1. Scalars and Vectors: The Language of Physics
In physics, we quantify the world using two types of quantities: scalars and vectors.
- A scalar is a quantity that is fully described by a single number—its magnitude. Examples include mass (5 kg), temperature (293 K), and energy (100 J).
- A vector is a quantity that requires both a magnitude and a direction to be fully described. Examples include velocity (15 m/s, North), force (10 N, 30° above horizontal), and electric field.
Geometrically, we represent a vector as an arrow. The length of the arrow corresponds to the vector's magnitude, and its orientation in space represents its direction.
2. Vector Representation
Component Form
While arrows are intuitive, they are not practical for calculations. We make vectors algebraic by placing them in a Cartesian coordinate system. Any vector can be uniquely expressed as the sum of its components along the coordinate axes.
We introduce unit vectors ($\hat{i}, \hat{j}, \hat{k}$) which are dimensionless vectors of length one pointing along the positive x, y, and z axes, respectively. The vector $\vec{A}$ can then be written as:
$$ \vec{A} = A_x\hat{i} + A_y\hat{j} + A_z\hat{k} $$The scalar components ($A_x, A_y, A_z$) are the projections of $\vec{A}$ onto the axes. The magnitude of the vector is found using the Pythagorean theorem:
$$ |\vec{A}| = A = \sqrt{A_x^2 + A_y^2 + A_z^2} $$3. Vector Algebra
Addition and Subtraction
Geometrically, vectors are added "head-to-tail". To find $\vec{R} = \vec{A} + \vec{B}$, you place the tail of $\vec{B}$ at the head of $\vec{A}$. The resultant vector $\vec{R}$ is drawn from the tail of $\vec{A}$ to the head of $\vec{B}$. Algebraically, addition is simpler: we just add the corresponding components.
$$ \vec{A} + \vec{B} = (A_x + B_x)\hat{i} + (A_y + B_y)\hat{j} + (A_z + B_z)\hat{k} $$Vector subtraction is defined as the addition of a negative vector: $\vec{A} - \vec{B} = \vec{A} + (-\vec{B})$, where $-\vec{B}$ is a vector with the same magnitude as $\vec{B}$ but opposite direction.
4. Vector Products
Multiplying vectors is more complex than adding them, as there are two distinct types of products that are useful in physics.
The Dot Product (Scalar Product)
The dot product of two vectors, $\vec{A}$ and $\vec{B}$, results in a scalar. It measures the extent to which two vectors point in the same direction.
Geometric Definition: $\vec{A} \cdot \vec{B} = |\vec{A}||\vec{B}|\cos\theta$, where $\theta$ is the angle between the two vectors.
Component-wise Definition: $\vec{A} \cdot \vec{B} = A_x B_x + A_y B_y + A_z B_z$.
The physical concept of work is a perfect application: $W = \vec{F} \cdot \vec{d}$. Only the component of force in the direction of displacement does work.
The Cross Product (Vector Product)
The cross product of two vectors, $\vec{A}$ and $\vec{B}$, results in a new vector that is perpendicular to the plane containing $\vec{A}$ and $\vec{B}$.
Geometric Definition: The magnitude is $|\vec{A} \times \vec{B}| = |\vec{A}||\vec{B}|\sin\theta$. The direction is given by the right-hand rule.
Component-wise Definition (Determinant Form):
$$ \vec{A} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ A_x & A_y & A_z \\ B_x & B_y & B_z \end{vmatrix} = (A_yB_z - A_zB_y)\hat{i} - (A_xB_z - A_zB_x)\hat{j} + (A_x B_y - A_y B_x)\hat{k} $$The physical concept of torque is a key application: $\vec{\tau} = \vec{r} \times \vec{F}$.
5. Introduction to Vector Calculus
When a vector quantity changes with respect to a parameter like time, we can use calculus.
