Multivariable Calculus: Comprehensive Study Guide on Vectors, Dot Products, Direction Angles, and Vector Projections

Fundamentals of Vectors and Vector Algebra

• Definition of Vectors:

  • A vector is a directed line segment characterized by two fundamental properties: magnitude (also referred to as length or norm) and direction.
  • A vector originates at an initial point and terminates at a terminal point.
  • Geometrically, vectors represent directional displacement or movement from one point in space to another.

• The Zero Vector:

  • The zero vector, denoted $\mathbf{0}$, represents zero displacement and has a magnitude of $0$.
  • In two dimensions, its component form is $\mathbf{0} = (0, 0)$; in three dimensions, it is $\mathbf{0} = (0, 0, 0)$.
  • The zero vector acts as a point that moves nowhere and lacks a specific direction.

• Physical Applications:

  • Vectors are widely used in physics and engineering to represent vector quantities such as force, displacement, acceleration, and velocity.
  • The magnitude of a velocity vector represents speed, which isolates rate of motion from directional information.

• Vector Notation and Representation:

  • Component form using angled brackets: $\mathbf{a} = \langle a_1, a_2 \rangle$ in 2D or $\mathbf{a} = \langle a_1, a_2, a_3 \rangle$ in 3D (also written in parenthetical notation as $(a_1, a_2, a_3)$).
  • Standard basis unit vectors:
  • $\mathbf{i} = (1, 0, 0)$ (or $(1, 0)$ in 2D): A unit vector pointing $1\,\text{unit}$ in the positive x-direction.
  • $\mathbf{j} = (0, 1, 0)$ (or $(0, 1)$ in 2D): A unit vector pointing $1\,\text{unit}$ in the positive y-direction.
  • $\mathbf{k} = (0, 0, 1)$: A unit vector pointing $1\,\text{unit}$ in the positive z-direction.
  • Linear combination form: $\mathbf{a} = a_1\mathbf{i} + a_2\mathbf{j} + a_3\mathbf{k}$, where $a_1$, $a_2$, and $a_3$ are scalar values that scale the respective standard unit vectors.
  • The vector sum $\mathbf{a} = 3\mathbf{i} + 4\mathbf{j}$ represents a resultant vector formed by combining a rightward movement of $3\,\text{units}$ and an upward movement of $4\,\text{units}$.

• Vector Translation and Equivalence:

  • Vectors are free quantities; they can be translated anywhere in space without altering their value, provided their magnitude and direction remain unchanged.
  • Two vectors are equivalent if and only if their corresponding components are strictly equal.
  • Example: Starting at the initial point $(1, 1)$ and applying the vector displacement $\langle 3, 4 \rangle$ yields the terminal point $(1 + 3, 1 + 4) = (4, 5)$.

Vector Magnitude, Operations, and Algebraic Properties

• Vector Magnitude (Norm):

  • The magnitude of a vector is calculated using the distance formula derived from the Pythagorean theorem.
  • For a 2D vector $\mathbf{a} = \langle a_1, a_2 \rangle$:     $\|\mathbf{a}\| = \sqrt{a_1^2 + a_2^2}$     Example: For $\mathbf{a} = \langle 3, 4 \rangle$, the magnitude is $\|\mathbf{a}\| = \sqrt{3^2 + 4^2} = \sqrt{25} = 5$.
  • For a 3D vector $\mathbf{a} = \langle a_1, a_2, a_3 \rangle$:     $\|\mathbf{a}\| = \sqrt{a_1^2 + a_2^2 + a_3^2}$     Example: For $\mathbf{a} = \langle 3, 4, 1 \rangle$, the magnitude is $\|\mathbf{a}\| = \sqrt{3^2 + 4^2 + 1^2} = \sqrt{26}$.

• Vector Addition and Subtraction:

  • Vectors are added or subtracted by performing component-wise operations.
  • For $\mathbf{u} = (u_1, u_2, u_3)$ and $\mathbf{v} = (v_1, v_2, v_3)$:     $\mathbf{u} + \mathbf{v} = (u_1 + v_1, u_2 + v_2, u_3 + v_3)$$\mathbf{u} - \mathbf{v} = (u_1 - v_1, u_2 - v_2, u_3 - v_3)$

• Scalar Multiplication and Parallel Vectors:

  • Multiplying a vector $\mathbf{a}$ by a scalar $c$ scales its magnitude by $|c|$:     $c\mathbf{a} = (c a_1, c a_2, c a_3)$     Example: Scaling $\mathbf{a} = \langle 3, 4 \rangle$ by $2$ gives $2\mathbf{a} = \langle 6, 8 \rangle = 6\mathbf{i} + 8\mathbf{j}$.
  • If $c > 0$, the direction is unchanged; if $c < 0$, the direction is reversed.
  • Parallelism Definition: Two non-zero vectors $\mathbf{u}$ and $\mathbf{v}$ are parallel if and only if one is a scalar multiple of the other ($\mathbf{u} = c\mathbf{v}$ for some real scalar $c$).

