Introduction to Vectors: Direction Matters!
Welcome to the chapter on Vectors! Vectors might seem abstract, but they are incredibly practical. Whenever you describe motion—like a plane flying at a certain speed in a certain direction, or a force pushing an object—you need vectors. Unlike scalars (which only have magnitude, like temperature or mass), vectors have both magnitude (size) and direction.
This chapter bridges pure geometry and algebra, giving you tools to solve complex 2D and 3D problems easily. Don't worry if visualizing 3D space seems tricky at first; we will use simple algebraic rules to handle the heavy lifting!
1. Defining and Describing Vectors
1.1 Notation and Representation
Vectors can be written in two common ways. You must be comfortable with both, though column vectors are often preferred in calculations.
Column Vector Notation (Preferred in exams)
In 2D or 3D, we list the components representing movement along the $x$, $y$, and $z$ axes:
$$ \mathbf{a} = \begin{pmatrix} 3 \\ -2 \end{pmatrix} \quad \text{or} \quad \mathbf{b} = \begin{pmatrix} 4 \\ 1 \\ 5 \end{pmatrix} $$
$\mathbf{i}, \mathbf{j}, \mathbf{k}$ Notation
These are unit vectors (vectors of magnitude 1) aligned with the positive $x$, $y$, and $z$ axes, respectively.
$$ \mathbf{a} = 3\mathbf{i} - 2\mathbf{j} \quad \text{or} \quad \mathbf{b} = 4\mathbf{i} + 1\mathbf{j} + 5\mathbf{k} $$
Analogy: Imagine a journey. The vector $\mathbf{b}$ above means "Go 4 units forward, 1 unit right, and 5 units up."
1.2 The Magnitude of a Vector
The magnitude of a vector is its length, often called its modulus. We use Pythagoras' Theorem to find it.
If $\mathbf{a} = \begin{pmatrix} x \\ y \end{pmatrix}$, the magnitude is denoted $|\mathbf{a}|$.
$$ |\mathbf{a}| = \sqrt{x^2 + y^2} $$
In 3D, if $\mathbf{b} = \begin{pmatrix} x \\ y \\ z \end{pmatrix}$, the magnitude is:
$$ |\mathbf{b}| = \sqrt{x^2 + y^2 + z^2} $$
Quick Review: The magnitude calculation is just an extension of the distance formula in 3D space.
Did you know? A unit vector is any vector whose magnitude is exactly 1. You can turn any vector $\mathbf{a}$ into a unit vector $\hat{\mathbf{a}}$ by dividing it by its magnitude: $$\hat{\mathbf{a}} = \frac{1}{|\mathbf{a}|}\mathbf{a}$$
2. Algebraic Operations and Geometric Interpretation
2.1 Vector Addition and Subtraction
Adding or subtracting vectors is easy: you simply add or subtract the corresponding components.
If $\mathbf{a} = \begin{pmatrix} 3 \\ 1 \\ 4 \end{pmatrix}$ and $\mathbf{b} = \begin{pmatrix} 2 \\ 5 \\ -1 \end{pmatrix}$:
Addition:
$$ \mathbf{a} + \mathbf{b} = \begin{pmatrix} 3+2 \\ 1+5 \\ 4+(-1) \end{pmatrix} = \begin{pmatrix} 5 \\ 6 \\ 3 \end{pmatrix} $$
Subtraction:
$$ \mathbf{a} - \mathbf{b} = \begin{pmatrix} 3-2 \\ 1-5 \\ 4-(-1) \end{pmatrix} = \begin{pmatrix} 1 \\ -4 \\ 5 \end{pmatrix} $$
Geometric Interpretation
Vector addition follows the Triangle Rule: to find $\mathbf{a} + \mathbf{b}$, place the tail of $\mathbf{b}$ at the head of $\mathbf{a}$. The resultant vector starts at the tail of $\mathbf{a}$ and ends at the head of $\mathbf{b}$.
2.2 Multiplication by a Scalar
Multiplying a vector by a scalar (a normal number) changes its magnitude, but not its direction (unless the scalar is negative, which reverses the direction).
If $\mathbf{a} = \begin{pmatrix} 3 \\ 1 \\ 4 \end{pmatrix}$ and $k=2$, then:
$$ 2\mathbf{a} = 2 \begin{pmatrix} 3 \\ 1 \\ 4 \end{pmatrix} = \begin{pmatrix} 6 \\ 2 \\ 8 \end{pmatrix} $$
Key Takeaway: If two vectors, $\mathbf{u}$ and $\mathbf{v}$, are parallel, then $\mathbf{u} = k\mathbf{v}$ for some scalar $k$.
