Vectors: 2D and 3D.
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Lesson Objectives
- Understand vector representation in 2D and 3D coordinate systems
- Perform vector addition, subtraction, and scalar multiplication
- Calculate the magnitude (length) of a vector
- Find the dot product and use it to determine angles between vectors
- Find the cross product of 3D vectors
- Understand vector projections and their applications
- Apply vector methods to solve geometric and physics problems
Introduction to Vectors
Vectors are mathematical quantities that have both magnitude (size) and direction. They are essential in physics, engineering, computer graphics, and many other fields. Unlike scalars (which have only magnitude), vectors require both a numerical value and a direction to be fully described. This section covers vector representation in 2D and 3D, vector operations, dot and cross products, and advanced vector concepts.
Vector Notation
A vector can be written as $\vec{v}$, $\mathbf{v}$, or $\overrightarrow{AB}$
In component form: $\vec{v} = \langle v_x, v_y \rangle$ (2D) or $\vec{v} = \langle v_x, v_y, v_z \rangle$ (3D)
Magnitude: $|\vec{v}| = \sqrt{v_x^2 + v_y^2}$ (2D) or $|\vec{v}| = \sqrt{v_x^2 + v_y^2 + v_z^2}$ (3D)
A vector can be written as $\vec{v}$, $\mathbf{v}$, or $\overrightarrow{AB}$
In component form: $\vec{v} = \langle v_x, v_y \rangle$ (2D) or $\vec{v} = \langle v_x, v_y, v_z \rangle$ (3D)
Magnitude: $|\vec{v}| = \sqrt{v_x^2 + v_y^2}$ (2D) or $|\vec{v}| = \sqrt{v_x^2 + v_y^2 + v_z^2}$ (3D)
Key Definitions:
• Vector: A quantity with both magnitude and direction.
• Scalar: A quantity with only magnitude (e.g., mass, temperature, speed).
• Position Vector: A vector from the origin to a point $P(x,y,z)$, written as $\vec{r} = \langle x, y, z \rangle$.
• Unit Vector: A vector with magnitude 1, denoted as $\hat{v} = \frac{\vec{v}}{|\vec{v}|}$.
• Zero Vector: A vector with all components zero, denoted $\vec{0}$.
• Vector: A quantity with both magnitude and direction.
• Scalar: A quantity with only magnitude (e.g., mass, temperature, speed).
• Position Vector: A vector from the origin to a point $P(x,y,z)$, written as $\vec{r} = \langle x, y, z \rangle$.
• Unit Vector: A vector with magnitude 1, denoted as $\hat{v} = \frac{\vec{v}}{|\vec{v}|}$.
• Zero Vector: A vector with all components zero, denoted $\vec{0}$.
Vector Representation in 2D and 3D
Vectors can be represented geometrically as directed line segments (arrows) or algebraically using components. In 2D, a vector is written as $\vec{v} = \langle v_x, v_y \rangle$ or $\vec{v} = v_x\hat{i} + v_y\hat{j}$. In 3D, we add a third component: $\vec{v} = \langle v_x, v_y, v_z \rangle$ or $\vec{v} = v_x\hat{i} + v_y\hat{j} + v_z\hat{k}$.
Standard Unit Vectors:
In 2D: $\hat{i} = \langle 1, 0 \rangle$, $\hat{j} = \langle 0, 1 \rangle$
In 3D: $\hat{i} = \langle 1, 0, 0 \rangle$, $\hat{j} = \langle 0, 1, 0 \rangle$, $\hat{k} = \langle 0, 0, 1 \rangle$
Any vector can be written as: $\vec{v} = x\hat{i} + y\hat{j} + z\hat{k}$
In 2D: $\hat{i} = \langle 1, 0 \rangle$, $\hat{j} = \langle 0, 1 \rangle$
In 3D: $\hat{i} = \langle 1, 0, 0 \rangle$, $\hat{j} = \langle 0, 1, 0 \rangle$, $\hat{k} = \langle 0, 0, 1 \rangle$
Any vector can be written as: $\vec{v} = x\hat{i} + y\hat{j} + z\hat{k}$
Example 1: 2D Vector Representation
Problem: Write the vector from point $A(2, 3)$ to $B(5, 7)$ in component form.
Solution:
$\overrightarrow{AB} = \langle x_B - x_A, y_B - y_A \rangle = \langle 5-2, 7-3 \rangle = \langle 3, 4 \rangle$
Answer: $\vec{v} = \langle 3, 4 \rangle$ or $3\hat{i} + 4\hat{j}$
Problem: Write the vector from point $A(2, 3)$ to $B(5, 7)$ in component form.
Solution:
$\overrightarrow{AB} = \langle x_B - x_A, y_B - y_A \rangle = \langle 5-2, 7-3 \rangle = \langle 3, 4 \rangle$
Answer: $\vec{v} = \langle 3, 4 \rangle$ or $3\hat{i} + 4\hat{j}$
Example 2: 3D Vector Representation
Problem: Write the vector from $P(1, -2, 4)$ to $Q(3, 0, -1)$ in component form.
Solution:
$\overrightarrow{PQ} = \langle 3-1, 0-(-2), -1-4 \rangle = \langle 2, 2, -5 \rangle$
Answer: $\vec{v} = \langle 2, 2, -5 \rangle$ or $2\hat{i} + 2\hat{j} - 5\hat{k}$
Problem: Write the vector from $P(1, -2, 4)$ to $Q(3, 0, -1)$ in component form.
Solution:
$\overrightarrow{PQ} = \langle 3-1, 0-(-2), -1-4 \rangle = \langle 2, 2, -5 \rangle$
Answer: $\vec{v} = \langle 2, 2, -5 \rangle$ or $2\hat{i} + 2\hat{j} - 5\hat{k}$
Example 3: Magnitude of a Vector
Problem: Find the magnitude of $\vec{v} = \langle 3, -4, 12 \rangle$.
Solution:
$|\vec{v}| = \sqrt{3^2 + (-4)^2 + 12^2} = \sqrt{9 + 16 + 144} = \sqrt{169} = 13$
Answer: $|\vec{v}| = 13$
Problem: Find the magnitude of $\vec{v} = \langle 3, -4, 12 \rangle$.
Solution:
$|\vec{v}| = \sqrt{3^2 + (-4)^2 + 12^2} = \sqrt{9 + 16 + 144} = \sqrt{169} = 13$
Answer: $|\vec{v}| = 13$
Example 4: Unit Vector
Problem: Find the unit vector in the direction of $\vec{v} = \langle 6, -8 \rangle$.