Derivatives: The derivative of a vector is found by differentiating each component. If $\vec{r}(t)$ is the position vector, then:
$$ \vec{v}(t) = \frac{d\vec{r}}{dt} = \frac{dx}{dt}\hat{i} + \frac{dy}{dt}\hat{j} + \frac{dz}{dt}\hat{k} $$The Gradient Operator ($\nabla$): In multivariable calculus, the vector differential operator $\nabla$ ("del") is fundamental. In Cartesian coordinates:
$$ \nabla = \hat{i}\frac{\partial}{\partial x} + \hat{j}\frac{\partial}{\partial y} + \hat{k}\frac{\partial}{\partial z} $$This operator can act on scalar and vector fields to give physically meaningful results:
- Gradient of a scalar field ($f$): $\nabla f$ is a vector field that points in the direction of the greatest rate of increase of $f$. (e.g., Force from Potential Energy: $\vec{F} = -\nabla U$).
- Divergence of a vector field ($\vec{V}$): $\nabla \cdot \vec{V}$ is a scalar field that measures the "outflow" or "source strength" at a point. (e.g., Gauss's Law for electricity: $\nabla \cdot \vec{E} = \rho/\epsilon_0$).
- Curl of a vector field ($\vec{V}$): $\nabla \times \vec{V}$ is a vector field that measures the "rotation" or "circulation" at a point. (e.g., Faraday's Law of Induction: $\nabla \times \vec{E} = -\frac{\partial \vec{B}}{\partial t}$).
6. Problem Set
Problem 1 (Intermediate):
Given vectors $\vec{A} = 2\hat{i} + 3\hat{j}$ and $\vec{B} = -1\hat{i} + 2\hat{j}$, find the magnitude and direction of the resultant vector $\vec{R} = \vec{A} + \vec{B}$.
Magnitude: $|\vec{R}| = \sqrt{1^2 + 5^2} = \sqrt{26} \approx 5.1$.
Direction: $\theta = \arctan(R_y/R_x) = \arctan(5/1) \approx 78.7^\circ$ above the positive x-axis.
Problem 2 (Intermediate):
Find the angle $\theta$ between the vectors $\vec{A} = 3\hat{i} + 4\hat{j}$ and $\vec{B} = 5\hat{i} + 12\hat{j}$.
$\vec{A} \cdot \vec{B} = (3)(5) + (4)(12) = 15 + 48 = 63$.
$|\vec{A}| = \sqrt{3^2+4^2} = 5$.
$|\vec{B}| = \sqrt{5^2+12^2} = 13$.
$\cos\theta = \frac{\vec{A} \cdot \vec{B}}{|\vec{A}||\vec{B}|} = \frac{63}{5 \cdot 13} = \frac{63}{65}$.
$\theta = \arccos(63/65) \approx 14.25^\circ$.
Problem 3 (Intermediate):
Find a unit vector perpendicular to both $\vec{A} = \hat{i} - \hat{j} + 2\hat{k}$ and $\vec{B} = 2\hat{i} + \hat{j} - 3\hat{k}$.
$\vec{C} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 2 \\ 2 & 1 & -3 \end{vmatrix} = \hat{i}(3-2) - \hat{j}(-3-4) + \hat{k}(1-(-2)) = \hat{i} + 7\hat{j} + 3\hat{k}$.
The magnitude is $|\vec{C}| = \sqrt{1^2+7^2+3^2} = \sqrt{1+49+9} = \sqrt{59}$.
The unit vector is $\hat{C} = \frac{\vec{C}}{|\vec{C}|} = \frac{1}{\sqrt{59}}(\hat{i} + 7\hat{j} + 3\hat{k})$.
Problem 4 (Intermediate):
Calculate the work done by a force $\vec{F} = (2\hat{i} - 4\hat{j})\,$N in moving an object through a displacement $\vec{d} = (3\hat{i} + 3\hat{j})\,$m.
$W = (2)(3) + (-4)(3) = 6 - 12 = -6\,$J.
Problem 5 (Intermediate):
Find the projection of vector $\vec{A} = 2\hat{i} - 3\hat{j} + 6\hat{k}$ onto vector $\vec{B} = \hat{i} + 2\hat{j} + 2\hat{k}$.
$\vec{A} \cdot \vec{B} = (2)(1) + (-3)(2) + (6)(2) = 2 - 6 + 12 = 8$.
$|\vec{B}| = \sqrt{1^2+2^2+2^2} = \sqrt{9} = 3$.