2.3 Position Vectors and Displacement
A position vector describes the location of a point relative to the origin $O=(0, 0, 0)$. The position vector of point $A$ is $\vec{OA}$, usually written as $\mathbf{a}$.
The vector representing the displacement from point $A$ to point $B$ (the vector $\vec{AB}$) is found using the crucial formula:
$$ \mathbf{AB} = \mathbf{b} - \mathbf{a} \quad \text{or} \quad \vec{AB} = \vec{OB} - \vec{OA} $$
Memory Aid: "End minus Start." To go from A to B, you take the position of B and subtract the position of A.
2.4 Distance Between Two Points
If points $A$ and $B$ have position vectors $\mathbf{a}$ and $\mathbf{b}$, the distance between them is simply the magnitude of the displacement vector $\vec{AB}$.
$$ \text{Distance } AB = |\vec{AB}| = |\mathbf{b} - \mathbf{a}| $$
Example: If $A(1, 0, 5)$ and $B(4, 3, 1)$, then $\vec{AB} = \begin{pmatrix} 4-1 \\ 3-0 \\ 1-5 \end{pmatrix} = \begin{pmatrix} 3 \\ 3 \\ -4 \end{pmatrix}$. The distance is $|\vec{AB}| = \sqrt{3^2 + 3^2 + (-4)^2} = \sqrt{9+9+16} = \sqrt{34}$.
3. The Vector Equation of a Line
In A-Level Mathematics, we describe a straight line not with $y=mx+c$, but using vectors. This works perfectly in 3D space, where the Cartesian form is much harder.
3.1 The General Form
The vector equation of a line $L$ passing through a fixed point $A$ (with position vector $\mathbf{a}$) and parallel to a direction vector $\mathbf{b}$ is:
$$ \mathbf{r} = \mathbf{a} + t\mathbf{b} $$
- $\mathbf{r}$: The position vector of any point on the line (the variable).
- $\mathbf{a}$: A fixed point on the line (the "starting point").
- $\mathbf{b}$: The direction vector (tells the line where to go).
- $t$: The parameter (a scalar that tells you how far you travel in direction $\mathbf{b}$).
Example: A line passing through $(1, 0, 2)$ in the direction $\begin{pmatrix} -1 \\ 2 \\ 3 \end{pmatrix}$.
$$ \mathbf{r} = \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 1 \\ 0 \\ 2 \end{pmatrix} + t \begin{pmatrix} -1 \\ 2 \\ 3 \end{pmatrix} $$
3.2 Intersection of Lines (2D and 3D)
To find if two lines, $L_1$ and $L_2$, intersect, we set their vector equations equal to each other.
If $L_1: \mathbf{r}_1 = \mathbf{a}_1 + t\mathbf{b}_1$ and $L_2: \mathbf{r}_2 = \mathbf{a}_2 + s\mathbf{b}_2$ (note: use a different parameter, like $s$, for the second line!), then at the intersection point, $\mathbf{r}_1 = \mathbf{r}_2$.
Step-by-Step for Intersection
- Equate Components: Set the $\mathbf{i}$, $\mathbf{j}$, and $\mathbf{k}$ components equal to generate three simultaneous equations involving the two parameters, $t$ and $s$.
- Solve for Parameters: Use any two of the equations (e.g., the $x$ and $y$ components) to solve for $t$ and $s$.
- Check Consistency (The Crucial Step): Substitute the calculated values of $t$ and $s$ into the third equation (the $z$ component).
- If the third equation holds true, the lines intersect.
- If the third equation gives a contradiction (e.g., $5=7$), the lines do not intersect.
- Find Point: Substitute the value of $t$ back into $\mathbf{r}_1$ (or $s$ into $\mathbf{r}_2$) to find the coordinates of the intersection point.
Common Mistake: Students often solve the first two equations for $t$ and $s$, and assume the lines intersect, forgetting to verify those values in the third equation. Always check the third component!
3.3 Parallel and Skew Lines (3D)
For two lines $L_1$ and $L_2$ with direction vectors $\mathbf{b}_1$ and $\mathbf{b}_2$:
- Parallel: $L_1$ and $L_2$ are parallel if $\mathbf{b}_1 = k\mathbf{b}_2$ for some non-zero scalar $k$.
- Skew (3D Only): Lines are skew if they are not parallel AND they do not intersect. They pass near each other but never touch, which is unique to 3D space.
Key Takeaway for Lines: Use different parameters for different lines, and always check the third equation when solving for intersection.
4. The Scalar Product (Dot Product)
The scalar product, also called the dot product, is a mathematical tool that takes two vectors and returns a scalar (a number). It tells us about the relationship between their directions.
