MATH-283 (Calculus III) Curriculum

Session 03 - Dot and Cross Product

Handout Worksheet

Fall Semester 2026

Session Objectives & Overview

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

Compute algebraic and geometric dot products in 2D and 3D, determine orthogonality, calculate the angle between vectors, and evaluate scalar and vector projections.
Evaluate $3 \times 3$ matrix determinants to compute cross products, apply the right-hand rule, generate orthogonal unit vectors, and calculate geometric areas.
Model and solve real-world UNLV engineering systems involving mechanical work ( $W = \vec{F} \cdot \vec{d}$ ) and rotational torque ( $\vec{\tau} = \vec{r} \times \vec{F}$ ) while maintaining strict fluency in scalar versus vector outputs.

Micro-Lecture

1. Multiplying Vectors: The Fundamental Distinction (Scalars vs. Vectors)

Mathematical Foundation: In vector calculus, there is no single generic operation called "vector multiplication." Instead, we define two distinct operations with entirely different mathematical outputs and physical meanings:
1. The Dot Product (Scalar Product, $\vec{u} \cdot \vec{v}$ ): Multiplies corresponding components to return a single real number (scalar). It quantifies directional alignment: $\vec{u} \cdot \vec{v} = |\vec{u}| |\vec{v}| \cos\theta$ . If two non-zero vectors are perpendicular ( $\theta = 90^\circ$ ), their dot product is strictly $0$ .
2. The Cross Product (Vector Product, $\vec{u} \times \vec{v}$ ): Evaluates a $3 \times 3$ matrix determinant to return a brand new 3D vector that is strictly orthogonal (perpendicular) to both input vectors: $|\vec{u} \times \vec{v}| = |\vec{u}| |\vec{v}| \sin\theta$ . If two vectors are parallel ( $\theta = 0^\circ$ or $180^\circ$ ), their cross product is the zero vector $\vec{0}$ .
Engineering Context: A common misconception is providing a vector as an answer to a dot product. In physics and engineering, the dot product collapses force and motion into scalar energy or power (Joules, Watts), whereas the cross product constructs directional rotational axes, torque moments ( $\text{N}\cdot\text{m}$ ), and aerodynamic lift vectors.

2. Projections: Decomposing Loads onto Structural Axes

Mathematical Foundation: Vector Projections allow us to split any vector $\vec{u}$ into a component parallel to a target vector $\vec{v}$ and a component perpendicular to $\vec{v}$ . The scalar projection $\text{comp}_{\vec{v}}\vec{u} = \frac{\vec{u} \cdot \vec{v}}{|\vec{v}|}$ is a signed number giving the length of the shadow. The vector projection $\text{proj}_{\vec{v}}\vec{u} = \left(\frac{\vec{u} \cdot \vec{v}}{|\vec{v}|^2}\right)\vec{v}$ is the actual vector pointing along $\vec{v}$ .
Engineering Context: Essential for civil and structural engineers isolating axial tension and compression forces along individual diagonal truss members in bridge designs.

3. Physical Modeling: Work ( $W$ ) and Torque ( $\vec{\tau}$ )

Mathematical Foundation: Mechanical Work is calculated via the dot product: $W = \vec{F} \cdot \vec{d} = |\vec{F}| |\vec{d}| \cos\theta$ (scalar in Joules). Torque is calculated via the cross product: $\vec{\tau} = \vec{r} \times \vec{F}$ (vector in $\text{N}\cdot\text{m}$ defining the axis and magnitude of rotational turning effort).
Engineering Context: Used across mechanical conveyors, wind turbine blade aerodynamics, industrial valve operation, and robotic joint actuators.

Skill Block 1

Worked Example A: Algebraic Dot Product and Orthogonality

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

Worked Solution:

Identify Components: Let $\vec{u} = \langle 6, 7 \rangle$ with $u_x = 6, u_y = 7$ and $\vec{v} = \langle 7, -6 \rangle$ with $v_x = 7, v_y = -6$ .
Apply Algebraic Dot Product Formula: $\vec{u} \cdot \vec{v} = u_x v_x + u_y v_y = (6)(7) + (7)(-6)$
Compute the Scalar Value: $\vec{u} \cdot \vec{v} = 42 + (-42) = 0$
Interpret Result: Since $\vec{u} \cdot \vec{v} = 0$ and both vectors are non-zero, $\vec{u}$ and $\vec{v}$ are strictly orthogonal (perpendicular). Notice the result is the scalar number $0$ , not a vector.

