Vector Algebra: Fundamentals & Geometric Applications - Lecture Notes

Vector Algebra and Scalar Quantities

• Definition of Physical Quantities:
  • Scalar Quantities: Physical quantities that have only magnitude.
  • Vector Quantities: Physical quantities that have both magnitude and direction and obey the laws of vector algebra (specifically vector addition) are called vectors.
  • The note emphasizes that having magnitude and direction alone is not sufficient; the laws of algebra must be followed ($\vec{v}_1 + \vec{v}_2$).

Representation of a Vector

• Visual Representation: A vector is represented by a directed line segment (an arrow).
• Components of the Representation:
  • Tail (Initial Point): The starting point of the vector arrow.
  • Head (Terminal Point): The ending point or tip of the arrow.
  • Length of Arrow: Directly represents the magnitude of the vector.
  • Direction of Arrow: Represents the direction of the vector.
• Symbolic Representation: A vector is typically denoted as $\vec{a}$. Its magnitude is denoted by $|\vec{a}|$

Special Types of Vectors

• Zero Vector (Null Vector):

  • Definition: A vector with zero magnitude where the initial and terminal points coincide.
  • Notation: Denoted by $\vec{0}$.
  • Magnitude: $|\vec{0}| = 0$.
  • Direction: The direction of a zero vector is considered arbitrary.
  • Properties:
    • $x \times 0 = 0$ is analogous to $k \cdot \vec{0} = \vec{0}$.
    • $x + 0 = x$ is analogous to $\vec{a} + \vec{0} = \vec{a}$.
    • $x - 0 = x$ is analogous to $\vec{a} - \vec{0} = \vec{a}$.
  • Example: A boy throwing a ball straight up and catching it back; the total displacement of the ball is a null vector.

• Unit Vector:

  • Definition: A vector whose magnitude is exactly one unit.
  • Purpose: The primary concept of a unit vector is to impart direction to a physical quantity or to write a vector of a specific magnitude in a specific direction.
  • Formula: $\hat{a} = \frac{\vec{a}}{|\vec{a}|}$.
  • Magnitude: $|\hat{a}| = 1$.
  • Usage: To write a vector of magnitude 10 along $\vec{a}$, compute $10\hat{a}$.

• Equal Vectors:

  • Two vectors are equal ($\vec{a} = \vec{b}$) if and only if they meet three criteria:
    1. Same Magnitude: $|\vec{a}| = |\vec{b}|$
    2. Same Direction: The arrows must point the same way.
    3. Same Physical Quantity: They must represent the same type of physical phenomenon.
  • Line of Support: They must have the same or parallel lines of support.

Multiplication of Vectors by Scalars

• Scalar Multiplication: If $\vec{a}$ is a vector and $m$ is a scalar, then $m\vec{a}$ is a vector parallel to $\vec{a}$ with a magnitude $|m|$ times the magnitude of $\vec{a}$.
• Mathematical Properties:
  • $m(\vec{a}) = (\vec{a})m = m\vec{a}$
  • Associative: $m(n\vec{a}) = n(m\vec{a}) = (mn)\vec{a}$
  • Distributive over Vector Addition: $m(\vec{a} + \vec{b}) = m\vec{a} + m\vec{b}$
  • Distributive over Scalar Addition: $(m + n)\vec{a} = m\vec{a} + n\vec{a}$
• Important Rule: $|k\vec{a}| = |k| |\vec{a}|$
  • Example: For $-2\vec{a}$, the magnitude is $|-2| |\vec{a}| = 2|\vec{a}|$

Kaam ki Baat: Unit Vectors in Space

• Perpendicular to a Given Plane: There are exactly 2 distinct unit vectors perpendicular to a given plane (one pointing upward and one downward).
• Parallel to a Given Plane: There are infinitely many unit vectors parallel to a given plane.
• Perpendicular to a Given Line: In 3D space, there are infinitely many distinct unit vectors perpendicular to a given line (forming a disk/plane of vectors around the line).