4.1 Calculation of the Scalar Product
If $\mathbf{a} = \begin{pmatrix} a_1 \\ a_2 \\ a_3 \end{pmatrix}$ and $\mathbf{b} = \begin{pmatrix} b_1 \\ b_2 \\ b_3 \end{pmatrix}$, the scalar product is:
$$ \mathbf{a} \cdot \mathbf{b} = a_1 b_1 + a_2 b_2 + a_3 b_3 $$
You multiply corresponding components and add the results.
4.2 Geometric Definition and Angle Calculation
The scalar product is also defined geometrically:
$$ \mathbf{a} \cdot \mathbf{b} = |\mathbf{a}||\mathbf{b}| \cos \theta $$
Where $\theta$ is the angle between the two vectors when their tails are joined.
This formula is rearranged to find the angle $\theta$: $$ \cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}||\mathbf{b}|} $$
Step-by-Step for Finding the Angle Between Two Vectors
- Calculate the Dot Product: Find $\mathbf{a} \cdot \mathbf{b}$.
- Calculate Magnitudes: Find $|\mathbf{a}|$ and $|\mathbf{b}|$.
- Substitute: Plug the results into the cosine formula and solve for $\theta$.
4.3 The Perpendicular Condition
If two vectors $\mathbf{a}$ and $\mathbf{b}$ are perpendicular (orthogonal), the angle $\theta$ between them is $90^\circ$. Since $\cos(90^\circ) = 0$, this leads to a vital rule:
$$ \text{If } \mathbf{a} \text{ is perpendicular to } \mathbf{b} \text{, then } \mathbf{a} \cdot \mathbf{b} = 0 $$
Memory Trick: Think of the dot product being "zero" when the vectors are "squashed" flat against each other (at 90 degrees).
Key Takeaway for Dot Product: The scalar product is your primary tool for finding angles and proving perpendicularity in vector problems.
5. Perpendicular Distance from a Point to a Line
This is one of the most common and challenging application questions in the Vectors chapter. It uses the vector line equation and the perpendicular condition ($\mathbf{a} \cdot \mathbf{b} = 0$).
Goal: Find the shortest distance from a point $P$ to a line $L$.
The shortest distance is always along the line segment $PF$ where $F$ is a point on $L$ such that $PF$ is perpendicular to $L$. $F$ is called the foot of the perpendicular.
Step-by-Step Process
Assume line $L$ is $\mathbf{r} = \mathbf{a} + t\mathbf{b}$ and $P$ is a given point with position vector $\mathbf{p}$.
- Define the Foot of the Perpendicular ($F$): Since $F$ lies on the line $L$, its position vector $\mathbf{f}$ must be defined by the line equation using the parameter $t$. $$ \mathbf{f} = \mathbf{a} + t\mathbf{b} $$
- Find the Vector $\vec{PF}$: Use "End minus Start" to find the vector from $P$ to $F$. This vector will contain the parameter $t$. $$ \vec{PF} = \mathbf{f} - \mathbf{p} = (\mathbf{a} + t\mathbf{b}) - \mathbf{p} $$
- Apply the Perpendicular Condition: Since $\vec{PF}$ is perpendicular to the line $L$, it must be perpendicular to the direction vector $\mathbf{b}$. Therefore, their scalar product is zero. $$ \vec{PF} \cdot \mathbf{b} = 0 $$
- Solve for $t$: Expand the dot product equation. This will result in a simple linear equation in $t$. Solve for the value of $t$.
- Find Point $F$: Substitute this value of $t$ back into the expression for $\mathbf{f}$ (Step 1) to find the exact coordinates of $F$.
- Calculate the Distance: Find the magnitude of the vector $\vec{PF}$ (from Step 2, using the calculated $t$) or calculate the distance between $P$ and $F$ (Step 5). $$ \text{Distance } = |\vec{PF}| $$
Accessibility Tip: The algebra in Step 3 looks complicated, but remember you are substituting specific numbers for $\mathbf{a}, \mathbf{b},$ and $\mathbf{p}$. The equation will simplify quickly, usually leading to something like $14t = 28$. Be methodical and show your substitution clearly!
Summary: Key Takeaways for Vectors (P2)
Vectors provide the language for physics and 3D geometry. Master these core concepts:
- Displacement: $\vec{AB} = \mathbf{b} - \mathbf{a}$ (End minus Start).
- Magnitude: $|\mathbf{a}| = \sqrt{x^2 + y^2 + z^2}$.
- Line Equation: $\mathbf{r} = \mathbf{a} + t\mathbf{b}$.
- Scalar Product: Used for finding the angle: $\cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}||\mathbf{b}|}$.
- Perpendicularity Condition: $\mathbf{a} \cdot \mathbf{b} = 0$. This is the foundation for finding perpendicular distances.