Chapter 10 Vector Algebra (Concepts)

Introduction to Vectors

In our study of mathematics and physics, we encounter various types of quantities. These quantities can be broadly categorised into two main types based on whether they have direction in addition to magnitude: scalar quantities and vector quantities. Understanding the distinction between these two is fundamental to the study of Vector Algebra.

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Scalar Quantity

Definition: A quantity that is completely described by its magnitude alone is called a scalar quantity. Magnitude refers to the numerical value along with the appropriate unit. Scalar quantities do not have a direction associated with them.

Examples of scalar quantities include:

• Length (e.g., 5 metres)
• Mass (e.g., 10 kilograms)
• Time (e.g., 2 hours)
• Distance (e.g., 10 kilometres travelled)
• Speed (e.g., 60 kilometres per hour)
• Temperature (e.g., 30 degrees Celsius)
• Volume (e.g., 1 litre)
• Work (e.g., 500 Joules)
• Energy (e.g., 1000 calories)
• Density, Electric Charge, Potential, etc.

Scalar quantities can be added, subtracted, multiplied, and divided using the ordinary rules of arithmetic and algebra. For example, if you have two masses of 5 kg and 3 kg, the total mass is simply $5 + 3 = 8$ kg.

Vector Quantity

Definition: A quantity that has both magnitude and direction is called a vector quantity. To fully describe a vector quantity, we must specify both how much it is (magnitude) and in which direction it acts.

Examples of vector quantities include:

• Displacement (e.g., 5 metres towards East)
• Velocity (e.g., 60 kilometres per hour towards North)
• Acceleration (e.g., 9.8 metres per second squared downwards)
• Force (e.g., 10 Newtons applied to the right)
• Momentum, Weight, Electric Field Intensity, etc.

Vector quantities cannot be manipulated using ordinary arithmetic. Adding two vectors requires special rules that account for their directions. This is the basis of Vector Algebra. For example, if you walk 5 metres East and then 3 metres West, your total displacement is not $5 + 3 = 8$ metres, but $5 - 3 = 2$ metres East. The directions matter.

Representation of a Vector

A vector is geometrically represented by a directed line segment. A directed line segment has a starting point and an ending point, and its length and direction convey the magnitude and direction of the vector, respectively.

Consider two points A and B. If we draw a line segment from A to B and put an arrowhead at B, this represents a vector starting at A and ending at B.

• Point A is called the initial point or the tail of the vector.
• Point B is called the terminal point or the head of the vector.

The vector represented by the directed line segment from A to B is denoted by $\vec{\text{AB}}$. The arrow above indicates the direction from A to B.

The magnitude of the vector $\vec{\text{AB}}$ is the length of the line segment AB. It is denoted by $|\vec{\text{AB}}|$ or simply AB.

Alternatively, a vector can be denoted by a single boldface lowercase letter, like $\mathbf{a}$, or more commonly in handwriting, by a lowercase letter with an arrow above it, like $\vec{a}$. The magnitude of vector $\vec{a}$ is denoted by $|\vec{a}|$ or simply $a$.

The vector from B to A, denoted by $\vec{\text{BA}}$, has the same magnitude as $\vec{\text{AB}}$ but the opposite direction. Thus, $\vec{\text{BA}} = -\vec{\text{AB}}$.

Direction Cosines and Direction Ratios

In a three-dimensional Cartesian coordinate system (with x, y, and z axes), the direction of a vector can be uniquely determined by the angles it makes with the positive directions of the coordinate axes.

Consider a non-zero vector $\vec{r}$ in 3D space. Let this vector make angles $\alpha, \beta,$ and $\gamma$ with the positive x, y, and z axes respectively. These angles $\alpha, \beta, \gamma$ are called the direction angles of the vector $\vec{r}$. The values of these angles are usually taken in the range $[0, \pi]$.

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

For any vector in 3D space, the sum of the squares of its direction cosines is always equal to 1.

This identity, $\cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1$, is a fundamental property of direction cosines.

Any three numbers $a, b,$ and $c$ that are proportional to the direction cosines $l, m,$ and $n$ of a vector are called the direction ratios of that vector.

This means that there exists a non-zero constant $k$ such that:

If we are given the direction ratios $a, b, c$ of a vector, we can find the direction cosines $l, m, n$ using the relationship $l^2 + m^2 + n^2 = 1$. Substitute $l=ak, m=bk, n=ck$ into equation (2):

Substitute this value of $k$ back into equation (3) to find the direction cosines in terms of the direction ratios:

The $\pm$ sign indicates the two possible directions along a line defined by the direction ratios. For a specific directed vector, the sign is determined by which direction the vector is pointing.

Vector Algebra

Vector algebra is the system of mathematical operations specifically designed for vector quantities. Unlike scalar algebra, which deals with magnitudes only, vector algebra incorporates both magnitude and direction. Basic operations include the addition and subtraction of vectors, and the multiplication of a vector by a scalar. More advanced operations include the scalar product (dot product) and the vector product (cross product) of two vectors, which we will study later.

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Types of Vectors

To understand vector operations, it is helpful to be familiar with different types of vectors:

• Zero Vector (Null Vector): A vector whose initial and terminal points coincide is called a zero vector. Since the initial and terminal points are the same, its length (magnitude) is zero. A zero vector has no specific direction, or its direction is considered arbitrary or indeterminate. It is denoted by $\vec{0}$. The magnitude of the zero vector is $|\vec{0}| = 0$. The zero vector plays a role similar to the number zero in scalar arithmetic.
• Unit Vector: A vector whose magnitude is exactly 1 unit is called a unit vector. Unit vectors are primarily used to specify direction. The unit vector in the same direction as a given non-zero vector $\vec{a}$ is obtained by dividing the vector $\vec{a}$ by its magnitude $|\vec{a}|$. This unit vector is denoted by $\hat{a}$ (read as 'a cap' or 'a hat').

$\hat{a} = \frac{\vec{a}}{|\vec{a}|}$

... (1)

Here, $|\hat{a}| = \left|\frac{\vec{a}}{|\vec{a}|}\right| = \frac{|\vec{a}|}{|| \vec{a} ||} = \frac{|\vec{a}|}{|\vec{a}|} = 1$.

• Coinitial Vectors: Two or more vectors are said to be coinitial if they all have the same initial point.
• Collinear Vectors: Two or more vectors are said to be collinear if they are parallel to the same line, irrespective of their magnitude and direction. If $\vec{a}$ and $\vec{b}$ are two non-zero collinear vectors, then one vector can be expressed as a scalar multiple of the other. That is, $\vec{a} = k \vec{b}$ for some non-zero scalar $k$. Conversely, if $\vec{a} = k \vec{b}$ for some scalar $k$, then $\vec{a}$ and $\vec{b}$ are collinear. If $k > 0$, they are in the same direction; if $k < 0$, they are in opposite directions. If $k=0$, one vector is the zero vector, which is considered collinear with any vector.
• Equal Vectors: Two vectors $\vec{a}$ and $\vec{b}$ are said to be equal if and only if they have the exact same magnitude and the exact same direction. This means that if $\vec{a} = \vec{b}$, then $|\vec{a}| = |\vec{b}|$ and they point in the same direction. Note that equal vectors do not necessarily have the same initial point. Any two vectors with the same magnitude and direction are considered equal (this is the concept of free vectors).
• Negative of a Vector: The negative of a vector $\vec{a}$, denoted by $-\vec{a}$, is a vector that has the same magnitude as $\vec{a}$ but points in the opposite direction. If $\vec{\text{AB}}$ represents the vector $\vec{a}$, then the vector $\vec{\text{BA}}$ represents the negative of $\vec{a}$, so $\vec{\text{BA}} = -\vec{\text{AB}} = -\vec{a}$. The magnitude of $-\vec{a}$ is $|-\vec{a}| = |\vec{a}|$.

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Addition of Vectors

Vector addition is the operation of combining two or more vectors to produce a single equivalent vector called the resultant or sum. Vector addition follows specific geometric laws that take direction into account.

1. Triangle Law of Vector Addition: This law is used to add two vectors. If two vectors are represented in magnitude and direction by the two sides of a triangle taken in order (i.e., the terminal point of the first vector is the initial point of the second vector), then their sum (resultant) is represented in magnitude and direction by the third side of the triangle taken in the opposite order.

