Lesson Notes By Weeks and Term v3 - Senior Secondary 2
Vectors in three dimension
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Vectors in three dimension
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Subject: Further Mathematics
Class: Senior Secondary 2
Term: 1st Term
Week: 8
Theme: Mechanics
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Find dot or scalar product of vectors. Solve problems in volving application of dot products. Find the vector or cross product of two vectors. Solve simple problems on application of vector product.
A vector in three dimensions is typically represented as a directed line segment in a 3D coordinate system (x, y, z). It can be expressed in component form using unit vectors $\mathbf{i}$, $\mathbf{j}$, and $\mathbf{k}$ along the x, y, and z axes respectively. For a vector $\mathbf{a}$, its component form is $\mathbf{a} = a_1\mathbf{i} + a_2\mathbf{j} + a_3\mathbf{k}$ or as a column vector $\begin{pmatrix} a_1 \\ a_2 \\ a_3 \end{pmatrix}$. The magnitude (or length) of vector $\mathbf{a}$ is given by $|\mathbf{a}| = \sqrt{a_1^2 + a_2^2 + a_3^2}$.
Example of a 3D Vector: A vector representing the displacement from a point (1, 2, 0) to another point (4, 6, 5) can be written as: $\mathbf{a} = (4-1)\mathbf{i} + (6-2)\mathbf{j} + (5-0)\mathbf{k} = 3\mathbf{i} + 4\mathbf{j} + 5\mathbf{k}$. Its magnitude is $|\mathbf{a}| = \sqrt{3^2 + 4^2 + 5^2} = \sqrt{9 + 16 + 25} = \sqrt{50} = 5\sqrt{2}$ units. The dot product of two vectors, $\mathbf{a}$ and $\mathbf{b}$, results in a scalar quantity. It is a measure of how much two vectors point in the same direction.
A. Geometric Definition: If $\mathbf{a}$ and $\mathbf{b}$ are two non-zero vectors and $\theta$ is the angle between them ($0 \le \theta \le \pi$), then their dot product is defined as: $\mathbf{a} \cdot \mathbf{b} = |\mathbf{a}||\mathbf{b}|\cos\theta$
B. Algebraic Definition: Given $\mathbf{a} = a_1\mathbf{i} + a_2\mathbf{j} + a_3\mathbf{k}$ and $\mathbf{b} = b_1\mathbf{i} + b_2\mathbf{j} + b_3\mathbf{k}$, their dot product is calculated as: $\mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3$ Derivation of the Algebraic Formula: Consider the unit vectors $\mathbf{i}$, $\mathbf{j}$, $\mathbf{k}$. $\mathbf{i} \cdot \mathbf{i} = |\mathbf{i}||\mathbf{i}|\cos(0^\circ) = (1)(1)(1) = 1$ Similarly, $\mathbf{j} \cdot \mathbf{j} = 1$ and $\mathbf{k} \cdot \mathbf{k} = 1$. Also, $\mathbf{i} \cdot \mathbf{j} = |\mathbf{i}||\mathbf{j}|\cos(90^\circ) = (1)(1)(0) = 0$ Similarly, $\mathbf{j} \cdot \mathbf{k} = 0$ and $\mathbf{k} \cdot \mathbf{i} = 0$. Now, expand the dot product using the distributive property: $\mathbf{a} \cdot \mathbf{b} = (a_1\mathbf{i} + a_2\mathbf{j} + a_3\mathbf{k}) \cdot (b_1\mathbf{i} + b_2\mathbf{j} + b_3\mathbf{k})$ $= a_1\mathbf{i} \cdot (b_1\mathbf{i} + b_2\mathbf{j} + b_3\mathbf{k}) + a_2\mathbf{j} \cdot (b_1\mathbf{i} + b_2\mathbf{j} + b_3\mathbf{k}) + a_3\mathbf{k} \cdot (b_1\mathbf{i} + b_2\mathbf{j} + b_3\mathbf{k})$ $= a_1b_1(\mathbf{i} \cdot \mathbf{i}) + a_1b_2(\mathbf{i} \cdot \mathbf{j}) + a_1b_3(\mathbf{i} \cdot \mathbf{k})$ $+ a_2b_1(\mathbf{j} \cdot \mathbf{i}) + a_2b_2(\mathbf{j} \cdot \mathbf{j}) + a_2b_3(\mathbf{j} \cdot \mathbf{k})$ $+ a_3b_1(\mathbf{k} \cdot \mathbf{i}) + a_3b_2(\mathbf{k} \cdot \mathbf{j}) + a_3b_3(\mathbf{k} \cdot \mathbf{k})$ Substituting the values of the dot products of unit vectors: $= a_1b_1(1) + a_1b_2(0) + a_1b_3(0)$ $+ a_2b_1(0) + a_2b_2(1) + a_2b_3(0)$ $+ a_3b_1(0) + a_3b_2(0) + a_3b_3(1)$ $= a_1b_1 + a_2b_2 + a_3b_3$. Thus, the formula $\mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3$ is derived.
