A Level Pure Mathematics 3 (9709) Study Notes: VECTORS (3.7)

Welcome to the world of Vectors! This chapter is one of the most useful and conceptually satisfying parts of Pure Mathematics 3. Why? Because vectors give us the power to describe and solve problems involving position, movement, and forces in 2D and 3D space. Think of it as upgrading your coordinate geometry skills to handle the real world, where everything doesn't just happen on a flat graph!

Don't worry if 3D space feels confusing at first. We will break down every concept step-by-step, using clear notation and lots of visual explanations.

1. Vector Basics: Definition and Notation

1.1 Scalars vs. Vectors

In mathematics, quantities fall into two main categories:

  • Scalar Quantity: Only has magnitude (size).
    Examples: Mass (5 kg), Speed (60 km/h), Temperature (20°C).
  • Vector Quantity: Has both magnitude (size) and direction.
    Examples: Velocity (60 km/h North), Force (10 N downwards), Displacement (5 m East).

Analogy: A scalar tells you how much. A vector tells you how much and where to go.

1.2 Standard Notation

We use special notation to distinguish a vector from a scalar. In print, vectors are usually written in bold, $\mathbf{a}$, or when handwritten, underlined, $\underline{a}$.

Vectors are typically expressed in two main ways in P3:

  1. Column Vector Form (Component Form): Best for calculations.

    In 2D: \(\mathbf{a} = \begin{pmatrix} x \\ y \end{pmatrix}\)
    In 3D: \(\mathbf{a} = \begin{pmatrix} x \\ y \\ z \end{pmatrix}\)

  2. \(\mathbf{i}, \mathbf{j}, \mathbf{k}\) Form (Unit Vector Form): Also excellent for calculations, especially showing direction. \(\mathbf{i}\), \(\mathbf{j}\), and \(\mathbf{k}\) are unit vectors (magnitude 1) in the direction of the \(x\), \(y\), and \(z\) axes, respectively.

    In 3D: \(\mathbf{a} = x\mathbf{i} + y\mathbf{j} + z\mathbf{k}\)

Quick Review: Notation

The vector \(\mathbf{a}\) can be written as \(\begin{pmatrix} 2 \\ 5 \\ -1 \end{pmatrix}\) or \(2\mathbf{i} + 5\mathbf{j} - \mathbf{k}\).

2. Position and Displacement Vectors

2.1 Position Vectors

A Position Vector ($\mathbf{r}$ or $\mathbf{a}$) specifies the location of a point $A$ relative to the origin $O$.

If point \(A\) has coordinates \((x, y, z)\), its position vector is \(\vec{OA} = \mathbf{a} = \begin{pmatrix} x \\ y \\ z \end{pmatrix}\).

2.2 Displacement Vectors (Vector between Two Points)

The vector representing the journey from point $A$ to point $B$ is the displacement vector, \(\vec{AB}\).

To find \(\vec{AB}\), you subtract the position vector of the starting point ($A$) from the position vector of the end point ($B$):

Formula: \(\vec{AB} = \mathbf{b} - \mathbf{a}\)

Memory Aid: The displacement vector is always "End minus Start".

2.3 Vector Arithmetic and Geometry

Addition and Subtraction: Performed component-wise. This follows the triangle rule (or parallelogram rule) geometrically.

If \(\mathbf{a} = \begin{pmatrix} a_1 \\ a_2 \\ a_3 \end{pmatrix}\) and \(\mathbf{b} = \begin{pmatrix} b_1 \\ b_2 \\ b_3 \end{pmatrix}\), then:

\(\mathbf{a} + \mathbf{b} = \begin{pmatrix} a_1 + b_1 \\ a_2 + b_2 \\ a_3 + b_3 \end{pmatrix}\)

Geometrical Interpretation (Parallelogram Rule):

If $OABC$ is a parallelogram, then the diagonal vector \(\vec{OB}\) is the sum of the adjacent sides: \(\vec{OB} = \vec{OA} + \vec{OC}\). This is a crucial geometric interpretation you must be comfortable using in proofs!

Scalar Multiplication: Multiplying a vector $\mathbf{a}$ by a scalar $k$ changes its magnitude (length) but not its direction (if $k > 0$).

\(k\mathbf{a} = k \begin{pmatrix} a_1 \\ a_2 \\ a_3 \end{pmatrix} = \begin{pmatrix} k a_1 \\ k a_2 \\ k a_3 \end{pmatrix}\)

If $k$ is negative, the direction of the vector is reversed (it points in the opposite direction).

2.4 Finding the Midpoint

The position vector of the midpoint $M$ of a line segment $AB$ is the average of the position vectors of $A$ and $B$.

Midpoint Formula: \(\vec{OM} = \mathbf{m} = \frac{1}{2}(\mathbf{a} + \mathbf{b})\)

Key Takeaway: All vector arithmetic (add, subtract, scalar multiply) is done by combining the individual components (\(i\), \(j\), \(k\)). The displacement vector is always "End minus Start".