If we have vector $\vec{a} = \vec{\text{AB}}$ and vector $\vec{b} = \vec{\text{BC}}$, then their sum $\vec{a} + \vec{b}$ is represented by the vector $\vec{\text{AC}}$.

2. Parallelogram Law of Vector Addition: This is another geometric method for adding two vectors. If two vectors are represented in magnitude and direction by the two adjacent sides of a parallelogram drawn from a common initial point, then their sum (resultant) is represented in magnitude and direction by the diagonal of the parallelogram drawn from the same common initial point.

If we have vector $\vec{a} = \vec{\text{OA}}$ and vector $\vec{b} = \vec{\text{OB}}$, and we complete the parallelogram OACB, then their sum $\vec{a} + \vec{b}$ is represented by the diagonal $\vec{\text{OC}}$.

The parallelogram law is equivalent to the triangle law. In parallelogram OACB, $\vec{\text{OA}} = \vec{\text{a}}$, $\vec{\text{OB}} = \vec{\text{b}}$. Since opposite sides of a parallelogram are parallel and equal in length, $\vec{\text{AC}} = \vec{\text{OB}} = \vec{\text{b}}$. Applying the triangle law to triangle OAC, we have $\vec{\text{OA}} + \vec{\text{AC}} = \vec{\text{OC}}$, which is $\vec{a} + \vec{b} = \vec{\text{OC}}$.

Properties of Vector Addition

Vector addition satisfies several important properties, similar to scalar addition:

• Commutative Law: The order of addition does not matter. For any two vectors $\vec{a}$ and $\vec{b}$, $\vec{a} + \vec{b} = \vec{b} + \vec{a}$. This can be seen from the parallelogram law, where $\vec{\text{OA}} + \vec{\text{OB}} = \vec{\text{OC}}$ and $\vec{\text{OB}} + \vec{\text{OA}} = \vec{\text{OC}}$ (considering the other triangle OBC).
• Associative Law: When adding three or more vectors, the grouping of vectors does not affect the sum. For any three vectors $\vec{a}, \vec{b}, \vec{c}$, $(\vec{a} + \vec{b}) + \vec{c} = \vec{a} + (\vec{b} + \vec{c})$.
• Existence of Additive Identity: The zero vector acts as the additive identity. For any vector $\vec{a}$, $\vec{a} + \vec{0} = \vec{0} + \vec{a} = \vec{a}$.
• Existence of Additive Inverse: For every vector $\vec{a}$, there exists a vector $-\vec{a}$ (its negative) such that their sum is the zero vector. $\vec{a} + (-\vec{a}) = (-\vec{a}) + \vec{a} = \vec{0}$.

Subtraction of Vectors

Subtraction of vectors is defined in terms of vector addition and the negative of a vector. Subtracting vector $\vec{b}$ from vector $\vec{a}$ is the same as adding the negative of vector $\vec{b}$ to vector $\vec{a}$.

Geometrically, if $\vec{a} = \vec{\text{OA}}$ and $\vec{b} = \vec{\text{OB}}$, then $\vec{a} - \vec{b}$ can be represented by the vector $\vec{\text{BA}}$. This can be seen by applying the triangle law to $\vec{\text{OB}} + \vec{\text{BA}} = \vec{\text{OA}}$, which implies $\vec{b} + \vec{\text{BA}} = \vec{a}$, so $\vec{\text{BA}} = \vec{a} - \vec{b}$.

Multiplication of a Vector by a Scalar

Multiplying a vector $\vec{a}$ by a scalar $k$ produces a new vector, denoted by $k\vec{a}$. The properties of the resulting vector are:

• Magnitude: The magnitude of $k\vec{a}$ is the absolute value of the scalar $k$ multiplied by the magnitude of vector $\vec{a}$. $|k\vec{a}| = |k| |\vec{a}|$.
• Direction:
  • If $k > 0$, the vector $k\vec{a}$ has the same direction as $\vec{a}$.
  • If $k < 0$, the vector $k\vec{a}$ has the direction opposite to $\vec{a}$.
  • If $k = 0$, the vector $0\vec{a}$ is the zero vector, $\vec{0}$, with zero magnitude and no specific direction.

Scalar multiplication is closely related to the concept of collinearity. Two non-zero vectors $\vec{a}$ and $\vec{b}$ are collinear if and only if one is a scalar multiple of the other, i.e., $\vec{a} = k\vec{b}$ for some non-zero scalar $k$.

Properties of Scalar Multiplication

Scalar multiplication interacts with vector addition according to distributive and associative laws:

• Associativity of Scalar Multiplication: $k(l\vec{a}) = (kl)\vec{a}$, where $k, l$ are scalars and $\vec{a}$ is a vector.
• Distributivity over Vector Addition: $k(\vec{a} + \vec{b}) = k\vec{a} + k\vec{b}$, where $k$ is a scalar and $\vec{a}, \vec{b}$ are vectors.
• Distributivity over Scalar Addition: $(k+l)\vec{a} = k\vec{a} + l\vec{a}$, where $k, l$ are scalars and $\vec{a}$ is a vector.
• Multiplication by 1 and -1: $1 \cdot \vec{a} = \vec{a}$ and $(-1) \cdot \vec{a} = -\vec{a}$.
• Multiplication by 0: $0 \cdot \vec{a} = \vec{0}$ (the zero vector).
• Magnitude Property: $|k\vec{a}| = |k| |\vec{a}|$.

Theorem on Two Non-Zero Non-Collinear Vectors

This theorem is a fundamental result in vector algebra that describes how any vector lying in a specific plane can be represented in terms of two given vectors that define that plane. It lays the groundwork for the concept of representing vectors using components in two dimensions.

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Statement of the Theorem

Theorem: If $\vec{a}$ and $\vec{b}$ are two given non-zero and non-collinear vectors, then any vector $\vec{r}$ that is coplanar with $\vec{a}$ and $\vec{b}$ can be uniquely expressed as a linear combination of $\vec{a}$ and $\vec{b}$.

In other words, for any vector $\vec{r}$ lying in the plane determined by $\vec{a}$ and $\vec{b}$, there exist unique scalars $x$ and $y$ such that:

Here, $x\vec{a} + y\vec{b}$ is called a linear combination of vectors $\vec{a}$ and $\vec{b}$. The theorem guarantees that this representation is unique for any given coplanar vector $\vec{r}$ and given non-collinear vectors $\vec{a}$ and $\vec{b}$.

Geometric Interpretation and Proof Idea

Let O be an arbitrary point in space. Consider the vectors $\vec{a}$ and $\vec{b}$ originating from O. Since $\vec{a}$ and $\vec{b}$ are non-collinear and non-zero, they lie on distinct lines passing through O, and these lines are not parallel. Two such intersecting lines uniquely define a plane. Let this plane be $\Pi$. Any vector $\vec{r}$ that is coplanar with $\vec{a}$ and $\vec{b}$ can also be considered as a vector lying in this plane $\Pi$.

Let the vector $\vec{r}$ originate from O, so $\vec{\text{OP}} = \vec{r}$ for some point P in the plane $\Pi$.

We can draw a line through P parallel to the vector $\vec{b}$, and let it intersect the line containing vector $\vec{a}$ (passing through O) at point Q. Similarly, draw a line through P parallel to the vector $\vec{a}$, and let it intersect the line containing vector $\vec{b}$ (passing through O) at point R.

This construction forms a parallelogram OQPR (unless P lies on the line of $\vec{a}$ or $\vec{b}$, in which case it's a degenerate parallelogram). By the parallelogram law of vector addition (or the triangle law applied to $\triangle$ OQP), we have:

Since point Q lies on the line containing the vector $\vec{a}$ originating from O, the vector $\vec{\text{OQ}}$ must be parallel to $\vec{a}$. As $\vec{a}$ is non-zero, $\vec{\text{OQ}}$ must be a scalar multiple of $\vec{a}$. Let $\vec{\text{OQ}} = x\vec{a}$ for some scalar $x$. The scalar $x$ represents how many "units" of vector $\vec{a}$ (or its negative) are needed to get from O to Q.