• Fundamental Algebraic Properties of Vector Operations:   For any vectors $\mathbf{u}$, $\mathbf{v}$, $\mathbf{w}$ and scalars $m$, $n$:

  • Commutative Property of Addition: $\mathbf{u} + \mathbf{v} = \mathbf{v} + \mathbf{u}$
  • Associative Property of Addition: $(\mathbf{u} + \mathbf{v}) + \mathbf{w} = \mathbf{u} + (\mathbf{v} + \mathbf{w})$
  • Additive Identity Property: $\mathbf{u} + \mathbf{0} = \mathbf{u}$
  • Additive Inverse Property: $\mathbf{u} + (-\mathbf{u}) = \mathbf{0}$
  • Distributive Property over Vector Addition: $m(\mathbf{u} + \mathbf{v}) = m\mathbf{u} + m\mathbf{v}$
  • Distributive Property over Scalar Addition: $(m + n)\mathbf{u} = m\mathbf{u} + n\mathbf{u}$
  • Associative Property of Scalar Multiplication: $m(n\mathbf{u}) = (mn)\mathbf{u}$
  • Scalar Identity Properties: $1\mathbf{u} = \mathbf{u}$, $0\mathbf{u} = \mathbf{0}$

• Proof of the Associative Property of Vector Addition:

  • Objective: Prove that $\mathbf{a} + (\mathbf{b} + \mathbf{c}) = (\mathbf{a} + \mathbf{b}) + \mathbf{c}$ for 2D vectors $\mathbf{a} = (a_1, a_2)$, $\mathbf{b} = (b_1, b_2)$, and $\mathbf{c} = (c_1, c_2)$.
  • Step 1: Evaluate the inner addition: $\mathbf{b} + \mathbf{c} = (b_1 + c_1, b_2 + c_2)$.
  • Step 2: Add vector $\mathbf{a}$ to the result: $\mathbf{a} + (\mathbf{b} + \mathbf{c}) = (a_1 + (b_1 + c_1), a_2 + (b_2 + c_2))$
  • Step 3: Apply the associative property of addition for real numbers inside each component:     $(a_1 + (b_1 + c_1), a_2 + (b_2 + c_2)) = ((a_1 + b_1) + c_1, (a_2 + b_2) + c_2)$
  • Step 4: Decompose the combined component vector into a sum of two vectors:     $((a_1 + b_1) + c_1, (a_2 + b_2) + c_2) = (a_1 + b_1, a_2 + b_2) + (c_1, c_2)$
  • Step 5: Decompose the first vector into $\mathbf{a}$ and $\mathbf{b}$:     $(a_1 + b_1, a_2 + b_2) + (c_1, c_2) = ((a_1, a_2) + (b_1, b_2)) + (c_1, c_2) = (\mathbf{a} + \mathbf{b}) + \mathbf{c}$

Trigonometric Representation and Unit Vectors

• Trigonometric Form of 2D Vectors:

  • Let $\theta$ be the angle measured counterclockwise from the positive x-axis ($\theta = 0$).
  • Any 2D vector $\mathbf{v}$ with magnitude $\|\mathbf{v}\|$ can be expressed as:     $\mathbf{v} = \|\mathbf{v}\| \langle \cos(\theta), \sin(\theta) \rangle = \langle \|\mathbf{v}\| \cos(\theta), \|\mathbf{v}\| \sin(\theta) \rangle$
  • Example: For $\mathbf{v} = \langle 1, 1 \rangle$:
  • Magnitude: $\|\mathbf{v}\| = \sqrt{1^2 + 1^2} = \sqrt{2}$.
  • Angle: $\theta = 45^\circ$ or $\frac{\pi}{4}\,\text{radians}$.
  • Unit circle coordinates: $\left(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right)$.
  • Reconstruction: \mathbf{v} = \sqrt{2} \langle \cos(45^\circ), \sin(45^\circ) \rangle = \sqrt{2} \left\langle \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \right\rangle = \langle 1, 1 \rangle$.\n\n* **Unit Vector Normalization**:\n * A unit vector is a vector with a magnitude of exactly 1, isolating direction without carrying magnitude scaling.\n * To normalize a non-zero vector \mathbf{v}, divide it by its magnitude:\n    \mathbf{u} = \frac{\mathbf{v}}{|\mathbf{v}|}\n * Example 1: Find the unit vector parallel to \mathbf{v} = \langle 2, 5 \rangle:\n * Magnitude: |\mathbf{v}| = \sqrt{2^2 + 5^2} = \sqrt{29}.\n * Unit vector: \mathbf{u} = \left\langle \frac{2}{\sqrt{29}}, \frac{5}{\sqrt{29}} \right\rangle$.
  • Example 2: Find a unit vector in the opposite direction of $\mathbf{v} = (-1, 1, 1)$:
  • Magnitude: $\|\mathbf{v}\| = \sqrt{(-1)^2 + 1^2 + 1^2} = \sqrt{3}$.
  • Opposite unit vector: $\mathbf{u} = -\frac{\mathbf{v}}{\|\mathbf{v}\|} = \left(\frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}}\right)$.
  • Rationalized form: $\left(\frac{\sqrt{3}}{3}, -\frac{\sqrt{3}}{3}, -\frac{\sqrt{3}}{3}\right)$.

• Vector Construction Problems:

  • Directed Line Segment Vector: Find vector $\mathbf{a}$ from point $A(3, 1, -4)$ to point $B(-2, 5, 6)$:
  • Movement in x: $-2 - 3 = -5$
  • Movement in y: $5 - 1 = 4$
  • Movement in z: $6 - (-4) = 10$
  • Vector: $\mathbf{a} = (-5, 4, 10)$.
  • Magnitude: $\|\mathbf{a}\| = \sqrt{(-5)^2 + 4^2 + 10^2} = \sqrt{25 + 16 + 100} = \sqrt{141}$.
  • Linear Combination Calculation: Given $\mathbf{a} = (-5, 4, 10)$ and $\mathbf{b} = (2, 3, -1)$, evaluate $5\mathbf{a} - 2\mathbf{b}$.     $5(-5, 4, 10) - 2(2, 3, -1) = (-25, 20, 50) - (4, 6, -2) = (-29, 14, 52)$
  • Specified Magnitude in Given Direction: Find a vector in the same direction as $\vec{BC}$ with magnitude $\sqrt{7}$, where $B(2, 4, 0)$ and $C(0, 0, 5)$:
  • Direction vector: $\vec{BC} = (0 - 2, 0 - 4, 5 - 0) = (-2, -4, 5)$.
  • Magnitude: $\|\vec{BC}\| = \sqrt{(-2)^2 + (-4)^2 + 5^2} = \sqrt{4 + 16 + 25} = \sqrt{45}$.
  • Unit vector: $\mathbf{u} = \left(-\frac{2}{\sqrt{45}}, -\frac{4}{\sqrt{45}}, \frac{5}{\sqrt{45}}\right)$.
  • Scale by $\sqrt{7}$: $\mathbf{v} = \left(-\frac{2\sqrt{7}}{\sqrt{45}}, -\frac{4\sqrt{7}}{\sqrt{45}}, \frac{5\sqrt{7}}{\sqrt{45}}\right)$.
  • Simplify radicals ($\sqrt{45} = 3\sqrt{5}$):       $\mathbf{v} = \left(-\frac{2\sqrt{35}}{15}, -\frac{4\sqrt{35}}{15}, \frac{\sqrt{35}}{3}\right)$
  • Specified Length at Specified Angle: Find a vector of length $4$ at angle $\theta = \frac{\pi}{6}\,\text{radians}$ ($30^\circ$) with the positive x-axis:
  • Unit circle point at $\frac{\pi}{6}$: $\left(\frac{\sqrt{3}}{2}, \frac{1}{2}\right)$.
  • Scale by length $4$: $4 \left(\frac{\sqrt{3}}{2}, \frac{1}{2}\right) = (2\sqrt{3}, 2)$.
  • Angle Identification: Find the angle made by $\mathbf{v} = (-1, \sqrt{3})$ with the positive x-axis:
  • Magnitude: $\|\mathbf{v}\| = \sqrt{(-1)^2 + (\sqrt{3})^2} = \sqrt{4} = 2$.
  • Normalized point: $\left(-\frac{1}{2}, \frac{\sqrt{3}}{2}\right)$.
  • Reference angle: $\frac{\pi}{3}\,\text{radians}$ ($60^\circ$).
  • Located in Quadrant II ($x < 0, y > 0$): $\theta = \pi - \frac{\pi}{3} = \frac{2\pi}{3}\,\text{radians}$ ($120^\circ$).