3. Magnitude and Unit Vectors

3.1 Calculating the Magnitude

The magnitude (or length) of a vector \(\mathbf{a}\) is denoted by \(|\mathbf{a}|\) (or sometimes \(\|\mathbf{a}\|\)). Since the components form a right-angled triangle (or box in 3D), we use Pythagoras' theorem.

If \(\mathbf{a} = \begin{pmatrix} a_1 \\ a_2 \\ a_3 \end{pmatrix}\), then:

Magnitude Formula: \(|\mathbf{a}| = \sqrt{a_1^2 + a_2^2 + a_3^2}\)

Example: If \(\mathbf{a} = 3\mathbf{i} - 2\mathbf{j} + 6\mathbf{k}\), then \(|\mathbf{a}| = \sqrt{3^2 + (-2)^2 + 6^2} = \sqrt{9 + 4 + 36} = \sqrt{49} = 7\).

3.2 Unit Vectors

A unit vector is a vector that has a magnitude of exactly 1. It is often used purely to define a direction.

To find the unit vector in the direction of \(\mathbf{a}\), you simply divide the vector \(\mathbf{a}\) by its own magnitude:

Unit Vector \(\hat{\mathbf{a}}\) Formula: \(\hat{\mathbf{a}} = \frac{\mathbf{a}}{|\mathbf{a}|}\)

Tip: Normalising a vector means converting it into a unit vector.

Common Mistake to Avoid: When calculating the magnitude, remember that squaring a negative number always results in a positive number. E.g., \((-2)^2 = 4\).

4. The Equation of a Straight Line

In P3, we describe a straight line not with \(y=mx+c\), but using vectors. This allows us to work in 3D space easily.

4.1 The Vector Equation of a Line

A line is uniquely defined by:

  1. A position vector of a fixed point on the line, \(\mathbf{a}\).
  2. A direction vector that the line follows, \(\mathbf{b}\).

The position vector \(\mathbf{r}\) of any general point on the line can be written as:

Vector Equation: \(\mathbf{r} = \mathbf{a} + t\mathbf{b}\)

Here, \(t\) is a scalar parameter (any real number). By changing the value of \(t\), you move along the line starting from point $\mathbf{a}$ in the direction $\mathbf{b}$.

Example: A line passing through point $A$ with position vector \(\mathbf{a} = \mathbf{i} + 2\mathbf{j}\) and parallel to the direction vector \(\mathbf{b} = 3\mathbf{i} - \mathbf{k}\) has the equation:
\(\mathbf{r} = (\mathbf{i} + 2\mathbf{j}) + t(3\mathbf{i} - \mathbf{k})\)

4.2 Writing in Component Form

If we write the components separately, it often makes solving simultaneous equations easier:

If \(\mathbf{r} = \begin{pmatrix} x \\ y \\ z \end{pmatrix}\), \(\mathbf{a} = \begin{pmatrix} a_1 \\ a_2 \\ a_3 \end{pmatrix}\), and \(\mathbf{b} = \begin{pmatrix} b_1 \\ b_2 \\ b_3 \end{pmatrix}\), then the line equation splits into three separate equations:

Component Form (Parametric Equations):
\(x = a_1 + t b_1\)
\(y = a_2 + t b_2\)
\(z = a_3 + t b_3\)

Step-by-Step: Finding the Equation of a Line Through Two Points

  1. Identify the position vectors of the two points, $\mathbf{A}$ and $\mathbf{B}$.
  2. Choose one position vector as your fixed point $\mathbf{a}$. (E.g., \(\mathbf{a} = \vec{OA}\)).
  3. Find the direction vector \(\mathbf{b}\) by calculating the displacement between the points: \(\mathbf{b} = \vec{AB} = \mathbf{b} - \mathbf{a}\).
  4. Substitute these into \(\mathbf{r} = \mathbf{a} + t\mathbf{b}\).

Key Takeaway: The vector equation of a line needs two things: a fixed point to start at, and a direction vector to follow. The parameter \(t\) is the map scale.

5. Relationships Between Two Lines

Given two lines, \(L_1: \mathbf{r} = \mathbf{a} + t\mathbf{b}\) and \(L_2: \mathbf{r} = \mathbf{c} + s\mathbf{d}\), we must determine if they are parallel, intersecting, or skew. (Note: We use different parameters, \(t\) and \(s\), for the two lines—this is critical!)

5.1 Parallel Lines

Two lines are parallel if their direction vectors are scalar multiples of each other.

\(L_1\) is parallel to \(L_2\) if \(\mathbf{b} = k\mathbf{d}\) for some scalar \(k\).

If they are parallel, check if they are the same line by seeing if point $\mathbf{c}$ lies on $L_1$ (i.e., can you find a value of \(t\) such that \(\mathbf{c} = \mathbf{a} + t\mathbf{b}\)?).

5.2 Intersecting Lines

If the direction vectors are not parallel, the lines might intersect at a single point.