Solution:
$|\vec{v}| = \sqrt{6^2 + (-8)^2} = \sqrt{36 + 64} = \sqrt{100} = 10$
$\hat{v} = \frac{\vec{v}}{|\vec{v}|} = \langle \frac{6}{10}, \frac{-8}{10} \rangle = \langle 0.6, -0.8 \rangle$
Answer: $\hat{v} = \langle 0.6, -0.8 \rangle$
Problem: Find the unit vector in the direction of $\vec{v} = \langle 6, -8 \rangle$.
Solution:
$|\vec{v}| = \sqrt{6^2 + (-8)^2} = \sqrt{36 + 64} = \sqrt{100} = 10$
$\hat{v} = \frac{\vec{v}}{|\vec{v}|} = \langle \frac{6}{10}, \frac{-8}{10} \rangle = \langle 0.6, -0.8 \rangle$
Answer: $\hat{v} = \langle 0.6, -0.8 \rangle$
Watch Out!
A vector has both magnitude and direction. Two vectors are equal only if they have the same magnitude AND the same direction. Position matters for location, but free vectors can be translated.
A vector has both magnitude and direction. Two vectors are equal only if they have the same magnitude AND the same direction. Position matters for location, but free vectors can be translated.
Practice for Vector Representation
- Find the vector from $A(3, -2)$ to $B(7, 1)$.
- Find the vector from $P(2, -1, 3)$ to $Q(5, 2, -4)$.
- Find the magnitude of $\vec{v} = \langle 5, 12 \rangle$.
- Find the magnitude of $\vec{v} = \langle 2, -3, 6 \rangle$.
- Find the unit vector in the direction of $\vec{v} = \langle 8, -15 \rangle$.
Vector Operations
Vector operations include addition, subtraction, and scalar multiplication. These operations follow component-wise rules and have important geometric interpretations.
Vector Operation Rules:
For vectors $\vec{u} = \langle u_x, u_y, u_z \rangle$, $\vec{v} = \langle v_x, v_y, v_z \rangle$, and scalar $k$:
• Addition: $\vec{u} + \vec{v} = \langle u_x + v_x, u_y + v_y, u_z + v_z \rangle$
• Subtraction: $\vec{u} - \vec{v} = \langle u_x - v_x, u_y - v_y, u_z - v_z \rangle$
• Scalar Multiplication: $k\vec{v} = \langle kv_x, kv_y, kv_z \rangle$
• Properties: Commutative, associative, distributive
For vectors $\vec{u} = \langle u_x, u_y, u_z \rangle$, $\vec{v} = \langle v_x, v_y, v_z \rangle$, and scalar $k$:
• Addition: $\vec{u} + \vec{v} = \langle u_x + v_x, u_y + v_y, u_z + v_z \rangle$
• Subtraction: $\vec{u} - \vec{v} = \langle u_x - v_x, u_y - v_y, u_z - v_z \rangle$
• Scalar Multiplication: $k\vec{v} = \langle kv_x, kv_y, kv_z \rangle$
• Properties: Commutative, associative, distributive
Example 1: Vector Addition
Problem: If $\vec{u} = \langle 2, -3, 5 \rangle$ and $\vec{v} = \langle 1, 4, -2 \rangle$, find $\vec{u} + \vec{v}$.
Solution:
$\vec{u} + \vec{v} = \langle 2+1, -3+4, 5+(-2) \rangle = \langle 3, 1, 3 \rangle$
Answer: $\langle 3, 1, 3 \rangle$
Problem: If $\vec{u} = \langle 2, -3, 5 \rangle$ and $\vec{v} = \langle 1, 4, -2 \rangle$, find $\vec{u} + \vec{v}$.
Solution:
$\vec{u} + \vec{v} = \langle 2+1, -3+4, 5+(-2) \rangle = \langle 3, 1, 3 \rangle$
Answer: $\langle 3, 1, 3 \rangle$
Example 2: Vector Subtraction
Problem: If $\vec{u} = \langle 5, -2, 3 \rangle$ and $\vec{v} = \langle 2, 4, -1 \rangle$, find $\vec{u} - \vec{v}$.
Solution:
$\vec{u} - \vec{v} = \langle 5-2, -2-4, 3-(-1) \rangle = \langle 3, -6, 4 \rangle$
Answer: $\langle 3, -6, 4 \rangle$
Problem: If $\vec{u} = \langle 5, -2, 3 \rangle$ and $\vec{v} = \langle 2, 4, -1 \rangle$, find $\vec{u} - \vec{v}$.
Solution:
$\vec{u} - \vec{v} = \langle 5-2, -2-4, 3-(-1) \rangle = \langle 3, -6, 4 \rangle$
Answer: $\langle 3, -6, 4 \rangle$
Example 3: Scalar Multiplication
Problem: If $\vec{v} = \langle 3, -2, 6 \rangle$, find $4\vec{v}$.
Solution:
$4\vec{v} = \langle 4 \times 3, 4 \times (-2), 4 \times 6 \rangle = \langle 12, -8, 24 \rangle$
Answer: $\langle 12, -8, 24 \rangle$
Problem: If $\vec{v} = \langle 3, -2, 6 \rangle$, find $4\vec{v}$.
Solution:
$4\vec{v} = \langle 4 \times 3, 4 \times (-2), 4 \times 6 \rangle = \langle 12, -8, 24 \rangle$
Answer: $\langle 12, -8, 24 \rangle$
Example 4: Linear Combination
Problem: If $\vec{u} = \langle 2, 1 \rangle$ and $\vec{v} = \langle -3, 4 \rangle$, find $2\vec{u} - 3\vec{v}$.
Solution:
$2\vec{u} = \langle 4, 2 \rangle$, $3\vec{v} = \langle -9, 12 \rangle$
$2\vec{u} - 3\vec{v} = \langle 4 - (-9), 2 - 12 \rangle = \langle 13, -10 \rangle$
Answer: $\langle 13, -10 \rangle$
Problem: If $\vec{u} = \langle 2, 1 \rangle$ and $\vec{v} = \langle -3, 4 \rangle$, find $2\vec{u} - 3\vec{v}$.
Solution:
$2\vec{u} = \langle 4, 2 \rangle$, $3\vec{v} = \langle -9, 12 \rangle$
$2\vec{u} - 3\vec{v} = \langle 4 - (-9), 2 - 12 \rangle = \langle 13, -10 \rangle$
Answer: $\langle 13, -10 \rangle$
Watch Out!
Vector addition is commutative ($\vec{u} + \vec{v} = \vec{v} + \vec{u}$) but vector subtraction is not ($\vec{u} - \vec{v} \neq \vec{v} - \vec{u}$).
