Chapter 10 Vector Algebra (Class 12 - Maths NCERT Concept Notes)

Introduction to Vectors

In our daily life, we frequently encounter physical quantities that can be fully described by a single numerical value along with a unit of measurement. Consider the following observations:

1. "The height of the Statue of Unity in Gujarat is $182\text{ metres}$."

2. "The capacity of a domestic overhead water tank is $1000\text{ litres}$."

3. "A train takes $4\text{ hours}$ to travel from New Delhi to Ludhiana."

In these instances, the values $182\text{ m}$, $1000\text{ L}$, and $4\text{ hours}$ represent the magnitude of the characteristics (height, volume, and time). However, there are many physical characteristics that cannot be described completely by using just one number. For example, if you are told that a wind is blowing at a speed of $30\text{ km/h}$ in the coastal regions of Odisha, this information is insufficient to predict which areas are at risk. To understand the impact, one must also know the direction in which the wind is moving.

Similarly, if a cricketer hits a ball with a force of $50\text{ N}$, the trajectory of the ball depends entirely on the direction in which the force was applied. In this chapter, we shall study vectors—quantities that possess both magnitude and direction—which are fundamental tools in physics, engineering, and navigation.

---

Basic Concepts

To differentiate between quantities that require direction and those that do not, we classify them into Scalars and Vectors.

Scalars

A quantity which has only magnitude and no direction is called a scalar quantity or simply a scalar. These quantities follow the ordinary laws of algebra for addition, subtraction, multiplication, and division.

Requirements to specify a Scalar:

(i) A unit, in terms of which the quantity is measured.

(ii) A real number (positive, negative, or zero) representing the magnitude.

Examples of Scalars: Length, distance, speed, mass, time, temperature, work done, energy, area, volume, density, and money (represented in Indian currency as $\textsf{₹}$). In the context of mathematics, every real number is considered a scalar.

Vectors

A quantity which has magnitude as well as direction is called a vector quantity or simply a vector. A vector is not completely defined unless its direction is specified along with its magnitude.

Requirements to specify a Vector:

(i) A unit of measurement.

(ii) A real number representing the magnitude (length of the vector).

(iii) A particular direction (e.g., North-East, $45^\circ$ to the $x$-axis, etc.).

Examples of Vectors: Displacement, velocity, acceleration, force, weight, and momentum.

Comparison Table: Scalars vs Vectors

Characteristic	Scalar Quantity	Vector Quantity
Components	Only Magnitude	Magnitude and Direction
Algebra	Ordinary Laws of Algebra	Vector Algebra Laws
Change	Changes with magnitude change	Changes with magnitude or direction change
Example	Distance ($5\text{ km}$)	Displacement ($5\text{ km}$ towards East)

---

Representation of Vectors

Since a vector possesses both magnitude and direction, it is represented geometrically by a directed line segment. This representation helps in visualizing the vector's orientation in space and its relative size.

Directed Line Segment

A portion of a straight line where the two end points are distinguished as initial and terminal is called a directed line segment. Let us consider a straight line in a specific direction.

1. Initial Point: The point $A$ from where the vector starts is called the initial point or the origin.

2. Terminal Point: The point $B$ where the vector ends is called the terminal point, tip, or end point.

The directed line segment with initial point $A$ and terminal point $B$ is denoted by the symbol $\vec{AB}$. It is important to note that the order of points is crucial. The segments $\vec{AB}$ and $\vec{BA}$ represent different vectors because their directions are exactly opposite.

Length (Modulus) of a Vector

The distance between the initial point $A$ and the terminal point $B$ is called the length or magnitude or modulus of the directed line segment $\vec{AB}$. It represents the "size" of the physical quantity.

The modulus is denoted by $|\vec{AB}|$. Vectors are also frequently represented by single lowercase letters with an arrow or a bar over them, such as $\vec{a}$ or $\bar{a}$. In such cases, the modulus is denoted as $|\vec{a}|$ or simply $a$.

A fundamental property of the modulus is that it remains the same regardless of the direction of the segment. Thus, the length of the segment from $A$ to $B$ is equal to the length of the segment from $B$ to $A$.

Line of Support and Sense

To fully define the geometry of a vector, we consider the following terms:

1. Support: The infinite straight line of which the directed line segment $\vec{AB}$ is a part is called its line of support or simply the support.

2. Sense: The sense of a directed line segment refers to its direction from the initial point to the terminal point. For example, the sense of $\vec{AB}$ is from $A$ to $B$, whereas the sense of $\vec{BA}$ is from $B$ to $A$.

In conclusion, a vector is determined by three factors: its length (magnitude), its support (the line it lies on), and its sense (its direction along that line).

Note on Physical Quantities:

Any physical quantity that can be represented by a directed line segment is a vector. For example, if a flight travels from New Delhi (Point $A$) to Bengaluru (Point $B$), the displacement is represented by $\vec{AB}$, where the length of the segment is the direct distance in kilometres and the direction is Southwards.

---

Position Vector of a Point

In three-dimensional geometry, the location of any object or point in space is defined relative to a fixed reference point. In vector algebra, this reference point is the origin, and the directed distance from this origin to the point is known as the Position Vector (abbreviated as P.V.).

Consider a point $P$ in space having coordinates $(x, y, z)$ with respect to a fixed point $O(0, 0, 0)$ as the origin. The vector $\vec{OP}$, which has $O$ as its initial point and $P$ as its terminal point, is called the position vector of point $P$ with respect to $O$.

The position vector $\vec{OP}$ is conventionally denoted by the symbol $\vec{r}$. The points $A, B, C...$ in space are similarly represented by position vectors $\vec{a}, \vec{b}, \vec{c}...$ relative to the origin.

Derivation of the Magnitude of a Position Vector

To find the length (magnitude) of the position vector $\vec{OP}$, we apply the three-dimensional distance formula between the origin $O(0, 0, 0)$ and the point $P(x, y, z)$.

The distance $d$ between any two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ is given by:

Substituting the coordinates of $O(0, 0, 0)$ and $P(x, y, z)$ into the formula:

If we denote the magnitude of $\vec{r}$ simply as $r$, we obtain the fundamental relation:

Terminology and Visual Representation

The point $P$ is often referred to as the tip or the head of the vector, while the origin $O$ is referred to as the tail or the initial point. The coordinates $x, y,$ and $z$ are known as the scalar components of the position vector along the respective axes.

Example

In maritime navigation, if a ship is located at a point $P$ whose coordinates relative to a lighthouse (origin $O$) are $(30, 40, 0)$ kilometres, the displacement of the ship from the lighthouse is represented by the position vector $\vec{OP}$. The direct distance $r$ is calculated as:

---

Direction Cosines and Direction Ratios

To specify the orientation of a vector in three-dimensional space, we use specific angles that the vector makes with the coordinate axes. These angles and their trigonometric ratios are fundamental in both geometry and vector analysis.

Direction Angles and Direction Cosines (D.C.s)

Let a position vector $\vec{OP}$ (denoted by $\vec{r}$) make angles $\alpha, \beta,$ and $\gamma$ with the positive directions of the $x, y,$ and $z$-axes respectively. These angles $\alpha, \beta,$ and $\gamma$ are known as the direction angles of the vector $\vec{r}$.

The cosines of these angles, i.e., $\cos \alpha, \cos \beta,$ and $\cos \gamma$, are called the direction cosines of the vector $\vec{OP}$. They are standardly denoted by the letters $l, m,$ and $n$.

Derivation of the Relationship between D.C.s

To establish the relationship between the direction cosines, consider a point $P(x, y, z)$ in three-dimensional space. Let $\vec{OP} = \vec{r}$ be the position vector of point $P$ with respect to the origin $O(0, 0, 0)$, and let its magnitude be $|\vec{r}| = r$.

If we drop a perpendicular from point $P$ to the $x$-axis, it meets the axis at point $A$. This forms a right-angled triangle $OAP$, where $\angle OAP = 90^\circ$ and $\angle POA = \alpha$.

In $\triangle OAP$:

By symmetry, if we project the vector onto the $y$ and $z$ axes by dropping perpendiculars to those axes, we obtain similar relations for the other coordinates:

We know from the distance formula that for any point $P(x, y, z)$ in space, the square of the magnitude $r$ is given by:

Substituting the values of $x, y,$ and $z$ in terms of $l, m,$ and $n$ into above equation:

Dividing both sides of the equation by $r^2$ (noting that $r \neq 0$ for a proper vector):

This identity proves that the sum of the squares of the direction cosines of any vector is always unity (one). This is a fundamental property used to verify if a given set of values can represent the direction cosines of a vector.

Direction Ratios (D.R.s)

While direction cosines are unique for a given vector, it is often convenient to use numbers that are simply proportional to them. Any three real numbers $a, b,$ and $c$ that are proportional to the direction cosines $l, m,$ and $n$ of a vector are called its Direction Ratios (or direction numbers).

Mathematically, if $a, b,$ and $c$ are the direction ratios, then the relationship with direction cosines $l, m,$ and $n$ is expressed as:

From our previous derivation of position vectors, we established that $x = lr$, $y = mr$, and $z = nr$. Comparing this with the definition above, we can see that the coordinates $(x, y, z)$ of the terminal point of a position vector are themselves a set of direction ratios for that vector. This is because:

Why are the Ratios Equal to $k$?

By definition, any three numbers $a, b,$ and $c$ are called Direction Ratios if they satisfy the property of being proportional to the Direction Cosines $l, m,$ and $n$. In mathematics, proportionality between two sequences $(l, m, n)$ and $(a, b, c)$ is expressed by saying there exists some non-zero constant $k$ such that:

If we rearrange these individual equations to solve for $k$, we get:

Since all three expressions are equal to the same value $k$, we can equate them to form a continued proportion:

The Physical Meaning of $k$ in terms of Coordinates

To understand why the coordinates $(x, y, z)$ of a point $P$ act as direction ratios and what the "constant" represents in that case, let us look at the projection of the position vector $\vec{r}$.

From the geometry of a position vector in 3D space, we established that:

If we solve for the magnitude $r$ (the distance of the point from the origin) in each case:

Therefore:

In this specific case, the constant of proportionality is the magnitude of the vector, $r$. If we compare equation (i) and equation (ii), we can see that when we use coordinates as direction ratios ($a=x, b=y, c=z$), the constant $k$ becomes $\frac{1}{r}$.

Deriving the Value of $k$

We can find the general value of $k$ for any set of D.R.s $(a, b, c)$ using the identity of D.C.s:

Step 1: Square and add the D.C.s expressed in terms of $k$.

Step 2: Use the identity $l^2 + m^2 + n^2 = 1$.

Step 3: Solve for $k$.

Conversion from Direction Ratios to Direction Cosines

Given the direction ratios $a, b,$ and $c$, we can determine the unique direction cosines $l, m,$ and $n$. The derivation is as follows:

From the above equation, we have:

We know the fundamental identity for direction cosines:

Substituting the values of $l, m,$ and $n$ in terms of $k$:

Substituting this value of $k$ back into the expressions for $l, m,$ and $n$, we get the conversion formulas:

Key Differences and Observations

1. Uniqueness

Direction cosines ($l, m, n$) are unique for a specific vector. However, a vector can have infinitely many sets of direction ratios, as any three numbers in the same proportion will serve the purpose.

2. The Unit Vector Connection

For a unit vector $\hat{a}$, the magnitude is $1$. This implies that for a unit vector, the direction cosines are exactly equal to the scalar components of the vector.

3. Coordinate Representation

If we are given a point $P(x, y, z)$ in space, its coordinates are its direction ratios relative to the origin. To find the direction cosines, we divide each coordinate by the distance $r = \sqrt{x^2 + y^2 + z^2}$.

Summary Table:

Property	Direction Cosines (D.C.s)	Direction Ratios (D.R.s)
Notation	$l, m, n$	$a, b, c$
Sum of Squares	$l^2 + m^2 + n^2 = 1$	$a^2 + b^2 + c^2 \neq 1$ (usually)
Nature	Unique set of values	Infinite possible sets

---

Types of Vectors

In vector algebra, vectors are classified into several types based on their magnitude, direction, and their relationship with other vectors. Understanding these classifications is essential for performing vector operations and solving physical problems.

Classification of Vectors

1. Zero Vector (or Null Vector)

A vector whose magnitude is zero is called a zero vector or a null vector. It is denoted by the symbol $\vec{0}$.

In this case, the initial and terminal points of the vector coincide. For example, if a car starts at point $A$ and does not move, its displacement is represented by the zero vector $\vec{AA}$.

Key Properties:

(i) Its magnitude is $|\vec{0}| = 0$.

(ii) It has no specific direction, or it can be regarded as having any arbitrary direction.

(iii) It is represented by directed line segments like $\vec{AA}, \vec{BB},$ or $\vec{PP}$.

2. Proper Vector

Any vector whose magnitude is non-zero is called a proper vector. Mathematically, a vector $\vec{a}$ is a proper vector if and only if:

3. Unit Vector

A vector whose magnitude is exactly unity ($1$ unit) is called a unit vector. It is used to specify a direction in space without affecting the magnitude of other quantities.

If $\vec{a}$ is a proper vector with magnitude $a = |\vec{a}|$, then the unit vector in the direction of $\vec{a}$ is denoted by $\hat{a}$ (read as "a-cap").

Derivation of the Unit Vector Formula:

By definition, a unit vector $\hat{a}$ has the same direction as $\vec{a}$ but a magnitude of $1$. To obtain this, we divide the vector $\vec{a}$ by its own magnitude:

Rearranging this, any vector can be expressed as the product of its magnitude and its direction (unit vector):

4. Equal Vectors

Two vectors $\vec{a}$ and $\vec{b}$ are said to be equal (written as $\vec{a} = \vec{b}$) if and only if they satisfy three conditions simultaneously:

(i) They have equal magnitudes ($|\vec{a}| = |\vec{b}|$).

(ii) They have the same or parallel supports.

(iii) They have the same direction (sense).

It is important to note that equality is independent of the position of their initial points (for free vectors).

5. Negative of a Vector

A vector having the same magnitude as that of a given vector $\vec{a}$, but directed in the exactly opposite sense, is called the negative of $\vec{a}$. It is denoted by $-\vec{a}$.

If a directed line segment $\vec{AB}$ represents vector $\vec{a}$, then the segment $\vec{BA}$ represents $-\vec{a}$.

6. Collinear or Parallel Vectors

Two or more vectors are called collinear if they have the same or parallel supports, irrespective of their magnitudes and directions. Collinear vectors can be classified into two types:

(i) Like Vectors: Collinear vectors having the same direction.

(ii) Unlike Vectors: Collinear vectors having opposite directions.

Mathematical Condition for Collinearity:

Two vectors $\vec{a}$ and $\vec{b}$ are collinear if and only if there exists a non-zero scalar $\lambda$ such that:

7. Coplanar Vectors

Three or more vectors are called coplanar if they lie in the same plane or are parallel to the same plane. An important result in vector geometry is that any two vectors are always coplanar, as a plane can always be defined passing through two intersecting or parallel lines.