$A_B = 8/3$. The vector projection is this scalar times the unit vector of B: $\vec{A}_B = A_B \hat{B} = \frac{8}{3} \frac{\vec{B}}{|\vec{B}|} = \frac{8}{9}(\hat{i} + 2\hat{j} + 2\hat{k})$.
Problem 6 (Advanced):
A particle's position is given by $\vec{r}(t) = (R\cos(\omega t))\hat{i} + (R\sin(\omega t))\hat{j}$. Show that the acceleration vector is always directed towards the origin and has a magnitude of $\omega^2 R$.
Velocity: $\vec{v}(t) = \frac{d\vec{r}}{dt} = (-R\omega\sin(\omega t))\hat{i} + (R\omega\cos(\omega t))\hat{j}$.
Acceleration: $\vec{a}(t) = \frac{d\vec{v}}{dt} = (-R\omega^2\cos(\omega t))\hat{i} - (R\omega^2\sin(\omega t))\hat{j}$.
We can factor out $-\omega^2$: $\vec{a}(t) = -\omega^2 (R\cos(\omega t)\hat{i} + R\sin(\omega t)\hat{j}) = -\omega^2\vec{r}(t)$.
This shows the acceleration vector $\vec{a}$ is always in the opposite direction to the position vector $\vec{r}$, meaning it always points towards the origin.
Its magnitude is $|\vec{a}| = |-\omega^2\vec{r}| = \omega^2 |\vec{r}| = \omega^2 \sqrt{(R\cos\omega t)^2 + (R\sin\omega t)^2} = \omega^2 R$.
Problem 7 (Advanced):
Prove the vector triple product identity, also known as the BAC-CAB rule: $\vec{A} \times (\vec{B} \times \vec{C}) = \vec{B}(\vec{A} \cdot \vec{C}) - \vec{C}(\vec{A} \cdot \vec{B})$.
First, $\vec{B} \times \vec{C} = (B_yC_z - B_zC_y)\hat{i} + \dots$
Then, the x-component of $\vec{A} \times (\vec{B} \times \vec{C})$ is $A_y(\vec{B} \times \vec{C})_z - A_z(\vec{B} \times \vec{C})_y$.
$[\vec{A} \times (\vec{B} \times \vec{C})]_x = A_y(B_x C_y - B_y C_x) - A_z(B_z C_x - B_x C_z)$.
$= A_yB_xC_y - A_yB_yC_x - A_zB_zC_x + A_zB_xC_z$.
Now for the right side: $[\vec{B}(\vec{A} \cdot \vec{C}) - \vec{C}(\vec{A} \cdot \vec{B})]_x = B_x(A_x C_x + A_y C_y + A_z C_z) - C_x(A_x B_x + A_y B_y + A_z B_z)$.
$= A_xB_xC_x + A_yB_xC_y + A_zB_xC_z - A_xB_xC_x - A_yB_yC_x - A_zB_zC_x$.
$= A_yB_xC_y + A_zB_xC_z - A_yB_yC_x - A_zB_zC_x$.
The expressions match. The y and z components can be shown to match similarly.
Problem 8 (Advanced):
If a vector $\vec{v}(t)$ has a constant magnitude, show that its derivative $\frac{d\vec{v}}{dt}$ is perpendicular to $\vec{v}$.
This also means $|\vec{v}|^2 = \vec{v} \cdot \vec{v} = c^2$.
Now, differentiate this equation with respect to time $t$: $\frac{d}{dt}(\vec{v} \cdot \vec{v}) = \frac{d}{dt}(c^2) = 0$.
Using the product rule for dot products: $\frac{d\vec{v}}{dt} \cdot \vec{v} + \vec{v} \cdot \frac{d\vec{v}}{dt} = 0$.
Since the dot product is commutative, this is $2\vec{v} \cdot \frac{d\vec{v}}{dt} = 0$.
Therefore, $\vec{v} \cdot \frac{d\vec{v}}{dt} = 0$.
The dot product of two non-zero vectors is zero if and only if they are perpendicular. Thus, the derivative of a vector of constant magnitude is always perpendicular to the vector itself. This is why centripetal acceleration is perpendicular to velocity in uniform circular motion.
