1. Scalars and Vectors
Scalar Quantity
A Scalar quantity is defined only by its magnitude (a number and a unit). It follows ordinary arithmetic.
Examples: Mass ($\text{kg}$), Temperature ($\text{K}$), Time ($\text{s}$).
Vector Quantity
A Vector quantity is defined by both its magnitude and direction in space. It follows specialized vector algebra.
Examples: Force ($\vec{F}$), Velocity ($\vec{v}$), Displacement ($\Delta \vec{r}$).
2. Vector Algebra and Representation
Vector Component Representation
A vector $\vec{A}$ can be expressed in Cartesian coordinates using unit vectors ($\hat{i}$, $\hat{j}$, $\hat{k}$).
The magnitude of $\vec{A}$ is found using the Pythagorean theorem:
Vector Addition and Subtraction
Vectors are added by combining their respective components:
$$\vec{R} = \vec{A} + \vec{B} = (A_x + B_x)\hat{i} + (A_y + B_y)\hat{j} + (A_z + B_z)\hat{k}$$Vector subtracted by $\vec{B}$ is equivalent to adding $-\vec{B}$:
$$\vec{D} = \vec{A} - \vec{B} = (A_x - B_x)\hat{i} + (A_y - B_y)\hat{j} + (A_z - B_z)\hat{k}$$3. Vector Products
3.1. The Dot Product (Scalar Product)
The dot product of two vectors, $\vec{A}$ and $\vec{B}$, yields a scalar quantity. It measures the extent to which the vectors are parallel. The geometric definition involves the angle $\theta$ between them.
The dot product is used to find the angle between two vectors.
Physics Application: Work
Work is a scalar quantity that is the dot product of Force and Displacement.
$$W = \vec{F} \cdot \vec{d} = |\vec{F}| |\vec{d}| \cos \theta$$3.2. The Cross Product (Vector Product)
The cross product of two vectors, $\vec{A}$ and $\vec{B}$, yields a new vector $\vec{C}$, which is perpendicular to both $\vec{A}$ and $\vec{B}$. The direction of $\vec{C}$ is determined by the Right-Hand Rule.
Physics Application: Torque
Torque is a vector quantity that is the cross product of the position vector ($\vec{r}$) and the Force ($\vec{F}$).
$$\vec{\tau} = \vec{r} \times \vec{F}$$4. The $\nabla$ Operator (Del or Nabla)
The Del operator, $\nabla$ (nabla), is a differential vector operator used in vector calculus. It acts on scalar and vector fields.
4.1. The Gradient of a Scalar Field ($\text{grad } f$)
The gradient of a scalar function $f(x, y, z)$ (like Temperature $T$) is a vector that points in the direction of the greatest rate of increase of the function at that point.
Physics Application: Force
The Force vector ($\vec{F}$) is the negative gradient of the potential energy scalar field ($U$): $\vec{F} = -\nabla U$.
4.2. The Divergence of a Vector Field ($\text{div } \vec{F}$)
The divergence of a vector field $\vec{F}$ is a scalar quantity that measures the magnitude of the vector field's source or sink (flow or flux per unit volume).
Physics Application: Gauss's Law
In its differential form, Gauss's Law relates the divergence of the electric field ($\vec{E}$) to the charge density ($\rho$): $\nabla \cdot \vec{E} = \frac{\rho}{\epsilon_0}$.
4.3. The Curl of a Vector Field ($\text{curl } \vec{F}$)
The curl of a vector field $\vec{F}$ is a vector quantity that measures the rotation (or vorticity) of the field at a given point.
Physics Application: Magnetic Field
The curl of the magnetic field ($\vec{B}$) is related to the current density ($\vec{J}$) in Ampère's Law (ignoring displacement current): $\nabla \times \vec{B} = \mu_0 \vec{J}$.