8. Co-initial and Co-terminal Vectors

1. Co-initial Vectors: Vectors having the same initial point are called co-initial vectors.

2. Co-terminal Vectors: Vectors having the same terminal point are called co-terminal vectors.

9. Free and Localised Vectors

1. Free Vector: If the effect of a vector depends only on its magnitude and direction and is independent of the position of its origin, it is called a free vector. In pure mathematics and this chapter, we primarily deal with free vectors.

2. Localised Vector: When the initial point of a vector is restricted to a certain specific point, it is called a localised vector. For example, a force acting on a rigid body is a localised vector (or sliding vector) because the point of application changes the resulting torque.

Summary Comparison Table

Vector Type	Magnitude	Direction
Zero Vector	Zero ($0$)	Indeterminate
Unit Vector	Unity ($1$)	Specific direction
Equal Vectors	Must be equal	Must be identical
Negative Vector	Must be equal	Exactly opposite ($180^\circ$)
Collinear	Can be different	Same or Opposite

---

Answer:

---

Answer:

---

Answer:

---

Answer:

Vector Algebra

In previous sections, we learned about the nature of vectors. Now, we shall explore Vector Algebra, which involves mathematical operations such as addition, subtraction, and multiplication of vectors by scalars. These operations follow specific geometric laws rather than simple arithmetic rules.

---

Addition of Vectors

When two vectors act on a single point or represent successive displacements, their combined effect is represented by a single vector called the Resultant. This resultant is determined by two fundamental geometric laws.

1. Triangle Law of Vector Addition

The Triangle Law is the most basic rule for adding two vectors. It is often referred to as the "Head-to-Tail" method.

Statement:

If two vectors are represented in magnitude and direction by two sides of a triangle taken in order (meaning the terminal point of the first is the initial point of the second), then their sum or resultant is represented by the third side of the triangle taken in the opposite order (from the initial point of the first to the terminal point of the second).

Geometrical Illustration:

Let $\vec{a}$ and $\vec{b}$ be two vectors. To find their sum:

1. Position vector $\vec{a}$ such that it is represented by $\vec{OA}$.

2. Position vector $\vec{b}$ such that its initial point coincides with the terminal point $A$ of $\vec{a}$, represented by $\vec{AB}$.

3. The vector joining the starting point $O$ to the final point $B$ is the resultant vector $\vec{c}$.

2. Parallelogram Law of Vector Addition

The Parallelogram Law is used when two vectors are co-initial (originating from the same point). This is particularly useful in physics when representing two forces acting simultaneously on a single particle.

Statement:

If two vectors $\vec{a}$ and $\vec{b}$ are represented in magnitude and direction by the two adjacent sides of a parallelogram directed away from a common vertex, then their sum is represented by the diagonal of the parallelogram passing through that same vertex.

Geometrical Illustration:

Let $\vec{OA} = \vec{a}$ and $\vec{OB} = \vec{b}$ be two vectors acting from the origin $O$.

1. Complete the parallelogram $OACB$ by drawing $AC$ parallel to $OB$ and $BC$ parallel to $OA$.

2. Since $AC$ is parallel and equal to $OB$, it represents the same vector $\vec{b}$.

3. The diagonal $\vec{OC}$ starting from $O$ represents the resultant.

Relationship Between Triangle and Parallelogram Laws

It is important to understand that these two laws are mathematically equivalent. In the parallelogram $OACB$ shown in Fig 10.6, we can observe that:

Applying the Triangle Law in $\triangle OAC$:

Thus, the Parallelogram Law is essentially a double application of the Triangle Law. The Triangle Law is preferred for vectors occurring in a sequence (like steps in a journey), while the Parallelogram Law is preferred for vectors occurring simultaneously at a point (like forces acting on a cricket ball).

Comparison of Addition Methods

Feature	Triangle Law	Parallelogram Law
Vector Arrangement	Head of first to Tail of second	Tail to Tail (Co-initial)
Resultant	Closing side of the triangle	Diagonal from the common vertex
Primary Use	Consecutive displacements	Concurrent forces or velocities

---

Properties of Vector Addition

Vector addition is not just a geometric operation; it possesses several algebraic properties that allow us to manipulate vector equations much like we do with real numbers. These properties ensure consistency in calculations involving multiple forces or displacements.

(i) Commutative Law

The commutative law states that the order in which two vectors are added does not affect the resultant vector. Whether we add vector $\vec{b}$ to $\vec{a}$ or $\vec{a}$ to $\vec{b}$, the final vector remains identical in magnitude and direction.

Proof:

Consider a parallelogram $OACB$ where $\vec{OA} = \vec{a}$ and $\vec{OB} = \vec{b}$. Since opposite sides of a parallelogram are equal and parallel, we have $\vec{BC} = \vec{OA} = \vec{a}$ and $\vec{AC} = \vec{OB} = \vec{b}$.

In $\triangle OAC$, by the Triangle Law:

In $\triangle OBC$, by the Triangle Law:

From equations (i) and (ii), we conclude:

(ii) Associative Law

The associative law states that when three or more vectors are added, the way they are grouped does not change the result. This allows us to add vectors in any sequence.

Proof:

Let $\vec{OA} = \vec{a}$, $\vec{AB} = \vec{b}$, and $\vec{BC} = \vec{c}$.

1. To find $(\vec{a} + \vec{b}) + \vec{c}$:

First add $\vec{a}$ and $\vec{b}$ in $\triangle OAB$ to get $\vec{OB} = \vec{a} + \vec{b}$. Then add $\vec{c}$ to this resultant in $\triangle OBC$.

2. To find $\vec{a} + (\vec{b} + \vec{c})$:

First add $\vec{b}$ and $\vec{c}$ in $\triangle ABC$ to get $\vec{AC} = \vec{b} + \vec{c}$. Then add this to $\vec{a}$ in $\triangle OAC$.

From (i) and (ii), we get:

(iii) Existence of Additive Identity

For any vector $\vec{a}$, there exists a unique vector $\vec{0}$ (zero vector) such that adding it to $\vec{a}$ results in the original vector $\vec{a}$.

If a flight from Mumbai to Delhi ($\vec{a}$) has a ground delay of $0\text{ km}$ displacement ($\vec{0}$), its total displacement remains $\vec{a}$.

(iv) Existence of Additive Inverse

For every vector $\vec{a}$, there exists a negative vector $-\vec{a}$ (same magnitude but opposite direction) such that their sum is the zero vector.

Example: If a person walks $5\text{ metres}$ East ($\vec{a}$) and then $5\text{ metres}$ West ($-\vec{a}$), the final displacement is zero.

Summary Table of Properties

Property	Mathematical Expression	Significance
Commutative	$\vec{a} + \vec{b} = \vec{b} + \vec{a}$	Order of operations is flexible.
Associative	$(\vec{a} + \vec{b}) + \vec{c} = \vec{a} + (\vec{b} + \vec{c})$	Grouping of vectors is flexible.
Identity	$\vec{a} + \vec{0} = \vec{a}$	Zero vector does not change the original vector.
Inverse	$\vec{a} + (-\vec{a}) = \vec{0}$	Opposite vectors cancel each other out.

---

Addition of More Than Two Vectors (Vector Polygon Law)

The Triangle Law of vector addition is restricted to the addition of only two vectors. When we need to find the resultant of three or more vectors, we extend this principle into what is known as the Polygon Law of Vector Addition. This law is essentially the result of repeated applications of the Triangle Law.

Statement of Polygon Law

If a number of vectors are represented in magnitude and direction by the sides of an open polygon taken in order, then their resultant is represented in magnitude and direction by the closing side of the polygon taken in the opposite order (from the initial point of the first vector to the terminal point of the last vector).

Derivation and Geometric Proof

Let us consider four vectors $\vec{a}, \vec{b}, \vec{c},$ and $\vec{d}$ represented by the sides of an open polygon. We wish to find their sum $\vec{r} = \vec{a} + \vec{b} + \vec{c} + \vec{d}$.

Construction:

1. Draw $\vec{OA} = \vec{a}$.

2. From the terminal point $A$, draw $\vec{AB} = \vec{b}$.

3. From the terminal point $B$, draw $\vec{BC} = \vec{c}$.

4. From the terminal point $C$, draw $\vec{CD} = \vec{d}$.

Proof:

Applying the Triangle Law in $\triangle OAB$:

Now, applying the Triangle Law in $\triangle OBC$:

Finally, applying the Triangle Law in $\triangle OCD$:

Thus, the vector $\vec{OD}$ represents the resultant vector $\vec{r}$. It closes the polygon by connecting the starting point $O$ to the final terminal point $D$.

Important Property: The Closed Polygon

If the terminal point of the last vector coincides with the initial point of the first vector, the polygon becomes closed. In such a case, the resultant vector is a zero vector.

Example

Consider a traveler exploring Rajasthan. Suppose the traveler makes the following consecutive displacements:

1. From Jaipur to Ajmer ($\vec{a}$)

2. From Ajmer to Jodhpur ($\vec{b}$)

3. From Jodhpur to Jaisalmer ($\vec{c}$)

The total displacement of the traveler from the starting point (Jaipur) to the final destination (Jaisalmer) is the resultant vector $\vec{r} = \vec{a} + \vec{b} + \vec{c}$. According to the Polygon Law, this is represented by the straight-line vector directed from Jaipur to Jaisalmer.

Summary Table: Application of Vector Addition Laws

Law	Number of Vectors	Configuration
Triangle Law	2 Vectors	Head-to-Tail arrangement.
Parallelogram Law	2 Vectors	Tail-to-Tail (Co-initial) arrangement.
Polygon Law	More than 2 Vectors	Successive Head-to-Tail arrangement forming a chain.

---

Subtraction of Two Vectors

In vector algebra, the operation of subtraction is not defined as a separate process but is treated as a specific case of vector addition. To subtract a vector $\vec{b}$ from a vector $\vec{a}$, we simply add the negative of $\vec{b}$ to $\vec{a}$.

Mathematical Definition

For any two vectors $\vec{a}$ and $\vec{b}$, the difference $\vec{a} - \vec{b}$ is defined as follows:

Geometric Construction

To find the difference geometrically, follow these steps:

1. Represent the vector $\vec{a}$ by $\vec{OA}$.

2. Represent the vector $\vec{b}$ by $\vec{OB}$.

3. To obtain $-\vec{b}$, reverse the direction of $\vec{OB}$ to get $\vec{OB'}$.

4. Now, apply the Triangle Law to add $\vec{a}$ and $-\vec{b}$. The resultant vector $\vec{r} = \vec{a} + (-\vec{b})$ represents the subtraction.

From the figure, it is evident that if $\vec{a}$ and $\vec{b}$ are represented by the adjacent sides of a parallelogram, one diagonal represents the sum ($\vec{a} + \vec{b}$), while the other diagonal represents the difference ($\vec{a} - \vec{b}$).

---

Representation of a Vector in terms of Position Vectors

In analytical geometry, it is often necessary to define a displacement vector between two arbitrary points in space. If we know the location of these points relative to a fixed origin (their position vectors), we can easily determine the vector connecting them. This concept is fundamental for navigating 3D space, whether in engineering designs or satellite tracking.

Proof of the Relation

Let $P$ and $Q$ be two points in space. Let $O$ be the origin $(0, 0, 0)$. The vectors $\vec{OP}$ and $\vec{OQ}$ are the position vectors (P.V.) of points $P$ and $Q$ respectively. We wish to find the vector $\vec{PQ}$ which starts at $P$ and ends at $Q$.

According to the Triangle Law of Vector Addition, in $\triangle OPQ$, the sides taken in order are $\vec{OP}$ and $\vec{PQ}$, while the closing side in the opposite order is $\vec{OQ}$. Therefore:

To find the expression for $\vec{PQ}$, we subtract $\vec{OP}$ from both sides of the equation:

This result implies that any vector is equal to the Position Vector of its Terminal point minus the Position Vector of its Initial point. For any two points $A$ and $B$, this is generalized as:

Practical Application

Consider the calculation of a direct flight path between New Delhi and Bengaluru. If we set a coordinate system with its origin at the center of the Earth, let the position of New Delhi be $\vec{d}$ and Bengaluru be $\vec{b}$.

The displacement vector $\vec{DB}$ representing the journey from New Delhi to Bengaluru is given by:

This allows pilots to determine the exact heading (direction) and distance (magnitude) required for the flight. If New Delhi is at coordinates $(x_1, y_1, z_1)$ and Bengaluru is at $(x_2, y_2, z_2)$, the magnitude of the journey in kilometres is found as:

---

Solving a Vector Equation

A vector equation is an algebraic expression where the unknowns are vectors. Solving such an equation involves finding a vector that satisfies the equality. For instance, consider the basic equation $\vec{a} + \vec{x} = \vec{b}$, where $\vec{a}$ and $\vec{b}$ are known vectors and $\vec{x}$ is the unknown vector.

Proof of the Unique Solution

To solve for $\vec{x}$ in the equation $\vec{a} + \vec{x} = \vec{b}$, we add the additive inverse of $\vec{a}$ (which is $-\vec{a}$) to both sides:

Using the Associative Property of vector addition:

Since $(-\vec{a} + \vec{a}) = \vec{0}$ (Additive Inverse property):

Using the Additive Identity property ($\vec{0} + \vec{x} = \vec{x}$):

Proof of Uniqueness:

To prove that this solution is unique, let us assume there are two solutions, $\vec{x_1}$ and $\vec{x_2}$, such that:

Equating the two expressions for $\vec{b}$:

Adding $-\vec{a}$ to both sides:

Hence, the solution $\vec{x} = \vec{b} - \vec{a}$ is unique. This confirms that vector equations follow a logic very similar to scalar linear equations.

---

Multiplication of a Vector by a Scalar

Multiplication of a vector by a scalar (a real number) is an operation that alters the magnitude of the vector and may also reverse its direction, but it does not change its line of support (except in the case of a zero scalar). This process is also known as scaling a vector.

Let $\vec{a}$ be a given vector and $\lambda$ be a scalar. The product $\lambda \vec{a}$ is a new vector defined by the following characteristics:

1. Magnitude of the Scaled Vector

The magnitude of the new vector $\lambda \vec{a}$ is the product of the absolute value of the scalar and the magnitude of the original vector.

2. Direction of the Scaled Vector

The direction of $\lambda \vec{a}$ depends entirely on the sign of the scalar $\lambda$:

(i) If $\lambda > 0$, the direction of $\lambda \vec{a}$ is the same as that of $\vec{a}$.

(ii) If $\lambda < 0$, the direction of $\lambda \vec{a}$ is exactly opposite to that of $\vec{a}$.

(iii) If $\lambda = 0$, the result is the zero vector ($\vec{0}$), which has no specific direction.

Important Remarks and Special Cases

Identity and Negative Vectors

If the scalar is $1$, the vector remains unchanged. If the scalar is $-1$, we obtain the negative of the vector.

