Comprehensive Guide to Two-Dimensional Vectors and Foundations for 3D Vector Calculus

Overview and Study Recommendations for Two-Dimensional Vectors

• Two-dimensional ($2\text{D}$) vector operations serve as the primary conceptual foundation required for transitioning into three-dimensional ($3\text{D}$) vector applications.
• Recommended student workflow when working through practice problems:
  • Pause before attempting each problem.
  • Solve the problem independently.
  • Review the step-by-step solution to confirm both the final result and the conceptual approach.
  • Pay close attention to conceptual remarks and structural techniques, as these extend directly to higher dimensions where visual diagrams are no longer feasible.

Displacement Vectors and Position Vectors Between Points

• Position Vectors:
  • The position vector of a point $A(x, y)$ relative to the origin $O(0, 0)$ is denoted as $OA = \begin{pmatrix} x \\ y \end{pmatrix}$.
• Displacement Vector Between Two Points Formula:
  • Given two points $A$ and $B$ with position vectors $OA$ and $OB$:     $AB = OB - OA$
  • Rule of thumb: Subtract the position vector of the first point (start point) from the position vector of the second point (end point) — "second letter minus first letter".
• Vector Reversal Property:
  • Reversing the direction of a displacement vector negates its components:     $BA = -AB$$CO = -OC$
• Comprehensive Worked Example with Given Points:
  • Given points: $O(0, 0)$, $A(1, -2)$, $B(3, 0)$, and $C(2, -3)$.
  • Position vector $OA$:     $OA = \begin{pmatrix} 1 \\ -2 \end{pmatrix}$
  • Finding vector $AB$:     $AB = OB - OA = \begin{pmatrix} 3 \\ 0 \end{pmatrix} - \begin{pmatrix} 1 \\ -2 \end{pmatrix} = \begin{pmatrix} 3 - 1 \\ 0 - (-2) \end{pmatrix} = \begin{pmatrix} 2 \\ 2 \end{pmatrix}$
  • Visual Verification: Moving from point $A(1, -2)$ to point $B(3, 0)$ requires moving $2\,\text{units}$ right ($+2$) and $2\,\text{units}$ up ($+2$).
  • Finding vector $BC$:     $BC = OC - OB = \begin{pmatrix} 2 \\ -3 \end{pmatrix} - \begin{pmatrix} 3 \\ 0 \end{pmatrix} = \begin{pmatrix} 2 - 3 \\ -3 - 0 \end{pmatrix} = \begin{pmatrix} -1 \\ -3 \end{pmatrix}$
  • Visual Verification: Moving from point $B(3, 0)$ to point $C(2, -3)$ requires moving $1\,\text{unit}$ left ($-1$) and $3\,\text{units}$ down ($-3$).
  • Finding vector $CO$:     $CO = -OC = -\begin{pmatrix} 2 \\ -3 \end{pmatrix} = \begin{pmatrix} -2 \\ 3 \end{pmatrix}$
  • Finding vector $CB$:
  • Method 1: $CB = OB - OC = \begin{pmatrix} 3 \\ 0 \end{pmatrix} - \begin{pmatrix} 2 \\ -3 \end{pmatrix} = \begin{pmatrix} 1 \\ 3 \end{pmatrix}$
  • Method 2: $CB = -BC = -\begin{pmatrix} -1 \\ -3 \end{pmatrix} = \begin{pmatrix} 1 \\ 3 \end{pmatrix}$