Vector addition is commutative ($\vec{u} + \vec{v} = \vec{v} + \vec{u}$) but vector subtraction is not ($\vec{u} - \vec{v} \neq \vec{v} - \vec{u}$).
Practice for Vector Operations
- If $\vec{u} = \langle 4, -2, 7 \rangle$ and $\vec{v} = \langle 3, 5, -1 \rangle$, find $\vec{u} + \vec{v}$.
- If $\vec{u} = \langle 6, -3, 2 \rangle$ and $\vec{v} = \langle 4, 1, -5 \rangle$, find $\vec{u} - \vec{v}$.
- If $\vec{v} = \langle -2, 5, -3 \rangle$, find $3\vec{v}$.
- If $\vec{u} = \langle 1, 4 \rangle$ and $\vec{v} = \langle 2, -3 \rangle$, find $2\vec{u} + 5\vec{v}$.
- If $\vec{a} = \langle 3, -2, 1 \rangle$ and $\vec{b} = \langle -1, 4, 2 \rangle$, find $3\vec{a} - 2\vec{b}$.
The Dot Product (Scalar Product)
The dot product of two vectors produces a scalar (a number). It is used to find the angle between vectors, determine orthogonality, and calculate projections.
Dot Product Formulas
For $\vec{u} = \langle u_x, u_y, u_z \rangle$ and $\vec{v} = \langle v_x, v_y, v_z \rangle$:
$\vec{u} \cdot \vec{v} = u_x v_x + u_y v_y + u_z v_z$
$\vec{u} \cdot \vec{v} = |\vec{u}| |\vec{v}| \cos \theta$, where $\theta$ is the angle between them
$\cos \theta = \frac{\vec{u} \cdot \vec{v}}{|\vec{u}| |\vec{v}|}$
For $\vec{u} = \langle u_x, u_y, u_z \rangle$ and $\vec{v} = \langle v_x, v_y, v_z \rangle$:
$\vec{u} \cdot \vec{v} = u_x v_x + u_y v_y + u_z v_z$
$\vec{u} \cdot \vec{v} = |\vec{u}| |\vec{v}| \cos \theta$, where $\theta$ is the angle between them
$\cos \theta = \frac{\vec{u} \cdot \vec{v}}{|\vec{u}| |\vec{v}|}$
Properties of Dot Product:
1. $\vec{u} \cdot \vec{v} = \vec{v} \cdot \vec{u}$ (commutative)
2. $\vec{u} \cdot (\vec{v} + \vec{w}) = \vec{u} \cdot \vec{v} + \vec{u} \cdot \vec{w}$ (distributive)
3. $k(\vec{u} \cdot \vec{v}) = (k\vec{u}) \cdot \vec{v} = \vec{u} \cdot (k\vec{v})$
4. $\vec{u} \cdot \vec{u} = |\vec{u}|^2$
5. If $\vec{u} \cdot \vec{v} = 0$, then $\vec{u}$ and $\vec{v}$ are perpendicular (orthogonal).
1. $\vec{u} \cdot \vec{v} = \vec{v} \cdot \vec{u}$ (commutative)
2. $\vec{u} \cdot (\vec{v} + \vec{w}) = \vec{u} \cdot \vec{v} + \vec{u} \cdot \vec{w}$ (distributive)
3. $k(\vec{u} \cdot \vec{v}) = (k\vec{u}) \cdot \vec{v} = \vec{u} \cdot (k\vec{v})$
4. $\vec{u} \cdot \vec{u} = |\vec{u}|^2$
5. If $\vec{u} \cdot \vec{v} = 0$, then $\vec{u}$ and $\vec{v}$ are perpendicular (orthogonal).
Example 1: Calculating Dot Product
Problem: Find $\vec{u} \cdot \vec{v}$ if $\vec{u} = \langle 2, -3, 4 \rangle$ and $\vec{v} = \langle 5, 1, -2 \rangle$.
Solution:
$\vec{u} \cdot \vec{v} = (2)(5) + (-3)(1) + (4)(-2) = 10 - 3 - 8 = -1$
Answer: $-1$
Problem: Find $\vec{u} \cdot \vec{v}$ if $\vec{u} = \langle 2, -3, 4 \rangle$ and $\vec{v} = \langle 5, 1, -2 \rangle$.
Solution:
$\vec{u} \cdot \vec{v} = (2)(5) + (-3)(1) + (4)(-2) = 10 - 3 - 8 = -1$
Answer: $-1$
Example 2: Finding Angle Between Vectors
Problem: Find the angle between $\vec{u} = \langle 1, 0 \rangle$ and $\vec{v} = \langle 0, 1 \rangle$.
Solution:
$\vec{u} \cdot \vec{v} = 1(0) + 0(1) = 0$
$|\vec{u}| = 1$, $|\vec{v}| = 1$
$\cos \theta = \frac{0}{1 \times 1} = 0$ → $\theta = 90^\circ$
Answer: $90^\circ$ (vectors are perpendicular)
Problem: Find the angle between $\vec{u} = \langle 1, 0 \rangle$ and $\vec{v} = \langle 0, 1 \rangle$.
Solution:
$\vec{u} \cdot \vec{v} = 1(0) + 0(1) = 0$
$|\vec{u}| = 1$, $|\vec{v}| = 1$
$\cos \theta = \frac{0}{1 \times 1} = 0$ → $\theta = 90^\circ$
Answer: $90^\circ$ (vectors are perpendicular)
Example 3: Determining Orthogonality
Problem: Are $\vec{u} = \langle 3, -2, 1 \rangle$ and $\vec{v} = \langle 2, 3, 0 \rangle$ perpendicular?
Solution:
$\vec{u} \cdot \vec{v} = 3(2) + (-2)(3) + 1(0) = 6 - 6 + 0 = 0$
Yes, they are perpendicular.
Answer: Yes
Problem: Are $\vec{u} = \langle 3, -2, 1 \rangle$ and $\vec{v} = \langle 2, 3, 0 \rangle$ perpendicular?
Solution:
$\vec{u} \cdot \vec{v} = 3(2) + (-2)(3) + 1(0) = 6 - 6 + 0 = 0$
Yes, they are perpendicular.
Answer: Yes
Example 4: Dot Product with Itself
Problem: If $\vec{u} = \langle 3, -4 \rangle$, find $\vec{u} \cdot \vec{u}$.
Solution:
$\vec{u} \cdot \vec{u} = 3(3) + (-4)(-4) = 9 + 16 = 25 = |\vec{u}|^2$
Answer: $25$
Problem: If $\vec{u} = \langle 3, -4 \rangle$, find $\vec{u} \cdot \vec{u}$.