Collinearity

The vectors $\vec{a}$ and $\lambda \vec{a}$ are always collinear because they lie on the same or parallel lines of support. Conversely, if two vectors $\vec{a}$ and $\vec{b}$ are collinear, there must exist a scalar $\lambda$ such that:

Unit Vector Representation

If $\vec{a}$ is a proper vector (non-zero), we can define a unit vector $\hat{a}$ in its direction by scaling the vector $\vec{a}$ by the reciprocal of its magnitude:

---

Properties of Multiplication of a Vector by a Scalar

Scalar multiplication distributes over both scalar addition and vector addition, and it follows an associative-like property with multiple scalars. Let $\vec{a}$ and $\vec{b}$ be vectors and $m, n$ be scalars:

(i) Distributive Law over Scalar Addition

Multiplying a vector by the sum of two scalars is the same as adding the results of multiplying the vector by each scalar individually.

(ii) Associativity of Scalars

The product of two scalars with a vector can be calculated by multiplying the scalars first or by multiplying the vector by scalars one after the other.

(iii) Distributive Law over Vector Addition

Multiplying a scalar by the sum of two vectors is the same as multiplying each vector by the scalar and then adding the results.

Theorem on Two Non-Zero Non-Collinear Vectors

In this section, we explore the fundamental relationship between vectors that are not parallel (non-collinear). This theorem is the basis for understanding Linear Independence in vector spaces and allows us to uniquely represent vectors in a plane.

---

Theorem: Linear Independence of Non-Collinear Vectors

Statement:

If $\vec{a}$ and $\vec{b}$ are two non-zero and non-collinear vectors, and $x, y$ are two scalars, then the linear combination $x\vec{a} + y\vec{b} = \vec{0}$ holds true if and only if $x = 0$ and $y = 0$.

Given:

1. $\vec{a} \neq \vec{0}$ and $\vec{b} \neq \vec{0}$.

2. $\vec{a}$ and $\vec{b}$ are non-collinear (they are not parallel).

To Prove:

$x\vec{a} + y\vec{b} = \vec{0} \iff x = 0, y = 0$.

Proof:

Part 1: The Direct Proof

Let us assume that $x\vec{a} + y\vec{b} = \vec{0}$. We will use the method of contradiction to prove $x$ and $y$ must be zero.

If possible, let $x \neq 0$. Then we can rearrange the equation as follows:

In the above equation, we see that vector $\vec{a}$ is expressed as a scalar multiple of vector $\vec{b}$. By the definition of collinearity, this implies that $\vec{a}$ and $\vec{b}$ are collinear.

However, this contradicts our given hypothesis that $\vec{a}$ and $\vec{b}$ are non-collinear. Therefore, our assumption that $x \neq 0$ must be false. Hence, $x = 0$.

Similarly, by assuming $y \neq 0$, we can show that $\vec{b} = (-\frac{x}{y})\vec{a}$, which again leads to a contradiction. Thus, $y$ must also be $0$.

Part 2: The Converse Proof

Conversely, let $x = 0$ and $y = 0$. Then:

Equality of Components

If $\vec{a}$ and $\vec{b}$ are two non-zero non-collinear vectors, then:

---

Theorem: Resolution of Vectors in a Plane

Statement:

If $\vec{a}$ and $\vec{b}$ are two non-zero non-collinear vectors in a plane, then every vector $\vec{r}$ coplanar with $\vec{a}$ and $\vec{b}$ can be uniquely expressed as a linear combination:

Proof of Uniqueness

Suppose the vector $\vec{r}$ can be expressed in two different ways:

Equating (i) and (ii):

Since $\vec{a}$ and $\vec{b}$ are non-collinear, by Theorem 10.2.9, the coefficients must be zero:

This confirms that the representation is unique.

Condition for Coplanarity

Three vectors $\vec{a}, \vec{b},$ and $\vec{c}$ are coplanar if and only if one of them (say $\vec{c}$) can be expressed as a linear combination of the other two:

Components of a Vector in a Plane

In a two-dimensional rectangular coordinate system, any vector can be expressed as a linear combination of two mutually perpendicular unit vectors. This representation is fundamental in physics and engineering as it simplifies vector addition, subtraction, and other algebraic operations.

Position Vector of a Point in a Plane

Consider a rectangular coordinate system $XOY$ with the origin at $O(0, 0)$. Let $P(x, y)$ be any point in this plane. We define two fixed vectors of unit length, denoted by $\hat{i}$ and $\hat{j}$, which point in the positive directions of the $x$-axis and $y$-axis, respectively.

Step-by-Step Representation

Given: A point $P(x, y)$ and unit vectors $\hat{i}$ and $\hat{j}$.

To Find: The position vector $\vec{OP}$ in terms of its components.

Construction: From point $P(x, y)$, draw perpendiculars $PM$ and $PN$ to the $x$-axis and $y$-axis respectively. Join $OP$.

Derivation of the Position Vector

From the geometry of the coordinate system, the distance $OM$ is the $x$-coordinate of $P$, and the distance $ON$ is the $y$-coordinate of $P$. Thus:

Since $\hat{i}$ and $\hat{j}$ are unit vectors along $OX$ and $OY$, we can write the segments as vectors:

By the Parallelogram Law of Vector Addition, since $OMPN$ is a rectangle (and thus a parallelogram), the resultant vector $\vec{OP}$ is the diagonal:

Substituting the values from (i) and (ii), we get the standard form of a vector in a plane:

Distinction between Scalar and Vector Components

It is crucial to differentiate between the magnitude of the projections and the projections themselves as vectors.

Term	Along x-axis	Along y-axis
Scalar Components	$x$	$y$
Vector Components	$x\hat{i}$	$y\hat{j}$
Nature	Real Numbers	Directed Segments

Magnitude (Length) of the Vector

The magnitude of the position vector $\vec{r}$ corresponds to the actual length of the line segment $OP$.

Proof: In the right-angled triangle $\triangle OMP$, the angle $\angle OMP = 90^\circ$. Applying the Pythagoras Theorem:

Taking the square root on both sides to find the magnitude (represented as $|\vec{r}|$ or $r$):

Direction of the Vector

If the vector $\vec{r}$ makes an angle $\theta$ with the positive direction of the $x$-axis, then from $\triangle OMP$:

This shows that the scalar components $x$ and $y$ depend on the magnitude of the vector and its orientation in the plane.

---

Vector Joining Two Points

When we are given two distinct points in a Cartesian plane, we can define a directed line segment representing the vector from one point to another. This is particularly useful in finding displacement and distance between two locations in space.

Given: Two points $P(x_1, y_1)$ and $Q(x_2, y_2)$ in the $XOY$ plane.

To Find: The vector $\vec{PQ}$ and its magnitude $|\vec{PQ}|$.

Construction: Draw perpendiculars $PM$ and $QN$ from points $P$ and $Q$ respectively to the $x$-axis ($OX$). Also, draw $PR$ perpendicular to $QN$ ($PR \perp QN$).

Derivation of the Vector $\vec{PQ}$

From the geometric construction, we observe the following horizontal and vertical lengths:

Now, we represent these directed segments as vectors using the unit vectors $\hat{i}$ and $\hat{j}$:

According to the Triangle Law of Vector Addition in $\triangle PRQ$:

Position Vector Method (Alternate Proof)

We can also derive this using the position vectors of $P$ and $Q$ with respect to the origin $O$:

By the Triangle Law in $\triangle OPQ$:

Magnitude of $\vec{PQ}$ (Distance Formula)

The magnitude $|\vec{PQ}|$ represents the distance between points $P$ and $Q$. In the right-angled triangle $PRQ$, we apply the Pythagoras Theorem:

Summary of Components of $\vec{PQ}$

Direction	Scalar Component	Vector Component
Along x-axis	$(x_2 - x_1)$	$(x_2 - x_1)\hat{i}$
Along y-axis	$(y_2 - y_1)$	$(y_2 - y_1)\hat{j}$

---

Theorem: Algebraic Operations on Vectors in Component Form

Let $\vec{a} = a_1\hat{i} + a_2\hat{j}$ and $\vec{b} = b_1\hat{i} + b_2\hat{j}$ be any two vectors in the $XOY$ plane, and let $m$ be a scalar. Then the following properties hold true:

(i) Addition of Vectors: The sum of two vectors is obtained by adding their corresponding scalar components.

(ii) Equality of Vectors: Two vectors are equal if and only if their respective components are identical.

(iii) Multiplication by a Scalar: When a vector is multiplied by a scalar, each of its components is multiplied by that scalar.

Proof of the Theorem:

(i) Addition of Vectors

Consider the two vectors in their component form. To find their sum, we apply the Commutative and Associative laws of vector addition:

By applying the distributive property of scalar multiplication over vector addition:

(ii) Equality of Vectors

This proof relies on a fundamental property of basis vectors: if $\hat{i}$ and $\hat{j}$ are non-zero and non-collinear (orthogonal), then any linear combination $x\hat{i} + y\hat{j} = \vec{0}$ implies $x = 0$ and $y = 0$.

Since $\hat{i}$ and $\hat{j}$ are along different axes, they are non-collinear. Thus, the coefficients must be zero:

(iii) Multiplication by a Scalar

Using the distributive property of a scalar over the sum of vectors:

Summary Table of Operations

The following table summarizes the component-wise operations for vectors $\vec{a}$ and $\vec{b}$ in 2D space:

Operation	Resulting Vector Formulation
Addition ($\vec{a} + \vec{b}$)	$(a_1 + b_1)\hat{i} + (a_2 + b_2)\hat{j}$
Subtraction ($\vec{a} - \vec{b}$)	$(a_1 - b_1)\hat{i} + (a_2 - b_2)\hat{j}$
Scalar Multiplication ($k\vec{a}$)	$(ka_1)\hat{i} + (ka_2)\hat{j}$
Equality ($\vec{a} = \vec{b}$)	$a_1 = b_1$ AND $a_2 = b_2$
Magnitude ($|\vec{a}|$)	$\sqrt{a_1^2 + a_2^2}$

Example: Subtraction of Vectors

Subtraction is simply the addition of a negative vector. If we need to find $\vec{a} - \vec{b}$:

Components of a Vector in Space

In three-dimensional space, any vector can be resolved into three mutually perpendicular components along the $x$, $y$, and $z$ axes. This is an extension of the 2D concept, introducing a third unit vector $\hat{k}$ along the $z$-axis.

Position Vector of a Point in Space

Consider a rectangular coordinate system $OXYZ$ with origin $O(0, 0, 0)$. Let $P(x, y, z)$ be any point in space. The position vector $\vec{OP}$ (represented as $\vec{r}$) defines the position of point $P$ relative to the origin.

Given: Point $P(x, y, z)$ and origin $O(0, 0, 0)$.

To Prove: $\vec{OP} = x\hat{i} + y\hat{j} + z\hat{k}$

Construction and Geometrical Representation

To visualize the components, imagine a rectangular parallelepiped (a box) where one vertex is at the origin $O$ and the diagonally opposite vertex is at $P$. The edges of this box lie along the $x$, $y$, and $z$ axes.

1. Let $A$ be a point on the $x$-axis such that $OA = x$. Then $\vec{OA} = x\hat{i}$.

2. Let $B$ be a point in the $xy$-plane such that $AB$ is parallel to the $y$-axis and $AB = y$. Then $\vec{AB} = y\hat{j}$.

3. From $B$, move parallel to the $z$-axis to reach point $P$ such that $BP = z$. Then $\vec{BP} = z\hat{k}$.

Derivation:

Using the triangle law of vector addition in $\triangle OAB$ (in the $xy$-plane):

Now, applying the triangle law in $\triangle OBP$:

Substituting the value of $\vec{OB}$ from equation (i):

Classification of Components

The vector $\vec{r}$ is composed of three parts, each corresponding to an axis. These are categorized into scalar and vector forms.

Type	X-component	Y-component	Z-component
Scalar Components	$x$	$y$	$z$
Vector Components	$x\hat{i}$	$y\hat{j}$	$z\hat{k}$

Note: The scalar components $x, y, z$ are also known as the rectangular components of the vector $\vec{r}$.

Magnitude of the Vector in Space

The magnitude $|\vec{r}|$ represents the straight-line distance from the origin $O(0, 0, 0)$ to the point $P(x, y, z)$.

Derivation of Magnitude:

In the right-angled triangle $\triangle OAB$ (lying in the $xy$-plane), using Pythagoras Theorem:

Now, consider $\triangle OBP$. Since the $z$-axis (and thus $BP$) is perpendicular to the $xy$-plane, $BP$ is perpendicular to $OB$. Therefore, $\triangle OBP$ is a right-angled triangle at $B$.

Substituting the value of $OB^2$ from equation (i):

Thus, the magnitude of vector $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$ is given by:

Direction Cosines of a Vector

If $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$ makes angles $\alpha, \beta,$ and $\gamma$ with the positive $x, y,$ and $z$ axes respectively, then:

$\cos \alpha = \frac{x}{|\vec{r}|}, \quad \cos \beta = \frac{y}{|\vec{r}|}, \quad \cos \gamma = \frac{z}{|\vec{r}|}$

These are called Direction Cosines, usually denoted by $l, m,$ and $n$. A key property in 3D space is:

---

Vector Joining Two Points in Space

If we are given two points in a three-dimensional space, the vector starting from one point and ending at another is called the vector joining two points. This is fundamentally based on the concept of position vectors relative to a fixed origin $O(0, 0, 0)$.

Given: Two points $P(x_1, y_1, z_1)$ and $Q(x_2, y_2, z_2)$ in the rectangular coordinate system.

To Find: The vector $\vec{PQ}$ and its magnitude $|\vec{PQ}|$.

Geometrical Construction

1. Let $O$ be the origin $(0,0,0)$.

2. Join $OP$ and $OQ$ to represent the position vectors of points $P$ and $Q$ respectively.

3. The position vector of $P$ is $\vec{OP} = x_1\hat{i} + y_1\hat{j} + z_1\hat{k}$.

4. The position vector of $Q$ is $\vec{OQ} = x_2\hat{i} + y_2\hat{j} + z_2\hat{k}$.

Derivation of the Vector $\vec{PQ}$

According to the Triangle Law of Vector Addition in $\triangle OPQ$:

To find $\vec{PQ}$, we can rearrange the equation as follows:

Substituting the values of the position vectors $\vec{OQ}$ and $\vec{OP}$:

By grouping the components along the $\hat{i}$, $\hat{j}$, and $\hat{k}$ directions, we get:

Magnitude and Distance

The magnitude of the vector $\vec{PQ}$ represents the displacement or the distance between points $P$ and $Q$. Using the formula for the magnitude of a vector in space:

This formula is identical to the Distance Formula used in 3D Coordinate Geometry.