Worked Example B: Angle Between 3D Vectors

Find the angle $\theta$ between $\vec{a} = 5\hat{i} + 3\hat{j}$ and $\vec{b} = 4\hat{i} + 2\hat{j}$ .

Worked Solution:

Calculate Dot Product: $\vec{a} \cdot \vec{b} = (5)(4) + (3)(2) = 20 + 6 = 26$
Calculate Vector Magnitudes: $|\vec{a}| = \sqrt{5^2 + 3^2} = \sqrt{25 + 9} = \sqrt{34} \approx 5.831$ $|\vec{b}| = \sqrt{4^2 + 2^2} = \sqrt{16 + 4} = \sqrt{20} \approx 4.472$
Solve for $\cos\theta$ and $\theta$ : $\cos\theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|} = \frac{26}{\sqrt{34}\sqrt{20}} = \frac{26}{\sqrt{680}} \approx \frac{26}{26.077} \approx 0.9970$ $\theta = \cos^{-1}(0.9970) \approx 4.40^\circ \quad (\text{or } 0.0768\text{ rad})$

Worked Example C: Scalar and Vector Projection

Given $\vec{u} = \langle 1, 2, 3 \rangle$ and $\vec{v} = \langle -2, 0, 4 \rangle$ , compute the scalar projection $\text{comp}_{\vec{v}}\vec{u}$ and the vector projection $\text{proj}_{\vec{v}}\vec{u}$ .

Worked Solution:

Compute $\vec{u} \cdot \vec{v}$ and $|\vec{v}|^2$ : $\vec{u} \cdot \vec{v} = (1)(-2) + (2)(0) + (3)(4) = -2 + 0 + 12 = 10$ $|\vec{v}|^2 = (-2)^2 + 0^2 + 4^2 = 4 + 0 + 16 = 20 \implies |\vec{v}| = \sqrt{20} = 2\sqrt{5}$
Compute Scalar Projection (Scalar Output): $\text{comp}_{\vec{v}}\vec{u} = \frac{\vec{u} \cdot \vec{v}}{|\vec{v}|} = \frac{10}{\sqrt{20}} = \frac{10}{2\sqrt{5}} = \sqrt{5} \approx 2.236$
Compute Vector Projection (Vector Output): $\text{proj}_{\vec{v}}\vec{u} = \left(\frac{\vec{u} \cdot \vec{v}}{|\vec{v}|^2}\right)\vec{v} = \frac{10}{20} \langle -2, 0, 4 \rangle = \frac{1}{2} \langle -2, 0, 4 \rangle = \langle -1, 0, 2 \rangle$

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, what type of mathematical object (scalar or vector) is the output of a dot product? Why is writing $\langle 12, 10 \rangle$ as the result of a dot product incorrect?
(b) Under Key Equations & Structural Rules, what is the formula for the scalar projection $\text{comp}_{\vec{v}}\vec{u}$ , and how does it differ in output type from the vector projection $\text{proj}_{\vec{v}}\vec{u}$ ?
(c) Review the linked engineering application Structural Load Decomposition (Civil Engineering). How is the dot product used by civil engineers to determine if a bridge truss member is under compression or tension?

Solution Key:

(a) As emphasized in the Conceptual Framework section, the dot product (scalar product) collapses two vectors into a single real number (scalar). The individual component products are added together ( $u_1 v_1 + u_2 v_2$ ). Writing vector brackets like $\langle 12, 10 \rangle$ incorrectly retains vector structure instead of computing the final scalar sum ( $12 + 10 = 22$ ).
(b) According to the Key Equations & Structural Rules section, the scalar projection is given by $\text{comp}_{\vec{v}}\vec{u} = \frac{\vec{u}\cdot\vec{v}}{|\vec{v}|}$ , which yields a scalar representing the signed magnitude of the shadow of $\vec{u}$ on $\vec{v}$ . In contrast, the vector projection $\text{proj}_{\vec{v}}\vec{u} = \left(\frac{\vec{u}\cdot\vec{v}}{|\vec{v}|^2}\right)\vec{v}$ multiplies this scalar by a unit vector along $\vec{v}$ , producing an actual vector quantity.
(c) In Structural Load Decomposition (Civil Engineering), civil engineers compute $\text{comp}_{\vec{a}}\vec{F}_{\text{load}} = \vec{F}_{\text{load}} \cdot \hat{u}_a$ to isolate the axial force pushing or pulling along the truss member. A negative scalar value indicates the force opposes the member direction vector, placing the steel beam under compressive stress.