Localized (Bound) vs. Free Vectors

• In Mathematics:
  • A vector is viewed as a "directed line segment."
  • We primarily deal with Free Vectors. A free vector can be translated parallel to itself in space without changing its identity or effect. Example: Velocity.
• In Physics:
  • Localized or Bound Vectors are common. These are vectors where the point of application matters. Moving the vector parallel to itself changes its physical effect. Example: Force.

Collinear (Parallel) Vectors

• Definition: Two vectors are collinear if their directed line segments are parallel, regardless of their individual directions.
• Classification:
  • Like Parallel Vectors: Same direction ($K > 0$).
  • Unlike Parallel Vectors: Opposite directions ($K < 0$).
• Mathematical Condition: Two non-zero vectors $\vec{a}$ and $\vec{b}$ are collinear if and only if $\vec{a} = K\vec{b}$ for some scalar $K \in \mathbb{R}$.

Golden Result (G1): Non-Collinear Vectors

• Theorem: If $\vec{a}$ and $\vec{b}$ are two non-zero, non-collinear vectors, then:     $x\vec{a} + y\vec{b} = 0 \Rightarrow x = 0 \text{ and } y = 0$
• Proof by Contradiction:
  • Assume $x \neq 0$.
  • Then $x\vec{a} = -y\vec{b} \Rightarrow \vec{a} = (-\frac{y}{x})\vec{b}$.
  • This implies $\vec{a} = \lambda\vec{b}$, meaning $\vec{a}$ and $\vec{b}$ are collinear, which contradicts the initial condition.
• General Application: If $x_1\vec{a} + y_1\vec{b} = x_2\vec{a} + y_2\vec{b}$, then $x_1 = x_2$ and $y_1 = y_2$.

Golden Result (G2): The Median Vector

• Theorem: In triangle $ABC$, if $AD$ is the median from vertex $A$ to side $BC$, then:     $\vec{AB} + \vec{AC} = 2\vec{AD}$
• Parallelogram Law Connection: The sum of two vectors emanating from a vertex ($\vec{AB}, \vec{AC}$) is equal to twice the median vector drawn to the opposite side.
• Extensions:
  • $\vec{BA} + \vec{BC} = 2\vec{BE}$ (where $BE$ is the median to $AC$).
  • $\vec{CA} + \vec{CB} = 2\vec{CF}$ (where $CF$ is the median to $AB$).

Position Vectors and Coordinate Geometry

• Definition: The position vector of a point $P$ relative to a fixed origin $O$ is denoted as $\vec{OP}$, or simplified as $\vec{p}$.
• Difference Vector: For any two points $A$ and $B$ in space:     $\vec{AB} = \text{Position Vector of } B - \text{Position Vector of } A$$\vec{PQ} = \vec{OQ} - \vec{OP}$
• Coordinates in 3D: If point $P$ has coordinates $(x, y, z)$, its position vector is:     $\vec{OP} = x\hat{i} + y\hat{j} + z\hat{k}$
• Magnitude in 3D: The length of the vector $\vec{OP}$ is:     $|\vec{OP}| = \sqrt{x^2 + y^2 + z^2}$

Section Formula

• Internal Division: If point $P$ divides the line segment joining $A(\vec{a})$ and $B(\vec{b})$ internally in the ratio $m:n$:     $\vec{r} = \frac{m\vec{b} + n\vec{a}}{m+n}$
  • Midpoint Formula (when $m=n$): $\vec{r} = \frac{\vec{a} + \vec{b}}{2}$
• External Division: If point $P$ divides the segment externally in the ratio $m:n$:     $\vec{r} = \frac{m\vec{b} - n\vec{a}}{m - n}$
• General Coordinate Relation: If $\vec{r} = \alpha\hat{i} + \beta\hat{j} + \gamma\hat{k}$, then:     $\alpha = \frac{mx_2 + nx_1}{m+n}, \quad \beta = \frac{my_2 + ny_1}{m+n}, \quad \gamma = \frac{mz_2 + nz_1}{m+n}$