C. Properties of the Dot Product: Commutative: $\mathbf{a} \cdot \mathbf{b} = \mathbf{b} \cdot \mathbf{a}$ Distributive: $\mathbf{a} \cdot (\mathbf{b} + \mathbf{c}) = \mathbf{a} \cdot \mathbf{b} + \mathbf{a} \cdot \mathbf{c}$ Scalar Multiplication: $(c\mathbf{a}) \cdot \mathbf{b} = \mathbf{a} \cdot (c\mathbf{b}) = c(\mathbf{a} \cdot \mathbf{b})$ $\mathbf{a} \cdot \mathbf{a} = |\mathbf{a}|^2$
D. Applications of the Dot Product: Finding the angle between two vectors: From $\mathbf{a} \cdot \mathbf{b} = |\mathbf{a}||\mathbf{b}|\cos\theta$, we can derive: $\cos\theta = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}||\mathbf{b}|}$ Checking for Perpendicularity: If two non-zero vectors $\mathbf{a}$ and $\mathbf{b}$ are perpendicular (orthogonal), then the angle between them is $90^\circ$. $\mathbf{a} \cdot \mathbf{b} = |\mathbf{a}||\mathbf{b}|\cos(90^\circ) = |\mathbf{a}||\mathbf{b}|(0) = 0$. Thus, $\mathbf{a} \cdot \mathbf{b} = 0$ is a necessary and sufficient condition for two non-zero vectors to be perpendicular.
Worked Example 1 (Dot Product): Given vectors $\mathbf{p} = 2\mathbf{i} - \mathbf{j} + 3\mathbf{k}$ and $\mathbf{q} = 4\mathbf{i} + 2\mathbf{j} - \mathbf{k}$. (a) Find $\mathbf{p} \cdot \mathbf{q}$. (b) Find the angle between $\mathbf{p}$ and $\mathbf{q}$.
Solution: (a) $\mathbf{p} \cdot \mathbf{q} = (2)(4) + (-1)(2) + (3)(-1) = 8 - 2 - 3 = 3$. (b) First, find the magnitudes of $\mathbf{p}$ and $\mathbf{q}$: $|\mathbf{p}| = \sqrt{2^2 + (-1)^2 + 3^2} = \sqrt{4 + 1 + 9} = \sqrt{14}$. $|\mathbf{q}| = \sqrt{4^2 + 2^2 + (-1)^2} = \sqrt{16 + 4 + 1} = \sqrt{21}$. Now, use the formula for the angle: $\cos\theta = \frac{\mathbf{p} \cdot \mathbf{q}}{|\mathbf{p}||\mathbf{q}|} = \frac{3}{\sqrt{14}\sqrt{21}} = \frac{3}{\sqrt{294}}$. $\sqrt{294} = \sqrt{49 \times 6} = 7\sqrt{6}$. So, $\cos\theta = \frac{3}{7\sqrt{6}} = \frac{3\sqrt{6}}{7 \times 6} = \frac{\sqrt{6}}{14}$. $\theta = \cos^{-1}\left(\frac{\sqrt{6}}{14}\right) \approx \cos^{-1}(0.1749) \approx 79.9^\circ$ (to 1 decimal place). The cross product of two vectors, $\mathbf{a}$ and $\mathbf{b}$, results in a new vector that is perpendicular to both $\mathbf{a}$ and $\mathbf{b}$.