Similarly, since point R lies on the line containing the vector $\vec{b}$ originating from O, the vector $\vec{\text{OR}}$ must be parallel to $\vec{b}$. As $\vec{b}$ is non-zero, $\vec{\text{OR}}$ must be a scalar multiple of $\vec{b}$. Let $\vec{\text{OR}} = y\vec{b}$ for some scalar $y$.

Substituting these into the equation $\vec{\text{OP}} = \vec{\text{OQ}} + \vec{\text{OR}}$:

This shows that any vector $\vec{r}$ coplanar with $\vec{a}$ and $\vec{b}$ can be expressed as a linear combination of $\vec{a}$ and $\vec{b}$.

Proof of Uniqueness: To prove that this representation is unique, assume that the same vector $\vec{r}$ can be expressed in two different ways as a linear combination of $\vec{a}$ and $\vec{b}$:

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

Rearrange the terms:

We are given that $\vec{a}$ and $\vec{b}$ are non-zero and non-collinear vectors. The only way a linear combination of two non-collinear vectors can be equal to the zero vector is if both scalar coefficients in the linear combination are zero.

Thus, from equation (A), we must have:

This proves that the scalars $x$ and $y$ are unique for any given vector $\vec{r}$ coplanar with $\vec{a}$ and $\vec{b}$.

This theorem is important because it shows that any vector in a plane can be represented as a combination of just two vectors in that plane, provided they are not pointing in the same or opposite direction (non-collinear). These two vectors form a basis for the plane.

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Extension to Three Non-Zero Non-Coplanar Vectors in Space

The concept extends to three dimensions. To represent any vector in three-dimensional space, we need three vectors that do not lie in the same plane (non-coplanar).

Theorem: Any vector $\vec{r}$ in space can be uniquely expressed as a linear combination of three given non-zero, non-coplanar vectors $\vec{a}, \vec{b},$ and $\vec{c}$.

That is, for any vector $\vec{r}$ in space, there exist unique scalars $x, y,$ and $z$ such that:

Here, $x\vec{a} + y\vec{b} + z\vec{c}$ is a linear combination of vectors $\vec{a}, \vec{b},$ and $\vec{c}$. The scalars $x, y, z$ are unique.

Geometric Interpretation and Proof Idea: Let O be the origin. Consider the vectors $\vec{a}, \vec{b}, \vec{c}$ originating from O. Since they are non-coplanar, they define the three dimensions of space. Any point P in space corresponds to a position vector $\vec{\text{OP}} = \vec{r}$ originating from O. We can form a parallelepiped with $\vec{\text{OP}}$ as its diagonal and edges parallel to $\vec{a}, \vec{b}, \vec{c}$. The edge vectors originating from O will be scalar multiples of $\vec{a}, \vec{b}, \vec{c}$, say $x\vec{a}, y\vec{b}, z\vec{c}$. Then, by the vector addition rule for parallelepipeds, $\vec{r} = x\vec{a} + y\vec{b} + z\vec{c}$. The uniqueness is proved similarly by assuming two different linear combinations are equal to $\vec{r}$, leading to $(x_1-x_2)\vec{a} + (y_1-y_2)\vec{b} + (z_1-z_2)\vec{c} = \vec{0}$. Since $\vec{a}, \vec{b}, \vec{c}$ are non-coplanar, this is only possible if all coefficients are zero, proving $x_1=x_2, y_1=y_2, z_1=z_2$.

This theorem is fundamental for coordinate geometry in three dimensions. The most common choice for the three non-coplanar vectors are the standard unit vectors $\hat{i}, \hat{j}, \hat{k}$ along the x, y, and z axes respectively, which are mutually orthogonal (perpendicular) and have magnitude 1. Any vector in space can be uniquely written as $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$.

Components of a Vector in a Plane

The theorem on two non-zero, non-collinear vectors in a plane provides the basis for representing any vector in that plane using a coordinate system. When we use a rectangular Cartesian coordinate system, the non-collinear vectors are chosen to be the standard unit vectors along the axes. This representation is called the component form of a vector.

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Component Form in 2D (Rectangular Coordinates)

Consider a rectangular Cartesian coordinate system in a plane with origin O. Let the x-axis and the y-axis be perpendicular to each other. We define unit vectors along the positive x and y axes as $\hat{i}$ and $\hat{j}$ respectively.

• $\hat{i}$ is a vector of magnitude 1 pointing along the positive x-axis.
• $\hat{j}$ is a vector of magnitude 1 pointing along the positive y-axis.

These unit vectors $\hat{i}$ and $\hat{j}$ are non-zero and non-collinear. According to the theorem on two non-collinear vectors, any vector $\vec{r}$ lying in this plane can be uniquely expressed as a linear combination of $\hat{i}$ and $\hat{j}$.

Let P be any point in the plane with coordinates $(x, y)$ relative to the origin O. The position vector of point P is the vector $\vec{\text{OP}} = \vec{r}$, which originates from the origin O and terminates at P. We can reach point P from the origin by moving $x$ units along the x-axis and $y$ units along the y-axis. The vector representing the movement along the x-axis is $x\hat{i}$ (a scalar $x$ times the unit vector $\hat{i}$), and the vector representing the movement along the y-axis is $y\hat{j}$.

By the triangle law of vector addition, the position vector $\vec{r} = \vec{\text{OP}}$ is the sum of these two component vectors:

The scalars $x$ and $y$ are called the scalar components or rectangular components of the vector $\vec{r}$ along the x and y axes respectively. The vectors $x\hat{i}$ and $y\hat{j}$ are called the vector components of $\vec{r}$ along the x and y axes.

A vector in a plane can also be conveniently represented by its ordered pair of scalar components $(x, y)$.

Magnitude of a Vector in 2D Component Form

If a vector $\vec{r}$ is given in component form as $\vec{r} = x\hat{i} + y\hat{j}$, its magnitude $|\vec{r}|$ is the length of the line segment from the origin O(0, 0) to the point P(x, y). This length can be found using the distance formula, which is derived from the Pythagorean theorem in the right-angled triangle formed by the vector $\vec{r}$ and its components $x\hat{i}$ and $y\hat{j}$.

Vector Operations in Component Form (2D)

Expressing vectors in component form simplifies vector operations, allowing us to perform them algebraically on the corresponding components.

Let $\vec{a} = a_1\hat{i} + a_2\hat{j}$ and $\vec{b} = b_1\hat{i} + b_2\hat{j}$ be two vectors in a plane.

• Addition: The sum of two vectors is found by adding their corresponding components.

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

• Subtraction: The difference of two vectors is found by subtracting their corresponding components.

  $\vec{a} - \vec{b} = (a_1\hat{i} + a_2\hat{j}) - (b_1\hat{i} + b_2\hat{j}) = (a_1-b_1)\hat{i} + (a_2-b_2)\hat{j}$

• Scalar Multiplication: Multiplying a vector by a scalar involves multiplying each component by the scalar. For a scalar $k$:

  k$\vec{a} = k(a_1\hat{i} + a_2\hat{j}) = (ka_1)\hat{i} + (ka_2)\hat{j}$

• Equality of Vectors: Two vectors are equal if and only if their corresponding components are equal.

  $\vec{a} = \vec{b} \iff a_1 = b_1 \text{ and } a_2 = b_2$

• Collinearity: Two non-zero vectors $\vec{a} = a_1\hat{i} + a_2\hat{j}$ and $\vec{b} = b_1\hat{i} + b_2\hat{j}$ are collinear if and only if there exists a non-zero scalar $k$ such that $\vec{a} = k\vec{b}$. In component form, this means $a_1\hat{i} + a_2\hat{j} = k(b_1\hat{i} + b_2\hat{j}) = (kb_1)\hat{i} + (kb_2)\hat{j}$. By equality of vectors, this implies $a_1 = kb_1$ and $a_2 = kb_2$. If $b_1 \neq 0$ and $b_2 \neq 0$, this is equivalent to the ratio of corresponding components being equal: $\frac{a_1}{b_1} = \frac{a_2}{b_2} = k$. (Special care needed if components are zero).