Component Analysis

The vector $\vec{PQ}$ can be broken down into its respective components as shown in the table below:

Component Type	Expression
X-Scalar Component	$(x_2 - x_1)$
Y-Scalar Component	$(y_2 - y_1)$
Z-Scalar Component	$(z_2 - z_1)$
Vector Form	$(x_2 - x_1)\hat{i} + (y_2 - y_1)\hat{j} + (z_2 - z_1)\hat{k}$

Note: The values $(x_2 - x_1)$, $(y_2 - y_1)$, and $(z_2 - z_1)$ are the direction ratios of the line segment joining $P$ and $Q$.

---

Operations and Properties of Vectors in Three-Dimensional Space

When vectors are expressed in their rectangular component forms, mathematical operations such as addition, subtraction, and scalar multiplication become straightforward algebraic manipulations of their respective components along the $x$, $y$, and $z$ axes.

Given: Let there be two vectors $\vec{a}$ and $\vec{b}$ represented in space as:

$\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}$

$\vec{b} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k}$

Let $m$ and $n$ be any scalars (real numbers).

(i) Addition of Vectors

The sum of two vectors $\vec{a}$ and $\vec{b}$ is a vector obtained by adding their corresponding scalar components. This follows the parallelogram law or triangle law of vector addition applied to each dimension independently.

Derivation:

By using the distributive and associative properties of vectors, we group the components of $\hat{i}$, $\hat{j}$, and $\hat{k}$ together:

Similarly, the subtraction of two vectors is given by:

(ii) Equality of Vectors

Two vectors $\vec{a}$ and $\vec{b}$ are said to be equal if and only if they have the same magnitude and the same direction. In terms of their rectangular components, two vectors are equal if their corresponding scalar components are identical.

This property is extensively used to solve for unknown variables in vector equations by comparing the coefficients of $\hat{i}$, $\hat{j}$, and $\hat{k}$ on both sides.

(iii) Multiplication of a Vector by a Scalar

When a vector $\vec{a}$ is multiplied by a scalar $m$, the magnitude of the vector is scaled by $|m|$ and its direction remains the same if $m > 0$, or becomes opposite if $m < 0$.

Derivation:

Applying the distributive law of scalar multiplication over vector addition:

From this, we can also observe the distributive property for multiple scalars:

Summary of Vector Operations

The following table summarizes the resultant components for different operations between vectors $\vec{a}$ and $\vec{b}$:

Operation	Resultant Component Form
Addition ($\vec{a} + \vec{b}$)	$(a_1+b_1)\hat{i} + (a_2+b_2)\hat{j} + (a_3+b_3)\hat{k}$
Subtraction ($\vec{a} - \vec{b}$)	$(a_1-b_1)\hat{i} + (a_2-b_2)\hat{j} + (a_3-b_3)\hat{k}$
Scalar Multiplication ($m\vec{a}$)	$(ma_1)\hat{i} + (ma_2)\hat{j} + (ma_3)\hat{k}$
Condition for $\vec{a} = \vec{b}$	$a_1=b_1, \ a_2=b_2, \ a_3=b_3$

Important Note: These algebraic operations are only valid when the vectors are referred to the same coordinate system and same set of unit vectors $\hat{i}$, $\hat{j}$, and $\hat{k}$.

---

Direction Ratios and Collinearity of Vectors

In three-dimensional space, the orientation and relationship between two vectors can be mathematically determined using their components. Two fundamental concepts in this regard are Direction Ratios and the condition of Collinearity (or parallelism).

Direction Ratios

If $\vec{r} = a\hat{i} + b\hat{j} + c\hat{k}$ is any vector in space, the scalar components $a, b,$ and $c$ are called the direction ratios of the vector $\vec{r}$.

Unlike direction cosines ($l, m, n$), which are unique for a given direction, direction ratios are not unique. Any set of numbers proportional to the direction cosines can serve as direction ratios. For a vector $\vec{r}$, the scalar components $a, b, c$ are the simplest set of direction ratios.

Collinearity (Parallelism) of Vectors

Two vectors $\vec{a}$ and $\vec{b}$ are said to be collinear (or parallel) if and only if they act along the same line or parallel lines. Mathematically, one vector must be a scalar multiple of the other.

Given: Two vectors $\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}$ and $\vec{b} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k}$.

Condition: For $\vec{a}$ and $\vec{b}$ to be collinear, there must exist a non-zero scalar $\lambda$ such that:

Derivation of the Proportionality Condition

Substituting the component forms into the condition $\vec{b} = \lambda \vec{a}$:

Expanding the right-hand side using the distributive law of scalar multiplication:

By the property of equality of vectors, we equate the corresponding coefficients of $\hat{i}$, $\hat{j}$, and $\hat{k}$:

From the above equations, we can express $\lambda$ as the ratio of the components:

Conclusion: Two vectors are collinear if and only if their respective direction ratios are proportional.

---

Distinction between Direction Ratios and Direction Cosines

Feature	Direction Ratios (a, b, c)	Direction Cosines (l, m, n)
Definition	Components of the vector $\vec{r}$.	Cosines of angles $\alpha, \beta, \gamma$.
Uniqueness	Infinite sets (multiples of each other).	Unique for a given direction.
Relationship	$a, b, c \propto l, m, n$	$l = \frac{a}{|\vec{r}|}, m = \frac{b}{|\vec{r}|}, n = \frac{c}{|\vec{r}|}$
Square Sum	$a^2 + b^2 + c^2 = |\vec{r}|^2$	$l^2 + m^2 + n^2 = 1$

Note: In many competitive exams and textbooks, direction ratios are also referred to as direction numbers.

Section Formulae

The section formula is used to find the position vector of a point that divides a line segment joining two given points in a specific ratio, either internally or externally.

Section Formula: Internal Division

The Section Formula is a mathematical tool used to determine the position vector of a point that divides a line segment into two parts according to a given ratio. When the point lies between the two endpoints of the segment, the division is said to be internal.

Given:

1. Two points $A$ and $B$ in space with position vectors $\vec{OA} = \vec{a}$ and $\vec{OB} = \vec{b}$ relative to an origin $O$.

2. A point $P$ with position vector $\vec{OP} = \vec{r}$ that lies on the line segment $AB$.

3. The point $P$ divides $AB$ internally in the ratio $m : n$, such that $\frac{AP}{PB} = \frac{m}{n}$.

To Find:

The position vector $\vec{r}$ in terms of $\vec{a}$, $\vec{b}$, $m$, and $n$.

Geometrical Representation

In the rectangular coordinate system, we represent the vectors originating from $O(0,0,0)$. The points $A, P,$ and $B$ are collinear, meaning they lie on the same straight line.

Proof and Derivation

Since the point $P$ divides the line segment $AB$ internally in the ratio $m : n$, the ratio of the lengths of the segments $AP$ and $PB$ is given by:

By cross-multiplying the magnitudes, we get:

Because $A, P,$ and $B$ lie on a straight line and $P$ is between $A$ and $B$, the vectors $\vec{AP}$ and $\vec{PB}$ are in the same direction. Thus, we can write the vector equality:

We know that a vector joining two points is the difference of their position vectors. Therefore:

Substituting these vector expressions into our equality:

Expanding both sides using the distributive property of scalar multiplication:

To solve for $\vec{r}$, we group all terms containing $\vec{r}$ on the left-hand side and the remaining terms on the right-hand side:

Factoring out the vector $\vec{r}$:

Dividing by the scalar $(m + n)$, we arrive at the final formula:

Coordinate Form (In $R^3$)

If we express the position vectors in terms of their rectangular components, where $A(x_1, y_1, z_1)$, $B(x_2, y_2, z_2)$, and $P(x, y, z)$, the coordinates of $P$ are given by:

Coordinate	Value
X-coordinate ($x$)	$\frac{mx_2 + nx_1}{m + n}$
Y-coordinate ($y$)	$\frac{my_2 + ny_1}{m + n}$
Z-coordinate ($z$)	$\frac{mz_2 + nz_1}{m + n}$

---

Section Formula: External Division

When a point $P$ divides a line segment $AB$ externally, it means that the point $P$ lies on the line passing through $A$ and $B$, but not between them. Instead, it lies on the line segment produced. The ratio of the distances is measured from the endpoints to this external point $P$.

Given:

1. Two points $A$ and $B$ with position vectors $\vec{OA} = \vec{a}$ and $\vec{OB} = \vec{b}$ respectively.

2. A point $P$ with position vector $\vec{OP} = \vec{r}$ that lies on the line $AB$ produced.

3. The point $P$ divides the segment $AB$ externally in the ratio $m : n$, such that $AP : BP = m : n$.

To Find:

The position vector $\vec{r}$ of point $P$.

Derivation of the Formula

Since the point $P$ divides the line segment $AB$ externally in the ratio $m : n$, we have the following ratio of lengths:

By cross-multiplying the scalar lengths:

Since points $A, B,$ and $P$ are collinear and $P$ lies outside $AB$, the vectors $\vec{AP}$ and $\vec{BP}$ are in the same direction. Thus, we can represent this as a vector equation:

Now, we express these vectors in terms of their respective position vectors from the origin $O$:

Substituting these into the vector equation:

Expanding both sides using the distributive property:

To isolate the vector $\vec{r}$, we rearrange the terms:

Factoring out $\vec{r}$ from the right-hand side:

Finally, dividing by the scalar $(m - n)$, we obtain the position vector of point $P$:

Coordinate Comparison in Space

The following table illustrates the coordinates of point $P(x, y, z)$ when it divides the line joining $A(x_1, y_1, z_1)$ and $B(x_2, y_2, z_2)$ externally.

Coordinate	External Division Formula
X-coordinate ($x$)	$\frac{mx_2 - nx_1}{m - n}$
Y-coordinate ($y$)	$\frac{my_2 - ny_1}{m - n}$
Z-coordinate ($z$)	$\frac{mz_2 - nz_1}{m - n}$

Key Points to Remember

1. Sign Convention: The external division formula is identical to the internal division formula, except that $n$ is replaced by $-n$.

2. Validity: The formula for external division is valid only when $m \neq n$. If $m = n$, the point $P$ cannot divide the segment externally as it would be at infinity.

3. Direction: If $m > n$, the point $P$ lies on the side of $B$ (segment $AB$ produced). If $m < n$, the point $P$ lies on the side of $A$ (segment $BA$ produced).

---

The Mid-point Formula

The Mid-point Formula is a specific case of the internal section formula. It is used to find the position vector of a point $P$ that divides the line segment $AB$ into two equal halves.

Given:

1. Two points $A$ and $B$ with position vectors $\vec{a}$ and $\vec{b}$ respectively.

2. Point $P$ is the mid-point of the segment $AB$.

Condition:

Since $P$ is the mid-point, it divides $AB$ in the ratio $1 : 1$. Thus, we set the values of $m$ and $n$ as:

Derivation:

By substituting these values into the internal section formula $\vec{r} = \frac{m\vec{b} + n\vec{a}}{m + n}$, we get:

Simplifying the expression, we obtain the position vector $\vec{r}$ of the mid-point:

Component Form (Coordinates)

In a 3D rectangular coordinate system, if the coordinates of the endpoints are $A(x_1, y_1, z_1)$ and $B(x_2, y_2, z_2)$, the coordinates of the mid-point $P(x, y, z)$ are calculated as follows:

Axis	Mid-point Coordinate
X-axis	$x = \frac{x_1 + x_2}{2}$
Y-axis	$y = \frac{y_1 + y_2}{2}$
Z-axis	$z = \frac{z_1 + z_2}{2}$

Geometrical Significance

The mid-point vector $\vec{r}$ represents the average of the position vectors of the two endpoints. Geometrically, it is the vector directed from the origin to the exact center of the line segment $AB$.

Note: This formula is extensively used in finding the median of a triangle or the center of a line in vector algebra and 3D geometry.

---

Centroid of a Triangle

The Centroid of a triangle is the point of concurrency where all three medians of the triangle intersect. A median is a line segment joining a vertex to the mid-point of the opposite side. A key property of the centroid ($G$) is that it divides each median internally in the ratio $2 : 1$.

Given:

Let $ABC$ be a triangle where the position vectors of the vertices $A$, $B$, and $C$ with respect to an origin $O$ are $\vec{a}$, $\vec{b}$, and $\vec{c}$ respectively.

Derivation of the Centroid Formula

To find the position vector of the centroid $G$, we follow these steps:

Step 1: Finding the mid-point of side $BC$

Let $D$ be the mid-point of the side $BC$. By applying the mid-point formula, the position vector of $D$ ($\vec{OD}$) is given by:

Step 2: Applying the Section Formula on Median $AD$

The centroid $G$ lies on the median $AD$ and divides it internally in the ratio $2 : 1$ (where $m = 2$ and $n = 1$). Using the section formula for internal division:

Step 3: Substitution and Simplification

Now, we substitute the value of $\vec{OD}$ from equation (xi) into the expression for $\vec{OG}$:

By canceling the scalar $2$ in the numerator:

Rearranging the vectors in alphabetical order, we get the final position vector $\vec{g}$:

Properties and Concurrency

It can be mathematically shown that if we calculate the point dividing the other medians (say $BE$ or $CF$) in the ratio $2 : 1$, we arrive at the exact same position vector. This proves that:

1. All three medians of a triangle are concurrent (they pass through the same point).

2. The point of concurrency is the centroid $G$ with position vector $\frac{\vec{a} + \vec{b} + \vec{c}}{3}$.

Coordinate Form in 3D Space

If the vertices are given as coordinates, the centroid $G(x, y, z)$ is calculated as the arithmetic mean of the coordinates of the vertices:

Axis	Centroid Coordinate Formula
X-coordinate	$x = \frac{x_1 + x_2 + x_3}{3}$
Y-coordinate	$y = \frac{y_1 + y_2 + y_3}{3}$
Z-coordinate	$z = \frac{z_1 + z_2 + z_3}{3}$

Note: This formula is applicable for any triangle in a 2D plane as well, by simply omitting the $z$ component.

---

Section and Centroid Formulas

For efficient problem-solving in vector algebra, it is essential to have a consolidated view of the formulas used to find the position vectors of specific points on line segments or within geometric figures like triangles.

Variables used:

1. $\vec{a}$ and $\vec{b}$ represent the position vectors of the endpoints $A$ and $B$.

2. $m : n$ represents the ratio in which the point divides the segment.

3. $\vec{r}$ (or $\vec{g}$) represents the position vector of the resultant point (Section point or Centroid).

Tabular Summary of Vector Formulae

Case / Condition	Position Vector Formula
Internal Division (Ratio $m : n$)	$\vec{r} = \frac{m\vec{b} + n\vec{a}}{m + n}$
External Division (Ratio $m : n$)	$\vec{r} = \frac{m\vec{b} - n\vec{a}}{m - n}$
Mid-point of Segment $AB$	$\vec{r} = \frac{\vec{a} + \vec{b}}{2}$
Centroid of Triangle $ABC$	$\vec{g} = \frac{\vec{a} + \vec{b} + \vec{c}}{3}$
Centroid of Tetrahedron	$\vec{g} = \frac{\vec{a} + \vec{b} + \vec{c} + \vec{d}}{4}$

Key Points for Application

1. Consistency: In the internal division formula, the numerator uses an addition sign ($+$), whereas in external division, it uses a subtraction sign ($-$).