The Dot Product and Angle Calculations

• Definition of the Dot Product:

  • The dot product (also known as the inner product) combines component values of two vectors to yield a single scalar value.
  • For 3D vectors $\mathbf{u} = (u_1, u_2, u_3)$ and $\mathbf{v} = (v_1, v_2, v_3)$:     $\mathbf{u} \cdot \mathbf{v} = u_1 v_1 + u_2 v_2 + u_3 v_3$
  • Crucial distinction: The output of a dot product is strictly a scalar, not a vector (contains no brackets, commas, or basis vectors $\mathbf{i}, \mathbf{j}, \mathbf{k}$).

• Geometric Formula for the Dot Product:

  • The dot product relates vector magnitudes and the angle $\theta$ between them ($0 \le \theta \le \pi$):     $\mathbf{a} \cdot \mathbf{b} = \|\mathbf{a}\| \|\mathbf{b}\| \cos(\theta)$
  • Solving for $\cos(\theta)$ (assuming non-zero vectors):     $\cos(\theta) = \frac{\mathbf{a} \cdot \mathbf{b}}{\|\mathbf{a}\| \|\mathbf{b}\|}$
  • Scaling by magnitudes ensures the value stays in the range $[-1, 1]$, allowing inverse cosine calculation.

• Geometric Interpretation of the Dot Product Sign:

  • Positive Dot Product ($\mathbf{a} \cdot \mathbf{b} > 0$): The angle $\theta$ is acute ($0^\circ \le \theta < 90^\circ$). Vectors point in generally the same direction.
  • Zero Dot Product ($\mathbf{a} \cdot \mathbf{b} = 0$): The angle is $\theta = 90^\circ$ ($\frac{\pi}{2}\,\text{radians}$). Non-zero vectors are orthogonal.
  • Negative Dot Product ($\mathbf{a} \cdot \mathbf{b} < 0$): The angle $\theta$ is obtuse ($90^\circ < \theta \le 180^\circ$). Vectors point in generally opposite directions.
  • Terminology Note: The term "perpendicular" is typically applied to intersecting lines in geometry, whereas "orthogonal" is the standard term for vectors in multivariable calculus and physics.

• Algebraic Properties of the Dot Product:

  • Commutative Property: $\mathbf{a} \cdot \mathbf{b} = \mathbf{b} \cdot \mathbf{a}$
  • Scalar Associativity/Distributivity: $c(\mathbf{a} \cdot \mathbf{b}) = (c\mathbf{a}) \cdot \mathbf{b} = \mathbf{a} \cdot (c\mathbf{b})$
  • Dot Product with Self: $\mathbf{a} \cdot \mathbf{a} = \|\mathbf{a}\|^2$ (since $a_1^2 + a_2^2 + a_3^2 = \|\mathbf{a}\|^2$)
  • Dot Product with Zero Vector: $\mathbf{a} \cdot \mathbf{0} = 0$

• Worked Numerical Example:

  • Given vectors $\mathbf{a} = (3, -4, 2)$ and $\mathbf{b} = (1, 2, -5)$:
  • Vector Addition: $\mathbf{a} + \mathbf{b} = (3+1, -4+2, 2-5) = (4, -2, -3) = 4\mathbf{i} - 2\mathbf{j} - 3\mathbf{k}$
  • Vector Combination: $5\mathbf{a} - 4\mathbf{b} = 5(3, -4, 2) - 4(1, 2, -5) = (15, -20, 10) - (4, 8, -20) = (11, -28, 30)$
  • Dot Product Computation:     $\mathbf{a} \cdot \mathbf{b} = 3(1) + (-4)(2) + 2(-5) = 3 - 8 - 10 = -15$
  • Magnitude Computations:     $\|\mathbf{a}\| = \sqrt{3^2 + (-4)^2 + 2^2} = \sqrt{9 + 16 + 4} = \sqrt{29}$$\|\mathbf{b}\| = \sqrt{1^2 + 2^2 + (-5)^2} = \sqrt{1 + 4 + 25} = \sqrt{30}$
  • Cosine of Inter-Vector Angle:     $\cos(\theta) = \frac{-15}{\sqrt{29} \sqrt{30}}$
  • Since the dot product is negative ($-15$), the angle $\theta$ between $\mathbf{a}$ and $\mathbf{b}$ is obtuse.