A. Geometric Definition: If $\mathbf{a}$ and $\mathbf{b}$ are two non-zero vectors and $\theta$ is the angle between them ($0 \le \theta \le \pi$), then their cross product is defined as: $\mathbf{a} \times \mathbf{b} = |\mathbf{a}||\mathbf{b}|\sin\theta \ \mathbf{\hat{n}}$ where $\mathbf{\hat{n}}$ is a unit vector perpendicular to both $\mathbf{a}$ and $\mathbf{b}$, whose direction is given by the right-hand rule. (If the fingers of the right hand curl from $\mathbf{a}$ to $\mathbf{b}$, the thumb points in the direction of $\mathbf{\hat{n}}$).
B. Algebraic Definition (using Determinants): Given $\mathbf{a} = a_1\mathbf{i} + a_2\mathbf{j} + a_3\mathbf{k}$ and $\mathbf{b} = b_1\mathbf{i} + b_2\mathbf{j} + b_3\mathbf{k}$, their cross product is calculated using a determinant of a 3x3 matrix: $\mathbf{a} \times \mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix}$ Expanding the determinant along the first row: $\mathbf{a} \times \mathbf{b} = \mathbf{i}(a_2b_3 - a_3b_2) - \mathbf{j}(a_1b_3 - a_3b_1) + \mathbf{k}(a_1b_2 - a_2b_1)$
C. Properties of the Cross Product: Anti-commutative: $\mathbf{a} \times \mathbf{b} = -(\mathbf{b} \times \mathbf{a})$ Distributive: $\mathbf{a} \times (\mathbf{b} + \mathbf{c}) = \mathbf{a} \times \mathbf{b} + \mathbf{a} \times \mathbf{c}$ Scalar Multiplication: $(c\mathbf{a}) \times \mathbf{b} = \mathbf{a} \times (c\mathbf{b}) = c(\mathbf{a} \times \mathbf{b})$ Parallelism: If two non-zero vectors $\mathbf{a}$ and $\mathbf{b}$ are parallel, then the angle between them is $0^\circ$ or $180^\circ$. In either case, $\sin\theta = 0$. Thus, $\mathbf{a} \times \mathbf{b} = \mathbf{0}$ (the zero vector) is a necessary and sufficient condition for two non-zero vectors to be parallel. $\mathbf{a} \times \mathbf{a} = \mathbf{0}$ The magnitude of the cross product, $|\mathbf{a} \times \mathbf{b}| = |\mathbf{a}||\mathbf{b}|\sin\theta$, represents the area of the parallelogram formed by vectors $\mathbf{a}$ and $\mathbf{b}$ as adjacent sides.
D. Applications of the Cross Product: Finding a vector perpendicular to two given vectors: The result of $\mathbf{a} \times \mathbf{b}$ is a vector perpendicular to both $\mathbf{a}$ and $\mathbf{b}$.
Calculating the area of a parallelogram: The area of a parallelogram with adjacent sides formed by vectors $\mathbf{a}$ and $\mathbf{b}$ is $|\mathbf{a} \times \mathbf{b}|$.
Calculating the area of a triangle: The area of a triangle with sides formed by vectors $\mathbf{a}$ and $\mathbf{b}$ is $\frac{1}{2}|\mathbf{a} \times \mathbf{b}|$.
Worked Example 2 (Cross Product): Given vectors $\mathbf{u} = \mathbf{i} - 2\mathbf{j} + 3\mathbf{k}$ and $\mathbf{v} = 4\mathbf{i} + 2\mathbf{k}$. (a) Find $\mathbf{u} \times \mathbf{v}$. (b) Find the area of the parallelogram formed by $\mathbf{u}$ and $\mathbf{v}$.
Solution: (a) Write $\mathbf{v}$ as $4\mathbf{i} + 0\mathbf{j} + 2\mathbf{k}$. $\mathbf{u} \times \mathbf{v} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & -2 & 3 \\ 4 & 0 & 2 \end{vmatrix}$ $= \mathbf{i}((-2)(2) - (3)(0)) - \mathbf{j}((1)(2) - (3)(4)) + \mathbf{k}((1)(0) - (-2)(4))$ $= \mathbf{i}(-4 - 0) - \mathbf{j}(2 - 12) + \mathbf{k}(0 - (-8))$ $= -4\mathbf{i} - \mathbf{j}(-10) + \mathbf{k}(8)$ $= -4\mathbf{i} + 10\mathbf{j} + 8\mathbf{k}$. (b) The area of the parallelogram is the magnitude of the cross product: $|\mathbf{u} \times \mathbf{v}| = |-4\mathbf{i} + 10\mathbf{j} + 8\mathbf{k}|$ $= \sqrt{(-4)^2 + 10^2 + 8^2}$ $= \sqrt{16 + 100 + 64}$ $= \sqrt{180} = \sqrt{36 \times 5} = 6\sqrt{5}$ square units.