Position Vector of a Point (2D)

The position vector of a point P with coordinates $(x, y)$ relative to the origin O(0, 0) is the vector $\vec{\text{OP}}$ originating from the origin and terminating at P. In component form, the position vector of P(x, y) is:

The magnitude of the position vector $\vec{\text{OP}}$ is the distance of the point P from the origin: $|\vec{\text{OP}}| = \sqrt{x^2 + y^2}$.

Vector Joining Two Points (2D)

Let P$_1$ be a point with coordinates $(x_1, y_1)$ and P$_2$ be a point with coordinates $(x_2, y_2)$. The vector joining P$_1$ to P$_2$ is the directed line segment $\vec{\text{P}_1\text{P}_2}$.

We can express this vector in terms of the position vectors of P$_1$ and P$_2$. Let $\vec{r}_1 = \vec{\text{OP}_1} = x_1\hat{i} + y_1\hat{j}$ be the position vector of P$_1$, and $\vec{r}_2 = \vec{\text{OP}_2} = x_2\hat{i} + y_2\hat{j}$ be the position vector of P$_2$.

Consider the triangle OP$_1$P$_2$. By the triangle law of vector addition, $\vec{\text{OP}_1} + \vec{\text{P}_1\text{P}_2} = \vec{\text{OP}_2}$.

Rearranging this equation to find $\vec{\text{P}_1\text{P}_2}$:

Substitute the component forms of $\vec{r}_1$ and $\vec{r}_2$:

The components of the vector joining P$_1(x_1, y_1)$ to P$_2(x_2, y_2)$ are $(x_2 - x_1)$ and $(y_2 - y_1)$.

The magnitude of the vector $\vec{\text{P}_1\text{P}_2}$ is the distance between points P$_1$ and P$_2$, given by:

Answer:

Components of a Vector in Space

Just as we represent vectors in a plane using two components, we can represent any vector in three-dimensional space using three components. This representation is based on extending the concept of a basis to three mutually perpendicular directions.

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Component Form in 3D (Rectangular Coordinates)

Consider a rectangular Cartesian coordinate system in space with the origin at O. We define three mutually perpendicular unit vectors along the positive x, y, and z axes as $\hat{i}$, $\hat{j}$, and $\hat{k}$ respectively.

• $\hat{i}$ is a unit vector along the positive x-axis.
• $\hat{j}$ is a unit vector along the positive y-axis.
• $\hat{k}$ is a unit vector along the positive z-axis.

These three unit vectors $\hat{i}, \hat{j}, \hat{k}$ are non-zero and non-coplanar. According to the theorem on three non-coplanar vectors, any vector $\vec{r}$ in space can be uniquely expressed as a linear combination of these three vectors.

Let P be any point in space with coordinates $(x, y, z)$ relative to the origin O. The position vector of point P is $\vec{\text{OP}} = \vec{r}$. We can reach point P from the origin by moving $x$ units parallel to the x-axis, $y$ units parallel to the y-axis, and $z$ units parallel to the z-axis. The vector representing the movement along the x-direction is $x\hat{i}$, along the y-direction is $y\hat{j}$, and along the z-direction is $z\hat{k}$.

By vector addition (specifically, the rule for adding vectors in three dimensions, which can be viewed as an extension of the parallelogram law), the position vector $\vec{r} = \vec{\text{OP}}$ is the sum of these three component vectors:

The scalars $x, y,$ and $z$ are called the scalar components or rectangular components of the vector $\vec{r}$ along the x, y, and z axes respectively. The vectors $x\hat{i}, y\hat{j},$ and $z\hat{k}$ are called the vector components of $\vec{r}$ along the axes.

A vector in space can also be represented by the ordered triplet of its scalar components $(x, y, z)$.

Magnitude of a Vector in 3D Component Form

If a vector $\vec{r}$ is given in component form as $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$, its magnitude $|\vec{r}|$ is the length of the vector from the origin O(0, 0, 0) to the point P(x, y, z). This length is the distance from the origin to P, which can be calculated using the distance formula in 3D space, derived from the Pythagorean theorem applied twice.

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

Vector Operations in Component Form (3D)

Similar to the 2D case, vector operations in 3D become algebraic operations on their corresponding components.

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

• Addition: The sum of two vectors is found by adding their corresponding components.

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

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

• Subtraction: The difference of two vectors is found by subtracting their corresponding components.

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

  $= (a_1-b_1)\hat{i} + (a_2-b_2)\hat{j} + (a_3-b_3)\hat{k}$

• Scalar Multiplication: Multiplying a vector by a scalar involves multiplying each component by the scalar. For a scalar $k$:

  k$\vec{a} = k(a_1\hat{i} + a_2\hat{j} + a_3\hat{k}) = (ka_1)\hat{i} + (ka_2)\hat{j} + (ka_3)\hat{k}$

• Equality of Vectors: Two vectors are equal if and only if their corresponding components are equal.

  $\vec{a} = \vec{b} \iff a_1 = b_1, a_2 = b_2, \text{ and } a_3 = b_3$

• Collinearity: Two non-zero vectors $\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}$ and $\vec{b} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k}$ are collinear if and only if there exists a non-zero scalar $k$ such that $\vec{a} = k\vec{b}$. In component form, this means $a_1 = kb_1$, $a_2 = kb_2$, and $a_3 = kb_3$. If $b_1, b_2, b_3$ are all non-zero, this is equivalent to the ratio of corresponding components being equal: $\frac{a_1}{b_1} = \frac{a_2}{b_2} = \frac{a_3}{b_3} = k$.

Direction Ratios and Direction Cosines from Components

If a vector $\vec{r}$ is expressed in component form $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$, then its scalar components $x, y,$ and $z$ are the direction ratios of the vector. Any triplet $(a, b, c)$ proportional to $(x, y, z)$ will also be direction ratios of $\vec{r}$.

If $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$ is a non-zero vector, and $R = |\vec{r}| = \sqrt{x^2 + y^2 + z^2}$ is its magnitude, the direction cosines $l, m, n$ are given by the ratios of the components to the magnitude:

The unit vector in the direction of $\vec{r}$ is $\hat{r} = \frac{\vec{r}}{|\vec{r}|} = \frac{x}{R}\hat{i} + \frac{y}{R}\hat{j} + \frac{z}{R}\hat{k} = l\hat{i} + m\hat{j} + n\hat{k}$. The components of the unit vector are precisely the direction cosines. As shown before, $l^2 + m^2 + n^2 = (\frac{x}{R})^2 + (\frac{y}{R})^2 + (\frac{z}{R})^2 = \frac{x^2+y^2+z^2}{R^2} = \frac{R^2}{R^2} = 1$.

Position Vector of a Point (3D)

The position vector of a point P with coordinates $(x, y, z)$ relative to the origin O(0, 0, 0) is the vector $\vec{\text{OP}}$ originating from the origin and terminating at P. In component form, the position vector of P(x, y, z) is:

The magnitude of the position vector $\vec{\text{OP}}$ is the distance of the point P from the origin: $|\vec{\text{OP}}| = \sqrt{x^2 + y^2 + z^2}$.

Vector Joining Two Points (3D)

Let P$_1$ be a point with coordinates $(x_1, y_1, z_1)$ and P$_2$ be a point with coordinates $(x_2, y_2, z_2)$. The vector joining P$_1$ to P$_2$ is the directed line segment $\vec{\text{P}_1\text{P}_2}$.

Using position vectors, $\vec{\text{P}_1\text{P}_2} = \vec{\text{OP}_2} - \vec{\text{OP}_1}$. Let $\vec{r}_1 = \vec{\text{OP}_1} = x_1\hat{i} + y_1\hat{j} + z_1\hat{k}$ be the position vector of P$_1$. Let $\vec{r}_2 = \vec{\text{OP}_2} = x_2\hat{i} + y_2\hat{j} + z_2\hat{k}$ be the position vector of P$_2$.

The components of the vector joining P$_1(x_1, y_1, z_1)$ to P$_2(x_2, y_2, z_2)$ are $(x_2 - x_1), (y_2 - y_1),$ and $(z_2 - z_1)$.

