In day-to-day life, quantities are classified into two categories: scalars (having only magnitude) and vectors (having both magnitude and direction).
Consider a point $P(x, y, z)$ in a three-dimensional rectangular coordinate system with origin $O(0, 0, 0)$. The vector $\vec{OP}$ is called the position vector of point $P$ with respect to $O$, denoted as $\vec{r}$.
Using the distance formula, the magnitude of $\vec{OP}$ (or $\vec{r}$) is given by:
$$|\vec{OP}| = |\vec{r}| = \sqrt{x^2 + y^2 + z^2}$$Let $\alpha, \beta, \gamma$ be the angles (called direction angles ) made by the position vector $\vec{r}$ with the positive directions of the $x, y,$ and $z$-axes respectively.
Vectors are classified based on their magnitude, direction, and initial/terminal points:
| Vector Type | Mathematical Definition & Properties | Notation / Representation |
|---|---|---|
| Zero Vector (Null Vector) | Initial and terminal points coincide. Magnitude is zero. Direction is indeterminate (can be regarded as having any direction). | $\vec{0}$ or $\vec{AA}$, $\vec{BB}$ |
| Unit Vector | A vector whose magnitude is unity (1 unit). | $\hat{a} = \frac{\vec{a}}{|\vec{a}|}$ |
| Coinitial Vectors | Two or more vectors having the same initial point. | Vectors starting from a common point $A$. |
| Collinear Vectors | Two or more vectors parallel to the same line, irrespective of their magnitudes and directions. | $\vec{a} = \lambda \vec{b}$ for some scalar $\lambda$ |
| Equal Vectors | Vectors having the same magnitude and direction, regardless of the positions of their initial points. | $\vec{a} = \vec{b}$ |
| Negative of a Vector | A vector having the same magnitude as a given vector $\vec{a}$ but opposite direction. | $-\vec{a}$ or $\vec{BA} = -\vec{AB}$ |
| Free Vectors | Vectors that can be subject to parallel displacement without changing magnitude and direction. | Standard vectors used in this chapter. |
Mathematical Problem Statement: Classify the following measures as scalars and vectors: (i) 5 seconds, (ii) $1000\text{ cm}^3$, (iii) 10 Newton, (iv) $30\text{ km/hr}$, (v) $10\text{ g/cm}^3$, (vi) $20\text{ m/s}$ towards north.
Step-by-Step Algorithmic Solution Trace:
- Identify if the quantity has only magnitude or both magnitude and direction.
- (i) Time (5 seconds) has only magnitude $\rightarrow$ Scalar .
- (ii) Volume ($1000\text{ cm}^3$) has only magnitude $\rightarrow$ Scalar .
- (iii) Force (10 Newton) has both magnitude and direction $\rightarrow$ Vector .
- (iv) Speed ($30\text{ km/hr}$) has only magnitude $\rightarrow$ Scalar .
- (v) Density ($10\text{ g/cm}^3$) has only magnitude $\rightarrow$ Scalar .
- (vi) Velocity ($20\text{ m/s}$ towards north) has both magnitude and direction $\rightarrow$ Vector .
Final Evaluation Output: Scalars: (i), (ii), (iv), (v); Vectors: (iii), (vi).
Vector addition is defined geometrically using two fundamental laws:
If two vectors are represented by two sides of a triangle taken in order, then their sum (resultant) is represented by the third side of the triangle taken in the opposite order.
$$\vec{AB} + \vec{BC} = \vec{AC}$$
When the sides of a triangle are taken in order, the resultant is zero: $\vec{AB} + \vec{BC} + \vec{CA} = \vec{AA} = \vec{0}$.
If two vectors $\vec{a}$ and $\vec{b}$ are represented by two adjacent sides of a parallelogram in magnitude and direction, then their sum $\vec{a} + \vec{b}$ is represented by the diagonal of the parallelogram passing through their common point.
$$\vec{OA} + \vec{OB} = \vec{OC} \quad \text{(since } \vec{AC} = \vec{OB}\text{)}$$
Let $\vec{a}$ be a vector and $\lambda$ be a scalar. The product $\lambda \vec{a}$ is a vector collinear with $\vec{a}$.
Let $\hat{i}, \hat{j}, \hat{k}$ be the unit vectors along the positive $x, y, z$ axes respectively.