Direction Angles and Direction Cosines

• Definitions:

  • In a 3D Cartesian coordinate system, the orientation of a vector $\mathbf{a} = (a_1, a_2, a_3)$ relative to the coordinate axes is defined by three direction angles:
  • $\alpha$: Angle made with the positive x-axis.
  • $\beta$: Angle made with the positive y-axis.
  • $\gamma$: Angle made with the positive z-axis.
  • The cosines of these angles are known as direction cosines.

• Derivation via Standard Basis Vectors:

  • To find $\alpha$, consider the positive x-axis unit vector $\mathbf{i} = (1, 0, 0)$ (with magnitude $\|\mathbf{i}\| = 1$).
  • Dot product: $\mathbf{a} \cdot \mathbf{i} = a_1(1) + a_2(0) + a_3(0) = a_1$.
  • Applying the angle formula:     $\cos(\alpha) = \frac{\mathbf{a} \cdot \mathbf{i}}{\|\mathbf{a}\| \|\mathbf{i}\|} = \frac{a_1}{\|\mathbf{a}\|}$
  • Similarly, using $\mathbf{j} = (0, 1, 0)$ and $\mathbf{k} = (0, 0, 1)$:     $\cos(\beta) = \frac{a_2}{\|\mathbf{a}\|}$$\cos(\gamma) = \frac{a_3}{\|\mathbf{a}\|}$

• Comprehensive Numerical Example:

  • Consider the 3D vector $\mathbf{a} = (-2, 1, 5)$:
  • Magnitude: $\|\mathbf{a}\| = \sqrt{(-2)^2 + 1^2 + 5^2} = \sqrt{4 + 1 + 25} = \sqrt{30}$.
  • Positive x-axis Angle ($\alpha$):
  • Direction cosine: $\cos(\alpha) = \frac{-2}{\sqrt{30}}$
  • The negative value indicates an obtuse angle ($\alpha > 90^\circ$).
  • Angle calculation: $\alpha = \arccos\left(\frac{-2}{\sqrt{30}}\right) \approx 111.42^\circ$
  • Positive y-axis Angle ($\beta$):
  • Direction cosine: $\cos(\beta) = \frac{1}{\sqrt{30}}$
  • The positive value indicates an acute angle.
  • Angle calculation: $\beta = \arccos\left(\frac{1}{\sqrt{30}}\right) \approx 79.48^\circ$
  • Positive z-axis Angle ($\gamma$):
  • Direction cosine: $\cos(\gamma) = \frac{5}{\sqrt{30}}$
  • Angle calculation: $\gamma = \arccos\left(\frac{5}{\sqrt{30}}\right) \approx 24.09^\circ$

Vector Projections and Scalar Components

• Conceptual Foundations:

  • Consider two vectors $\mathbf{a}$ and $\mathbf{b}$ sharing the same initial point.
  • Imagine a light source positioned straight above the terminal point of $\mathbf{b}$, shining perpendicularly down onto the line containing $\mathbf{a}$.
  • The shadow cast by $\mathbf{b}$ onto $\mathbf{a}$ forms a new vector laying directly on top of $\mathbf{a}$ (parallel to $\mathbf{a}$).
  • This shadow vector is the vector projection of $\mathbf{b}$ onto $\mathbf{a}$, denoted $\text{proj}_{\mathbf{a}}\mathbf{b}$.
  • The signed length/magnitude of this shadow vector is the scalar component of $\mathbf{b}$ along $\mathbf{a}$, denoted $\text{comp}_{\mathbf{a}}\mathbf{b}$.