5. Key Identities and Proofs
5.1. The Dot Product is Commutative
Proof:
From the geometric definition, the angle $\theta$ between $\vec{A}$ and $\vec{B}$ is the same as the angle between $\vec{B}$ and $\vec{A}$. Since $\cos \theta = \cos (-\theta)$, the order does not matter:
$$\vec{A} \cdot \vec{B} = |\vec{A}| |\vec{B}| \cos \theta$$ $$\vec{B} \cdot \vec{A} = |\vec{B}| |\vec{A}| \cos \theta$$Therefore, $\vec{A} \cdot \vec{B} = \vec{B} \cdot \vec{A}$.
5.2. The Curl of a Gradient is Zero (Conservative Field Proof)
Proof:
From Clairaut's Theorem (equality of mixed partial derivatives), if $f$ has continuous second partial derivatives, then:
$$\frac{\partial^2 f}{\partial x \partial y} = \frac{\partial^2 f}{\partial y \partial x}$$The curl of the gradient is given by $\nabla \times (\nabla f)$:
The x-component of the curl is:
$$\text{Curl}_x = \frac{\partial}{\partial y}\left(\frac{\partial f}{\partial z}\right) - \frac{\partial}{\partial z}\left(\frac{\partial f}{\partial y}\right) = \frac{\partial^2 f}{\partial y \partial z} - \frac{\partial^2 f}{\partial z \partial y}$$Since $\frac{\partial^2 f}{\partial y \partial z} = \frac{\partial^2 f}{\partial z \partial y}$, the x-component is zero. This applies to the y and z components as well.
Using $\nabla \times (\nabla f) = \vec{0}$ is a fundamental test for a conservative field (e.g., a gravitational or electrostatic field).
6. Practice Problems: Statements
Attempt all problems below before checking the solutions on the following pages.
Intermediate Level (5 Problems)
- Given $\vec{A} = \hat{i} - 2\hat{j} + 3\hat{k}$ and $\vec{B} = 2\hat{i} + 4\hat{j} - \hat{k}$. Calculate the resultant vector $\vec{R} = 2\vec{A} + 3\vec{B}$.
- A displacement vector $\vec{d}$ has an x-component of $12\text{ m}$ and a magnitude of $15\text{ m}$. Find the possible values for the y-component $d_y$ and the angle $\theta$ $\text{(in degrees)}$ the vector makes with the positive x-axis.
- A force $\vec{F} = (5\hat{i} - 2\hat{j})\text{ N}$ causes a displacement $\vec{d} = (3\hat{i} + 4\hat{j})\text{ m}$. Calculate the work done $W = \vec{F} \cdot \vec{d}$ (a scalar quantity).
- Find the magnitude of the torque vector $\vec{\tau} = \vec{r} \times \vec{F}$, where the position vector is $\vec{r} = 4\hat{i} + \hat{k}$ and the force vector is $\vec{F} = 2\hat{j} - 3\hat{k}$.
- Calculate the scalar projection of vector $\vec{A} = 6\hat{i} + 2\hat{j}$ onto vector $\vec{B} = \hat{i} - 3\hat{j}$. The projection is given by $\text{proj}_{\vec{B}} \vec{A} = \frac{\vec{A} \cdot \vec{B}}{|\vec{B}|}$.
Advanced Level (3 Problems)
- Three points in space are given by $P(1, 0, 0)$, $Q(0, 2, 0)$, and $R(0, 0, 3)$. Calculate the area of the triangle formed by these three points using the cross product.
- The velocity of a fluid is described by the vector field $\vec{v}(x, y, z) = (x^2 + y)\hat{i} + (z^2 + c x)\hat{j} + (y^2 + z)\hat{k}$. Find the value of the constant $c$ such that the field is irrotational (meaning $\nabla \times \vec{v} = \vec{0}$).
- The temperature $T$ (in $\text{K}$) in a room is given by the scalar field $T(x, y, z) = 2xy^2 + \sin(\pi z)$. Find the vector direction in which the temperature increases most rapidly at the point $P(2, 1, 0)$, and state the maximum rate of increase.