Solution:
$\vec{u} \cdot \vec{u} = 3(3) + (-4)(-4) = 9 + 16 = 25 = |\vec{u}|^2$
Answer: $25$
Practice for Dot Product
- Find $\vec{u} \cdot \vec{v}$ if $\vec{u} = \langle 4, -2, 3 \rangle$ and $\vec{v} = \langle 1, 5, -2 \rangle$.
- Find the angle between $\vec{u} = \langle 2, 3 \rangle$ and $\vec{v} = \langle 4, -1 \rangle$.
- Determine if $\vec{u} = \langle 1, 2, -1 \rangle$ and $\vec{v} = \langle 2, -1, 0 \rangle$ are perpendicular.
- If $\vec{u} = \langle 5, 12 \rangle$, find $\vec{u} \cdot \vec{u}$.
- Find the value of $k$ such that $\langle 2, k, 3 \rangle$ is perpendicular to $\langle 1, 2, -2 \rangle$.
The Cross Product (Vector Product)
The cross product of two 3D vectors produces a vector that is perpendicular to both original vectors. It is used to find normal vectors, area of parallelograms, and torque in physics.
Cross Product Formula
For $\vec{u} = \langle u_x, u_y, u_z \rangle$ and $\vec{v} = \langle v_x, v_y, v_z \rangle$ in 3D:
$\vec{u} \times \vec{v} = \langle u_y v_z - u_z v_y, u_z v_x - u_x v_z, u_x v_y - u_y v_x \rangle$
Or using determinant: $\vec{u} \times \vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ u_x & u_y & u_z \\ v_x & v_y & v_z \end{vmatrix}$
$|\vec{u} \times \vec{v}| = |\vec{u}| |\vec{v}| \sin \theta$ (magnitude gives area of parallelogram)
For $\vec{u} = \langle u_x, u_y, u_z \rangle$ and $\vec{v} = \langle v_x, v_y, v_z \rangle$ in 3D:
$\vec{u} \times \vec{v} = \langle u_y v_z - u_z v_y, u_z v_x - u_x v_z, u_x v_y - u_y v_x \rangle$
Or using determinant: $\vec{u} \times \vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ u_x & u_y & u_z \\ v_x & v_y & v_z \end{vmatrix}$
$|\vec{u} \times \vec{v}| = |\vec{u}| |\vec{v}| \sin \theta$ (magnitude gives area of parallelogram)
Properties of Cross Product:
1. $\vec{u} \times \vec{v} = -(\vec{v} \times \vec{u})$ (anti-commutative)
2. $\vec{u} \times \vec{u} = \vec{0}$
3. $\vec{u} \times \vec{v}$ is perpendicular to both $\vec{u}$ and $\vec{v}$
4. $\hat{i} \times \hat{j} = \hat{k}$, $\hat{j} \times \hat{k} = \hat{i}$, $\hat{k} \times \hat{i} = \hat{j}$
5. The magnitude $|\vec{u} \times \vec{v}|$ equals the area of the parallelogram formed by $\vec{u}$ and $\vec{v}$.
1. $\vec{u} \times \vec{v} = -(\vec{v} \times \vec{u})$ (anti-commutative)
2. $\vec{u} \times \vec{u} = \vec{0}$
3. $\vec{u} \times \vec{v}$ is perpendicular to both $\vec{u}$ and $\vec{v}$
4. $\hat{i} \times \hat{j} = \hat{k}$, $\hat{j} \times \hat{k} = \hat{i}$, $\hat{k} \times \hat{i} = \hat{j}$
5. The magnitude $|\vec{u} \times \vec{v}|$ equals the area of the parallelogram formed by $\vec{u}$ and $\vec{v}$.
Example 1: Calculating Cross Product
Problem: Find $\vec{u} \times \vec{v}$ if $\vec{u} = \langle 1, 2, 3 \rangle$ and $\vec{v} = \langle 4, 5, 6 \rangle$.
Solution:
$u_y v_z - u_z v_y = 2(6) - 3(5) = 12 - 15 = -3$
$u_z v_x - u_x v_z = 3(4) - 1(6) = 12 - 6 = 6$
$u_x v_y - u_y v_x = 1(5) - 2(4) = 5 - 8 = -3$
$\vec{u} \times \vec{v} = \langle -3, 6, -3 \rangle$
Answer: $\langle -3, 6, -3 \rangle$
Problem: Find $\vec{u} \times \vec{v}$ if $\vec{u} = \langle 1, 2, 3 \rangle$ and $\vec{v} = \langle 4, 5, 6 \rangle$.
Solution:
$u_y v_z - u_z v_y = 2(6) - 3(5) = 12 - 15 = -3$
$u_z v_x - u_x v_z = 3(4) - 1(6) = 12 - 6 = 6$
$u_x v_y - u_y v_x = 1(5) - 2(4) = 5 - 8 = -3$
$\vec{u} \times \vec{v} = \langle -3, 6, -3 \rangle$
Answer: $\langle -3, 6, -3 \rangle$
Example 2: Cross Product of Unit Vectors
Problem: Find $\hat{i} \times \hat{j}$.
Solution:
$\hat{i} = \langle 1, 0, 0 \rangle$, $\hat{j} = \langle 0, 1, 0 \rangle$
$\hat{i} \times \hat{j} = \langle (0)(0) - (0)(1), (0)(0) - (1)(0), (1)(1) - (0)(0) \rangle = \langle 0, 0, 1 \rangle = \hat{k}$
Answer: $\hat{k}$
Problem: Find $\hat{i} \times \hat{j}$.
Solution:
$\hat{i} = \langle 1, 0, 0 \rangle$, $\hat{j} = \langle 0, 1, 0 \rangle$
$\hat{i} \times \hat{j} = \langle (0)(0) - (0)(1), (0)(0) - (1)(0), (1)(1) - (0)(0) \rangle = \langle 0, 0, 1 \rangle = \hat{k}$
Answer: $\hat{k}$
Example 3: Area of Parallelogram
Problem: Find the area of the parallelogram formed by $\vec{u} = \langle 2, 0, 0 \rangle$ and $\vec{v} = \langle 0, 3, 0 \rangle$.
Solution:
$\vec{u} \times \vec{v} = \langle 0, 0, 6 \rangle$
Area = $|\vec{u} \times \vec{v}| = \sqrt{0^2 + 0^2 + 6^2} = 6$
Answer: 6 square units
Problem: Find the area of the parallelogram formed by $\vec{u} = \langle 2, 0, 0 \rangle$ and $\vec{v} = \langle 0, 3, 0 \rangle$.