Show your steps and calculations here.

Practice Problem 2

Find the dot product of $\vec{u} = 3\hat{i} - 2\hat{j}$ and $\vec{v} = 4\hat{i} - 5\hat{j}$ , and compute the exact angle $\theta$ between them.

Solution Key:

Compute Dot Product: $\vec{u} \cdot \vec{v} = (3)(4) + (-2)(-5) = 12 + 10 = 22$ (Note: The result is the scalar $22$ .)
Compute Magnitudes: $|\vec{u}| = \sqrt{3^2 + (-2)^2} = \sqrt{9 + 4} = \sqrt{13}$ $|\vec{v}| = \sqrt{4^2 + (-5)^2} = \sqrt{16 + 25} = \sqrt{41}$
Calculate Angle: $\cos\theta = \frac{\vec{u} \cdot \vec{v}}{|\vec{u}| |\vec{v}|} = \frac{22}{\sqrt{13}\sqrt{41}} = \frac{22}{\sqrt{533}} \approx 0.9529$ $\theta = \cos^{-1}(0.9529) \approx 17.65^\circ \quad (\approx 0.308\text{ rad})$

Show your steps and calculations here.

Practice Problem 3

Consider the 3D vectors $\vec{a} = \langle 2, -1, 3 \rangle$ and $\vec{b} = \langle 1, 4, -2 \rangle$ .

Compute $\vec{a} \cdot \vec{b}$ . Are the vectors orthogonal, parallel, or neither?
Find the scalar projection $\text{comp}_{\vec{b}}\vec{a}$ of $\vec{a}$ onto $\vec{b}$ .
Find the vector projection $\text{proj}_{\vec{b}}\vec{a}$ of $\vec{a}$ onto $\vec{b}$ .

Solution Key:

Compute Dot Product and Analyze Relationship: $\vec{a} \cdot \vec{b} = (2)(1) + (-1)(4) + (3)(-2) = 2 - 4 - 6 = -6$ Since $\vec{a} \cdot \vec{b} = -6 \ne 0$ , the vectors are not orthogonal. Since $\vec{a}$ is not a scalar multiple of $\vec{b}$ , they are not parallel. Because $\vec{a} \cdot \vec{b} < 0$ , the angle between them is obtuse ( $90^\circ < \theta \le 180^\circ$ ).
Compute Scalar Projection: $|\vec{b}| = \sqrt{1^2 + 4^2 + (-2)^2} = \sqrt{1 + 16 + 4} = \sqrt{21}$ $\text{comp}_{\vec{b}}\vec{a} = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|} = \frac{-6}{\sqrt{21}} = -\frac{2\sqrt{21}}{7} \approx -1.310$
Compute Vector Projection: $\text{proj}_{\vec{b}}\vec{a} = \left(\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|^2}\right)\vec{b} = \frac{-6}{21}\langle 1, 4, -2 \rangle = -\frac{2}{7}\langle 1, 4, -2 \rangle = \left\langle -\frac{2}{7}, -\frac{8}{7}, \frac{4}{7} \right\rangle$

Show your steps and calculations here.

Regroup 1

Emphasize that the dot product is always a scalar number ( $c \in \mathbb{R}$ ), never a vector.
Clarify the geometric meaning of the sign of the dot product: positive = acute angle, zero = orthogonal, negative = obtuse angle.
Reinforce that scalar projection is a signed number, whereas vector projection is a directed vector pointing parallel to the reference vector.