2. Ratio Convention: If a point $P$ divides $AB$ in the ratio $k : 1$, the formula simplifies to:

3. Centroid Property: The centroid is effectively the mean position vector of all the vertices of the triangle.

4. Negative Ratio: In some competitive exam contexts, an external division in ratio $m : n$ is sometimes treated as an internal division in ratio $m : -n$.

5. Mid-point: The mid-point is the specific case of internal division where the weights assigned to both position vectors are equal ($1:1$).

Scalar (or Dot) Product of Two Vectors

In elementary algebra, multiplication is a simple operation between two real numbers. However, in Vector Algebra, the process is more complex because vectors possess both magnitude and direction. Unlike scalars, two vectors cannot be multiplied by a single unique method.

Based on the nature of the resultant quantity, vector multiplication is classified into two distinct types, which are widely used in Physics and Engineering:

Type of Product	Resultant Quantity	Physical Example
Scalar Product (Dot Product)	A Scalar (Real Number)	Work Done ($\text{Work} = \vec{F} \cdot \vec{d}$)
Vector Product (Cross Product)	A Vector	Torque ($\vec{\tau} = \vec{r} \times \vec{F}$)

The definition of these products is mathematically framed to satisfy physical laws. For instance, the dot product helps in finding the component of one vector along another, while the cross product helps in determining rotational effects.

---

Angle between Two Vectors

The concept of the angle between two vectors is fundamental to defining their products. Since vectors in this context are considered free vectors, they can be shifted parallel to themselves without changing their magnitude or direction.

Definition:

To find the angle between two non-zero vectors $\vec{a}$ and $\vec{b}$, we translate them such that their initial points (tails) coincide at a common point, say origin $O$. The smaller of the two angles formed between the vectors is known as the angle between them, usually denoted by $\theta$.

Range of the Angle ($\theta$):

In vector algebra, the angle $\theta$ is always restricted to the interval between $0$ and $\pi$ radians (or $0^\circ$ and $180^\circ$).

If the angle between two directions is greater than $\pi$ (e.g., $210^\circ$), we always consider the shorter path (reflex angle is ignored), which would be $360^\circ - 210^\circ = 150^\circ$.

Geometric Interpretation of the Angle

The angle $\theta$ determines the orientation of one vector relative to another:

1. If $\theta = 0^\circ$, the vectors are like-parallel (pointing in the same direction).

2. If $\theta = 180^\circ$ ($\pi$ rad), the vectors are unlike-parallel or anti-parallel (pointing in opposite directions).

3. If $\theta = 90^\circ$ ($\pi/2$ rad), the vectors are orthogonal or perpendicular to each other.

Note: The angle between a vector and itself is always $0^\circ$, and the angle between a vector $\vec{a}$ and its negative $-\vec{a}$ is always $180^\circ$.

---

Definition of Scalar (or Dot) Product

The Scalar Product of two non-zero vectors $\vec{a}$ and $\vec{b}$ is a mathematical operation that takes two vectors and returns a single real number (a scalar). It is frequently referred to as the Dot Product because of the dot symbol "$\cdot$" used to denote the operation.

It is defined as the product of the magnitudes of the two vectors and the cosine of the angle $\theta$ between them. Mathematically, it is expressed as:

Where:

1. $|\vec{a}|$ is the magnitude (length) of vector $\vec{a}$.

2. $|\vec{b}|$ is the magnitude (length) of vector $\vec{b}$.

3. $\theta$ is the angle between $\vec{a}$ and $\vec{b}$, such that $0 \leq \theta \leq \pi$.

Key Characteristic: The result of the dot product is always a scalar quantity. This means it has magnitude but no direction, even though it is derived from two vector quantities.

The Nature of the Scalar Result

A common question in vector algebra is how the multiplication of two directed quantities (vectors) can result in a quantity with no direction (a scalar). This happens because of the mathematical structure of the definition and the way components interact.

Mathematical Reason: Multiplication of Real Numbers

By definition, the dot product is calculated as:

Let us analyze the nature of each term in this product:

1. $|\vec{a}|$: This is the magnitude or length of vector $\vec{a}$, which is a real number (scalar).

2. $|\vec{b}|$: This is the magnitude or length of vector $\vec{b}$, which is also a real number (scalar).

3. $\cos \theta$: This is a trigonometric ratio. In mathematics, all trigonometric ratios are pure numbers with no units or direction (scalars).

Conclusion: Since the product of three scalars (Real Number $\times$ Real Number $\times$ Real Number) is always a Real Number, the result of the dot product is inherently a scalar.

Component-wise Reason: The "Disappearance" of Unit Vectors

When we multiply two vectors in their component forms, we see exactly "how" the directional unit vectors ($\hat{i}, \hat{j}, \hat{k}$) are eliminated.

Let:

$\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}$

$\vec{b} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k}$

When we perform $\vec{a} \cdot \vec{b}$, we multiply every term of $\vec{a}$ with every term of $\vec{b}$. However, due to the orthogonality of the unit vectors:

1. Expanding the Product:

We write the dot product as the product of two trinomials:

2. Full Expansion:

Expanding the brackets further, we get:

Applying Orthogonality Rules

The coordinate axes in the Cartesian system are mutually perpendicular. This gives rise to the orthogonality of the unit vectors $\hat{i}, \hat{j},$ and $\hat{k}$. We apply the following rules to the nine terms above:

Rule A: Self-Dot Product (Angle = $0^\circ$)

Rule B: Mutual-Dot Product (Angle = $90^\circ$)

Final Simplification

When we substitute the values from above equations into our expanded expression, six terms vanish (become zero) and three terms survive (multiplied by 1):

This leaves us with the clean algebraic sum:

As seen in the above equation, the final expression contains only the products of the scalar components. All directional unit vectors ($\hat{i}, \hat{j}, \hat{k}$) have been reduced to the number 1 or 0. This is the algebraic reason why the product becomes a scalar.

Geometric Reason: Measure of "Overlap"

Geometrically, the dot product does not create a new direction; instead, it measures the extent of alignment between two vectors. It answers the question: "How much of vector $\vec{b}$ is pointing in the direction of vector $\vec{a}$?"

The result is a magnitude that tells us about the relationship between the two vectors:

1. A Positive result means they point in generally the same direction.

2. A Negative result means they point in generally opposite directions.

3. A Zero result means they have no common direction (perpendicular).

The Geometric Origin: Where does $\cos \theta$ come from?

The presence of $\cos \theta$ in the scalar product formula is not arbitrary. It arises from the concept of projections. In vector algebra, we often need to determine how much of one vector acts in the direction of another. This "effective part" of a vector is called its scalar projection.

Consider two vectors $\vec{a}$ and $\vec{b}$ with an angle $\theta$ between them. If we drop a perpendicular from the tip of $\vec{b}$ onto the line containing $\vec{a}$, the length of the segment formed on $\vec{a}$ is the projection of $\vec{b}$ onto $\vec{a}$.

In the right-angled triangle formed:

The Scalar Product is defined as the magnitude of the first vector ($|\vec{a}|$) multiplied by the projection of the second vector onto the first ($|\vec{b}| \cos \theta$). Hence:

Mathematical Derivation (Using Law of Cosines)

To prove why $\vec{a} \cdot \vec{b}$ equals $|\vec{a}| |\vec{b}| \cos \theta$ formally, we use the Law of Cosines from trigonometry. Consider $\triangle OAB$ where $\vec{OA} = \vec{a}$, $\vec{OB} = \vec{b}$, and $\vec{AB} = \vec{a} - \vec{b}$.

Proof:

According to the Law of Cosines in $\triangle OAB$:

Substituting the magnitudes of the vectors:

From the property of the square of a vector ($|\vec{v}|^2 = \vec{v} \cdot \vec{v}$):

Expanding using the distributive law:

Since $\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a}$ (Commutative Property):

Comparing equation (i) and equation (ii):

Canceling $|\vec{a}|^2$ and $|\vec{b}|^2$ from both sides:

Special Cases based on the Angle ($\theta$)

The value of the scalar product varies significantly depending on the orientation of the vectors. Since $|\vec{a}|$ and $|\vec{b}|$ are always non-negative, the sign and magnitude of the dot product are determined by the cosine of the angle between them.

Case 1: Parallel Vectors (Like Vectors)

When two vectors are parallel and act in the same direction, the angle between them is $\theta = 0^\circ$.

Using the definition:

Since $\cos 0^\circ = 1$:

Case 2: Anti-parallel Vectors (Unlike Vectors)

When two vectors act in exactly opposite directions, the angle between them is $\theta = \pi$ (or $180^\circ$).

Using the definition:

Since $\cos \pi = -1$:

Case 3: Perpendicular Vectors (Orthogonal Vectors)

When two vectors are perpendicular to each other, the angle between them is $\theta = \frac{\pi}{2}$ (or $90^\circ$).

Using the definition:

Since $\cos \frac{\pi}{2} = 0$:

This is the Condition of Orthogonality. In 3D geometry and vector problems, if the dot product of two non-zero vectors is zero, they are confirmed to be perpendicular.

Summary Table of Scalar Product Values

Angle ($\theta$)	$\cos \theta$	Value of $\vec{a} \cdot \vec{b}$	Interpretation
$0^\circ$	$1$	$|\vec{a}| |\vec{b}|$	Parallel (Same direction)
$0 < \theta < 90^\circ$	Positive	$> 0$	Acute Angle
$90^\circ$	$0$	$0$	Perpendicular
$90^\circ < \theta < 180^\circ$	Negative	$< 0$	Obtuse Angle
$180^\circ$	$-1$	$-|\vec{a}| |\vec{b}|$	Anti-parallel (Opposite direction)

Note: If either $\vec{a} = 0$ or $\vec{b} = 0$, the angle $\theta$ is not defined, and in such cases, the dot product is defined as $0$.

Condition of Perpendicularity

Two non-zero vectors $\vec{a}$ and $\vec{b}$ are perpendicular (orthogonal) if and only if their scalar product is zero.

Conversely, if $\vec{a} \cdot \vec{b} = 0$, then either $\vec{a} = 0$, $\vec{b} = 0$, or the vectors are perpendicular to each other.

Sign of the Scalar Product

The sign of $\vec{a} \cdot \vec{b}$ is determined by the nature of the angle $\theta$:

Condition of $\theta$	Nature of Angle	Value of $\vec{a} \cdot \vec{b}$
$0 \leq \theta < \frac{\pi}{2}$	Acute Angle	$\vec{a} \cdot \vec{b} > 0$ (Positive)
$\theta = \frac{\pi}{2}$	Right Angle	$\vec{a} \cdot \vec{b} = 0$ (Zero)
$\frac{\pi}{2} < \theta \leq \pi$	Obtuse Angle	$\vec{a} \cdot \vec{b} < 0$ (Negative)

Square of a Vector

The scalar product of a vector $\vec{a}$ with itself is known as the square of the vector. In this case, the angle $\theta$ is $0^\circ$.

Derivation:

Thus, the square of a vector is equal to the square of its magnitude. The length of a vector can be recovered using:

Scalar Products of Unit Vectors $\hat{i}, \hat{j}, \hat{k}$

Since $\hat{i}, \hat{j},$ and $\hat{k}$ are mutually perpendicular unit vectors along the coordinate axes, their dot products follow specific rules:

1. Dot product with themselves:

Since $|\hat{i}| = |\hat{j}| = |\hat{k}| = 1$ and the angle is $0^\circ$:

2. Dot product with each other:

Since they are mutually perpendicular ($\theta = 90^\circ$):

$\cdot$	$\hat{i}$	$\hat{j}$	$\hat{k}$
$\hat{i}$	1	0	0
$\hat{j}$	0	1	0
$\hat{k}$	0	0	1

Note: The scalar product is commutative, meaning $\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a}$.

---

Geometrical Interpretation of Scalar Product of Two Vectors

Beyond the algebraic definition, the scalar product has a deep geometrical significance. It is essentially a measure of how much one vector "overlaps" or "projects" onto another.

The projection of a vector $\vec{a}$ on another vector $\vec{b}$ is the scalar component of $\vec{a}$ in the direction of $\vec{b}$. This concept is vital in Physics for calculating the component of a force acting along the direction of displacement.

Geometrical Construction

To visualize this, let us represent two non-zero vectors $\vec{a}$ and $\vec{b}$ starting from a common origin $O$.

Step 1: Let $\vec{OA} = \vec{a}$ and $\vec{OB} = \vec{b}$. Let the angle between these two vectors be $\theta$, where $0 \leq \theta \leq \pi$.

Step 2: From the terminal point $A$ of vector $\vec{a}$, drop a perpendicular $AM$ onto the line containing vector $\vec{b}$ (the $OB$ line). The line segment $OM$ is the projection of $\vec{a}$ on $\vec{b}$.

Step 3: Similarly, from the terminal point $B$ of vector $\vec{b}$, drop a perpendicular $BN$ onto the line containing vector $\vec{a}$ (the $OA$ line). The line segment $ON$ is the projection of $\vec{b}$ on $\vec{a}$.

Derivation of Projection Lengths

We can use basic trigonometry to find the lengths of these projections. Consider the right-angled triangles formed in our construction.

1. Finding the Projection of $\vec{a}$ on $\vec{b}$ ($OM$):

In the right-angled triangle $\triangle OAM$, the side $OM$ is the adjacent side and $OA$ is the hypotenuse.

Since $OA = |\vec{a}|$, we get:

2. Finding the Projection of $\vec{b}$ on $\vec{a}$ ($ON$):

In the right-angled triangle $\triangle OBN$, the side $ON$ is the adjacent side and $OB$ is the hypotenuse.

Since $OB = |\vec{b}|$, we get:

Comparison of the Two Projections

The following table summarizes the two types of projections for easy reference during exam preparation:

Target	Projection Length	Trigonometric Form
$\vec{a}$ on $\vec{b}$	$OM$	$|\vec{a}| \cos \theta$
$\vec{b}$ on $\vec{a}$	$ON$	$|\vec{b}| \cos \theta$

Important Observations

1. Scalar Nature: The projection is a magnitude (a scalar). It represents a length along a specific line.

2. Effect of Angle:

    • If $\theta$ is acute ($< 90^\circ$), the projection is positive and points in the direction of the target vector.

    • If $\theta$ is obtuse ($> 90^\circ$), $\cos \theta$ is negative. Geometrically, the projection falls on the line produced in the opposite direction.

    • If $\theta = 90^\circ$, the projection is zero because there is no overlap.

3. The "Dot" Connection: These lengths $OM$ and $ON$ are the building blocks of the Scalar Product formula. Since $\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta$, we can say that the dot product is the magnitude of one vector multiplied by the projection of the other on it.