Solution:
$\vec{u} \times \vec{v} = \langle 0, 0, 6 \rangle$
Area = $|\vec{u} \times \vec{v}| = \sqrt{0^2 + 0^2 + 6^2} = 6$
Answer: 6 square units
Example 4: Finding a Perpendicular Vector
Problem: Find a vector perpendicular to both $\vec{u} = \langle 1, 0, -1 \rangle$ and $\vec{v} = \langle 2, 1, 0 \rangle$.
Solution:
$\vec{u} \times \vec{v} = \langle (0)(0) - (-1)(1), (-1)(2) - (1)(0), (1)(1) - (0)(2) \rangle = \langle 1, -2, 1 \rangle$
Answer: $\langle 1, -2, 1 \rangle$ (or any scalar multiple)
Problem: Find a vector perpendicular to both $\vec{u} = \langle 1, 0, -1 \rangle$ and $\vec{v} = \langle 2, 1, 0 \rangle$.
Solution:
$\vec{u} \times \vec{v} = \langle (0)(0) - (-1)(1), (-1)(2) - (1)(0), (1)(1) - (0)(2) \rangle = \langle 1, -2, 1 \rangle$
Answer: $\langle 1, -2, 1 \rangle$ (or any scalar multiple)
Watch Out!
The cross product is only defined in 3 dimensions. It is anti-commutative: $\vec{u} \times \vec{v} = -(\vec{v} \times \vec{u})$. Also, the cross product of two parallel vectors is the zero vector.
The cross product is only defined in 3 dimensions. It is anti-commutative: $\vec{u} \times \vec{v} = -(\vec{v} \times \vec{u})$. Also, the cross product of two parallel vectors is the zero vector.
Practice for Cross Product
- Find $\vec{u} \times \vec{v}$ if $\vec{u} = \langle 2, -1, 3 \rangle$ and $\vec{v} = \langle 1, 4, -2 \rangle$.
- Find $\hat{j} \times \hat{k}$.
- Find a vector perpendicular to both $\langle 1, 2, 3 \rangle$ and $\langle 4, 5, 6 \rangle$.
- Find the area of the parallelogram formed by $\vec{u} = \langle 1, 2, 0 \rangle$ and $\vec{v} = \langle 3, 4, 0 \rangle$.
- Show that $\vec{u} \times \vec{v}$ is perpendicular to $\vec{u}$ by computing the dot product.
Advanced Vector Concepts
Advanced vector concepts include vector projections, scalar triple products, and applications in geometry such as finding distances and equations of lines and planes.
Vector Projection:
The projection of $\vec{u}$ onto $\vec{v}$ is: $\text{proj}_{\vec{v}} \vec{u} = \frac{\vec{u} \cdot \vec{v}}{|\vec{v}|^2} \vec{v}$
The scalar component (magnitude of projection) is: $\frac{\vec{u} \cdot \vec{v}}{|\vec{v}|}$
Scalar Triple Product: $\vec{u} \cdot (\vec{v} \times \vec{w})$ gives the volume of the parallelepiped formed by the three vectors.
The projection of $\vec{u}$ onto $\vec{v}$ is: $\text{proj}_{\vec{v}} \vec{u} = \frac{\vec{u} \cdot \vec{v}}{|\vec{v}|^2} \vec{v}$
The scalar component (magnitude of projection) is: $\frac{\vec{u} \cdot \vec{v}}{|\vec{v}|}$
Scalar Triple Product: $\vec{u} \cdot (\vec{v} \times \vec{w})$ gives the volume of the parallelepiped formed by the three vectors.
Example 1: Vector Projection
Problem: Find the projection of $\vec{u} = \langle 3, 4 \rangle$ onto $\vec{v} = \langle 1, 0 \rangle$.
Solution:
$\vec{u} \cdot \vec{v} = 3(1) + 4(0) = 3$, $|\vec{v}|^2 = 1$
$\text{proj}_{\vec{v}} \vec{u} = \frac{3}{1} \langle 1, 0 \rangle = \langle 3, 0 \rangle$
Answer: $\langle 3, 0 \rangle$
Problem: Find the projection of $\vec{u} = \langle 3, 4 \rangle$ onto $\vec{v} = \langle 1, 0 \rangle$.
Solution:
$\vec{u} \cdot \vec{v} = 3(1) + 4(0) = 3$, $|\vec{v}|^2 = 1$
$\text{proj}_{\vec{v}} \vec{u} = \frac{3}{1} \langle 1, 0 \rangle = \langle 3, 0 \rangle$
Answer: $\langle 3, 0 \rangle$
Example 2: Scalar Triple Product (Volume)
Problem: Find the volume of the parallelepiped formed by $\vec{u} = \langle 1, 0, 0 \rangle$, $\vec{v} = \langle 0, 1, 0 \rangle$, $\vec{w} = \langle 0, 0, 1 \rangle$.
Solution:
$\vec{v} \times \vec{w} = \hat{i}$
$\vec{u} \cdot (\vec{v} \times \vec{w}) = 1$
Volume = $|1| = 1$ cubic unit
Answer: 1
Problem: Find the volume of the parallelepiped formed by $\vec{u} = \langle 1, 0, 0 \rangle$, $\vec{v} = \langle 0, 1, 0 \rangle$, $\vec{w} = \langle 0, 0, 1 \rangle$.
Solution:
$\vec{v} \times \vec{w} = \hat{i}$
$\vec{u} \cdot (\vec{v} \times \vec{w}) = 1$
Volume = $|1| = 1$ cubic unit
Answer: 1
Example 3: Equation of a Line in 3D (Vector Form)
Problem: Find the vector equation of the line through $P(2, -1, 3)$ in the direction $\vec{d} = \langle 1, 4, -2 \rangle$.
Solution:
$\vec{r}(t) = \vec{r}_0 + t\vec{d} = \langle 2, -1, 3 \rangle + t\langle 1, 4, -2 \rangle = \langle 2 + t, -1 + 4t, 3 - 2t \rangle$
Answer: $\vec{r}(t) = \langle 2 + t, -1 + 4t, 3 - 2t \rangle$
Problem: Find the vector equation of the line through $P(2, -1, 3)$ in the direction $\vec{d} = \langle 1, 4, -2 \rangle$.
Solution:
$\vec{r}(t) = \vec{r}_0 + t\vec{d} = \langle 2, -1, 3 \rangle + t\langle 1, 4, -2 \rangle = \langle 2 + t, -1 + 4t, 3 - 2t \rangle$
Answer: $\vec{r}(t) = \langle 2 + t, -1 + 4t, 3 - 2t \rangle$
Example 4: Equation of a Plane
Problem: Find the equation of the plane through point $P(1, 2, 3)$ with normal vector $\vec{n} = \langle 2, -1, 4 \rangle$.