Skill Block 2

Worked Example A: 2D Vectors Embedded in 3D Cross Product

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

Worked Solution:

Embed Vectors in 3D: Set $z$ -components to zero: $\vec{p} = \langle 2, 4, 0 \rangle$ and $\vec{q} = \langle -4, -5, 0 \rangle$ .
Set Up Determinant: $\vec{p} \times \vec{q} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 4 & 0 \\ -4 & -5 & 0 \end{vmatrix}$
Expand Along Top Row: $\vec{p} \times \vec{q} = \hat{i}(4\cdot 0 - 0\cdot (-5)) - \hat{j}(2\cdot 0 - 0\cdot (-4)) + \hat{k}(2\cdot (-5) - 4\cdot (-4))$ $\vec{p} \times \vec{q} = 0\hat{i} - 0\hat{j} + (-10 + 16)\hat{k} = 6\hat{k} = \langle 0, 0, 6 \rangle$ (Note: The result is a 3D vector pointing straight out of the $xy$ -plane along the positive $z$ -axis.)

Worked Example B: Full 3D Vector Cross Product

Calculate $\vec{p} \times \vec{q}$ given $\vec{p} = 2\hat{i} + 4\hat{j} - 4\hat{k}$ and $\vec{q} = -4\hat{i} - 5\hat{j} - 2\hat{k}$ .

Worked Solution:

Set Up Determinant: $\vec{p} \times \vec{q} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 4 & -4 \\ -4 & -5 & -2 \end{vmatrix}$
Expand Along Top Row: $\vec{p} \times \vec{q} = \hat{i}\begin{vmatrix} 4 & -4 \\ -5 & -2 \end{vmatrix} - \hat{j}\begin{vmatrix} 2 & -4 \\ -4 & -2 \end{vmatrix} + \hat{k}\begin{vmatrix} 2 & 4 \\ -4 & -5 \end{vmatrix}$
Evaluate $2 \times 2$ Minors:
- $\hat{i}\text{ component:} (4)(-2) - (-4)(-5) = -8 - 20 = -28$
- $\hat{j}\text{ component:} -[(2)(-2) - (-4)(-4)] = -[-4 - 16] = -[-20] = +20$
- $\hat{k}\text{ component:} (2)(-5) - (4)(-4) = -10 - (-16) = -10 + 16 = +6$
Assemble Vector: $\vec{p} \times \vec{q} = -28\hat{i} + 20\hat{j} + 6\hat{k} = \langle -28, 20, 6 \rangle$

Worked Example C: Area of Parallelogram

Find the area of the parallelogram formed by the vectors $\vec{a} = \langle 3, 2, 0 \rangle$ and $\vec{b} = \langle -2, 5, 0 \rangle$ .

Worked Solution:

Compute Cross Product: $\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 2 & 0 \\ -2 & 5 & 0 \end{vmatrix} = \hat{k}(3\cdot 5 - 2\cdot (-2)) = \hat{k}(15 + 4) = 19\hat{k} = \langle 0, 0, 19 \rangle$
Compute Magnitude: $\text{Area} = |\vec{a} \times \vec{b}| = \sqrt{0^2 + 0^2 + 19^2} = 19$

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, what type of mathematical object does the cross product produce, and in what dimension is it defined?
(b) Trace the step-by-step expansion in Worked Example B. What is the critical sign rule that must be applied to the middle ( $\hat{j}$ ) component during the $3 \times 3$ determinant expansion?
(c) Review the linked engineering application Airfoil Lift and Circulation (Aerospace Engineering). How does the cross product of horizontal velocity and spanwise circulation generate vertical lift according to the right-hand rule?

Solution Key:

(a) As stated in the Conceptual Framework section, the cross product (vector product) takes two vectors and produces a new VECTOR that is orthogonal to both. It is defined strictly in three dimensions ( $\mathbb{R}^3$ ).
(b) Tracing Worked Example B, when expanding a $3 \times 3$ matrix determinant across the top row, the cofactor pattern alternating signs requires a negative sign in front of the $\hat{j}$ minor: $-\hat{j}(a_x b_z - a_z b_x)$ . Forgetting this negative sign is the most frequent computational error in cross products.
(c) In Airfoil Lift and Circulation (Aerospace Engineering), the Kutta-Joukowski theorem states $\vec{F}_L = \rho(\vec{v} \times \vec{\Gamma})$ . Pointing the right index finger along the horizontal flight velocity $\vec{v} = \langle 150, 0, 0 \rangle$ (the $+x$ axis) and curling fingers toward the spanwise circulation vector $\vec{\Gamma} = \langle 0, 4, 0 \rangle$ (the $+y$ axis) causes the thumb to point straight up along $+z$ , producing upward aerodynamic lift.