The magnitude of the vector $\vec{\text{P}_1\text{P}_2}$ is the distance between points P$_1$ and P$_2$, given by:

The direction ratios of $\vec{\text{P}_1\text{P}_2}$ are $(x_2 - x_1), (y_2 - y_1), (z_2 - z_1)$. The direction cosines are $\frac{x_2-x_1}{|\vec{\text{P}_1\text{P}_2}|}, \frac{y_2-y_1}{|\vec{\text{P}_1\text{P}_2}|}, \frac{z_2-z_1}{|\vec{\text{P}_1\text{P}_2}|}$.

Answer:

Answer:

Section formulae

The section formula in vector algebra is used to determine the position vector of a point that divides a line segment joining two given points in a specific ratio. This formula has two cases: internal division and external division.

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Internal Division

Consider two points A and B in space. Let their position vectors relative to a fixed origin O be $\vec{a} = \vec{\text{OA}}$ and $\vec{b} = \vec{\text{OB}}$, respectively. Let P be a point that lies on the line segment AB and divides it internally in the ratio $m:n$. This means that the ratio of the length of segment AP to the length of segment PB is $m/n$, and P is located between A and B.

The position vector of the point P, denoted by $\vec{r} = \vec{\text{OP}}$, is given by the formula:

where $m$ and $n$ are positive scalars representing the ratio $m:n$. Note that the position vector of the point corresponding to the ratio $m$ is multiplied by $n$, and vice versa, and the sum of the ratios is in the denominator.

Derivation of the Internal Division Formula

Given that point P divides the line segment AB internally in the ratio $m:n$, we can write the relationship between the lengths of the segments AP and PB as $\frac{\text{AP}}{\text{PB}} = \frac{m}{n}$.

This implies $n \cdot \text{AP} = m \cdot \text{PB}$. Since P lies between A and B, the vector $\vec{\text{AP}}$ is in the same direction as the vector $\vec{\text{PB}}$. Therefore, the vector relationship is:

Now, express the vectors $\vec{\text{AP}}$ and $\vec{\text{PB}}$ in terms of the position vectors of points A, B, and P relative to the origin O.

$\vec{\text{AP}} = \text{Position vector of P} - \text{Position vector of A} = \vec{\text{OP}} - \vec{\text{OA}} = \vec{r} - \vec{a}$.

$\vec{\text{PB}} = \text{Position vector of B} - \text{Position vector of P} = \vec{\text{OB}} - \vec{\text{OP}} = \vec{b} - \vec{r}$.

Substitute these expressions into equation (A):

Distribute the scalars:

Collect terms involving $\vec{r}$ on one side and other terms on the other side:

Factor out $\vec{r}$ from the left side:

Since $m$ and $n$ are positive, $m+n \neq 0$. Divide by $(m+n)$ to solve for $\vec{r}$:

This confirms the internal division formula.

Midpoint Formula

A special case of the internal division formula occurs when P is the midpoint of the line segment AB. The midpoint divides the segment in the ratio $1:1$. So, we set $m=1$ and $n=1$ in the internal division formula:

The position vector of the midpoint of a line segment is the average of the position vectors of its endpoints.

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External Division

Let A and B be two points with position vectors $\vec{a}$ and $\vec{b}$ respectively. Let P be a point that lies on the line containing segment AB, but outside the segment AB. If P divides the line segment AB externally in the ratio $m:n$, it means that the ratio of the distance from A to P to the distance from B to P is $m/n$.

The position vector of the point P, denoted by $\vec{r} = \vec{\text{OP}}$, is given by the formula:

where $m$ and $n$ are scalars and $m \neq n$. If $m>n$, P lies on the extension of AB beyond B. If $n>m$, P lies on the extension of BA beyond A.

Derivation of the External Division Formula

Given that P divides the line segment AB externally in the ratio $m:n$, we have $\frac{\text{AP}}{\text{BP}} = \frac{m}{n}$. This implies $n \cdot \text{AP} = m \cdot \text{BP}$.

Let's consider the case where P is on the extension of AB beyond B (i.e., $m>n$). In this case, the vectors $\vec{\text{AP}}$ and $\vec{\text{BP}}$ are in the same direction. The vector relationship reflecting the magnitude ratio is:

Express $\vec{\text{AP}}$ and $\vec{\text{BP}}$ in terms of position vectors:

$\vec{\text{AP}} = \vec{r} - \vec{a}$

$\vec{\text{BP}} = \vec{r} - \vec{b}$

Substitute these into equation (C):

$n\vec{r}$ - n$\vec{a}$ = m$\vec{r}$ - m$\vec{b}$

Collect terms with $\vec{r}$ on one side:

Solve for $\vec{r}$ (since $m \neq n$, $m-n \neq 0$):

This derivation holds for the case $m>n$. If $n>m$, P lies on the extension of BA beyond A. In this case, $\vec{\text{AP}}$ and $\vec{\text{BP}}$ are in opposite directions, so the vector relationship is $n\vec{\text{AP}} = -m\vec{\text{BP}}$, or $n(\vec{r}-\vec{a}) = -m(\vec{r}-\vec{b})$. This leads to $(n-m)\vec{r} = n\vec{a} - m\vec{b}$, giving $\vec{r} = \frac{n\vec{a} - m\vec{b}}{n-m} = \frac{m\vec{b} - n\vec{a}}{m-n}$. So the formula is consistent for both sub-cases of external division.

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Scalar (or Dot) Product of Two Vectors

In vector algebra, there are different ways to multiply vectors. The scalar product, also known as the dot product, is one such operation. It takes two vectors and produces a scalar quantity. The dot product is a fundamental tool for concepts like work done in physics, projection of one vector onto another, and finding the angle between two vectors.

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Definition of Scalar Product

Let $\vec{a}$ and $\vec{b}$ be two non-zero vectors, and let $\theta$ be the angle between them, such that $0 \leq \theta \leq \pi$. The scalar product of $\vec{a}$ and $\vec{b}$, denoted by $\vec{a} \cdot \vec{b}$, is defined as the product of their magnitudes and the cosine of the angle between them.

If either $\vec{a}$ or $\vec{b}$ (or both) is the zero vector ($\vec{0}$), the angle $\theta$ is not defined. In this case, the scalar product is defined to be zero:

The result of the dot product is a scalar, hence the name "scalar product". The unit of the dot product is the product of the units of the two vectors.

Geometric Interpretation of the Dot Product

The dot product has a useful geometric interpretation related to projections. From the definition $\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos\theta$, we can group the terms in two ways:

• $\vec{a} \cdot \vec{b} = |\vec{a}| (|\vec{b}| \cos\theta)$: Here, $(|\vec{b}| \cos\theta)$ is the scalar projection of vector $\vec{b}$ onto vector $\vec{a}$. It represents the length of the projection of $\vec{b}$ onto the line containing $\vec{a}$, with a sign depending on whether $\theta$ is acute ($>0$) or obtuse ($<0$). So, the dot product is the magnitude of $\vec{a}$ multiplied by the scalar projection of $\vec{b}$ onto $\vec{a}$.
• $\vec{a} \cdot \vec{b} = |\vec{b}| (|\vec{a}| \cos\theta)$: Similarly, $(|\vec{a}| \cos\theta)$ is the scalar projection of vector $\vec{a}$ onto vector $\vec{b}$. The dot product is the magnitude of $\vec{b}$ multiplied by the scalar projection of $\vec{a}$ onto $\vec{b}$.

The vector projection of $\vec{b}$ onto $\vec{a}$ is a vector in the direction of $\vec{a}$ (or $\hat{a}$) with magnitude equal to the scalar projection. The unit vector in the direction of $\vec{a}$ is $\hat{a} = \frac{\vec{a}}{|\vec{a}|}$. Vector projection of $\vec{b}$ onto $\vec{a}$ = $(|\vec{b}| \cos\theta) \hat{a} = \left(\frac{|\vec{a}| |\vec{b}| \cos\theta}{|\vec{a}|}\right) \frac{\vec{a}}{|\vec{a}|} = \left(\frac{\vec{a} \cdot \vec{b}}{|\vec{a}|^2}\right) \vec{a}$. Similarly, the vector projection of $\vec{a}$ onto $\vec{b}$ is $\left(\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|^2}\right) \vec{b}$.