• Derivation of the Scalar Component Formula:

  • In the right triangle formed by vector $\mathbf{b}$, the projection shadow, and the perpendicular light ray:     $\cos(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{\text{comp}_{\mathbf{a}}\mathbf{b}}{\|\mathbf{b}\|}$
  • Solving for the scalar component:     $\text{comp}_{\mathbf{a}}\mathbf{b} = \|\mathbf{b}\| \cos(\theta)$
  • Substituting $\cos(\theta) = \frac{\mathbf{a} \cdot \mathbf{b}}{\|\mathbf{a}\| \|\mathbf{b}\|}$$\text{comp}_{\mathbf{a}}\mathbf{b} = \|\mathbf{b}\| \left( \frac{\mathbf{a} \cdot \mathbf{b}}{\|\mathbf{a}\| \|\mathbf{b}\|} \right) = \frac{\mathbf{a} \cdot \mathbf{b}}{\|\mathbf{a}\|}$

• Derivation of the Vector Projection Formula:

  • The projection vector $\text{proj}_{\mathbf{a}}\mathbf{b}$ has length $\text{comp}_{\mathbf{a}}\mathbf{b}$ and points in the direction of the unit vector parallel to $\mathbf{a}$, which is $\frac{\mathbf{a}}{\|\mathbf{a}\|}$.
  • Multiplying length by direction:     $\text{proj}_{\mathbf{a}}\mathbf{b} = \left(\text{comp}_{\mathbf{a}}\mathbf{b}\right) \frac{\mathbf{a}}{\|\mathbf{a}\|} = \left(\frac{\mathbf{a} \cdot \mathbf{b}}{\|\mathbf{a}\|}\right) \frac{\mathbf{a}}{\|\mathbf{a}\|} = \frac{\mathbf{a} \cdot \mathbf{b}}{\|\mathbf{a}\|^2} \mathbf{a}$

• Detailed Step-by-Step Examples:

  • Example 1: Projecting $\mathbf{x}$ onto $\mathbf{y}$:
  • Given $\mathbf{x} = (3, 1, -4)$ and $\mathbf{y} = (2, 3, -4)$:
  • Dot product: $\mathbf{x} \cdot \mathbf{y} = 3(2) + 1(3) + (-4)(1) = 6 + 3 - 4 = 5$
  • Magnitude of $\mathbf{y}$: $\|\mathbf{y}\| = \sqrt{2^2 + 3^2 + (-4)^2} = \sqrt{4 + 9 + 16} = \sqrt{29}$
  • Scalar component of $\mathbf{x}$ along $\mathbf{y}$:       $\text{comp}_{\mathbf{y}}\mathbf{x} = \frac{5}{\sqrt{29}}$
  • Example 2: Projecting $\mathbf{y}$ onto $\mathbf{x}$:
  • Let $\mathbf{x} = (3, 1, 1)$ and $\mathbf{y}$ such that \mathbf{x} \cdot \mathbf{y} = 5$.\n * Magnitude of \mathbf{x}$:$|\mathbf{x}| = \sqrt{3^2 + 1^2 + 1^2} = \sqrt{9 + 1 + 1} = \sqrt{11}\n * Scalar component of \mathbf{y}$along$\mathbf{x}:\n      \text{comp}{\mathbf{x}}\mathbf{y} = \frac{5}{\sqrt{11}}\n * Vector projection of \mathbf{y}$onto$\mathbf{x}:\n      \text{proj}{\mathbf{x}}\mathbf{y} = \frac{\mathbf{x} \cdot \mathbf{y}}{|\mathbf{x}|^2} \mathbf{x} = \frac{5}{(\sqrt{11})^2} (3, 1, 1) = \frac{5}{11} (3, 1, 1) = \left( \frac{15}{11}, \frac{5}{11}, \frac{5}{11} \right)\n * Notice that \left(\frac{15}{11}, \frac{5}{11}, \frac{5}{11}\right)$is parallel to$\mathbf{x} = (3, 1, 1)$scaled by$\frac{5}{11}.\n\n* **Conditions for Equality of Projections**:\n * Equality of Scalar Components (\text{comp}{\mathbf{a}}\mathbf{b} = \text{comp}{\mathbf{b}}\mathbf{a}):\n * Formula equivalence: \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}|} = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{b}|}.\n * Since \mathbf{a} \cdot \mathbf{b} = \mathbf{b} \cdot \mathbf{a}, equality holds for non-zero vectors if and only if their magnitudes are equal:\n      |\mathbf{a}| = |\mathbf{b}|\n * Equality of Projection Vectors (\text{proj}{\mathbf{a}}\mathbf{b} = \text{proj}{\mathbf{b}}\mathbf{a}):\n * Case 1: Vectors \mathbf{a}$and$\mathbf{b}$are **orthogonal** ($\mathbf{a} \cdot \mathbf{b} = 0$). Both projection vectors equal the zero vector$\mathbf{0}.\n * Case 2: Vectors \mathbf{a}$and$\mathbf{b}$are **identical** ($\mathbf{a} = \mathbf{b}$$). They possess identical magnitudes and directions.