5\hat{j}$ .

Formula

When vectors are written in unit vector notation ( $\hat{i}, \hat{j}$ ), the dot product multiplies the corresponding components together:

\vec{u} \cdot \vec{v} = (u_x \cdot v_x) + (u_y \cdot v_y)

Step-by-Step Calculation

Identify the components of each vector:

$\vec{u}$ : $u_x = 3$ , $u_y = -2$
$\vec{v}$ : $v_x = 4$ , $v_y = -5$

Substitute the values into the formula:

\vec{u} \cdot \vec{v} = (3 \cdot 4) + (-2 \cdot -5)

Multiply the components:

For the $\hat{i}$ components: $3 \cdot 4 = 12$
For the $\hat{j}$ components: $-2 \cdot -5 = 10$

Add the results together:

\vec{u} \cdot \vec{v} = 12 + 10

\vec{u} \cdot \vec{v} = 22

Final Answer

The dot product is $22$ .

Extra Practice & Extensions

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

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

Calculate $\vec{p} \times \vec{q}$ using a determinant if $\vec{p} = 2\hat{i} + 4\hat{j}$ and $\vec{q} = -4\hat{i} - 5\hat{j}$ .

💡 Reveal Worked Solution

Here is the step-by-step solution to the problem in the image:

Problem Statement

Calculate $\vec{p} \times \vec{q}$ using a determinant if $\vec{p} = 2\hat{i} + 4\hat{j}$ and $\vec{q} = -4\hat{i} - 5\hat{j}$ .

Formula

Even though these vectors are in 2D ( $\hat{i}$ and $\hat{j}$ components), the cross product requires a 3D space because the resulting vector points perpendicular to the $xy$ -plane (along the $\hat{k}$ axis). We treat the $\hat{k}$ components as $0$ :

\vec{p} \times \vec{q} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ p_x & p_y & p_z \\ q_x & q_y & q_z \end{vmatrix}

Step-by-Step Calculation

Set up the matrix determinant with $p_z = 0$ and $q_z = 0$ :

\vec{p} \times \vec{q} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 4 & 0 \\ -4 & -5 & 0 \end{vmatrix}

Expand the determinant along the top row:

\vec{p} \times \vec{q} = \hat{i}\begin{vmatrix} 4 & 0 \\ -5 & 0 \end{vmatrix} - \hat{j}\begin{vmatrix} 2 & 0 \\ -4 & 0 \end{vmatrix} + \hat{k}\begin{vmatrix} 2 & 4 \\ -4 & -5 \end{vmatrix}

Evaluate each 2x2 minor:

$\hat{i}$ component: $(4 \cdot 0) - (0 \cdot -5) = 0$
$\hat{j}$ component: $(2 \cdot 0) - (0 \cdot -4) = 0$
$\hat{k}$ component: $(2 \cdot -5) - (4 \cdot -4) = -10 - (-16) = -10 + 16 = 6$

Combine the components:

\vec{p} \times \vec{q} = 0\hat{i} - 0\hat{j} + 6\hat{k}

\vec{p} \times \vec{q} = 6\hat{k}

Final Answer

The cross product is $6\hat{k}$ (or $\langle 0, 0, 6 \rangle$ ).

Active Practice Problem

Calculate $\vec{r} \times \vec{F}$ where $\vec{r} = \langle 3, 2 \rangle$ and $\vec{F} = \langle -2, 5 \rangle$ .

Workspace

Solution

Here is the step-by-step solution to the problem in the image:

Problem Statement

Calculate $\vec{r} \times \vec{F}$ where $\vec{r} = \langle 3, 2 \rangle$ and $\vec{F} = \langle -2, 5 \rangle$ .

Formula

For 2D vectors lying in the $xy$ -plane, we extend them into 3D by setting their $z$ -components to $0$ ( $r_z = 0$ and $F_z = 0$ ). The cross product is computed using a $3 \times 3$ determinant:

\vec{r} \times \vec{F} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ r_x & r_y & r_z \\ F_x & F_y & F_z \end{vmatrix}

Step-by-Step Calculation

Set up the determinant matrix:

\vec{r} \times \vec{F} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 2 & 0 \\ -2 & 5 & 0 \end{vmatrix}

Expand the determinant along the top row:

\vec{r} \times \vec{F} = \hat{i}\begin{vmatrix} 2 & 0 \\ 5 & 0 \end{vmatrix} - \hat{j}\begin{vmatrix} 3 & 0 \\ -2 & 0 \end{vmatrix} + \hat{k}\begin{vmatrix} 3 & 2 \\ -2 & 5 \end{vmatrix}

Evaluate the 2x2 minors:

$\hat{i}$ component: $(2 \cdot 0) - (0 \cdot 5) = 0$
$\hat{j}$ component: $(3 \cdot 0) - (0 \cdot -2) = 0$
$\hat{k}$ component: $(3 \cdot 5) - (2 \cdot -2) = 15 - (-4) = 15 + 4 = 19$

Combine the components:

\vec{r} \times \vec{F} = 0\hat{i} - 0\hat{j} + 19\hat{k}

\vec{r} \times \vec{F} = 19\hat{k}

Final Answer

The cross product is $19\hat{k}$ (or written in component form as $\langle 0, 0, 19 \rangle$ ).

Extra Practice & Extensions

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

Segment 4 of 8

📅 Regroup 2

Reflect on the physical modeling applications and mathematical setups.

Segment 5 of 8

Skill Block 3

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

Explanations and derivations of the core concepts.