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Properties of the Dot Product

The scalar product satisfies the following properties:

• Commutative Property: The order of the vectors in the dot product does not matter. For any two vectors $\vec{a}$ and $\vec{b}$, $\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a}$. This is true because $|\vec{a}| |\vec{b}| \cos\theta = |\vec{b}| |\vec{a}| \cos\theta$.
• Distributive Property: The dot product distributes over vector addition. For any three vectors $\vec{a}, \vec{b}, \vec{c}$, $\vec{a} \cdot (\vec{b} + \vec{c}) = \vec{a} \cdot \vec{b} + \vec{a} \cdot \vec{c}$.
• Scalar Multiplication: A scalar factor can be moved outside the dot product. For any vectors $\vec{a}, \vec{b}$ and scalar $k$, $(k\vec{a}) \cdot \vec{b} = k(\vec{a} \cdot \vec{b}) = \vec{a} \cdot (k\vec{b})$.
• Dot Product of Orthogonal Vectors: If two non-zero vectors $\vec{a}$ and $\vec{b}$ are perpendicular (orthogonal) to each other, the angle between them is $\theta = \pi/2$ or $90^\circ$. In this case, $\cos(\pi/2) = 0$.

  $\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos(\pi/2) = |\vec{a}| |\vec{b}| \cdot 0 = 0$

  Conversely, if the dot product of two non-zero vectors is zero, then the cosine of the angle between them must be zero, implying the angle is $90^\circ$ and the vectors are orthogonal. Thus, $\vec{a} \cdot \vec{b} = 0 \iff \vec{a} \perp \vec{b}$ (for non-zero vectors).
• Dot Product of Parallel Vectors: If two non-zero vectors $\vec{a}$ and $\vec{b}$ are parallel, the angle between them is either $\theta = 0$ (same direction) or $\theta = \pi$ (opposite direction). If $\theta=0$, $\cos 0^\circ = 1$, so $\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cdot 1 = |\vec{a}| |\vec{b}|$. If $\theta=\pi$, $\cos \pi = -1$, so $\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cdot (-1) = -|\vec{a}| |\vec{b}|$.
• Dot Product of a Vector with Itself: The dot product of a vector with itself gives the square of its magnitude. The angle between a vector and itself is $0^\circ$.

  $\vec{a} \cdot \vec{a} = |\vec{a}| |\vec{a}| \cos 0^\circ = |\vec{a}|^2 \cdot 1 = |\vec{a}|^2$

  ... (7)

  This provides a convenient way to calculate the magnitude: $|\vec{a}| = \sqrt{\vec{a} \cdot \vec{a}}$.
• Dot Products of Standard Orthogonal Unit Vectors: For the unit vectors $\hat{i}, \hat{j}, \hat{k}$ along the mutually perpendicular axes, their magnitudes are 1, and the angle between any two distinct vectors is $90^\circ$.

  $\hat{i} \cdot \hat{i} = |\hat{i}| |\hat{i}| \cos 0^\circ = 1 \cdot 1 \cdot 1 = 1$

  $\hat{j} \cdot \hat{j} = 1$

  $\hat{k} \cdot \hat{k} = 1$

  ... (8)

  $\hat{i} \cdot \hat{j} = |\hat{i}| |\hat{j}| \cos 90^\circ = 1 \cdot 1 \cdot 0 = 0$

  $\hat{j} \cdot \hat{k} = 0$

  $\hat{k} \cdot \hat{i} = 0$

  ... (9)

  (Also $\hat{j} \cdot \hat{i} = 0$, $\hat{k} \cdot \hat{j} = 0$, $\hat{i} \cdot \hat{k} = 0$ due to commutative property).

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Dot Product in Component Form

The dot product is very easy to calculate when the vectors are given in component form. Let $\vec{a} = a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{k}$ and $\vec{b} = b_1 \hat{i} + b_2 \hat{j} + b_3 \hat{k}$. We can find their dot product using the distributive property and the dot products of the unit vectors:

Apply the distributive property (each term in the first vector is dotted with each term in the second vector):

Using the properties from (8) and (9), the terms with $\hat{i} \cdot \hat{j}$, $\hat{i} \cdot \hat{k}$, etc. are zero, and terms with $\hat{i} \cdot \hat{i}$, $\hat{j} \cdot \hat{j}$, $\hat{k} \cdot \hat{k}$ are one:

The dot product of two vectors in component form is the sum of the products of their corresponding components. This formula is extremely useful for calculation.

Angle Between Two Vectors using Dot Product

The definition of the dot product, $\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos\theta$, allows us to find the angle $\theta$ between two non-zero vectors $\vec{a}$ and $\vec{b}$. Rearranging the formula, we get:

If the vectors are given in component form $\vec{a} = a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{k}$ and $\vec{b} = b_1 \hat{i} + b_2 \hat{j} + b_3 \hat{k}$, we can use the component formulas for the dot product and magnitudes:

Substitute these into the formula for $\cos\theta$:

We can then find $\theta = \cos^{-1}\left(\frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|}\right)$.

Answer:

Answer:

Vector (or Cross) Product of Two Vectors

The vector product or cross product of two non-zero vectors, $\vec{a}$ and $\vec{b}$, is an operation that produces a new vector, denoted by $\vec{a} \times \vec{b}$. The direction of this new vector is perpendicular to the plane containing the original two vectors $\vec{a}$ and $\vec{b}$, and its magnitude is related to the area of the parallelogram formed by these vectors.

Definition of Vector Product

Let $\vec{a}$ and $\vec{b}$ be two non-zero vectors. The vector product $\vec{a} \times \vec{b}$ is defined as:

where:

• $|\vec{a}|$ and $|\vec{b}|$ are the magnitudes of vectors $\vec{a}$ and $\vec{b}$ respectively.
• $\theta$ is the angle between $\vec{a}$ and $\vec{b}$, such that $0 \leq \theta \leq \pi$.
• $\hat{n}$ is a unit vector perpendicular to the plane containing $\vec{a}$ and $\vec{b}$. Its direction is determined by the right-hand rule. To apply the right-hand rule, if you curl the fingers of your right hand from vector $\vec{a}$ towards vector $\vec{b}$ through the smaller angle $\theta$, then your thumb points in the direction of $\hat{n}$ (and thus $\vec{a} \times \vec{b}$).

Observations:

1. $\vec{a} \times \vec{b}$ is a vector quantity.
2. If $\vec{a} = \vec{0}$ or $\vec{b} = \vec{0}$, then $\theta$ is not defined, and in this case, we define $\vec{a} \times \vec{b} = \vec{0}$.
3. If $\vec{a}$ and $\vec{b}$ are non-zero and collinear (i.e., parallel, so $\theta = 0$ or $\theta = \pi$), then $\sin\theta = 0$. In this case, $\vec{a} \times \vec{b} = \vec{0}$. Conversely, if $\vec{a} \times \vec{b} = \vec{0}$ and $\vec{a}, \vec{b}$ are non-zero, then $\vec{a}$ and $\vec{b}$ are collinear.

The magnitude of the cross product is given by:

$|\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin\theta$

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Geometric Interpretation of the Cross Product

The magnitude $|\vec{a} \times \vec{b}|$ represents the area of the parallelogram whose adjacent sides are represented by the vectors $\vec{a}$ and $\vec{b}$.

Consider a parallelogram with adjacent sides $\vec{a}$ and $\vec{b}$. Let $\theta$ be the angle between them.

Area of parallelogram = Base $\times$ Height

$= |\vec{a}| (|\vec{b}| \sin\theta) = |\vec{a} \times \vec{b}|$

Consequently, the area of the triangle formed by the vectors $\vec{a}$ and $\vec{b}$ as adjacent sides is $\frac{1}{2} |\vec{a} \times \vec{b}|$.

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Properties of the Cross Product

Let $\vec{a}$, $\vec{b}$, and $\vec{c}$ be any three vectors, and $k$ be a scalar.

1. Non-Commutative Property: The cross product is not commutative. In fact,

$\vec{a} \times \vec{b} = -(\vec{b} \times \vec{a})$

This means that the order of vectors in a cross product matters. Reversing the order reverses the direction of the resulting vector, though its magnitude remains the same.