Problem 9 (Irodov-like):
Use vectors to prove that the diagonals of a rhombus are perpendicular.
A rhombus is a parallelogram with sides of equal length, so $|\vec{A}| = |\vec{B}|$.
One diagonal, $\vec{D}_1$, is the vector sum of the sides: $\vec{D}_1 = \vec{A} + \vec{B}$.
The other diagonal, $\vec{D}_2$, connects the heads of the two vectors, so it is their difference: $\vec{D}_2 = \vec{A} - \vec{B}$.
To check if the diagonals are perpendicular, we take their dot product:
$\vec{D}_1 \cdot \vec{D}_2 = (\vec{A} + \vec{B}) \cdot (\vec{A} - \vec{B})$.
Using the distributive property of the dot product: $\vec{D}_1 \cdot \vec{D}_2 = \vec{A}\cdot\vec{A} - \vec{A}\cdot\vec{B} + \vec{B}\cdot\vec{A} - \vec{B}\cdot\vec{B}$.
Since the dot product is commutative ($\vec{A}\cdot\vec{B} = \vec{B}\cdot\vec{A}$), the middle terms cancel.
$\vec{D}_1 \cdot \vec{D}_2 = \vec{A}\cdot\vec{A} - \vec{B}\cdot\vec{B} = |\vec{A}|^2 - |\vec{B}|^2$.
Because it is a rhombus, we know $|\vec{A}| = |\vec{B}|$, so $|\vec{A}|^2 - |\vec{B}|^2 = 0$.
Since $\vec{D}_1 \cdot \vec{D}_2 = 0$, the diagonals are perpendicular.
Problem 10 (Irodov-like):
A boat moves relative to the water with a velocity that is $n=2.0$ times less than the river flow velocity. At what angle to the stream direction must the boat move to minimize drifting?
Let $|\vec{v}_{flow}| = u$. Then $|\vec{v}_{boat,water}| = u/n$. Let the river flow along the x-axis, so $\vec{v}_{flow} = u\hat{i}$.
Let the boat move at an angle $\alpha$ to the y-axis (across the stream). So its angle to the flow (x-axis) is $90+\alpha$. Let's use the angle $\theta$ with the flow direction (x-axis) for generality. $\vec{v}_{boat,water} = \frac{u}{n}(\cos\theta\hat{i} + \sin\theta\hat{j})$.
$\vec{v}_{boat,ground} = (u + \frac{u}{n}\cos\theta)\hat{i} + (\frac{u}{n}\sin\theta)\hat{j}$.
Drift is the downstream displacement, which depends on the x-component of the ground velocity. Time to cross is $t = W/v_y$ where $W$ is width. Drift $D = v_x t$. We want to minimize drift. In this phrasing, "drifting" means being carried downstream. To minimize this, the boat must point upstream to counteract the flow as much as possible. The downstream velocity component is $v_x = u + \frac{u}{n}\cos\theta$. To minimize this, we need $\cos\theta$ to be as negative as possible, which is -1. This happens when $\theta = 180^\circ$. The boat must move directly upstream. This minimizes downstream velocity, hence drift. Let's re-read. "Minimize drifting" might mean "minimize downstream distance while crossing". Let's say the boat wants to cross a river of width $W$. Time to cross is $t = \frac{W}{v_y} = \frac{W}{(u/n)\sin\theta}$. The drift distance is $D = v_x t = (u + \frac{u}{n}\cos\theta) \frac{W}{(u/n)\sin\theta} = W(n\csc\theta + \cot\theta)$. To minimize $D$, we take the derivative with respect to $\theta$ and set it to 0. $\frac{dD}{d\theta} = W(-n\csc\theta\cot\theta - \csc^2\theta) = 0$. $-W\csc\theta(n\cot\theta + \csc\theta) = 0$. Since $\csc\theta \ne 0$: $n\frac{\cos\theta}{\sin\theta} + \frac{1}{\sin\theta} = 0 \implies n\cos\theta = -1 \implies \cos\theta = -1/n$. Given $n=2$, $\cos\theta = -1/2$. This means $\theta = 120^\circ$. The boat must move at an angle of 120° with respect to the downstream direction.