Scalar Product expressed in terms of Projection

The definition of the scalar product, $\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta$, allows us to view the operation as a product of a magnitude and a projection. By grouping the terms differently, we can interpret the dot product from two distinct perspectives.

Case 1: Projection of $\vec{a}$ on $\vec{b}$

We can group the magnitude of $\vec{a}$ and the cosine term together. This grouping highlights the component of vector $\vec{a}$ that lies along the direction of vector $\vec{b}$.

Since $|\vec{a}| \cos \theta$ is the scalar projection of $\vec{a}$ on $\vec{b}$, we can say:

Case 2: Projection of $\vec{b}$ on $\vec{a}$

Similarly, by grouping the magnitude of $\vec{b}$ with the cosine term, we highlight the component of vector $\vec{b}$ that lies along the direction of vector $\vec{a}$.

Since $|\vec{b}| \cos \theta$ is the scalar projection of $\vec{b}$ on $\vec{a}$, we can say:

Conclusion: Geometrically, the scalar product of two vectors is equal to the modulus (magnitude) of one vector multiplied by the projection of the other vector upon it.

Formulas for Projection

In competitive examinations, calculating the projection is a frequent requirement. The projection is essentially the dot product divided by the magnitude of the vector on which the projection is being taken.

Target Projection	Mathematical Formula
Projection of $\vec{a}$ on $\vec{b}$	$\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|} = \vec{a} \cdot \hat{b}$
Projection of $\vec{b}$ on $\vec{a}$	$\frac{\vec{a} \cdot \vec{b}}{|\vec{a}|} = \vec{b} \cdot \hat{a}$

Note: The projection is a scalar quantity. If the dot product is negative (meaning the angle $\theta$ is obtuse), the projection is technically negative, but for length of projection, we always consider the absolute value $|\vec{a} \cdot \hat{b}|$.

Connection with Direction Cosines

The concept of projection is directly linked to the Direction Cosines of a vector in three-dimensional space. If a non-zero vector $\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}$ makes angles $\alpha, \beta, \gamma$ with the positive $x, y,$ and $z$ axes respectively, then these angles determine its orientation.

Derivation using Dot Product

The direction cosines $l, m, n$ are defined as the dot product of the vector with the unit vectors along the axes, divided by the vector's magnitude.

For the $x$-axis ($\hat{i}$):

For the $y$-axis ($\hat{j}$):

For the $z$-axis ($\hat{k}$):

Components as Projections

By rearranging the equations above, we can express the scalar components of the vector as projections along the respective coordinate axes:

Conclusion: The scalar components $a_1, a_2,$ and $a_3$ are defined as the projections of vector $\vec{a}$ along the $x$-axis, $y$-axis, and $z$-axis respectively. This confirms that a vector is simply the sum of its projection components in 3D space.

---

Fundamental Properties of Scalar Product

The scalar product follows several algebraic laws that are similar to the multiplication of real numbers. Understanding these properties is essential for simplifying complex vector expressions in Engineering and Board examinations.

Theorem 1: Commutative Law

The scalar product of two vectors is commutative. This means the order in which the vectors are multiplied does not change the result.

Proof:

Let $\theta$ be the angle between non-zero vectors $\vec{a}$ and $\vec{b}$. By definition:

Since the angle between $\vec{b}$ and $\vec{a}$ is also $\theta$, and the magnitudes $|\vec{a}|$ and $|\vec{b}|$ are real numbers (which follow commutative multiplication):

Theorem 2: Multiplication by a Scalar

If $\vec{a}$ and $\vec{b}$ are any two vectors and $m$ is any real number (scalar), then the scalar can be associated with either vector or the entire product.

Also:

1. $(-\vec{a}) \cdot \vec{b} = -(\vec{a} \cdot \vec{b}) = \vec{a} \cdot (-\vec{b})$

2. $(-\vec{a}) \cdot (-\vec{b}) = \vec{a} \cdot \vec{b}$

Theorem 3: Distributive Law

The scalar product is distributive over vector addition. This is one of the most important properties for expanding vector brackets.

This property can be extended to any number of vectors:

Scalar Product of Sums of Vectors

When multiplying two binomial vector expressions, we use the distributive and commutative properties together.

Expression: $(\vec{a} + \vec{b}) \cdot (\vec{c} + \vec{d})$

Derivation:

Vector Identities

Just like algebraic identities $(a+b)^2$, vector magnitudes follow similar patterns.

(i) Square of Sum:

(ii) Square of Difference:

(iii) Difference of Squares:

(iv) Sum of Squares of Sum and Difference:

(v) Difference of Squares of Sum and Difference:

---

Angle Between Two Vectors

One of the most important applications of the Scalar Product is finding the angle between two non-zero vectors. Since the dot product is directly dependent on the cosine of the angle $\theta$, we can rearrange the formula to isolate the angle.

General Formula

From the definition of the scalar product, we have:

To find the angle $\theta$ (where $0 \leq \theta \leq \pi$), we rearrange the equation as follows:

Or, in terms of $\theta$ itself:

Angle in Component Form

In most problems, vectors are provided in their rectangular component form. Let us consider two vectors:

$\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}$

$\vec{b} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k}$

Substituting the component-wise dot product and the magnitudes into the formula:

1. Scalar Product: $\vec{a} \cdot \vec{b} = a_1b_1 + a_2b_2 + a_3b_3$

2. Magnitude of $\vec{a}$: $|\vec{a}| = \sqrt{a_1^2 + a_2^2 + a_3^2}$

3. Magnitude of $\vec{b}$: $|\vec{b}| = \sqrt{b_1^2 + b_2^2 + b_3^2}$

Combining these into above equation, we get the formula for the angle in component form:

Vector (or Cross) Product of Two Vectors

The Vector Product (also known as the Cross Product) of two non-zero vectors $\vec{a}$ and $\vec{b}$ is a mathematical operation that results in a vector quantity. Unlike the dot product, the result of a cross product has both magnitude and a specific direction.

It is denoted by $\vec{a} \times \vec{b}$ (read as $\vec{a}$ cross $\vec{b}$) and is defined as:

Where:

1. $|\vec{a}|$ and $|\vec{b}|$ are the magnitudes of vectors $\vec{a}$ and $\vec{b}$.

2. $\theta$ is the angle between $\vec{a}$ and $\vec{b}$, such that $0 \leq \theta \leq \pi$.

3. $\hat{n}$ is a unit vector perpendicular to both $\vec{a}$ and $\vec{b}$. Its direction is determined by the right-handed screw rule.

Direction of the Resultant Vector ($\hat{n}$)

The unit vector $\hat{n}$ is perpendicular to the plane containing $\vec{a}$ and $\vec{b}$. If you curl the fingers of your right hand from $\vec{a}$ to $\vec{b}$ through the smaller angle $\theta$, your thumb points in the direction of $\hat{n}$. This makes the set of vectors ($\vec{a}, \vec{b}, \vec{a} \times \vec{b}$) a right-handed system.

In terms of a right-handed screw, the direction of $\hat{n}$ is the direction of the translation (advancement) of the screw if it is rotated in such a way that the rotation makes the first vector $\vec{a}$ coincide with the second vector $\vec{b}$.

Difference Between Scalar and Vector Products

To understand why one results in a scalar and the other in a vector, we must look at what they represent physically and mathematically.

Feature	Scalar (Dot) Product ($\vec{a} \cdot \vec{b}$)	Vector (Cross) Product ($\vec{a} \times \vec{b}$)
Resultant Nature	Scalar: A pure number with no direction.	Vector: A quantity with both magnitude and direction.
Geometric Focus	Measures Projection (Shadow).	Measures Area and Orientation.
Trigonometric Ratio	Uses $\cos \theta$ (Alignment).	Uses $\sin \theta$ (Perpendicularity).
Physical Application	Work Done, Flux.	Torque, Angular Momentum, Lorentz Force.

Why is one a Scalar and the other a Vector?

1. Scalar Product Result: The dot product effectively "multiplies" the magnitude of one vector by the component of the second vector that is parallel to it. Since we are multiplying two magnitudes (scalars) in the same direction, the directional information is "consumed," leaving only a total magnitude (a scalar).

2. Vector Product Result: The cross product is designed to represent rotational effects or surface areas. In physics, rotation (like turning a wrench) happens around an axis. That axis is the direction of the resultant vector. Because the rotation has a specific orientation (Clockwise or Anti-clockwise), the result must be a vector to describe that axis and sense of rotation.

The Origin of $\sin \theta$ and $\hat{n}$

1. Where did $\sin \theta$ come from?

The magnitude of $\vec{a} \times \vec{b}$ is equal to the Area of the Parallelogram formed by vectors $\vec{a}$ and $\vec{b}$.

Consider a parallelogram with sides $\vec{a}$ and $\vec{b}$. Let the base be $|\vec{a}|$. The height ($h$) of this parallelogram is the perpendicular component of $\vec{b}$ relative to $\vec{a}$.

Since $\text{Area} = \text{Base} \times \text{Height}$:

This is why $\sin \theta$ is used—it extracts the perpendicular component necessary to calculate area or rotational leverage (torque).

2. Where did $\hat{n}$ come from?

The symbol $\hat{n}$ stands for the Normal Vector. In 3D geometry, two non-parallel vectors define a unique plane. The cross product must point in a direction that is "uniquely related" to both. The only such unique direction is perpendicular (normal) to that plane.

However, there are always two perpendicular directions (up and down). The Right-Hand Rule is the universal convention used to choose one of these two directions consistently, represented by the unit vector $\hat{n}$.

Formal Definition Elaboration

• Magnitude: Corresponds to the area of the parallelogram formed by the two vectors.

• Direction ($\hat{n}$): Defined by the Right-Hand Screw Rule. If you rotate $\vec{a}$ into $\vec{b}$, a right-handed screw would move in the direction of $\vec{a} \times \vec{b}$.

Important Perspective: We use "Right-Handed Cartesian Coordinates." This means $\hat{i} \times \hat{j} = \hat{k}$. If we used a left-handed system, all our cross product directions would be reversed.

---

Fundamental Remarks

The Vector Product is a binary operation on two vectors in three-dimensional space. Unlike the scalar product, which measures "how much" one vector lies along another, the cross product measures the extent of perpendicularity and determines the orientation of the plane containing the vectors.

Nature of the Resultant Vector

The Vector Product (or cross product) of two vectors possesses several distinct geometric and algebraic characteristics regarding its resultant.

1. Resultant as a Vector Quantity

The cross product of two vectors is always a vector quantity. Unlike the scalar product, which yields a single numerical value, the output of $\vec{a} \times \vec{b}$ has both a magnitude ($|\vec{a}| |\vec{b}| \sin \theta$) and a specific direction ($\hat{n}$).

2. Case of Zero Vectors

If either of the vectors is a null vector, or if both are null vectors, the magnitude of the product becomes zero. In such cases, the direction of the resultant is not defined.

3. Perpendicularity of the Unit Vector ($\hat{n}$)

The unit vector $\hat{n}$ is perpendicular to both $\vec{a}$ and $\vec{b}$ simultaneously. Consequently, $\hat{n}$ is normal (perpendicular) to the entire plane containing the vectors $\vec{a}$ and $\vec{b}$.

In mathematical terms, we verify this orthogonality by checking that the dot product of the resultant with either individual vector is zero:

This property allows us to determine a unit vector normal to the plane of $\vec{a}$ and $\vec{b}$ using the formula:

4. Right-Handed Triad for Perpendicular Vectors

If the two vectors $\vec{a}$ and $\vec{b}$ are mutually perpendicular ($\theta = 90^\circ$ or $\frac{\pi}{2}$), then the three vectors $\vec{a}$, $\vec{b}$, and the resultant $\vec{a} \times \vec{b}$ form a right-handed triad of mutually perpendicular vectors.

In this specific case, the magnitude of the cross product reaches its maximum value:

Condition of Parallelism

The cross product serves as the ultimate test for parallelism or collinearity of two non-zero vectors. If the cross product of two vectors is the zero vector ($\vec{0}$), the vectors must be parallel.

Reasoning:

The magnitude of the cross product is defined by $|\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin \theta$. For non-zero vectors $\vec{a}$ and $\vec{b}$, the product can only be zero if $\sin \theta = 0$.

This occurs when:

1. $\theta = 0^\circ$: The vectors are in the same direction (Like-parallel).

2. $\theta = 180^\circ$ ($\pi$ rad): The vectors are in the opposite direction (Anti-parallel).

Self-Cross and Negative-Cross Products

Based on the condition of parallelism, certain standard results are derived which are used for simplifying vector expressions:

A. Self-Cross Product: Since any vector is parallel to itself ($\theta = 0^\circ$), its cross product with itself is always a null vector.

B. Negative-Cross Product: A vector and its negative point in exactly opposite directions ($\theta = 180^\circ$). Since $\sin 180^\circ = 0$, their cross product is also zero.

Relationship	Dot Product ($\cdot$)	Cross Product ($\times$)
Parallel ($\theta = 0^\circ$)	Maximum ($ab$)	Zero ($\vec{0}$)
Perpendicular ($\theta = 90^\circ$)	Zero ($0$)	Maximum ($ab\hat{n}$)

Anti-Commutative Property

One of the most distinct features of the vector product is that it does not obey the commutative law. Unlike real numbers ($2 \times 3 = 3 \times 2$) or the dot product ($\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a}$), changing the order of vectors in a cross product reverses the direction of the resultant.

The Right-Hand Screw Rule Explanation:

The direction of $\vec{a} \times \vec{b}$ is determined by the Right-Hand Rule. If you curl the fingers of your right hand from the first vector ($\vec{a}$) toward the second vector ($\vec{b}$), your thumb points in the direction of the product.

1. When we calculate $\vec{a} \times \vec{b}$, the rotation is from $A$ to $B$. Let's say the thumb points Upward.

2. When we calculate $\vec{b} \times \vec{a}$, the rotation is reversed (from $B$ to $A$). Following the same rule, the thumb must now point Downward.

Since the magnitudes are the same but the directions are exactly opposite ($180^\circ$ apart), one is the negative of the other.

Note: In algebraic expansions, this means you must maintain the order. For example, $(\vec{a} + \vec{b}) \times \vec{c}$ must be written as $\vec{a} \times \vec{c} + \vec{b} \times \vec{c}$, and not $\vec{c} \times \vec{a} + \vec{c} \times \vec{b}$.

---

Vector Products of Orthogonal Unit Vectors $\hat{i}, \hat{j}, \hat{k}$

The unit vectors $\hat{i}$, $\hat{j}$, and $\hat{k}$ represent the fundamental directions along the $X$, $Y$, and $Z$ axes of a Right-Handed Cartesian Coordinate System. Because these vectors are mutually perpendicular and have a magnitude of unity, their cross products follow very specific and predictable rules.

Cross Product of a Unit Vector with Itself (Self-Product)

When any unit vector is crossed with itself, the angle between the two vectors is $\theta = 0^\circ$. Since the sine of zero is zero, the magnitude of the resulting vector vanishes.