2. Scalar Multiplication: For any scalar $k$,

$(k\vec{a}) \times \vec{b} = k(\vec{a} \times \vec{b}) = \vec{a} \times (k\vec{b})$

3. Distributive Property over Vector Addition: The cross product distributes over vector addition.

$\vec{a} \times (\vec{b} + \vec{c}) = (\vec{a} \times \vec{b}) + (\vec{a} \times \vec{c})$ (Left distributivity)

$(\vec{b} + \vec{c}) \times \vec{a} = (\vec{b} \times \vec{a}) + (\vec{c} \times \vec{a})$ (Right distributivity)

4. Cross Product of Parallel or Collinear Vectors: If $\vec{a}$ and $\vec{b}$ are two non-zero vectors, then $\vec{a} \times \vec{b} = \vec{0}$ if and only if $\vec{a}$ and $\vec{b}$ are parallel (or collinear), i.e., $\theta = 0$ or $\theta = \pi$. This also implies that for any non-zero vector $\vec{a}$,

$\vec{a} \times \vec{a} = |\vec{a}| |\vec{a}| \sin 0^\circ \hat{n} = |\vec{a}|^2 \cdot 0 \cdot \hat{n} = \vec{0}$

5. Cross Products of Standard Orthogonal Unit Vectors $\hat{i}, \hat{j}, \hat{k}$ (Right-handed system):

Since $\hat{i}, \hat{j}, \hat{k}$ are mutually perpendicular unit vectors:

$\hat{i} \times \hat{i} = \vec{0}, \quad \hat{j} \times \hat{j} = \vec{0}, \quad \hat{k} \times \hat{k} = \vec{0}$

And, using the right-hand rule:

$\hat{i} \times \hat{j} = |\hat{i}||\hat{j}|\sin 90^\circ \hat{k} = (1)(1)(1) \hat{k} = \hat{k}$

$\hat{j} \times \hat{k} = \hat{i}$

$\hat{k} \times \hat{i} = \hat{j}$

Using the non-commutative property ($\vec{a} \times \vec{b} = -(\vec{b} \times \vec{a})$):

$\hat{j} \times \hat{i} = -(\hat{i} \times \hat{j}) = -\hat{k}$

$\hat{k} \times \hat{j} = -(\hat{j} \times \hat{k}) = -\hat{i}$

$\hat{i} \times \hat{k} = -(\hat{k} \times \hat{i}) = -\hat{j}$

These can be remembered using a cyclic order: $\hat{i} \xrightarrow{} \hat{j} \xrightarrow{} \hat{k} \xrightarrow{} \hat{i}$. If we multiply in the direction of the arrows (clockwise), the result is positive. If we multiply against the direction of the arrows (counter-clockwise), the result is negative.

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Cross Product in Component Form

Let $\vec{a} = a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{k}$ and $\vec{b} = b_1 \hat{i} + b_2 \hat{j} + b_3 \hat{k}$. Using the distributive property and the cross products of $\hat{i}, \hat{j}, \hat{k}$:

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

$= a_1 b_1 (\hat{i} \times \hat{i}) + a_1 b_2 (\hat{i} \times \hat{j}) + a_1 b_3 (\hat{i} \times \hat{k})$

$+ a_2 b_1 (\hat{j} \times \hat{i}) + a_2 b_2 (\hat{j} \times \hat{j}) + a_2 b_3 (\hat{j} \times \hat{k})$

$+ a_3 b_1 (\hat{k} \times \hat{i}) + a_3 b_2 (\hat{k} \times \hat{j}) + a_3 b_3 (\hat{k} \times \hat{k})$

Substitute the cross products of unit vectors:

$= a_1 b_1 (\vec{0}) + a_1 b_2 (\hat{k}) + a_1 b_3 (-\hat{j})$

$+ a_2 b_1 (-\hat{k}) + a_2 b_2 (\vec{0}) + a_2 b_3 (\hat{i})$

$+ a_3 b_1 (\hat{j}) + a_3 b_2 (-\hat{i}) + a_3 b_3 (\vec{0})$

Collect terms with $\hat{i}, \hat{j}, \hat{k}$:

$= (a_2 b_3 - a_3 b_2) \hat{i} + (a_3 b_1 - a_1 b_3) \hat{j} + (a_1 b_2 - a_2 b_1) \hat{k}$

This expression can be conveniently written as a determinant:

Expanding this determinant along the first row gives the component form.

Answer:

Scalar Triple Product

The scalar triple product is an operation that combines three vectors to produce a scalar quantity. It involves both the cross product and the dot product.

Definition of Scalar Triple Product

Let $\vec{a}, \vec{b}, \vec{c}$ be three vectors. The scalar triple product of $\vec{a}, \vec{b}, \vec{c}$ is defined as $\vec{a} \cdot (\vec{b} \times \vec{c})$.

It is often denoted by $[\vec{a}, \vec{b}, \vec{c}]$ or $(\vec{a}, \vec{b}, \vec{c})$.

Note that the scalar triple product is a scalar quantity.

Geometric Interpretation

The magnitude of the scalar triple product $|\vec{a} \cdot (\vec{b} \times \vec{c})|$ represents the volume of the parallelepiped formed by the three vectors $\vec{a}$, $\vec{b}$, and $\vec{c}$ as coterminous edges.

Volume of parallelepiped = Area of base $\times$ Height.

If the base is formed by $\vec{b}$ and $\vec{c}$, its area is $|\vec{b} \times \vec{c}|$. The vector $\vec{b} \times \vec{c}$ is perpendicular to the base. The height of the parallelepiped is the scalar projection of $\vec{a}$ onto the normal to the base, which is the direction of $\vec{b} \times \vec{c}$.

Height = $|\vec{a}| |\cos\phi|$, where $\phi$ is the angle between $\vec{a}$ and $(\vec{b} \times \vec{c})$.

Volume = $|\vec{b} \times \vec{c}| \cdot |\vec{a}| |\cos\phi| = |\vec{a} \cdot (\vec{b} \times \vec{c})|$.

The signed value of $\vec{a} \cdot (\vec{b} \times \vec{c})$ indicates the orientation of the three vectors (right-handed or left-handed system). If positive, they form a right-handed system; if negative, a left-handed system.

The volume of the tetrahedron formed by the three vectors $\vec{a}, \vec{b}, \vec{c}$ as coterminous edges is $\frac{1}{6} |\vec{a} \cdot (\vec{b} \times \vec{c})|$.

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Scalar Triple Product in Component Form

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 $\vec{c} = c_1 \hat{i} + c_2 \hat{j} + c_3 \hat{k}$.

We know that the cross product $\vec{b} \times \vec{c}$ is given by:

$\vec{b} \times \vec{c} = (b_2 c_3 - b_3 c_2) \hat{i} + (b_3 c_1 - b_1 c_3) \hat{j} + (b_1 c_2 - b_2 c_1) \hat{k}$

Alternatively, using determinants:

$\vec{b} \times \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{vmatrix}$

Now, taking the dot product of $\vec{a}$ with $(\vec{b} \times \vec{c})$:

$\vec{a} \cdot (\vec{b} \times \vec{c}) = (a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{k}) \cdot ((b_2 c_3 - b_3 c_2) \hat{i} + (b_3 c_1 - b_1 c_3) \hat{j} + (b_1 c_2 - b_2 c_1) \hat{k})$

$= a_1 (b_2 c_3 - b_3 c_2) + a_2 (b_3 c_1 - b_1 c_3) + a_3 (b_1 c_2 - b_2 c_1)$

This expression is the expansion of a $3 \times 3$ determinant where the components of the vectors $\vec{a}, \vec{b}, \vec{c}$ form the rows (or columns) of the determinant:

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Properties of Scalar Triple Product

1. Cyclic Permutation: The value of the scalar triple product remains unchanged if the vectors are cyclically permuted.

i.e., $[\vec{a}, \vec{b}, \vec{c}] = [\vec{b}, \vec{c}, \vec{a}] = [\vec{c}, \vec{a}, \vec{b}]$.