Teacher Activities: Introduction (10 minutes): Begin by briefly reviewing vectors in 2D, including magnitude and component form. Introduce the concept of 3D vectors as an extension, explaining the x, y, z axes and the unit vectors $\mathbf{i}$, $\mathbf{j}$, $\mathbf{k}$. Use the corner of the classroom or a prepared model (e.g., three perpendicular sticks) to illustrate 3D axes. State the learning objectives for the lesson.
Explaining Dot Product (20 minutes): Define the dot product geometrically and algebraically. Derive the formula $\mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3$ step-by-step, emphasizing the dot products of unit vectors ($\mathbf{i} \cdot \mathbf{i} = 1$, $\mathbf{i} \cdot \mathbf{j} = 0$, etc.). Discuss properties of the dot product (commutative, distributive).
Explain applications: finding the angle between vectors and checking for perpendicularity. Provide a simple example, relating it to structural integrity (e.g., perpendicular beams). Solve Worked Example 1 on the board, explaining each step clearly.
Explaining Cross Product (25 minutes): Define the cross product geometrically, emphasizing the right-hand rule for direction (demonstrate with hand). Explain the algebraic definition using the 3x3 determinant. Emphasize the signs in the expansion. Discuss properties of the cross product (anti-commutative, distributive).
Explain applications: finding a vector perpendicular to two others, calculating the area of a parallelogram/triangle. Solve Worked Example 2 on the board, showing the determinant expansion step-by-step.
Guided Practice (20 minutes): Provide 2-3 guided practice questions (from Section 4) for students to attempt in pairs or small groups. Circulate among students, providing support, checking understanding, and correcting misconceptions. Select groups to present their solutions on the board, facilitating peer correction and discussion.
Conclusion (5 minutes): Summarize the key differences and uses of dot and cross products. Assign independent practice questions and homework.
Student Activities: Active Listening and Note-Taking: Students will listen attentively to explanations, ask clarifying questions, and take comprehensive notes.
Participation: Students will respond to teacher questions and contribute to class discussions.
Collaborative Problem Solving: Students will work in pairs or small groups to solve guided practice problems, discussing strategies and solutions among themselves.
Presentation: Selected students or groups will present their solutions to guided practice problems on the board, explaining their reasoning.
Independent Practice: Students will attempt independent practice questions (from Section 5) to consolidate their learning.
Structural Engineering and Architecture: Vectors in three dimensions are essential for analysing forces and stresses in multi-storey buildings, bridges, and other complex structures (e.g., the Eko Bridge in Lagos or the Third Mainland Bridge). Engineers use dot products to calculate the component of a force acting in a particular direction (e.g., load distribution on beams) and cross products to determine torque or moments, ensuring structural stability and safety in designs across Nigeria.
Oil and Gas Exploration and Surveying: In Nigeria's vital oil and gas sector, geologists and engineers use 3D vectors to model subsurface formations, track boreholes, and analyse seismic data. Vectors define the direction of oil reservoirs, the path of drilling operations, and the orientation of geological faults. The dot product can determine the angle between a drilling path and a rock layer, while the cross product can help define planes of geological structures, facilitating efficient and safe resource extraction. Surveyors also use 3D vectors with GPS technology to map out terrain, plan roads, and establish property boundaries across diverse Nigerian landscapes.
Robotics and Animation: The burgeoning tech and creative industries in Nigeria (e.g., Nollywood animations, tech start-ups developing robotics for agriculture or security) heavily rely on 3D vectors. Robotics engineers use dot and cross products to program robot arm movements, ensure precise tool orientation, and avoid collisions. In computer animation and game development, vectors define object positions, rotations, and camera angles, enabling realistic 3D modelling and dynamic movement of characters and environments, for instance, in creating 3D animated films or virtual reality experiences with Nigerian cultural elements.