Worked Example

Find $\vec{a} \cdot \vec{b}$ if $|\vec{a}| = 16\text{ N}$ , $|\vec{b}| = 5\text{ m}$ , and the angle between $\vec{a}$ and $\vec{b}$ is $77^\circ$ . Express the answer rounded to two decimal places.

💡 Reveal Worked Solution

Here is the step-by-step solution to the problem in the image:

Problem Statement

Find $\vec{a} \cdot \vec{b}$ if $|\vec{a}| = 16\text{ N}$ , $|\vec{b}| = 5\text{ m}$ , and the angle between $\vec{a}$ and $\vec{b}$ is $77^\circ$ . Express the answer rounded to two decimal places.

Formula

When given the magnitudes of two vectors and the angle $\theta$ between them, the geometric definition of the dot product is used:

\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta

Step-by-Step Calculation

Identify the given values:

Magnitude of $\vec{a}$ : $|\vec{a}| = 16$
Magnitude of $\vec{b}$ : $|\vec{b}| = 5$
Angle: $\theta = 77^\circ$

Substitute the values into the formula:

\vec{a} \cdot \vec{b} = 16 \cdot 5 \cdot \cos(77^\circ)

Simplify the multiplication:

16 \cdot 5 = 80

\vec{a} \cdot \vec{b} = 80 \cdot \cos(77^\circ)

Calculate the cosine value (ensure your calculator is in Degree mode):

\cos(77^\circ) \approx 0.224951

Multiply to find the dot product:

\vec{a} \cdot \vec{b} \approx 80 \cdot 0.224951 = 17.99608

Round to two decimal places:

\vec{a} \cdot \vec{b} \approx 18.00

Final Answer

The dot product is $18.00$ (or $18.00\text{ J}$ / $\text{N} \cdot \text{m}$ if keeping track of the physical work units).

Active Practice Problem

Find the work (in Joules) done by the force $\vec{F} = \langle 4, -4, 4 \rangle$ (in Newtons) acting on a particle as it moves from point $P(-7, -2, 5)$ to point $Q(8, -7, -5)$ along a straight line. The coordinates of $P$ and $Q$ are measured in meters.

Units:

$1\text{ N} = 1\text{ kg} \cdot \text{m/s}^2$
$1\text{ J} = 1\text{ N} \cdot \text{m}$

Workspace

Solution

Here is the step-by-step solution to the problem in the image:

Problem Statement

Find the work (in Joules) done by the force $\vec{F} = \langle 4, -4, 4 \rangle$ (in Newtons) acting on a particle as it moves from point $P(-7, -2, 5)$ to point $Q(8, -7, -5)$ along a straight line. The coordinates of $P$ and $Q$ are measured in meters.

Formula

Work ( $W$ ) done by a constant force vector $\vec{F}$ over a displacement vector $\vec{d}$ is given by the dot product:

W = \vec{F} \cdot \vec{d}

Where the displacement vector $\vec{d}$ from a starting point $P$ to an ending point $Q$ is calculated as:

\vec{d} = \vec{PQ} = \langle x_Q - x_P, \ y_Q - y_P, \ z_Q - z_P \rangle

Step-by-Step Calculation

Step 1: Find the displacement vector $\vec{d}$

Subtract the coordinates of the initial point $P(-7, -2, 5)$ from the final point $Q(8, -7, -5)$ :

$d_x = 8 - (-7) = 8 + 7 = 15$
$d_y = -7 - (-2) = -7 + 2 = -5$
$d_z = -5 - 5 = -10$

So, the displacement vector is:

\vec{d} = \langle 15, -5, -10 \rangle\text{ m}

Step 2: Calculate the work using the dot product

Now, take the dot product of the force vector $\vec{F} = \langle 4, -4, 4 \rangle$ and the displacement vector $\vec{d} = \langle 15, -5, -10 \rangle$ :

W = \vec{F} \cdot \vec{d} = (4 \cdot 15) + (-4 \cdot -5) + (4 \cdot -10)

Step 3: Simplify the terms

$4 \cdot 15 = 60$
$-4 \cdot -5 = 20$
$4 \cdot -10 = -40$

Combine the values:

W = 60 + 20 - 40

W = 40\text{ J}

Final Answer

The work done by the force is $40\text{ Joules}$ .

Extra Practice & Extensions

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

Segment 6 of 8

📅 Regroup 3

Verify calculations and mathematical reasoning.

Segment 7 of 8

📅 Synthesis Wrap-up

Core takeaways from Session 03 and overview of homework homework assignment: The Dot Product The Cross Product.

Segment 8 of 8

Dot and Cross Product

Draft Handout - Content Not Finalized

📅 Micro-Lecture

Engineering Context: Quantifying Force

Skill Block 1

Worked Example

Problem Statement

Formula

Step-by-Step Calculation

Final Answer

Active Practice Problem

Workspace

Solution

Problem Statement

Formula

Step-by-Step Calculation

Final Answer

Extra Practice & Extensions

📅 Regroup 1

Skill Block 2

Worked Example

Problem Statement

Formula

Step-by-Step Calculation

Final Answer

Active Practice Problem

Workspace

Solution

Problem Statement

Formula

Step-by-Step Calculation

Final Answer

Extra Practice & Extensions

📅 Regroup 2

Skill Block 3

Worked Example

Problem Statement

Formula

Step-by-Step Calculation

Final Answer

Active Practice Problem

Workspace

Solution

Problem Statement

Formula

Step-by-Step Calculation

Step 1: Find the displacement vector d⃗\vec{d}d

Step 2: Calculate the work using the dot product

Step 3: Simplify the terms

Final Answer

Extra Practice & Extensions

📅 Regroup 3

📅 Synthesis Wrap-up

Step 1: Find the displacement vector $\vec{d}$