Solution:
$\vec{n} \cdot (\vec{r} - \vec{r}_0) = 0$ → $\langle 2, -1, 4 \rangle \cdot \langle x-1, y-2, z-3 \rangle = 0$
$2(x-1) - 1(y-2) + 4(z-3) = 0$ → $2x - 2 - y + 2 + 4z - 12 = 0$ → $2x - y + 4z - 12 = 0$
Answer: $2x - y + 4z = 12$
Problem: Find the equation of the plane through point $P(1, 2, 3)$ with normal vector $\vec{n} = \langle 2, -1, 4 \rangle$.
Solution:
$\vec{n} \cdot (\vec{r} - \vec{r}_0) = 0$ → $\langle 2, -1, 4 \rangle \cdot \langle x-1, y-2, z-3 \rangle = 0$
$2(x-1) - 1(y-2) + 4(z-3) = 0$ → $2x - 2 - y + 2 + 4z - 12 = 0$ → $2x - y + 4z - 12 = 0$
Answer: $2x - y + 4z = 12$
Practice for Advanced Concepts
- Find the projection of $\vec{u} = \langle 5, 12 \rangle$ onto $\vec{v} = \langle 1, 0 \rangle$.
- Find the scalar triple product $\vec{u} \cdot (\vec{v} \times \vec{w})$ for $\vec{u} = \langle 1, 2, 3 \rangle$, $\vec{v} = \langle 4, 5, 6 \rangle$, $\vec{w} = \langle 7, 8, 9 \rangle$.
- Write the vector equation of the line through $(1, -2, 4)$ with direction $\langle 3, 1, -5 \rangle$.
- Find the equation of the plane through $(2, -1, 1)$ with normal $\langle 1, 3, -2 \rangle$.
- Find the distance from the point $(1, 2, 3)$ to the plane $x + 2y + 2z = 10$ using projection.
Methods & Techniques
Mastering vectors requires systematic approaches and verification strategies. Use these techniques to ensure accuracy.
Verification / Checking Strategy:
1. For dot product: Verify that $\vec{u} \cdot \vec{v} = |\vec{u}||\vec{v}|\cos \theta$ by calculating both sides.
2. For cross product: Verify that $(\vec{u} \times \vec{v}) \cdot \vec{u} = 0$ and $(\vec{u} \times \vec{v}) \cdot \vec{v} = 0$.
3. For magnitude: Square the components and take the square root; check reasonableness.
4. For projections: Verify that the projection is parallel to $\vec{v}$ and that $\vec{u} - \text{proj}$ is perpendicular to $\vec{v}$.
1. For dot product: Verify that $\vec{u} \cdot \vec{v} = |\vec{u}||\vec{v}|\cos \theta$ by calculating both sides.
2. For cross product: Verify that $(\vec{u} \times \vec{v}) \cdot \vec{u} = 0$ and $(\vec{u} \times \vec{v}) \cdot \vec{v} = 0$.
3. For magnitude: Square the components and take the square root; check reasonableness.
4. For projections: Verify that the projection is parallel to $\vec{v}$ and that $\vec{u} - \text{proj}$ is perpendicular to $\vec{v}$.
Example: Checking Cross Product Perpendicularity
Original: $\vec{u} = \langle 1, 2, 3 \rangle$, $\vec{v} = \langle 4, 5, 6 \rangle$, $\vec{u} \times \vec{v} = \langle -3, 6, -3 \rangle$
Check:
$(\vec{u} \times \vec{v}) \cdot \vec{u} = (-3)(1) + (6)(2) + (-3)(3) = -3 + 12 - 9 = 0$ ✓
$(\vec{u} \times \vec{v}) \cdot \vec{v} = (-3)(4) + (6)(5) + (-3)(6) = -12 + 30 - 18 = 0$ ✓
Original: $\vec{u} = \langle 1, 2, 3 \rangle$, $\vec{v} = \langle 4, 5, 6 \rangle$, $\vec{u} \times \vec{v} = \langle -3, 6, -3 \rangle$
Check:
$(\vec{u} \times \vec{v}) \cdot \vec{u} = (-3)(1) + (6)(2) + (-3)(3) = -3 + 12 - 9 = 0$ ✓
$(\vec{u} \times \vec{v}) \cdot \vec{v} = (-3)(4) + (6)(5) + (-3)(6) = -12 + 30 - 18 = 0$ ✓
Common Pitfalls & How to Avoid Them:
• Pitfall 1: Confusing dot product (scalar) with cross product (vector) → Solution: Remember dot gives a number, cross gives a vector.
• Pitfall 2: Forgetting the anti-commutative property of cross product → Solution: $\vec{u} \times \vec{v} = -(\vec{v} \times \vec{u})$.
• Pitfall 3: Using cross product in 2D (not defined) → Solution: Cross product is only for 3D vectors.
• Pitfall 4: Sign errors in determinant calculation → Solution: Practice the determinant method carefully.
• Pitfall 5: Forgetting that $\vec{u} \cdot \vec{v} = 0$ implies perpendicular (not just any angle) → Solution: Only zero dot product gives $90^\circ$.
• Pitfall 1: Confusing dot product (scalar) with cross product (vector) → Solution: Remember dot gives a number, cross gives a vector.
• Pitfall 2: Forgetting the anti-commutative property of cross product → Solution: $\vec{u} \times \vec{v} = -(\vec{v} \times \vec{u})$.
• Pitfall 3: Using cross product in 2D (not defined) → Solution: Cross product is only for 3D vectors.
• Pitfall 4: Sign errors in determinant calculation → Solution: Practice the determinant method carefully.
• Pitfall 5: Forgetting that $\vec{u} \cdot \vec{v} = 0$ implies perpendicular (not just any angle) → Solution: Only zero dot product gives $90^\circ$.
Summary of Vector Products
| Product | Input | Output | Formula | Geometric Meaning |
|---|---|---|---|---|
| Dot Product | 2 vectors | Scalar | $\vec{u} \cdot \vec{v} = |\vec{u}||\vec{v}|\cos \theta$ | Angle between vectors, projection |
| Cross Product | 2 vectors (3D) | Vector | $|\vec{u} \times \vec{v}| = |\vec{u}||\vec{v}|\sin \theta$ | Perpendicular vector, area |
| Scalar Triple | 3 vectors | Scalar | $\vec{u} \cdot (\vec{v} \times \vec{w})$ | Volume of parallelepiped |
Technique Practice
- Verify that $\langle 1, 2, 3 \rangle \cdot \langle 4, 5, 6 \rangle = |\langle 1, 2, 3 \rangle||\langle 4, 5, 6 \rangle|\cos \theta$ by finding $\theta$.