Derivation:

Similarly, this applies to $\hat{j}$ and $\hat{k}$. Thus, for self-products, we always obtain a null vector:

Cross Product of Distinct Unit Vectors (Mutual Product)

Since $\hat{i}$, $\hat{j}$, and $\hat{k}$ are mutually perpendicular, the angle between any two distinct unit vectors is $\theta = 90^\circ$ $(\frac{\pi}{2} \ rad)$. Since $\sin 90^\circ = 1$, the magnitude of the cross product will be $1 \cdot 1 \cdot 1 = 1$. The direction, however, is determined by the Right-Hand Rule.

Case Study: $\hat{i} \times \hat{j}$

If you curl the fingers of your right hand from the positive $X$-axis ($\hat{i}$) toward the positive $Y$-axis ($\hat{j}$), your thumb points along the positive $Z$-axis ($\hat{k}$).

The Cyclic Rule (The "Circle" Shortcut)

Students are taught a visual shortcut to remember these products without using the hand rule every time. Imagine the letters $i, j,$ and $k$ arranged in a clockwise circle.

Rule A: Standard Order (Clockwise)

If you move in the direction of the arrows ($i \to j \to k \to i$), the product of any two consecutive vectors is the positive third vector.

Rule B: Reverse Order (Anti-clockwise)

If you move against the direction of the arrows, the product of any two vectors is the negative third vector.

Summary Table of Unit Vector Cross Products

The following table serves as a ready-reckoner for calculating cross products in component form (using the determinant method):

$\times$	$\hat{i}$	$\hat{j}$	$\hat{k}$
$\hat{i}$	$\vec{0}$	$\hat{k}$	$-\hat{j}$
$\hat{j}$	$-\hat{k}$	$\vec{0}$	$\hat{i}$
$\hat{k}$	$\hat{j}$	$-\hat{i}$	$\vec{0}$

Note: Understanding these unit vector products is mandatory for deriving the Determinant Form of the cross product, which is the standard method for multiplying vectors like $\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}$ and $\vec{b} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k}$.

Exam Tip: Remember that the cross product of any two unit vectors always results in a vector that is perpendicular to the plane of the first two. For instance, $\hat{i}$ and $\hat{j}$ are in the $XY$-plane, so their product must lie on the $Z$-axis ($\hat{k}$).

---

Algebraic Properties of Vector Product

The Vector Product (Cross Product) is a non-linear operation that possesses unique algebraic properties. Unlike the multiplication of real numbers, it is sensitive to the order of operands and follows specific distributive patterns. These properties form the foundation for solving complex problems in 3D geometry and Physics.

Theorem 1: Scalar Multiplication Property

This property states that when a vector in a cross product is multiplied by a scalar $m$, the resulting magnitude of the cross product is scaled by $|m|$. If $m$ is positive, the direction remains the same; if $m$ is negative, the direction is reversed.

Mathematical Expression:

Consider $\vec{a}$ and $\vec{b}$ as two vectors. If we double the length of $\vec{a}$ (i.e., $m = 2$), the area of the parallelogram formed by them also doubles. This confirms the change in magnitude. In a Right-Handed System, moving the scalar $m$ between the vectors does not change the plane or the sense of rotation, thus maintaining the direction $\hat{n}$.

Note: If $m = -1$, then $(-\vec{a}) \times \vec{b} = -(\vec{a} \times \vec{b})$. This is consistent with the fact that reversing one vector flips the direction of the resultant vector by $180^\circ$.

Theorem 2: The Distributive Law

The vector product is distributive with respect to vector addition. This means the cross product of a vector with the sum of other vectors can be calculated by summing the individual cross products.

Geometrically, the cross product $\vec{a} \times \vec{b}$ can be viewed as a projection of $\vec{b}$ onto a plane perpendicular to $\vec{a}$, followed by a $90^\circ$ rotation. Since the projection of a sum $(\vec{b} + \vec{c})$ is equal to the sum of individual projections, the distributive property holds true.

Right Distributive Law

The distributive property also holds when the multiplier vector is on the right side of the expression.

Distribution over Subtraction

Since subtraction is essentially the addition of a negative vector ($\vec{b} - \vec{c} = \vec{b} + (-\vec{c})$), the distributive law applies naturally to differences.

Generalization to Multiple Terms

The property extends to any finite number of vectors within the sum.

Vector Product of Sums (Expansion of Binomials)

When multiplying two vector sums, such as $(\vec{a} + \vec{b}) \times (\vec{c} + \vec{d})$, we must expand the terms while strictly maintaining their relative order. Swapping vectors during expansion will result in sign errors due to anti-commutativity ($\vec{x} \times \vec{y} = -\vec{y} \times \vec{x}$).

Step-by-Step Derivation:

Let us treat the first binomial $(\vec{a} + \vec{b})$ as a single vector $\vec{U}$.

Now, substitute back $\vec{U} = (\vec{a} + \vec{b})$:

Apply the Right Distributive Law to both terms:

Thus, the final expanded form is:

General Case Expansion

For the product of two expressions containing multiple terms, the expansion is given by:

$(\vec{a} + \vec{b} + \vec{c} + \dots) \times (\vec{p} + \vec{q} + \vec{r} + \dots) = $$(\vec{a} \times \vec{p} + \vec{a} \times \vec{q} + \vec{a} \times \vec{r} + \dots) +$$(\vec{b} \times \vec{p} + \vec{b} \times \vec{q} + \vec{b} \times \vec{r} + \dots) +$$(\vec{c} \times \vec{p} + \vec{c} \times \vec{q} + \vec{c} \times \vec{r} + \dots) + \dots$

---

Vector Product in terms of Rectangular Components

To compute the cross product of two vectors in a three-dimensional Cartesian coordinate system, we represent the vectors in terms of their rectangular components along the $X, Y$ and $Z$ axes. This analytical method is much more practical for calculations than the geometric definition.

Representation of Vectors

Let there be two vectors $\vec{a}$ and $\vec{b}$ defined in terms of their components as:

Derivation of the Component Form

Using the distributive law of vector products, we expand $\vec{a} \times \vec{b}$ as follows:

$\vec{a} \times \vec{b} = (a_1\hat{i} + a_2\hat{j} + a_3\hat{k}) \times (b_1\hat{i} + b_2\hat{j} + b_3\hat{k})$

Expanding this expression gives us nine terms:

$\vec{a} \times \vec{b} = a_1b_1(\hat{i} \times \hat{i}) + a_1b_2(\hat{i} \times \hat{j}) + a_1b_3(\hat{i} \times \hat{k}) + $$ a_2b_1(\hat{j} \times \hat{i}) + a_2b_2(\hat{j} \times \hat{j}) + a_2b_3(\hat{j} \times \hat{k}) + $$ a_3b_1(\hat{k} \times \hat{i}) + a_3b_2(\hat{k} \times \hat{j}) + a_3b_3(\hat{k} \times \hat{k})$

Applying Unit Vector Properties

From the properties of orthogonal unit vectors, we know:

Simplifying the Expression

Substituting these values into our expanded expression, the terms with self-products vanish, and we are left with:

$\vec{a} \times \vec{b} = a_1b_2(\hat{k}) + a_1b_3(-\hat{j}) + a_2b_1(-\hat{k}) + a_2b_3(\hat{i}) + $$ a_3b_1(\hat{j}) + a_3b_2(-\hat{i})$

Grouping the terms according to $\hat{i}, \hat{j}$ and $\hat{k}$:

Determinant Form of Vector Product

The result obtained in the above equation can be very conveniently represented and remembered using a determinant of order $3 \times 3$. In this notation, the first row consists of the unit vectors $\hat{i}, \hat{j}, \hat{k}$, the second row consists of the components of the first vector $\vec{a}$, and the third row consists of the components of the second vector $\vec{b}$.

Expansion of the Determinant

To find the vector product, we expand the determinant along the first row:

$\vec{a} \times \vec{b} = \hat{i} \begin{vmatrix} a_2 & a_3 \\ b_2 & b_3 \end{vmatrix} - \hat{j} \begin{vmatrix} a_1 & a_3 \\ b_1 & b_3 \end{vmatrix} + \hat{k} \begin{vmatrix} a_1 & a_2 \\ b_1 & b_2 \end{vmatrix}$

Solving the $2 \times 2$ determinants:

$\vec{a} \times \vec{b} = \hat{i}(a_2b_3 - a_3b_2) - \hat{j}(a_1b_3 - a_3b_1) + \hat{k}(a_1b_2 - a_2b_1)$

Note that this expansion is identical to equation provided above. The negative sign for the $\hat{j}$ component is a standard part of determinant expansion, but it can be absorbed into the parentheses to match the $a_3b_1 - a_1b_3$ form if desired.

Answer:

---

Angle Between Two Vectors

The Vector Product provides a trigonometric method to determine the angle between two non-zero vectors. While the scalar (dot) product involves the cosine of the angle, the vector product involves the sine of the angle.

Theoretical Derivation

Let $\theta$ be the angle between two non-zero vectors $\vec{a}$ and $\vec{b}$, such that $0 \leq \theta \leq \pi$. Let $\hat{n}$ be a unit vector perpendicular to the plane of $\vec{a}$ and $\vec{b}$, such that $\vec{a}$, $\vec{b}$, and $\hat{n}$ form a right-handed system.

By the definition of the vector product, we have:

Taking the magnitude of both sides in equation provided above:

Since $\hat{n}$ is a unit vector, $|\hat{n}| = 1$. Also, for $0 \leq \theta \leq \pi$, the value of $\sin \theta$ is always non-negative ($\sin \theta \geq 0$). Thus:

Rearranging the terms to find the value of $\sin \theta$:

Analytical Expression (Component Form)

To use this formula in practical calculations, we express the magnitudes in terms of rectangular components.

Let $\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}$ and $\vec{b} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k}$.

1. Magnitude of the Cross Product

As derived previously, the components of the cross product are obtained via the determinant expansion. The magnitude of this resultant vector is:

2. Magnitudes of Individual Vectors

The magnitudes of vectors $\vec{a}$ and $\vec{b}$ are:

3. Final Formula for $\sin \theta$

Substituting these component values into Equation (i), we get the elaborate analytical formula:

Special Observations

1. Parallel Vectors: If $\vec{a}$ and $\vec{b}$ are parallel or collinear, $\theta = 0^\circ$ or $180^\circ$. In both cases, $\sin \theta = 0$, leading to $|\vec{a} \times \vec{b}| = 0$.

2. Perpendicular Vectors: If $\vec{a}$ and $\vec{b}$ are perpendicular, $\theta = 90^\circ$. Since $\sin 90^\circ = 1$, the magnitude is maximized: $|\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}|$.

3. Ambiguity of $\theta$: Note that $\sin \theta$ cannot distinguish between an acute angle $\theta$ and its supplementary obtuse angle $(180^\circ - \theta)$, as the sine value is positive for both. In contrast, the Dot Product ($\cos \theta$) is often preferred for finding the exact angle because the sign of the cosine clearly indicates whether the angle is acute (positive) or obtuse (negative).

Answer:

---

Unit Vector Perpendicular to Two Given Vectors

One of the most important applications of the Vector Product is finding a vector that is perpendicular to a plane defined by two other vectors. By definition, the cross product $\vec{a} \times \vec{b}$ results in a vector that is mutually perpendicular to both $\vec{a}$ and $\vec{b}$.

Derivation of the Unit Vector

Let $\vec{a}$ and $\vec{b}$ be two non-zero, non-parallel vectors. Let $\theta$ be the angle between them and $\hat{n}$ be a unit vector perpendicular to their plane such that ($\vec{a}, \vec{b}, \hat{n}$) form a right-handed system.

From the definition of the vector product, we have:

Taking the magnitude on both sides of equation (i):

Since $|\hat{n}| = 1$ (as it is a unit vector) and $\sin \theta \geq 0$ for $0 \leq \theta \leq \pi$, we get:

Substituting the value of $|\vec{a}| |\vec{b}| \sin \theta$ from equation (ii) into equation (i):

Thus, the unit vector $\hat{n}$ is given by:

The Concept of Two Perpendicular Unit Vectors

It is important to note that for any given plane, there are two unit vectors perpendicular to it—one pointing "upward" and the other "downward."

1. The vector $\frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|}$ points in the direction determined by the right-hand rule ($\vec{a}$ to $\vec{b}$).

2. The vector $\frac{\vec{b} \times \vec{a}}{|\vec{b} \times \vec{a}|}$ points in the opposite direction.

Therefore, the unit vectors perpendicular to the plane of $\vec{a}$ and $\vec{b}$ are represented as:

Vector of a Specific Magnitude

If we need to find a vector having a specific magnitude, say $p$, which is perpendicular to the plane of $\vec{a}$ and $\vec{b}$, we simply multiply the unit vector by the required magnitude.

Answer:

---

Lagrange's Identity

Lagrange’s Identity is a powerful relation that establishes a link between the Vector (Cross) Product and the Scalar (Dot) Product of two vectors. It allows us to calculate the magnitude of the cross product if the dot product and the magnitudes of the individual vectors are known, and vice-versa.

Statement of the Theorem

For any two vectors $\vec{a}$ and $\vec{b}$, the square of the magnitude of their vector product is equal to the product of the squares of their magnitudes minus the square of their scalar product.

Note: Here $(\vec{a} \times \vec{b})^2$ is synonymous with $|\vec{a} \times \vec{b}|^2$.

Proof of the Identity

Given: Two vectors $\vec{a}$ and $\vec{b}$ with an angle $\theta$ between them ($0 \leq \theta \leq \pi$).

We know from the definition of the vector product:

Squaring both sides of the above equation:

Since $\hat{n}$ is a unit vector, $\hat{n} \cdot \hat{n} = |\hat{n}|^2 = 1$. Therefore:

Using the trigonometric identity $\sin^2 \theta = 1 - \cos^2 \theta$:

We know that the scalar product is defined as $\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta$. Substituting this into the equation provided above:

This identity is often represented in the form of a Gram Determinant:

Alternative Forms

The identity can be rearranged to find the magnitude of the cross product directly:

Answer:

---

Area of a Parallelogram

The Vector Product provides a very elegant way to calculate the area of geometric figures. Specifically, the magnitude of the cross product of two vectors representing the adjacent sides of a parallelogram is exactly equal to the area of that parallelogram.

Theorem

If two vectors $\vec{a}$ and $\vec{b}$ represent the two adjacent sides of a parallelogram, then the Area of the Parallelogram is given by the magnitude of their vector product, i.e., $|\vec{a} \times \vec{b}|$.

Geometric Derivation

Given: A parallelogram $ABCD$ where the adjacent sides $AB$ and $AD$ are represented by vectors $\vec{a}$ and $\vec{b}$ respectively. Let $\theta$ be the angle between $\vec{a}$ and $\vec{b}$.