Position Vectors of Important Triangle Centers

• Centroid ($G$): The point of intersection (POI) of medians.     $\vec{g} = \frac{\vec{\alpha} + \vec{\beta} + \vec{\gamma}}{3}$
• Incentre ($I$): The POI of internal angle bisectors.     $\vec{i} = \frac{a\vec{\alpha} + b\vec{\beta} + c\vec{\gamma}}{a+b+c}$
  • where $a, b, c$ are the lengths of the sides opposite to vertices $A, B, C$.
• Circumcentre ($O$): The POI of perpendicular bisectors of sides.     $\vec{O}_{circum} = \frac{\vec{\alpha} \sin(2A) + \vec{\beta} \sin(2B) + \vec{\gamma} \sin(2C)}{\sin(2A) + \sin(2B) + \sin(2C)}$
• Orthocentre ($H$): The POI of altitudes.     $\vec{h} = \frac{\vec{\alpha} \tan(A) + \vec{\beta} \tan(B) + \vec{\gamma} \tan(C)}{\tan(A) + \tan(B) + \tan(C)}$
• Euler Line Property: In any triangle, the Orthocentre ($H$), Centroid ($G$), and Circumcentre ($O$) are collinear. The Centroid ($G$) divides the segment $HO$ in the ratio $2:1$.

Solved Classroom Problems

• Problem 1: Length of Median

  • Given sides: $\vec{AB} = 3\hat{i} + 4\hat{k}$ and $\vec{AC} = 5\hat{i} - 2\hat{j} + 4\hat{k}$.
  • Formula: $2\vec{AD} = \vec{AB} + \vec{AC} = (3+5)\hat{i} + (-2)\hat{j} + (4+4)\hat{k} = 8\hat{i} - 2\hat{j} + 8\hat{k}$.
  • $AD = 4\hat{i} - \hat{j} + 4\hat{k}$.
  • Length: $|\vec{AD}| = \sqrt{4^2 + (-1)^2 + 4^2} = \sqrt{16 + 1 + 16} = \sqrt{33}$.

• Problem 2: Unit Vectors Perpendicular to two given vectors

  • Given: $\vec{a}(1, 1, 0)$ and $\vec{b}(0, 1, 1)$.
  • The number of unit length vectors perpendicular to two non-parallel vectors in space is 2.

• Problem 3: Linear Independence/Collinearity

  • Given: $\vec{a}, \vec{b}, \vec{c}$ are non-zero, no two collinear. If $\vec{a} + \vec{b}$ is collinear with $\vec{c}$ and $\vec{b} + \vec{c}$ is collinear with $\vec{a}$.
  • Equations: $\vec{a} + \vec{b} = \lambda\vec{c}$ and $\vec{b} + \vec{c} = \mu\vec{a}$.
  • Through substitution and comparing non-collinear components: $\lambda = -1$ and $\mu = -1$.
  • Result: $\vec{a} + \vec{b} + \vec{c} = \vec{0}$.

• Problem 4: Collinear Points (JEE Mains 2023)

  • Given Points: $A(\alpha, 10, 13)$, $B(6, 11, 11)$, $C(\frac{9}{2}, \beta, -8)$.
  • Method: Vectors $\vec{AB}$ and $\vec{BC}$ must be collinear ($\vec{AB} = K\vec{BC}$).
  • $\vec{AB} = (6-\alpha)\hat{i} + \hat{j} - 2\hat{k}$.
  • $\vec{BC} = -\frac{3}{2}\hat{i} + (\beta-11)\hat{j} - 19\hat{k}$.
  • Comparing components: $\frac{6-\alpha}{-3/2} = \frac{1}{\beta-11} = \frac{-2}{-19}$.
  • Calculated values: $19\alpha = 117$ and $6\beta = 123$.
  • Result: $(19\alpha - 6\beta)^2 = (117 - 123)^2 = (-6)^2 = 36$.