- Check that $\langle 1, 0, 0 \rangle \times \langle 0, 1, 0 \rangle = \langle 0, 0, 1 \rangle$.
- Identify the error: A student wrote $\vec{u} \times \vec{v} = \vec{v} \times \vec{u}$. Correct this.
- For $\vec{u} = \langle 3, 4 \rangle$, verify that $\vec{u} \cdot \vec{u} = |\vec{u}|^2$.
Real-World Applications
Vectors are essential in physics, engineering, computer graphics, navigation, and many other fields.
Application 1: Physics - Forces
Scenario: Two forces act on an object: $\vec{F}_1 = \langle 10, 5 \rangle$ N and $\vec{F}_2 = \langle -3, 8 \rangle$ N. Find the resultant force.
Problem: Add the vectors.
Solution:
$\vec{F}_R = \langle 10 + (-3), 5 + 8 \rangle = \langle 7, 13 \rangle$ N
Magnitude: $|\vec{F}_R| = \sqrt{49 + 169} = \sqrt{218} \approx 14.76$ N
Scenario: Two forces act on an object: $\vec{F}_1 = \langle 10, 5 \rangle$ N and $\vec{F}_2 = \langle -3, 8 \rangle$ N. Find the resultant force.
Problem: Add the vectors.
Solution:
$\vec{F}_R = \langle 10 + (-3), 5 + 8 \rangle = \langle 7, 13 \rangle$ N
Magnitude: $|\vec{F}_R| = \sqrt{49 + 169} = \sqrt{218} \approx 14.76$ N
Application 2: Physics - Work
Scenario: A force $\vec{F} = \langle 5, 3 \rangle$ N moves an object along displacement $\vec{d} = \langle 4, 2 \rangle$ m. Find the work done.
Problem: Work = $\vec{F} \cdot \vec{d}$.
Solution:
Work = $5(4) + 3(2) = 20 + 6 = 26$ J
Answer: 26 Joules
Scenario: A force $\vec{F} = \langle 5, 3 \rangle$ N moves an object along displacement $\vec{d} = \langle 4, 2 \rangle$ m. Find the work done.
Problem: Work = $\vec{F} \cdot \vec{d}$.
Solution:
Work = $5(4) + 3(2) = 20 + 6 = 26$ J
Answer: 26 Joules
Application 3: Physics - Torque
Scenario: A force $\vec{F} = \langle 0, 0, 10 \rangle$ N is applied at position $\vec{r} = \langle 2, 3, 0 \rangle$ m from the pivot. Find torque.
Problem: Torque $\vec{\tau} = \vec{r} \times \vec{F}$.
Solution:
$\vec{\tau} = \langle 2, 3, 0 \rangle \times \langle 0, 0, 10 \rangle = \langle 3(10) - 0(0), 0(0) - 2(10), 2(0) - 3(0) \rangle = \langle 30, -20, 0 \rangle$ N·m
Answer: $\langle 30, -20, 0 \rangle$ N·m
Scenario: A force $\vec{F} = \langle 0, 0, 10 \rangle$ N is applied at position $\vec{r} = \langle 2, 3, 0 \rangle$ m from the pivot. Find torque.
Problem: Torque $\vec{\tau} = \vec{r} \times \vec{F}$.
Solution:
$\vec{\tau} = \langle 2, 3, 0 \rangle \times \langle 0, 0, 10 \rangle = \langle 3(10) - 0(0), 0(0) - 2(10), 2(0) - 3(0) \rangle = \langle 30, -20, 0 \rangle$ N·m
Answer: $\langle 30, -20, 0 \rangle$ N·m
Application 4: Navigation - Velocity
Scenario: A plane flies with velocity $\vec{v}_p = \langle 400, 0 \rangle$ km/h relative to air. Wind velocity is $\vec{v}_w = \langle 50, 30 \rangle$ km/h. Find ground velocity.
Problem: Ground velocity = $\vec{v}_p + \vec{v}_w$.
Solution:
$\vec{v}_g = \langle 400 + 50, 0 + 30 \rangle = \langle 450, 30 \rangle$ km/h
Speed = $\sqrt{450^2 + 30^2} = \sqrt{202500 + 900} = \sqrt{203400} \approx 451$ km/h
Scenario: A plane flies with velocity $\vec{v}_p = \langle 400, 0 \rangle$ km/h relative to air. Wind velocity is $\vec{v}_w = \langle 50, 30 \rangle$ km/h. Find ground velocity.
Problem: Ground velocity = $\vec{v}_p + \vec{v}_w$.
Solution:
$\vec{v}_g = \langle 400 + 50, 0 + 30 \rangle = \langle 450, 30 \rangle$ km/h
Speed = $\sqrt{450^2 + 30^2} = \sqrt{202500 + 900} = \sqrt{203400} \approx 451$ km/h
Cross-Curricular Connections
- Physics: Forces, velocity, acceleration, torque, work, magnetic fields
- Engineering: Structural analysis, robotics, fluid dynamics
- Computer Graphics: 3D modelling, lighting calculations, camera transformations
- Navigation: GPS coordinates, relative velocity, bearings
- Economics: Portfolio vectors, multi-dimensional data analysis
Cumulative Practice Exercises
Try these problems on your own. Show all working steps. Use the verification strategies to check your answers.
- Find the vector from $A(2, -3, 1)$ to $B(5, 0, -4)$.
- Find the magnitude of $\vec{v} = \langle 6, -8, 0 \rangle$.
- Find the unit vector in the direction of $\vec{v} = \langle 3, 4 \rangle$.
- If $\vec{u} = \langle 2, -5, 3 \rangle$ and $\vec{v} = \langle 4, 1, -2 \rangle$, find $\vec{u} + \vec{v}$.
- If $\vec{u} = \langle 3, -2, 1 \rangle$ and $\vec{v} = \langle 5, 4, -3 \rangle$, find $2\vec{u} - 3\vec{v}$.
- Find $\vec{u} \cdot \vec{v}$ for $\vec{u} = \langle 2, -3, 5 \rangle$ and $\vec{v} = \langle 1, 4, -2 \rangle$.
- Find the angle between $\vec{u} = \langle 1, 2 \rangle$ and $\vec{v} = \langle 3, 4 \rangle$.
- Determine if $\vec{u} = \langle 2, -1, 3 \rangle$ and $\vec{v} = \langle 1, 2, 0 \rangle$ are perpendicular.
- Find $\vec{u} \times \vec{v}$ for $\vec{u} = \langle 1, 0, 0 \rangle$ and $\vec{v} = \langle 0, 1, 0 \rangle$.