Proof: From the determinant form (ii), $[\vec{b}, \vec{c}, \vec{a}] = \begin{vmatrix} b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \\ a_1 & a_2 & a_3 \end{vmatrix}$. Interchanging rows twice ($R_1 \leftrightarrow R_2$, then $R_2 \leftrightarrow R_3$ effectively, or $R_1 \leftrightarrow R_3$ and then $R_1 \leftrightarrow R_2$ to get back to original order form) gives the original determinant. For example, $R_1 \leftrightarrow R_2$ gives $-\begin{vmatrix} c_1 & c_2 & c_3 \\ b_1 & b_2 & b_3 \\ a_1 & a_2 & a_3 \end{vmatrix}$. Then $R_2 \leftrightarrow R_3$ gives $\begin{vmatrix} c_1 & c_2 & c_3 \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix} = [\vec{c}, \vec{a}, \vec{b}]$. A single cyclic shift $R_1 \to R_2 \to R_3 \to R_1$ is equivalent to two row interchanges. For instance, to get from $\begin{vmatrix} a \\ b \\ c \end{vmatrix}$ to $\begin{vmatrix} b \\ c \\ a \end{vmatrix}$: $\begin{vmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{vmatrix} \xrightarrow{R_1 \leftrightarrow R_2} - \begin{vmatrix} b_1 & b_2 & b_3 \\ a_1 & a_2 & a_3 \\ c_1 & c_2 & c_3 \end{vmatrix} \xrightarrow{R_2 \leftrightarrow R_3} (-1)(-1) \begin{vmatrix} b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \\ a_1 & a_2 & a_3 \end{vmatrix} = [\vec{b}, \vec{c}, \vec{a}]$. So, $[\vec{a}, \vec{b}, \vec{c}] = [\vec{b}, \vec{c}, \vec{a}]$. Similarly, $[\vec{b}, \vec{c}, \vec{a}] = [\vec{c}, \vec{a}, \vec{b}]$.

2. Interchange of Two Vectors: If any two vectors in the scalar triple product are interchanged, the sign of the product changes (anti-cyclic).

i.e., $[\vec{a}, \vec{b}, \vec{c}] = - [\vec{b}, \vec{a}, \vec{c}] = - [\vec{a}, \vec{c}, \vec{b}] = - [\vec{c}, \vec{b}, \vec{a}]$.

Proof: This follows directly from the property of determinants that interchanging any two rows (or columns) changes the sign of the determinant. For example, $[\vec{b}, \vec{a}, \vec{c}] = \begin{vmatrix} b_1 & b_2 & b_3 \\ a_1 & a_2 & a_3 \\ c_1 & c_2 & c_3 \end{vmatrix} = - \begin{vmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{vmatrix} = - [\vec{a}, \vec{b}, \vec{c}]$.

3. Dot and Cross Interchange: The positions of dot and cross can be interchanged without changing the value of the scalar triple product, provided the cyclic order of the vectors is maintained.

$\vec{a} \cdot (\vec{b} \times \vec{c}) = (\vec{a} \times \vec{b}) \cdot \vec{c}$

Proof: $(\vec{a} \times \vec{b}) \cdot \vec{c} = \vec{c} \cdot (\vec{a} \times \vec{b})$ (since dot product is commutative) $= [\vec{c}, \vec{a}, \vec{b}]$. By cyclic permutation property (Property 1), $[\vec{c}, \vec{a}, \vec{b}] = [\vec{a}, \vec{b}, \vec{c}]$. Thus, $\vec{a} \cdot (\vec{b} \times \vec{c}) = (\vec{a} \times \vec{b}) \cdot \vec{c}$.

4. Condition for Coplanarity of Three Vectors: Three non-zero vectors $\vec{a}, \vec{b}, \vec{c}$ are coplanar if and only if their scalar triple product is zero.

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

Proof: If $\vec{a}, \vec{b}, \vec{c}$ are coplanar, then the vector $\vec{b} \times \vec{c}$ (which is perpendicular to the plane containing $\vec{b}$ and $\vec{c}$) must also be perpendicular to $\vec{a}$ (since $\vec{a}$ lies in the same plane). The dot product of two perpendicular vectors is zero. So, $\vec{a} \cdot (\vec{b} \times \vec{c}) = 0$. Conversely, if $[\vec{a}, \vec{b}, \vec{c}] = 0$, then $\vec{a} \cdot (\vec{b} \times \vec{c}) = 0$. This implies that $\vec{a}$ is perpendicular to $\vec{b} \times \vec{c}$. Since $\vec{b} \times \vec{c}$ is perpendicular to the plane containing $\vec{b}$ and $\vec{c}$, $\vec{a}$ must lie in the plane containing $\vec{b}$ and $\vec{c}$ (or $\vec{b} \times \vec{c} = \vec{0}$, which means $\vec{b}$ and $\vec{c}$ are collinear, and thus $\vec{a}, \vec{b}, \vec{c}$ are coplanar). Geometrically, if the volume of the parallelepiped formed by $\vec{a}, \vec{b}, \vec{c}$ is zero, the vectors must be coplanar.

5. Identical Vectors: If any two of the three vectors are identical (or collinear), the scalar triple product is zero.

e.g., $[\vec{a}, \vec{a}, \vec{c}] = 0$.

Proof: If $\vec{a} = \vec{b}$, then $[\vec{a}, \vec{a}, \vec{c}] = \begin{vmatrix} a_1 & a_2 & a_3 \\ a_1 & a_2 & a_3 \\ c_1 & c_2 & c_3 \end{vmatrix} = 0$, since two rows of the determinant are identical. Alternatively, $\vec{a} \times \vec{a} = \vec{0}$, so $(\vec{a} \times \vec{a}) \cdot \vec{c} = \vec{0} \cdot \vec{c} = 0$. If $\vec{b} = k\vec{a}$ (collinear), then $[\vec{a}, k\vec{a}, \vec{c}] = k[\vec{a}, \vec{a}, \vec{c}] = k \cdot 0 = 0$.

6. Scalar Multiple: If any vector is multiplied by a scalar $k$, the scalar triple product is multiplied by $k$.

$[k\vec{a}, \vec{b}, \vec{c}] = k[\vec{a}, \vec{b}, \vec{c}]$

$[\vec{a}, k\vec{b}, \vec{c}] = k[\vec{a}, \vec{b}, \vec{c}]$

$[\vec{a}, \vec{b}, k\vec{c}] = k[\vec{a}, \vec{b}, \vec{c}]$

Proof: This follows from the property of determinants that multiplying a row by a scalar $k$ multiplies the determinant by $k$. $[k\vec{a}, \vec{b}, \vec{c}] = \begin{vmatrix} ka_1 & ka_2 & ka_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{vmatrix} = k \begin{vmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{vmatrix} = k[\vec{a}, \vec{b}, \vec{c}]$.

7. Distributive Property: The scalar triple product is distributive with respect to vector addition.

$[\vec{a} + \vec{d}, \vec{b}, \vec{c}] = [\vec{a}, \vec{b}, \vec{c}] + [\vec{d}, \vec{b}, \vec{c}]$

Proof: $(\vec{a} + \vec{d}) \cdot (\vec{b} \times \vec{c}) = \vec{a} \cdot (\vec{b} \times \vec{c}) + \vec{d} \cdot (\vec{b} \times \vec{c})$ (by distributivity of dot product). So, $[\vec{a} + \vec{d}, \vec{b}, \vec{c}] = [\vec{a}, \vec{b}, \vec{c}] + [\vec{d}, \vec{b}, \vec{c}]$. This also follows from determinant properties: $\begin{vmatrix} a_1+d_1 & a_2+d_2 & a_3+d_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{vmatrix} = \begin{vmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{vmatrix} + \begin{vmatrix} d_1 & d_2 & d_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{vmatrix}$.

8. The scalar triple product of three mutually perpendicular unit vectors $\hat{i}, \hat{j}, \hat{k}$ forming a right-handed system is 1.

$[\hat{i}, \hat{j}, \hat{k}] = \hat{i} \cdot (\hat{j} \times \hat{k}) = \hat{i} \cdot \hat{i} = 1$.

Also, $[\hat{i}, \hat{j}, \hat{k}] = \begin{vmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{vmatrix} = 1$.

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