Show your steps and calculations here.

Practice Problem 5

Given the vectors $\vec{u} = \langle 1, 2, 3 \rangle$ and $\vec{v} = \langle 4, 5, 6 \rangle$ :

Compute $\vec{u} \times \vec{v}$ .
Verify that the resulting vector is orthogonal to both $\vec{u}$ and $\vec{v}$ using the dot product.
Find a unit vector orthogonal to the plane containing both $\vec{u}$ and $\vec{v}$ .

Solution Key:

Compute Cross Product via Determinant: $\vec{u} \times \vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ 4 & 5 & 6 \end{vmatrix}$ $\vec{u} \times \vec{v} = \hat{i}(2\cdot 6 - 3\cdot 5) - \hat{j}(1\cdot 6 - 3\cdot 4) + \hat{k}(1\cdot 5 - 2\cdot 4)$ $\vec{u} \times \vec{v} = \hat{i}(12 - 15) - \hat{j}(6 - 12) + \hat{k}(5 - 8) = -3\hat{i} + 6\hat{j} - 3\hat{k} = \langle -3, 6, -3 \rangle$
Verify Orthogonality:
- $(\vec{u} \times \vec{v}) \cdot \vec{u} = (-3)(1) + (6)(2) + (-3)(3) = -3 + 12 - 9 = 0$
- $(\vec{u} \times \vec{v}) \cdot \vec{v} = (-3)(4) + (6)(5) + (-3)(6) = -12 + 30 - 18 = 0$ Both dot products equal zero, confirming strict orthogonality.
Find Orthogonal Unit Vector: $|\vec{u} \times \vec{v}| = \sqrt{(-3)^2 + 6^2 + (-3)^2} = \sqrt{9 + 36 + 9} = \sqrt{54} = 3\sqrt{6}$ $\hat{n} = \frac{\langle -3, 6, -3 \rangle}{3\sqrt{6}} = \left\langle -\frac{1}{\sqrt{6}}, \frac{2}{\sqrt{6}}, -\frac{1}{\sqrt{6}} \right\rangle$

Show your steps and calculations here.

Practice Problem 6

Find the area of the 3D triangle whose vertices are $P(1, 0, 1)$ , $Q(-2, 4, 1)$ , and $R(0, 1, 5)$ .

Solution Key:

Form Edge Vectors: $\vec{u} = \vec{PQ} = \langle -2 - 1, 4 - 0, 1 - 1 \rangle = \langle -3, 4, 0 \rangle$ $\vec{v} = \vec{PR} = \langle 0 - 1, 1 - 0, 5 - 1 \rangle = \langle -1, 1, 4 \rangle$
Calculate Cross Product: $\vec{u} \times \vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -3 & 4 & 0 \\ -1 & 1 & 4 \end{vmatrix}$ $\vec{u} \times \vec{v} = \hat{i}(4\cdot 4 - 0\cdot 1) - \hat{j}(-3\cdot 4 - 0\cdot (-1)) + \hat{k}(-3\cdot 1 - 4\cdot (-1))$ $\vec{u} \times \vec{v} = 16\hat{i} - (-12)\hat{j} + (-3 + 4)\hat{k} = 16\hat{i} + 12\hat{j} + 1\hat{k} = \langle 16, 12, 1 \rangle$
Compute Magnitude and Triangle Area: $|\vec{u} \times \vec{v}| = \sqrt{16^2 + 12^2 + 1^2} = \sqrt{256 + 144 + 1} = \sqrt{401}$ $\text{Area} = \frac{1}{2} |\vec{u} \times \vec{v}| = \frac{\sqrt{401}}{2} \approx 10.01$

Show your steps and calculations here.

Regroup 2

Emphasize that the cross product results in a vector in 3D, never a scalar number.
Address the common negative sign mistake on the $\hat{j}$ term in determinant expansions.
Review how the right-hand rule visually confirms the sign and orientation of cross products.

Skill Block 3

Worked Example A: Geometric Work Calculation with Angle

Find the mechanical work done if $|\vec{F}| = 16\text{ N}$ , displacement $|\vec{d}| = 5\text{ m}$ , and the angle between the force and displacement vectors is $77^\circ$ . Express the answer rounded to two decimal places.