Vector Operations: Addition, Subtraction, and Scalar Multiplication

• Vector Addition:
  • Performed component-wise:     $\begin{pmatrix} a_1 \\ a_2 \end{pmatrix} + \begin{pmatrix} b_1 \\ b_2 \end{pmatrix} = \begin{pmatrix} a_1 + b_1 \\ a_2 + b_2 \end{pmatrix}$
  • Example: Given $a = \begin{pmatrix} 1 \\ 2 \end{pmatrix}$ and $b = \begin{pmatrix} 1 \\ -3 \end{pmatrix}$:     $a + b = \begin{pmatrix} 1 + 1 \\ 2 + (-3) \end{pmatrix} = \begin{pmatrix} 2 \\ -1 \end{pmatrix}$
• Scalar Multiplication and Subtraction Combined:
  • Given $a = \begin{pmatrix} 1 \\ 2 \end{pmatrix}$ and $b = \begin{pmatrix} -2 \\ 1 \end{pmatrix}$:
  • Evaluate $2b - a$:     $2b = 2 \begin{pmatrix} -2 \\ 1 \end{pmatrix} = \begin{pmatrix} -4 \\ 2 \end{pmatrix}$$2b - a = \begin{pmatrix} -4 \\ 2 \end{pmatrix} - \begin{pmatrix} 1 \\ 2 \end{pmatrix} = \begin{pmatrix} -4 - 1 \\ 2 - 2 \end{pmatrix} = \begin{pmatrix} -5 \\ 0 \end{pmatrix}$

Vector Geometry and Geometric Proofs in Two Dimensions

• Vector Addition and Midpoint Traversal:
  • Vectors can be expressed in terms of defined side vectors in geometric figures.
  • Opposite sides of a parallelogram are equal in both magnitude and direction, representing identical vectors.
• Worked Example:
  • Let quadrilateral $ABCD$ have side vectors $AB = a$ and $AD = b$.
  • Let $M$ be the midpoint of side $AB$.
  • Express vector $MD$ in terms of $a$ and $b$:
  • Traversal path from $M$ to $D$:
    • Travel from $M$ to $A$: Since $M$ is the midpoint of $AB$, vector $MA = -\frac{1}{2}a$
    • Travel from $A$ to $D$: Vector $AD = b$
  • Summing the path:       $MD = MA + AD = -\frac{1}{2}a + b$
  • Express vector $MN$ (where $N$ is the midpoint of side $DC$) in terms of $a$ and $b$:
  • Traversal path from $M$ to $N$:
    • Travel from $M$ to $A$: $-\frac{1}{2}a$
    • Travel from $A$ to $D$: $b$
    • Travel from $D$ to $N$: $+\frac{1}{2}a$ (since side $DC$ is parallel and equal to $AB = a$)
  • Summing and simplifying:       $MN = -\frac{1}{2}a + b + \frac{1}{2}a = b$

Component Form and Vector Magnitude

• Unit Vector Bracket Notation ($i, j$ Form):
  • Any $2\text{D}$ vector can be expressed in terms of orthogonal unit vectors $i$ ($1\,\text{unit}$ along $x$-axis) and $j$ ($1\,\text{unit}$ along $y$-axis):     $v = v_x i + v_y j$
  • Subtraction with Brackets Warning:
  • When computing $AB = OB - OA$, always enclose $OA$ in brackets to distribute the negative sign to both components.
  • Example: Given $OA = i + 2j$ and $OB = 3i - 5j$:       $AB = (3i - 5j) - (i + 2j) = 3i - 5j - i - 2j = 2i - 7j$
  • Verification: Starting at $x = 1$ to $x = 3$ is $+2\,\text{units}$; starting at $y = 2$ down to $y = -5$ is $-7\,\text{units}$.
• Vector Magnitude Formula:
  • The magnitude (length) of vector $v = v_x i + v_y j$ is denoted by $|v|$:     $|v| = \sqrt{v_x^2 + v_y^2}$
  • Important Calculator Caveat: Always enclose negative components inside parentheses prior to squaring (e.g., $(-5)^2$) to avoid evaluation errors.
• Worked Examples for Magnitude:
  • Example 1: Given vector $v = 12i - 5j$:     $|v| = \sqrt{12^2 + (-5)^2} = \sqrt{144 + 25} = \sqrt{169} = 13$
  • Example 2: Find $AB$ and $|AB|$ for $A(2, 3)$ and $B(5, 1)$:     $AB = \begin{pmatrix} 5 \\ 1 \end{pmatrix} - \begin{pmatrix} 2 \\ 3 \end{pmatrix} = \begin{pmatrix} 3 \\ -2 \end{pmatrix} = 3i - 2j$$|AB| = \sqrt{3^2 + (-2)^2} = \sqrt{9 + 4} = \sqrt{13}$
  • Note on Exact Values: Keep answers in exact surd form ($\sqrt{13}$) rather than converting to decimals.
  • Reasonableness Check: Since $\sqrt{9} < \sqrt{13} < \sqrt{16}$, the length must lie between $3$ and $4$, which matches a displacement of $3\,\text{units}$ across and $2\,\text{units}$ down.