• Problem 5: Component Rotation (JEE Mains 2021)

  • Vector comp in Rectangular system: $(3p, 1)$.
  • Vector comp in Rotated system: $(p+1, \sqrt{10})$.
  • Property: Magnitude remains invariant under rotation.
  • Equation: $(3p)^2 + 1^2 = (p+1)^2 + (\sqrt{10})^2 \Rightarrow 9p^2 + 1 = p^2 + 2p + 1 + 10$.
  • $8p^2 - 2p - 10 = 0 \Rightarrow 4p^2 - p - 5 = 0$.
  • Roots: $p = -1, \frac{5}{4}$.

Question & Discussion (TAH and KTK Solutions)

Differential Equations (Revision Practice)

• TAH 1(a) (Solution by Tushar Verma): Radium decay rate $\frac{dm}{dt} = -km$. Resulting law: $m = m_0(1 - \frac{\alpha}{100})^{t/t_0}$.
• TAH 1(b) (Solution by Rohit Patil): Solve $\frac{d^2y}{dx^2} = e^{2x} + e^x + 2014$.
  • Integrating once: $\frac{dy}{dx} = \frac{1}{2}e^{2x} + e^x + 2014x + C_1$.
  • Final solution: $y = \frac{1}{4}e^{2x} + e^x + 1007x^2 + C_1x + C_2$.
• TAH 1(c) (Solution by Tushar Verma/Rohit Patil): Solve $\frac{d^2y}{dx^2} = \sin^2(x) = \frac{1 - \cos(2x)}{2}$.
  • Integrations yield: $y = \frac{x^2}{4} + \frac{\cos(2x)}{8} + C_1x + C_2$.
• TAH 1(d) (Solution by Tushar Verma/Rohit Patil): Solve $\frac{d^2y}{dx^2} = \frac{1}{4\sqrt{y}}$.
  • Using identity $2\frac{dy}{dx}\frac{d^2y}{dx^2} = \frac{d}{dx}(\frac{dy}{dx})^2$.
  • $(\frac{dy}{dx})^2 = \sqrt{y} + C_1$.
• TAH 2(a) (Solution by Aditya Singh Tomar): $(1+e^{2x})(\frac{dy}{dx} + y) = 1$. Passing through $(0, 2)$.
  • Linear differential equation form leads to $e^x y = \arctan(e^x) + C$.
  • Limiting value: $\lim_{x \to \infty} e^x y = \frac{3\pi}{4}$.
• TAH 3 (Solution by Aditya Singh Tomar): $f'(x) = 2 - \frac{f(x)}{x}$.
  • Solving as linear differential equation gives $f(x) = x + \frac{c}{x}$.
  • Limits check: $\lim_{x \to 0^+} x^2 f'(x) = 0$.
• TAH 4 (Solution by Pranay Thakur): Solution of $\frac{dy}{dx} = -\frac{x^2 + 3y^2}{3x^2 + y^2}$ with $y(1) = 0$.
  • Result: $\log_e |x+y| + \frac{2xy}{(x+y)^2} = 0$.
• TAH 8 (Solution by Himanshu Saini): Solve $x \log(x) \frac{dy}{dx} + y = x^2 \log(x)$.
  • Integrating factor $I.F. = \log(x)$.
  • Solution: $y \log(x) = \frac{x^2}{4}(2 \log(x) - 1) + 1$.
  • Value at $x=e$: $y(e) = \frac{e^2+4}{4}$.
• KTK 1 (Solution by Prithvi Raj): $2x^2 y \frac{dy}{dx} - (1 - xy^2) = 0$.
  • Substitution $y^2 = z$. Solution relates to form $\alpha x = \exp(x^{\beta} y^{\gamma})$.
  • Result: $\alpha + \beta - \gamma = 1$.
• KTK 5 (Solution by Rohit Patil): $(1+x^2)dy = y(x-y)dx$, $y(0)=1$.
  • Bernoulli equation solved by setting $z = 1/y$.
  • Solution involves $e^{3\beta - 1} = e^{(3 + 2\sqrt{2})}$.