Worked Solution:

Apply Geometric Formula: $W = \vec{F} \cdot \vec{d} = |\vec{F}| |\vec{d}| \cos\theta$
Substitute Given Values: $W = (16\text{ N})(5\text{ m})\cos(77^\circ) = 80 \cos(77^\circ)$
Evaluate and Round: $\cos(77^\circ) \approx 0.22495 \implies W \approx 80(0.22495) \approx 18.00\text{ Joules}$ (Note: Work is a scalar quantity measured in Joules.)

Worked Example B: Work Along a 3D Coordinate Path

Find the work done by the force $\vec{F} = \langle 4, -4, 4 \rangle\text{ N}$ acting on a particle as it moves along a straight line from $P(-7, -2, 5)\text{ m}$ to $Q(8, -7, -5)\text{ m}$ .

Worked Solution:

Compute Displacement Vector $\vec{d} = \vec{PQ}$ : $\vec{d} = \langle 8 - (-7), -7 - (-2), -5 - 5 \rangle = \langle 15, -5, -10 \rangle\text{ m}$
Compute Work via Dot Product: $W = \vec{F} \cdot \vec{d} = (4)(15) + (-4)(-5) + (4)(-10)$
Simplify: $W = 60 + 20 - 40 = 40\text{ Joules}$

Worked Example C: 3D Torque Vector on a Lever Arm

A force $\vec{F} = \langle 0, 12, 0 \rangle\text{ N}$ is applied at point $P$ , where the position vector from origin $O$ to $P$ is $\vec{r} = 2\hat{i} + 1\hat{j} + 0\hat{k}\text{ m}$ . Find the torque vector $\vec{\tau}$ about $O$ .

Worked Solution:

Set Up Determinant: $\vec{\tau} = \vec{r} \times \vec{F} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & 0 \\ 0 & 12 & 0 \end{vmatrix}$
Expand Along Top Row: $\vec{\tau} = \hat{i}(1\cdot 0 - 0\cdot 12) - \hat{j}(2\cdot 0 - 0\cdot 0) + \hat{k}(2\cdot 12 - 1\cdot 0) = 24\hat{k}\text{ N}\cdot\text{m}$ The torque vector is $\langle 0, 0, 24 \rangle\text{ N}\cdot\text{m}$ (magnitude $24\text{ N}\cdot\text{m}$ along the positive $z$ -axis).

Practice Problem 7: Concept & Reading Check

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

(a) Compare the mathematical definitions and output types of Mechanical Work ( $W$ ) and Torque ( $\vec{\tau}$ ). Which one is a scalar and which one is a vector?
(b) Why is Mechanical Work measured in Joules ( $\text{J}$ ), while Torque is measured in Newton-meters ( $\text{N}\cdot\text{m}$ ) even though both have algebraic dimensions of force times distance?
(c) Review the linked engineering application Robotic Arm Joint Torque (Robotics & Computer Engineering). How does the cross product allow robotics engineers to determine which robot joints experience bending moments under heavy payloads?

Solution Key:

(a) As outlined in the Conceptual Framework section, Mechanical Work is defined by the dot product ( $W = \vec{F}\cdot\vec{d}$ ) and produces a scalar quantity measuring energy transferred along motion. Torque is defined by the cross product ( $\vec{\tau} = \vec{r}\times\vec{F}$ ) and produces a 3D vector quantity defining the axis and strength of rotational turning effort.
(b) Although both share units of $\text{N}\cdot\text{m}$ , standard scientific and engineering convention reserves the unit Joules ( $\text{J}$ ) strictly for scalar energy transfer ( $W$ ). Torque ( $\vec{\tau}$ ) represents a directional rotational moment; labeling torque in Joules is incorrect because torque is not scalar energy.
(c) In Robotic Arm Joint Torque (Robotics & Computer Engineering), the resulting cross product $\vec{\tau} = \langle 60, 0, -80 \rangle\text{ N}\cdot\text{m}$ isolates the moment experienced around each coordinate axis: $+60\text{ N}\cdot\text{m}$ about the $x$ -axis (pitch) and $-80\text{ N}\cdot\text{m}$ about the $z$ -axis (roll). This allows control software to command the exact motor torques needed at each joint actuator.

Show your steps and calculations here.