Unit Vectors

• Definition:
  • A unit vector is a vector scaled to have a magnitude of exactly $1\,\text{unit}$ while preserving its original direction.
  • Designated using a hat symbol: $\hat{b}$ or $\hat{v}$.
• Formula:   $\hat{b} = \frac{1}{|b|} b$
• Worked Examples:
  • Example 1: Find the unit vector for $b = 3i - j$:
  • Step 1: Calculate magnitude:       $|b| = \sqrt{3^2 + (-1)^2} = \sqrt{9 + 1} = \sqrt{10}$
  • Step 2: Apply unit vector formula:       $\hat{b} = \frac{1}{\sqrt{10}}(3i - j) = \frac{3}{\sqrt{10}}i - \frac{1}{\sqrt{10}}j$
  • Example 2: Find the unit vector for $b = 6i - 4j$:
  • Step 1: Calculate magnitude:       $|b| = \sqrt{6^2 + (-4)^2} = \sqrt{36 + 16} = \sqrt{52} = \sqrt{4 \times 13} = 2\sqrt{13}$
  • Step 2: Apply unit vector formula:       $\hat{b} = \frac{1}{2\sqrt{13}}(6i - 4j) = \frac{6}{2\sqrt{13}}i - \frac{4}{2\sqrt{13}}j$
  • Step 3: Simplify fractions:       $\hat{b} = \frac{3}{\sqrt{13}}i - \frac{2}{\sqrt{13}}j$