Any vector $\vec{r}$ can be expressed in its component form as:
$$\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$$Let $\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}$ and $\vec{b} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k}$:
For points $P_1(x_1, y_1, z_1)$ and $P_2(x_2, y_2, z_2)$, the vector $\vec{P_1P_2}$ is given by:
$$\vec{P_1P_2} = (x_2 - x_1)\hat{i} + (y_2 - y_1)\hat{j} + (z_2 - z_1)\hat{k}$$Its magnitude is:
$$|\vec{P_1P_2}| = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}$$Let $P$ and $Q$ be two points with position vectors $\vec{a}$ and $\vec{b}$. Let $R$ be a point dividing the line segment $PQ$ in the ratio $m : n$:
Multiplication of two vectors is defined in two ways: Scalar (dot) product and Vector (cross) product.
The scalar product of two non-zero vectors $\vec{a}$ and $\vec{b}$, denoted by $\vec{a} \cdot \vec{b}$, is defined as:
$$\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos\theta$$where $\theta$ is the angle between $\vec{a}$ and $\vec{b}$ ($0 \le \theta \le \pi$).
The vector product of two non-zero vectors $\vec{a}$ and $\vec{b}$, denoted by $\vec{a} \times \vec{b}$, is defined as:
$$\vec{a} \times \vec{b} = |\vec{a}| |\vec{b}| \sin\theta \hat{n}$$where $\theta$ is the angle between $\vec{a}$ and $\vec{b}$ ($0 \le \theta \le \pi$), and $\hat{n}$ is a unit vector perpendicular to both $\vec{a}$ and $\vec{b}$ such that $\vec{a}, \vec{b}, \hat{n}$ form a right-handed system.
Mathematical Problem Statement: Find the angle $\theta$ between the vectors $\vec{a} = \hat{i} + \hat{j} - \hat{k}$ and $\vec{b} = \hat{i} - \hat{j} + \hat{k}$.
Step-by-Step Algorithmic Solution Trace:
- Compute the dot product $\vec{a} \cdot \vec{b}$:$$\vec{a} \cdot \vec{b} = (1)(1) + (1)(-1) + (-1)(1) = 1 - 1 - 1 = -1$$
- Compute the magnitudes $|\vec{a}|$ and $|\vec{b}|$:$$|\vec{a}| = \sqrt{1^2 + 1^2 + (-1)^2} = \sqrt{3}$$$$|\vec{b}| = \sqrt{1^2 + (-1)^2 + 1^2} = \sqrt{3}$$
- Use the cosine formula:$$\cos\theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|} = \frac{-1}{\sqrt{3}\sqrt{3}} = -\frac{1}{3}$$
- Solve for $\theta$:$$\theta = \cos^{-1}\left(-\frac{1}{3}\right)$$
Final Evaluation Output: $\theta = \cos^{-1}\left(-\frac{1}{3}\right)$
Mathematical Problem Statement: Find the area of a parallelogram whose adjacent sides are given by the vectors $\vec{a} = 3\hat{i} + \hat{j} + 4\hat{k}$ and $\vec{b} = \hat{i} - \hat{j} + \hat{k}$.
Step-by-Step Algorithmic Solution Trace:
- Compute the cross product $\vec{a} \times \vec{b}$ using the determinant:$$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 1 & 4 \\ 1 & -1 & 1 \end{vmatrix}$$
- Expand the determinant:$$\vec{a} \times \vec{b} = \hat{i}(1(1) - 4(-1)) - \hat{j}(3(1) - 4(1)) + \hat{k}(3(-1) - 1(1))$$$$\vec{a} \times \vec{b} = \hat{i}(1 + 4) - \hat{j}(3 - 4) + \hat{k}(-3 - 1) = 5\hat{i} + \hat{j} - 4\hat{k}$$
- Compute the magnitude of the cross product:$$|\vec{a} \times \vec{b}| = \sqrt{5^2 + 1^2 + (-4)^2} = \sqrt{25 + 1 + 16} = \sqrt{42}$$
Final Evaluation Output: Area of the parallelogram is $\sqrt{42}$ square units.
Historical Chronicle / Epigraph: William Rowan Hamilton (1805-1865)