Irodov-Like / Conceptual (2 Problems)
- Prove the following vector identity for a scalar function $\phi$ and a vector field $\vec{A}$: $$\nabla \times (\phi \vec{A}) = (\nabla \phi) \times \vec{A} + \phi (\nabla \times \vec{A})$$
- Consider the position vector field $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$. Calculate the Divergence ($\nabla \cdot \vec{r}$) and the Curl ($\nabla \times \vec{r}$) of this field. Explain the physical meaning of the results.
6. Practice Problems: Solutions
Detailed solutions for all 10 problems.
Solution P1 (Intermediate)
Given $\vec{A} = \hat{i} - 2\hat{j} + 3\hat{k}$ and $\vec{B} = 2\hat{i} + 4\hat{j} - \hat{k}$.
1. Calculate $2\vec{A}$ and $3\vec{B}$:
$$2\vec{A} = 2\hat{i} - 4\hat{j} + 6\hat{k}$$ $$3\vec{B} = 6\hat{i} + 12\hat{j} - 3\hat{k}$$2. Calculate resultant $\vec{R} = 2\vec{A} + 3\vec{B}$:
$$\vec{R} = (2+6)\hat{i} + (-4+12)\hat{j} + (6-3)\hat{k} = 8\hat{i} + 8\hat{j} + 3\hat{k}$$Result: $\vec{R} = 8\hat{i} + 8\hat{j} + 3\hat{k}$
Solution P2 (Intermediate)
Given $d_x = 12\text{ m}$ and $|\vec{d}| = 15\text{ m}$.
1. Find $d_y$ using the magnitude formula $|\vec{d}| = \sqrt{d_x^2 + d_y^2}$:
$$15^2 = 12^2 + d_y^2 \implies 225 = 144 + d_y^2$$ $$d_y^2 = 81 \implies d_y = \pm 9\text{ m}$$2. Find the angle $\theta$: $\cos \theta = \frac{d_x}{|\vec{d}|} = \frac{12}{15} = 0.8$.
$$\theta = \arccos(0.8) \approx 36.87^\circ$$Result: $d_y = \pm 9\text{ m}$, $\theta \approx 36.87^\circ$.
Solution P3 (Intermediate)
Work is the dot product $W = \vec{F} \cdot \vec{d}$.
$$\vec{F} = 5\hat{i} - 2\hat{j}, \quad \vec{d} = 3\hat{i} + 4\hat{j}$$ $$W = (5)(3) + (-2)(4) = 15 - 8 = 7$$Result: $W = 7\text{ Joules}$ (The unit of work is the Joule, J).
Solution P4 (Intermediate)
Torque is $\vec{\tau} = \vec{r} \times \vec{F}$. $\vec{r} = 4\hat{i} + 0\hat{j} + 1\hat{k}$, $\vec{F} = 0\hat{i} + 2\hat{j} - 3\hat{k}$.
$$\vec{\tau} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & 0 & 1 \\ 0 & 2 & -3 \end{vmatrix}$$ $$\vec{\tau} = \hat{i}((0)(-3) - (1)(2)) - \hat{j}((4)(-3) - (1)(0)) + \hat{k}((4)(2) - (0)(0))$$ $$\vec{\tau} = -2\hat{i} - (-12)\hat{j} + 8\hat{k} = -2\hat{i} + 12\hat{j} + 8\hat{k}$$Magnitude $|\vec{\tau}| = \sqrt{(-2)^2 + 12^2 + 8^2} = \sqrt{4 + 144 + 64} = \sqrt{212}$$
Result: $|\vec{\tau}| = \sqrt{212} \approx 14.56$
Solution P5 (Intermediate)
The scalar projection is $\text{proj}_{\vec{B}} \vec{A} = \frac{\vec{A} \cdot \vec{B}}{|\vec{B}|}$.