Practice Problem 8

A constant force $\vec{F} = 3\hat{i} + 2\hat{j} + 4\hat{k}\text{ N}$ acts on a mechanical carriage moving it from point $A(1, 1, 1)\text{ m}$ to point $B(4, 3, 2)\text{ m}$ .

Find the displacement vector $\vec{d} = \vec{AB}$ .
Calculate the mechanical work done by the force on the carriage.
If an opposing friction force vector $\vec{f} = \langle -1, -1, 0 \rangle\text{ N}$ acts during the same displacement, compute the net work performed on the carriage.

Solution Key:

Find Displacement Vector: $\vec{d} = \vec{AB} = \langle 4 - 1, 3 - 1, 2 - 1 \rangle = \langle 3, 2, 1 \rangle\text{ m}$
Compute Applied Work: $W_{\text{applied}} = \vec{F} \cdot \vec{d} = (3)(3) + (2)(2) + (4)(1) = 9 + 4 + 4 = 17\text{ Joules}$
Compute Net Work: $\vec{F}_{\text{net}} = \vec{F} + \vec{f} = \langle 3 - 1, 2 - 1, 4 + 0 \rangle = \langle 2, 1, 4 \rangle\text{ N}$ $W_{\text{net}} = \vec{F}_{\text{net}} \cdot \vec{d} = (2)(3) + (1)(2) + (4)(1) = 6 + 2 + 4 = 12\text{ Joules}$

Show your steps and calculations here.

Practice Problem 9

A heavy valve handle has a position vector from the pivot to the grip of $\vec{r} = \langle 0.5, 0.2, 0 \rangle\text{ meters}$ . A technician applies a vertical force $\vec{F} = \langle 0, 0, 50 \rangle\text{ Newtons}$ .

Compute the torque vector $\vec{\tau} = \vec{r} \times \vec{F}$ .
Calculate the magnitude of the torque ( $|\vec{\tau}|$ ).
State the direction of the torque vector in terms of coordinate axes.

Solution Key:

Calculate Torque Vector: $\vec{\tau} = \vec{r} \times \vec{F} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0.5 & 0.2 & 0 \\ 0 & 0 & 50 \end{vmatrix}$ $\vec{\tau} = \hat{i}(0.2\cdot 50 - 0\cdot 0) - \hat{j}(0.5\cdot 50 - 0\cdot 0) + \hat{k}(0.5\cdot 0 - 0.2\cdot 0)$ $\vec{\tau} = 10\hat{i} - 25\hat{j} + 0\hat{k} = \langle 10, -25, 0 \rangle\text{ N}\cdot\text{m}$
Compute Torque Magnitude: $|\vec{\tau}| = \sqrt{10^2 + (-25)^2 + 0^2} = \sqrt{100 + 625} = \sqrt{725} = 5\sqrt{29} \approx 26.93\text{ N}\cdot\text{m}$
Direction of Torque: The torque vector lies entirely in the $xy$ -plane, pointing in the positive $x$ and negative $y$ directions ( $\langle 10, -25, 0 \rangle$ ), representing a compound bending moment about the valve stem.

Show your steps and calculations here.

Regroup 3

Reinforce that Work ( $W = \vec{F}\cdot\vec{d}$ ) is a scalar energy transfer (measured in Joules), whereas Torque ( $\vec{\tau} = \vec{r}\times\vec{F}$ ) is a vector turning moment (measured in $\text{N}\cdot\text{m}$ ).
Remind students that forces perpendicular to displacement do zero work ( $\cos 90^\circ = 0$ ), whereas forces perpendicular to lever arms produce maximum torque ( $\sin 90^\circ = 1$ ).

Synthesis Wrap-up

Core Summary:
- Dot Product ( $\vec{u} \cdot \vec{v}$ ): Produces a scalar. Evaluates directional alignment, orthogonality ( $\vec{u}\cdot\vec{v}=0$ ), angles, projections, and mechanical work ( $W = \vec{F}\cdot\vec{d}$ ).
- Cross Product ( $\vec{u} \times \vec{v}$ ): Produces a vector in 3D. Evaluates normal directions, anti-commutativity, parallel lines ( $\vec{u}\times\vec{v}=\vec{0}$ ), geometric areas, and rotational torque ( $\vec{\tau} = \vec{r}\times\vec{F}$ ).
Homework Assignment (Knewton Alta):
- The Dot Product
- The Cross Product