Polar Form and Component Form Conversions

• Converting Polar Form $(r \angle \theta)$ to Component Form ($i, j$ Form):
  • General Formula:     $u = |u|\cos(\theta)i + |u|\sin(\theta)j$
  • Derivation: Right-angled triangle trigonometry where the adjacent side along $x$ is $|u|\cos(\theta)$ and the opposite side along $y$ is $|u|\sin(\theta)$.
  • Advantage of Using Direct Rotational Angle: Substituting the full angle $\theta$ (including negative signs for clockwise rotation) automatically determines both the magnitude and correct sign ($+\text{ or }-$) for each component.
• Conversion Examples:
  • Example 1: Magnitude $|u| = 4$, angle $\theta = -120^\circ$:
  • Formula setup:       $u = 4\cos(-120^\circ)i + 4\sin(-120^\circ)j$
  • Reference angle analysis:
    • Quadrant III (both cosine and sine are negative). Reference angle $= 60^\circ$.
    • Standard $30^\circ-60^\circ-90^\circ$ triangle sides: $1, 2, \sqrt{3}$.       $\cos(-120^\circ) = -\cos(60^\circ) = -\frac{1}{2}$$\sin(-120^\circ) = -\sin(60^\circ) = -\frac{\sqrt{3}}{2}$
  • Component substitution:       $u = 4\left(-\frac{1}{2}\right)i + 4\left(-\frac{\sqrt{3}}{2}\right)j = -2i - 2\sqrt{3}j$
  • Example 2: Magnitude $|v| = 4$, angle $\theta = 150^\circ$:
  • Formula setup:       $v = 4\cos(150^\circ)i + 4\sin(150^\circ)j$
  • Reference angle analysis:
    • Quadrant II (cosine negative, sine positive). Reference angle $= 30^\circ$.       $\cos(150^\circ) = -\cos(30^\circ) = -\frac{\sqrt{3}}{2}$$\sin(150^\circ) = \sin(30^\circ) = \frac{1}{2}$
  • Component substitution:       $v = 4\left(-\frac{\sqrt{3}}{2}\right)i + 4\left(\frac{1}{2}\right)j = -2\sqrt{3}i + 2j$
• Converting Component Form to Polar Form:
  • Example 1: Convert $v = -2\sqrt{3}i + 2j$ into polar form:
  • Magnitude:       $|v| = \sqrt{(-2\sqrt{3})^2 + 2^2} = \sqrt{4 \times 3 + 4} = \sqrt{12 + 4} = \sqrt{16} = 4$
  • Angle determination:
    • Plotting components places vector in Quadrant II ($-x, +y$).       $\tan(\theta_{\text{ref}}) = \frac{|y|}{|x|} = \frac{2}{2\sqrt{3}} = \frac{1}{\sqrt{3}}$$\theta_{\text{ref}} = \tan^{-1}\left(\frac{1}{\sqrt{3}}\right) = 30^\circ$$\theta = 180^\circ - 30^\circ = 150^\circ$
  • Polar form result: $4 \angle 150^\circ$
  • Example 2: Convert $v = -5i - 12j$ into polar form:
  • Magnitude:       $|v| = \sqrt{(-5)^2 + (-12)^2} = \sqrt{25 + 144} = \sqrt{169} = 13$
  • Angle determination:
    • Vector lies in Quadrant III ($-x, -y$).       $\tan(\theta_{\text{ref}}) = \frac{12}{5}$$\theta_{\text{ref}} = \tan^{-1}\left(\frac{12}{5}\right) \approx 67.38^\circ$$\theta = 180^\circ - 67.38^\circ = 112.62^\circ$
  • Polar form result: $13 \angle 112.62^\circ$

The Dot Product (Scalar Product)