1. Dot Product: $\vec{A} \cdot \vec{B} = (6)(1) + (2)(-3) = 6 - 6 = 0$
2. Magnitude of $\vec{B}$: $|\vec{B}| = \sqrt{1^2 + (-3)^2} = \sqrt{10}$
3. Projection: $\text{proj}_{\vec{B}} \vec{A} = \frac{0}{\sqrt{10}} = 0$
Result: The scalar projection is $0$. (This means the vectors are orthogonal/perpendicular.)
Advanced Level Solutions
Solution P6 (Advanced)
Points $P(1, 0, 0)$, $Q(0, 2, 0)$, $R(0, 0, 3)$. The area is $\frac{1}{2} |\vec{PQ} \times \vec{PR}|$.
1. Find vectors $\vec{PQ}$ and $\vec{PR}$:
$$\vec{PQ} = Q - P = (0-1)\hat{i} + (2-0)\hat{j} + (0-0)\hat{k} = -\hat{i} + 2\hat{j}$$ $$\vec{PR} = R - P = (0-1)\hat{i} + (0-0)\hat{j} + (3-0)\hat{k} = -\hat{i} + 3\hat{k}$$2. Calculate the Cross Product $\vec{C} = \vec{PQ} \times \vec{PR}$:
$$\vec{C} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 2 & 0 \\ -1 & 0 & 3 \end{vmatrix} = \hat{i}(6-0) - \hat{j}(-3-0) + \hat{k}(0-(-2)) = 6\hat{i} + 3\hat{j} + 2\hat{k}$$3. Find the Magnitude and Area:
$$|\vec{C}| = \sqrt{6^2 + 3^2 + 2^2} = \sqrt{36 + 9 + 4} = \sqrt{49} = 7$$ $$\text{Area} = \frac{1}{2} |\vec{C}| = \frac{7}{2} = 3.5$$Result: Area $= 3.5\text{ units}^2$
Solution P7 (Advanced)
For a vector field $\vec{v} = v_x \hat{i} + v_y \hat{j} + v_z \hat{k}$ to be irrotational, its curl must be zero: $\nabla \times \vec{v} = \vec{0}$.
The components of $\nabla \times \vec{v}$ are:
$$(\nabla \times \vec{v})_z = \frac{\partial v_y}{\partial x} - \frac{\partial v_x}{\partial y}$$ $$(\nabla \times \vec{v})_y = \frac{\partial v_x}{\partial z} - \frac{\partial v_z}{\partial x}$$ $$(\nabla \times \vec{v})_x = \frac{\partial v_z}{\partial y} - \frac{\partial v_y}{\partial z}$$The curl is calculated as:
$$\nabla \times \vec{v} = \hat{i}\left(\frac{\partial (y^2+z)}{\partial y} - \frac{\partial (z^2+cx)}{\partial z}\right) - \hat{j}\left(\frac{\partial (y^2+z)}{\partial x} - \frac{\partial (x^2+y)}{\partial z}\right) + \hat{k}\left(\frac{\partial (z^2+cx)}{\partial x} - \frac{\partial (x^2+y)}{\partial y}\right)$$ $$\nabla \times \vec{v} = \hat{i}(2y - 2z) - \hat{j}(0 - 0) + \hat{k}(c - 1)$$For the field to be irrotational, all components must be zero. Therefore, $c - 1 = 0$.
Result: $c = 1$.
Solution P8 (Advanced)
The direction of maximum increase is given by the Gradient $\nabla T$ evaluated at $P(2, 1, 0)$. The maximum rate of increase is the magnitude $|\nabla T|$.