- Find the area of the parallelogram formed by $\vec{u} = \langle 2, 3 \rangle$ and $\vec{v} = \langle 4, 1 \rangle$ (treat as 2D vectors in the xy-plane).
- Find the projection of $\vec{u} = \langle 4, 3 \rangle$ onto $\vec{v} = \langle 1, 2 \rangle$.
- Find the vector equation of the line through $(1, -2, 3)$ in the direction $\langle 2, 1, -4 \rangle$.
- Find the equation of the plane through $(1, 2, -1)$ with normal $\langle 2, -3, 1 \rangle$.
- Error analysis: A student calculated $\vec{u} \times \vec{v}$ for $\vec{u} = \langle 1, 2, 3 \rangle$ and $\vec{v} = \langle 4, 5, 6 \rangle$ and got $\langle -3, 6, -3 \rangle$. Is this correct? Verify.
- A force $\vec{F} = \langle 6, 8 \rangle$ N moves an object along displacement $\vec{d} = \langle 5, 12 \rangle$ m. Find the work done.
Answers to Cumulative Exercises
- Problem: Vector from $A(2, -3, 1)$ to $B(5, 0, -4)$.
Answer: $\langle 3, 3, -5 \rangle$ - Problem: Magnitude of $\vec{v} = \langle 6, -8, 0 \rangle$.
Answer: $\sqrt{36 + 64} = \sqrt{100} = 10$ - Problem: Unit vector in direction of $\vec{v} = \langle 3, 4 \rangle$.
Answer: $\langle \frac{3}{5}, \frac{4}{5} \rangle$ - Problem: $\vec{u} + \vec{v}$ for $\vec{u} = \langle 2, -5, 3 \rangle$, $\vec{v} = \langle 4, 1, -2 \rangle$.
Answer: $\langle 6, -4, 1 \rangle$ - Problem: $2\vec{u} - 3\vec{v}$ for $\vec{u} = \langle 3, -2, 1 \rangle$, $\vec{v} = \langle 5, 4, -3 \rangle$.
Answer: $2\vec{u} = \langle 6, -4, 2 \rangle$, $3\vec{v} = \langle 15, 12, -9 \rangle$, $2\vec{u} - 3\vec{v} = \langle -9, -16, 11 \rangle$ - Problem: Dot product $\vec{u} \cdot \vec{v}$ for $\vec{u} = \langle 2, -3, 5 \rangle$, $\vec{v} = \langle 1, 4, -2 \rangle$.
Answer: $2(1) + (-3)(4) + 5(-2) = 2 - 12 - 10 = -20$ - Problem: Angle between $\vec{u} = \langle 1, 2 \rangle$ and $\vec{v} = \langle 3, 4 \rangle$.
Answer: $\cos \theta = \frac{1(3)+2(4)}{\sqrt{5}\sqrt{25}} = \frac{3+8}{5\sqrt{5}} = \frac{11}{5\sqrt{5}} \approx 0.9839$, $\theta \approx 10.3^\circ$ - Problem: Perpendicular? $\vec{u} = \langle 2, -1, 3 \rangle$, $\vec{v} = \langle 1, 2, 0 \rangle$.
Answer: $2(1) + (-1)(2) + 3(0) = 2 - 2 + 0 = 0$ → Yes, perpendicular. - Problem: $\vec{u} \times \vec{v}$ for $\vec{u} = \langle 1, 0, 0 \rangle$, $\vec{v} = \langle 0, 1, 0 \rangle$.
Answer: $\langle 0, 0, 1 \rangle$ - Problem: Area of parallelogram: $\vec{u} = \langle 2, 3 \rangle$, $\vec{v} = \langle 4, 1 \rangle$ (treat as 2D, area = $|u_x v_y - u_y v_x|$).
Answer: $|2(1) - 3(4)| = |2 - 12| = 10$ square units - Problem: Projection of $\vec{u} = \langle 4, 3 \rangle$ onto $\vec{v} = \langle 1, 2 \rangle$.
Answer: $\frac{4(1)+3(2)}{1^2+2^2} \langle 1, 2 \rangle = \frac{4+6}{5} \langle 1, 2 \rangle = 2\langle 1, 2 \rangle = \langle 2, 4 \rangle$ - Problem: Vector equation through $(1, -2, 3)$ direction $\langle 2, 1, -4 \rangle$.
Answer: $\vec{r}(t) = \langle 1 + 2t, -2 + t, 3 - 4t \rangle$ - Problem: Plane through $(1, 2, -1)$ with normal $\langle 2, -3, 1 \rangle$.
Answer: $2(x-1) - 3(y-2) + 1(z+1) = 0$ → $2x - 2 - 3y + 6 + z + 1 = 0$ → $2x - 3y + z + 5 = 0$ - Problem: Error analysis: $\vec{u} = \langle 1, 2, 3 \rangle$, $\vec{v} = \langle 4, 5, 6 \rangle$, student got $\langle -3, 6, -3 \rangle$.
Answer: Correct: $u_y v_z - u_z v_y = 2(6)-3(5)=12-15=-3$, $u_z v_x - u_x v_z = 3(4)-1(6)=12-6=6$, $u_x v_y - u_y v_x = 1(5)-2(4)=5-8=-3$. Student is correct. - Problem: Work done: $\vec{F} = \langle 6, 8 \rangle$ N, $\vec{d} = \langle 5, 12 \rangle$ m.
Answer: $W = \vec{F} \cdot \vec{d} = 6(5) + 8(12) = 30 + 96 = 126$ J
Conclusion & Summary
Vectors are fundamental mathematical tools for representing quantities with both magnitude and direction. Understanding vector operations, dot products, and cross products enables us to solve problems in geometry, physics, and engineering.
Key Takeaways:
1. Vector Representation: $\vec{v} = \langle v_x, v_y, v_z \rangle = v_x\hat{i} + v_y\hat{j} + v_z\hat{k}$
2. Magnitude: $|\vec{v}| = \sqrt{v_x^2 + v_y^2 + v_z^2}$
3. Dot Product: $\vec{u} \cdot \vec{v} = |\vec{u}||\vec{v}|\cos \theta$ — gives a scalar; zero means perpendicular.
4. Cross Product: $\vec{u} \times \vec{v}$ gives a vector perpendicular to both; magnitude gives area.
5. Projection: $\text{proj}_{\vec{v}} \vec{u} = \frac{\vec{u} \cdot \vec{v}}{|\vec{v}|^2} \vec{v}$
6. Applications: Forces, work, torque, navigation, 3D geometry.
Keep practising vector operations. Vectors are essential for calculus, physics, and engineering!
Video Resource
Watch this video for more examples of vector operations, dot product, and cross product.
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