• Algebraic Definition (Component Form):
  • For two vectors $a = a_x i + a_y j$ and $b = b_x i + b_y j$:     $a \cdot b = a_x b_x + a_y b_y$
  • The dot product always yields a scalar quantity (a real number), not a vector.
• Algebraic Examples:
  • Example 1: Given $a = 2i - j$ and $b = 3i - 2j$:     $a \cdot b = (2)(3) + (-1)(-2) = 6 + 2 = 8$
  • Example 2: Given $a = 2i - j$ and $c = -i + 3j$:     $a \cdot c = (2)(-1) + (-1)(3) = -2 + (-3) = -5$
• Geometric Definition:
  • Formula:     $a \cdot b = |a||b|\cos(\theta)$     where $\theta$ is the angle between vectors $a$ and b$.\n * Example: Given |a||b| = 30$and angle$\theta = 45^\circ:\n    a \cdot b = 30 \times \cos(45^\circ) = 30 \times \frac{\sqrt{2}}{2} = 15\sqrt{2}\n\n# Angles Between Vectors\n\n* Derivation:\n * Equating algebraic and geometric dot products:\n    u \cdot v = |u||v|\cos(\theta)\n    \cos(\theta) = \frac{u \cdot v}{|u||v|}\n    \theta = \cos^{-1}\left(\frac{u \cdot v}{|u||v|}\right)\n * This method is essential for 3\text{D} vector calculations where geometric drawing is impossible.\n* Worked Examples:\n * Example 1: Find angle \theta$between$u = i + 2j$and$v = i - 4j:\n * Step 1: Dot product:\n      u \cdot v = (1)(1) + (2)(-4) = 1 - 8 = -7\n * Step 2: Magnitudes:\n      |u| = \sqrt{1^2 + 2^2} = \sqrt{5}\n      |v| = \sqrt{1^2 + (-4)^2} = \sqrt{17}\n * Step 3: Cosine equation:\n      \cos(\theta) = \frac{-7}{\sqrt{5}\sqrt{17}} = \frac{-7}{\sqrt{85}}\n      \theta = \cos^{-1}\left(\frac{-7}{\sqrt{85}}\right) \approx 139.40^\circ\n * Example 2: Find angle \theta$between$u = -2i + j$and$v = -2i - 2j:\n * Step 1: Dot product:\n      u \cdot v = (-2)(-2) + (1)(-2) = 4 - 2 = 2\n * Step 2: Magnitudes:\n      |u| = \sqrt{(-2)^2 + 1^2} = \sqrt{5}\n      |v| = \sqrt{(-2)^2 + (-2)^2} = \sqrt{8}\n * Step 3: Cosine equation:\n      \cos(\theta) = \frac{2}{\sqrt{5}\sqrt{8}} = \frac{2}{\sqrt{40}}\n      \theta = \cos^{-1}\left(\frac{2}{\sqrt{40}}\right) \approx 71.57^\circ\n\n# Perpendicular Vectors and Orthogonality\n\n* Condition for Orthogonality:\n * Two non-zero vectors u$and$v are perpendicular (orthogonal) if and only if their dot product equals zero:\n    u \cdot v = 0 \iff u \perp v\n * Reason: \cos(90^\circ) = 0$, making$|u||v|\cos(90^\circ) = 0$.
• Worked Example:
  • Given $u = 2i + 3j$ and $v = xi + 18j$, find $x$ such that $u \perp v$:
  • Step 1: Set dot product to zero:       $u \cdot v = 2(x) + 3(18) = 0$
  • Step 2: Solve linear equation:       $2x + 54 = 0$$2x = -54$$x = -27$
  • Geometric check confirms that $v = -27i + 18j$ forms an exact $90^\circ$ angle with u = 2i + 3j$.\n\n# Vector Projections and Vector Resolutes\n\n* Definition and Formulas:\n * The vector resolute of a$in the direction of$b$(denoted$a_{\text{on } b}$) is the vector component of$a$parallel to$b$.
  • Primary Formula:     $\text{Resolute } a \text{ on } b = (a \cdot \hat{b})\hat{b}$     where $\hat{b} = \frac{b}{|b|}$ is the unit vector of b$.\n * Computational Simplified Form:\n    \text{Resolute } a \text{ on } b = \left(\frac{a \cdot b}{|b|^2}\right)b\n * Note: (a \cdot \hat{b})$produces a scalar magnitude, which is then multiplied by direction vector$\hat{b}.\n* Worked Examples:\n * Example 1: Find vector resolute of a = i + 3j$in the direction of$b = i - 4j:\n * Step 1: Find unit vector \hat{b}:\n      |b| = \sqrt{1^2 + (-4)^2} = \sqrt{17}\n      \hat{b} = \frac{1}{\sqrt{17}}(i - 4j)\n * Step 2: Calculate dot product factor (a \cdot b):\n      a \cdot b = (1)(1) + (3)(-4) = 1 - 12 = -11\n * Step 3: Combine scalar projection and unit vector:\n      \text{Resolute } a \text{ on } b = \left(\frac{-11}{\sqrt{17}}\right) \frac{1}{\sqrt{17}}(i - 4j) = -\frac{11}{17}(i - 4j)\n * Step 4: Expand components:\n      \text{Resolute } a \text{ on } b = -\frac{11}{17}i + \frac{44}{17}j\n * Example 2: Find vector resolute of a = 3i + j$in the direction of$b = i + j:\n * Step 1: Find magnitude and unit vector \hat{b}:\n      |b| = \sqrt{1^2 + 1^2} = \sqrt{2}\n      \hat{b} = \frac{1}{\sqrt{2}}(i + j)\n * Step 2: Calculate dot product (a \cdot b):\n      a \cdot b = (3)(1) + (1)(1) = 3 + 1 = 4\n * Step 3: Evaluate vector projection:\n      \text{Resolute } a \text{ on } b = \left(\frac{a \cdot b}{|b|^2}\right) b = \left(\frac{4}{2}\right)(i + j) = 2(i + j) = 2i + 2j$$