1. Calculate Gradient $\nabla T$:
$$\nabla T = \frac{\partial T}{\partial x}\hat{i} + \frac{\partial T}{\partial y}\hat{j} + \frac{\partial T}{\partial z}\hat{k}$$ $$\frac{\partial T}{\partial x} = 2y^2, \quad \frac{\partial T}{\partial y} = 4xy, \quad \frac{\partial T}{\partial z} = \pi \cos(\pi z)$$2. Evaluate $\nabla T$ at $P(2, 1, 0)$:
$$\nabla T |_{(2, 1, 0)} = 2(1)^2\hat{i} + 4(2)(1)\hat{j} + \pi \cos(0)\hat{k}$$ $$\nabla T |_{(2, 1, 0)} = 2\hat{i} + 8\hat{j} + \pi\hat{k}$$3. Maximum Rate (Magnitude):
$$|\nabla T| = \sqrt{2^2 + 8^2 + \pi^2} = \sqrt{4 + 64 + 9.87} \approx \sqrt{77.87} \approx 8.82$$Result: Direction is $2\hat{i} + 8\hat{j} + \pi\hat{k}$. Maximum rate is $\approx 8.82\text{ K}/\text{unit distance}$.
Irodov-Like / Conceptual Solutions
Solution P9 (Irodov-Like / Conceptual)
Prove $\nabla \times (\phi \vec{A}) = (\nabla \phi) \times \vec{A} + \phi (\nabla \times \vec{A})$. We look at the $x$-component:
The $x$-component of the LHS, $\nabla \times (\phi \vec{A})$, is:
$$(\text{LHS})_x = \frac{\partial}{\partial y}(\phi A_z) - \frac{\partial}{\partial z}(\phi A_y)$$Using the product rule on each term:
$$(\text{LHS})_x = \left(\frac{\partial \phi}{\partial y} A_z + \phi \frac{\partial A_z}{\partial y}\right) - \left(\frac{\partial \phi}{\partial z} A_y + \phi \frac{\partial A_y}{\partial z}\right)$$ $$\text{(LHS)}_x = \left(\frac{\partial \phi}{\partial y} A_z - \frac{\partial \phi}{\partial z} A_y\right) + \phi \left(\frac{\partial A_z}{\partial y} - \frac{\partial A_y}{\partial z}\right)$$The first parenthesis is the $x$-component of the cross product $(\nabla \phi) \times \vec{A}$. The second parenthesis is the $x$-component of $\nabla \times \vec{A}$.
Thus, $(\text{LHS})_x = [(\nabla \phi) \times \vec{A}]_x + [\phi (\nabla \times \vec{A})]_x$. Since this holds for the $x$-component, and similarly for the $y$ and $z$ components, the identity is proven.
Result: Identity Proven (Product Rule for Curl)
Solution P10 (Irodov-Like / Conceptual)
The position vector field is $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$.
1. **Divergence** ($\nabla \cdot \vec{r}$):
$$\nabla \cdot \vec{r} = \frac{\partial}{\partial x}(x) + \frac{\partial}{\partial y}(y) + \frac{\partial}{\partial z}(z) = 1 + 1 + 1 = 3$$Divergence Result: $\nabla \cdot \vec{r} = 3$
2. **Curl** ($\nabla \times \vec{r}$):
$$\nabla \times \vec{r} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ x & y & z \end{vmatrix}$$ $$\nabla \times \vec{r} = \hat{i}\left(\frac{\partial z}{\partial y} - \frac{\partial y}{\partial z}\right) - \hat{j}\left(\frac{\partial z}{\partial x} - \frac{\partial x}{\partial z}\right) + \hat{k}\left(\frac{\partial y}{\partial x} - \frac{\partial x}{\partial y}\right) = \hat{i}(0-0) - \hat{j}(0-0) + \hat{k}(0-0) = \vec{0}$$Curl Result: $\nabla \times \vec{r} = \vec{0}$
Physical Meaning:
- Divergence is $3$: A positive divergence indicates a source or outward flow from every point. For the position vector, this means the vector field constantly radiates away from the origin in all directions.
- Curl is $\vec{0}$: A zero curl indicates the field is irrotational (or conservative). There is no "swirling" or rotational component to the position vector field.