To Prove: $\text{Area of parallelogram } ABCD = |\vec{a} \times \vec{b}|$

Construction: Draw a perpendicular $DN$ from the vertex $D$ to the base $AB$.

Proof:

In the right-angled triangle $\Delta DAN$, we use trigonometric ratios to find the height $DN$:

Substituting the magnitude of vector $\vec{b}$ for $AD$:

We know that the area of a parallelogram is calculated as the product of its base and its height:

Substituting $AB = |\vec{a}|$ and $DN = |\vec{b}| \sin \theta$ from equation (i):

By the definition of the magnitude of a vector product:

Comparing this with equation (ii), we conclude:

1. Area of a Triangle

A diagonal of a parallelogram divides it into two triangles of equal area. Therefore, if $\vec{a}$ and $\vec{b}$ are the adjacent sides of a triangle, its area is half the area of the corresponding parallelogram.

2. Area in terms of Diagonals

If $\vec{d_1}$ and $\vec{d_2}$ are the diagonals of a parallelogram, the area can also be expressed as:

Answer:

Scalar Triple Product

When we deal with three vectors $\vec{a}$, $\vec{b}$, and $\vec{c}$, we can combine them in a way that the final result is a scalar quantity. This operation is known as the Scalar Triple Product.

The scalar product of two vectors, where one of the vectors is itself the result of a vector product of two other vectors, is called a scalar triple product. It is mathematically expressed as:

In this operation, we first find the cross product $(\vec{b} \times \vec{c})$, which results in a vector. We then take the dot product of vector $\vec{a}$ with this resultant vector. Since the dot product of any two vectors is a scalar, the final result is a number.

Notations and Symbols

The scalar triple product $\vec{a} \cdot (\vec{b} \times \vec{c})$ is most commonly denoted by the Box Product notation:

Critical Remarks

1. Nature of the Resultant

The result of a scalar triple product is always a scalar quantity (a real number). It does not have any direction.

2. Position of Brackets and Dots/Crosses

In the expression $\vec{a} \cdot \vec{b} \times \vec{c}$, the brackets are often omitted because the operation is unambiguous. The cross product must be performed first. If we tried to perform the dot product first ($\vec{a} \cdot \vec{b}$), we would get a scalar, and we cannot take the cross product of a scalar and a vector ($\vec{c}$).

3. Scalar Multiplication vs. Triple Product

The expression $\vec{a}(\vec{b} \cdot \vec{c})$ is not a scalar triple product. Here, $(\vec{b} \cdot \vec{c})$ is a scalar (let's call it $k$). Thus, $\vec{a}(k)$ is simply a vector parallel to $\vec{a}$.

4. Meaningless Expressions

Certain combinations of dots and crosses are mathematically meaningless and should be avoided:

• $(\vec{a} \cdot \vec{b}) \cdot \vec{c}$ : Meaningless, because $(\vec{a} \cdot \vec{b})$ is a scalar, and a dot product requires two vectors.

• $\vec{a} \times (\vec{b} \cdot \vec{c})$ : Meaningless, because $(\vec{b} \cdot \vec{c})$ is a scalar, and a cross product requires two vectors.

---

Geometrical Interpretation of Scalar Triple Product

The Scalar Triple Product of three vectors $\vec{a}, \vec{b}$ and $\vec{c}$, denoted as $\vec{a} \cdot (\vec{b} \times \vec{c})$ or $[\vec{a} \ \vec{b} \ \vec{c}]$, has a profound geometric significance. It represents the signed volume of a parallelepiped, a three-dimensional figure formed by three sets of parallel planes.

The Volume Theorem

The scalar triple product $\vec{a} \cdot (\vec{b} \times \vec{c})$ is the scalar $\pm V$, where $V$ is the volume of the parallelepiped having $\vec{a}, \vec{b}$ and $\vec{c}$ as its three coterminous edges (edges meeting at a single vertex).

The choice of positive or negative sign depends on the orientation of the vectors:

1. Positive ($+V$): When $\vec{a}, \vec{b}, \vec{c}$ form a right-handed triad (the angle $\alpha$ between $\vec{a}$ and $(\vec{b} \times \vec{c})$ is acute).

2. Negative ($-V$): When $\vec{a}, \vec{b}, \vec{c}$ form a left-handed triad (the angle $\alpha$ is obtuse).

Step-by-Step Geometrical Derivation

1. Construction and Setup

Consider a parallelepiped with vertex $O$ as the origin. Let the three edges meeting at $O$ be represented by vectors:

Let the parallelogram $OBDC$ form the base of this parallelepiped. Let $\vec{n}$ be a vector representing the cross product of the base edges, i.e., $\vec{n} = \vec{b} \times \vec{c}$.

2. Analyzing the Base Area

By the property of the vector product, the magnitude of $\vec{n}$ is equal to the area of the parallelogram $OBDC$ which forms the base.

The direction of $\vec{n}$ is perpendicular to the plane of the base $OBDC$.

3. Finding the Perpendicular Height

Let $\alpha$ be the angle between the third edge $\vec{OA}$ (vector $\vec{a}$) and the normal vector $\vec{n}$. The height ($h$) of the parallelepiped is the perpendicular distance of vertex $A$ from the base. This is the projection of $\vec{a}$ along the direction of $\vec{n}$.

4. Combining the Components

By the definition of the scalar (dot) product between $\vec{a}$ and the cross product result $\vec{n}$:

Rearranging the terms:

Since $\text{Volume} = \text{Base Area} \times \text{Height}$, we get:

Sign and Orientation Analysis

The result of the scalar triple product is a real number. Its sign provides information about the spatial orientation (handedness) of the three vectors.

Case 1: Acute Angle ($0 \leq \alpha < 90^\circ$)

If $\vec{a}$ lies on the same side of the base as the vector $\vec{b} \times \vec{c}$, the angle $\alpha$ is acute, $\cos \alpha$ is positive, and the product is $+V$. The vectors form a right-handed triad.

Case 2: Obtuse Angle ($90^\circ < \alpha \leq 180^\circ$)

If $\vec{a}$ lies on the opposite side, the angle $\alpha$ is obtuse, $\cos \alpha$ is negative, and the product is $-V$. The vectors form a left-handed triad.

Important Corollaries

1: Absolute Volume

In physical applications where the direction of orientation is not required, the volume is taken as the absolute value of the scalar triple product.

2: Volume of a Tetrahedron

A tetrahedron is a triangular pyramid. If $\vec{a}, \vec{b}$ and $\vec{c}$ are the three coterminous edges of a tetrahedron, its volume is exactly one-sixth of the volume of the parallelepiped formed by the same edges.

3: Condition of Coplanarity

If three non-zero vectors $\vec{a}, \vec{b}$ and $\vec{c}$ are coplanar (lie in the same plane), the height of the parallelepiped formed by them becomes zero. Consequently, the volume of the parallelepiped is zero.

---

Scalar Triple Product in terms of Rectangular Components

To calculate the Scalar Triple Product analytically, we express the three vectors in terms of their rectangular components along the axes. This method transforms the vector operation into a simple determinant expansion, making it the most practical approach for numerical problems.

Analytical Representation

Let three vectors $\vec{a}, \vec{b}$ and $\vec{c}$ be defined as follows:

Step-by-Step Derivation

Step 1: Evaluating the Vector Product ($\vec{a} \times \vec{b}$)

Using the rectangular components, the cross product $\vec{a} \times \vec{b}$ is given by the determinant expansion:

On expansion, we get:

Step 2: Evaluating the Scalar Triple Product $(\vec{a} \times \vec{b}) \cdot \vec{c}$

Now, we take the dot product of the result obtained in the equation provided above with the vector $\vec{c} = c_1\hat{i} + c_2\hat{j} + c_3\hat{k}$:

$(\vec{a} \times \vec{b}) \cdot \vec{c} = [(a_2b_3 - a_3b_2)\hat{i} - (a_1b_3 - a_3b_1)\hat{j} + $$ (a_1b_2 - a_2b_1)\hat{k}] \cdot (c_1\hat{i} + c_2\hat{j} + c_3\hat{k})$

Multiplying the corresponding components ($\hat{i} \cdot \hat{i} = 1$, etc.):

Step 3: Representation as a Determinant

The algebraic expression in equation obtained above is identical to the expansion of a $3 \times 3$ determinant where the rows are formed by the components of vectors $\vec{a}, \vec{b}$ and $\vec{c}$.

Key Observations

1. Order of Rows: The rows follow the order of vectors in the operation. If we calculate $[\vec{c} \ \vec{a} \ \vec{b}]$, the components of $\vec{c}$ would occupy the first row.

2. Symmetry: Because the determinant value remains unchanged if rows are shifted in a cyclic order, we can conclude that:

Answer:

---

Condition of Coplanarity of Three Vectors

The concept of Coplanarity is a fundamental aspect of vector algebra, specifically when determining if three vectors lie in the same geometric plane. The Scalar Triple Product serves as the definitive mathematical test for this condition.

Three non-zero vectors $\vec{a}$, $\vec{b}$, and $\vec{c}$ are coplanar if and only if their scalar triple product is equal to zero.

Formal Proof

Given: Three non-zero vectors $\vec{a}$, $\vec{b}$, and $\vec{c}$.

To Prove: $\vec{a}, \vec{b}, \vec{c}$ are coplanar $\iff [\vec{a} \ \vec{b} \ \vec{c}] = 0$.

Part 1: Necessary Condition (If vectors are coplanar)

Suppose the vectors $\vec{a}$, $\vec{b}$, and $\vec{c}$ are coplanar. This means they all lie in the same plane or are parallel to the same plane.

1. Consider the vector product $\vec{a} \times \vec{b}$. By definition, this resultant vector is perpendicular to the plane containing $\vec{a}$ and $\vec{b}$.

2. Since $\vec{c}$ also lies in the same plane (as they are coplanar), the vector $\vec{a} \times \vec{b}$ must be perpendicular to $\vec{c}$ as well.

3. We know that the dot product of two perpendicular vectors is zero.

Part 2: Sufficient Condition (If scalar triple product is zero)

Suppose $[\vec{a} \ \vec{b} \ \vec{c}] = 0$ for non-zero vectors $\vec{a}$, $\vec{b}$, and $\vec{c}$.

1. Geometrically, $|[\vec{a} \ \vec{b} \ \vec{c}]|$ represents the volume of the parallelepiped formed by these vectors as coterminous edges.

2. If the volume is 0, the three-dimensional figure "collapses." This can only happen if the vectors are coplanar, or if two vectors are parallel, or if any of the vectors are null vectors.

3. Since we have given that the vectors are non-zero and non-parallel, they must lie in the same plane to result in zero volume.

Analytical Condition (Determinant Form)

If the vectors are given in component form as $\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}$, $\vec{b} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k}$, and $\vec{c} = c_1\hat{i} + c_2\hat{j} + c_3\hat{k}$, then the condition for coplanarity is:

Properties derived from Coplanarity

Condition	Result
$[\vec{a} \ \vec{b} \ \vec{c}] \neq 0$	Vectors are Non-coplanar (form a parallelepiped).
$[\vec{a} \ \vec{b} \ \vec{c}] = 0$	Vectors are Coplanar.
Two vectors are parallel	The product is always zero.

Answer:

---

Properties of Scalar Triple Product

The Scalar Triple Product (STP), denoted as $[\vec{a} \ \vec{b} \ \vec{c}]$, possesses several algebraic properties derived from the properties of determinants and the nature of dot and cross products. These properties are essential for simplifying complex vector identities in 3D geometry.

Property (i): Cyclic Permutation

The value of the scalar triple product remains unchanged as long as the cyclic order of the vectors $\vec{a}, \vec{b},$ and $\vec{c}$ is maintained.

Proof:

Let $\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}$, $\vec{b} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k}$ and $\vec{c} = c_1\hat{i} + c_2\hat{j} + c_3\hat{k}$. Then:

In a determinant, if we shift a row over two other rows (which is equivalent to two successive adjacent row swaps), the value remains the same.

Swap $R_1$ with $R_2$, then $R_2$ with $R_3$ to bring $\vec{b}$ to the first row:

Thus, $[\vec{a} \ \vec{b} \ \vec{c}] = [\vec{b} \ \vec{c} \ \vec{a}]$. The same logic applies to show equality with $[\vec{c} \ \vec{a} \ \vec{b}]$.

Property (ii): Interchange of Dot and Cross

The positions of the dot ($\cdot$) and cross ($\times$) operators can be interchanged without changing the value of the scalar triple product, provided the relative cyclic order of vectors is preserved.

Proof:

By the commutative property of the scalar (dot) product, we have:

From Property (i), we know $[\vec{a} \ \vec{b} \ \vec{c}] = [\vec{b} \ \vec{c} \ \vec{a}]$. Expanding the second box product:

Due to the cyclic symmetry, we can observe that the dot and cross can "trade places." Specifically, $\vec{a} \cdot (\vec{b} \times \vec{c}) = (\vec{a} \times \vec{b}) \cdot \vec{c}$.

Property (iii): Scalar Multiplication

If any vector in the scalar triple product is multiplied by a scalar $\lambda$, the scalar can be taken outside the product.

Proof:

Using the property of scalar association in vector products:

Since the dot product also allows scalar extraction ($m\vec{x} \cdot \vec{y} = m(\vec{x} \cdot \vec{y})$):

Property (iv): Non-Cyclic Permutation (Sign Reversal)

If the cyclic order of the vectors is broken by swapping any two vectors, the magnitude remains the same, but the sign changes.

Proof:

This corresponds to swapping two rows in a determinant. We know that interchanging any two rows of a determinant multiplies its value by $-1$.

Property (v): Zero Product with Equal Vectors

The scalar triple product is zero if any two of the given vectors are equal.

Proof:

By definition, $[\vec{a} \ \vec{a} \ \vec{c}] = (\vec{a} \times \vec{a}) \cdot \vec{c}$.

Since the cross product of any vector with itself is the null vector ($\vec{a} \times \vec{a} = \vec{0}$):

Property (vi): Parallel or Collinear Vectors

The scalar triple product is zero if any two of the vectors are parallel or collinear.

If $\vec{a} \parallel \vec{b}$, then there exists a scalar $\lambda$ such that $\vec{a} = \lambda\vec{b}$.

Proof:

Using Property (iii):

Using Property (v), since two vectors are now equal:

---

Summary of Properties

Nature of Change	Algebraic Expression
Cyclic Shift	$[\vec{a} \ \vec{b} \ \vec{c}] = [\vec{b} \ \vec{c} \ \vec{a}] = [\vec{c} \ \vec{a} \ \vec{b}]$
Single Swap	$[\vec{a} \ \vec{b} \ \vec{c}] = -[\vec{b} \ \vec{a} \ \vec{c}]$
Equal Vectors	$[\vec{a} \ \vec{a} \ \vec{b}] = 0$
Parallel Vectors	If $\vec{a} = \lambda \vec{b}$, then $[\vec{a} \ \vec{b} \ \vec{c}] = 0$