Step-by-Step: Finding the Intersection Point

  1. Set the two line equations equal to each other: \(\mathbf{a} + t\mathbf{b} = \mathbf{c} + s\mathbf{d}\).
  2. Split this vector equation into three component equations (for $x$, $y$, and $z$).
  3. Use any two equations (e.g., $x$ and $y$) to form simultaneous equations and solve for the parameters \(t\) and \(s\).
  4. Crucial Verification Step: Substitute the found values of \(t\) and \(s\) into the *third* (unused, e.g., $z$) equation.
    • If the third equation balances (LHS = RHS), the lines intersect.
    • If the third equation does not balance, the lines are skew.
  5. If they intersect, substitute the value of \(t\) (or \(s\)) back into the original line equation \(L_1\) (or \(L_2\)) to find the position vector of the intersection point.
5.3 Skew Lines

Lines are skew if they are not parallel but also do not intersect. This can only happen in 3D space. They appear to cross if you view them from certain angles, but they never actually meet.

How to determine if lines are skew: Follow the intersection steps (5.2). If the parameters \(t\) and \(s\) found from the first two equations fail the verification in the third equation, the lines are skew.

Did you know? Imagine two airplanes flying straight in the sky. If their flight paths are neither parallel nor intersect (i.e., one passes over the other far away), they represent skew lines.

6. The Scalar (Dot) Product

The scalar product is a fundamental tool used to find the angle between two vectors and to check for perpendicularity.

6.1 Definition and Calculation

The scalar product (or dot product) of two vectors \(\mathbf{a}\) and \(\mathbf{b}\) results in a scalar quantity (a number).

There are two ways to calculate it, depending on the information given:

  1. Using Components:

    If \(\mathbf{a} = a_1\mathbf{i} + a_2\mathbf{j} + a_3\mathbf{k}\) and \(\mathbf{b} = b_1\mathbf{i} + b_2\mathbf{j} + b_3\mathbf{k}\):

    Formula 1: \(\mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3\)

    This is simply multiplying the corresponding components and summing the results.
  2. Using Magnitudes and Angle:

    If \(\theta\) is the angle between \(\mathbf{a}\) and \(\mathbf{b}\) (when they are joined tail-to-tail):

    Formula 2: \(\mathbf{a} \cdot \mathbf{b} = |\mathbf{a}||\mathbf{b}| \cos \theta\)

6.2 Finding the Angle Between Two Vectors (or Lines)

We combine the two formulae above to isolate \(\cos \theta\):

Angle Formula: \(\cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}||\mathbf{b}|}\)

Step-by-Step: Finding the Angle \(\theta\)

  1. Calculate the scalar product \(\mathbf{a} \cdot \mathbf{b}\) using Formula 1 (components).
  2. Calculate the magnitude of \(\mathbf{a}\), \(|\mathbf{a}|\).
  3. Calculate the magnitude of \(\mathbf{b}\), \(|\mathbf{b}|\).
  4. Substitute these results into the Angle Formula and solve for \(\theta = \arccos(\dots)\).

Note: When finding the angle between two *lines*, you must use the direction vectors (\(\mathbf{b}\) and \(\mathbf{d}\)) of the lines.

6.3 Testing for Perpendicularity (Orthogonality)

This is the most crucial application of the scalar product.

If two vectors are perpendicular (meet at a 90° angle), then \(\theta = 90^\circ\). Since \(\cos(90^\circ) = 0\), the scalar product must be zero.

Condition for Perpendicularity: \(\mathbf{a} \cdot \mathbf{b} = 0\)

6.4 Applications: Foot of the Perpendicular

A common P3 problem is finding the foot of the perpendicular $N$ from a point $P$ to a line \(L\).

Let the line \(L\) be \(\mathbf{r} = \mathbf{a} + t\mathbf{b}\). Since $N$ lies on $L$, its position vector \(\mathbf{n}\) is given by \(\mathbf{n} = \mathbf{a} + t\mathbf{b}\).

Step-by-Step:

  1. Express the vector \(\vec{PN}\) (which is the vector from $P$ to a general point $N$ on the line) in terms of the parameter \(t\):
    \(\vec{PN} = \mathbf{n} - \mathbf{p} = (\mathbf{a} + t\mathbf{b}) - \mathbf{p}\).
  2. Since \(\vec{PN}\) must be perpendicular to the line \(L\), \(\vec{PN}\) must be perpendicular to the line's direction vector \(\mathbf{b}\).
  3. Apply the perpendicularity condition: \(\vec{PN} \cdot \mathbf{b} = 0\).
  4. Solve the resulting scalar equation for the parameter \(t\).
  5. Substitute the value of \(t\) back into the expression for \(\mathbf{n}\) to find the position vector of $N$.

Key Takeaway: The scalar product is your tool for angles and perpendicularity. Remember: If vectors are perpendicular, their dot product is zero. Use the dot product to find the foot of the perpendicular or the angle between lines.

Summary of Exclusions (What NOT to Study for 9709 P3 Vectors)

To keep your focus sharp, remember that the following advanced vector concepts are NOT required for the 9709 Pure Mathematics 3 syllabus:

  • Vector Product (Cross Product): This operation, which results in a vector perpendicular to the original two, is not required.
  • General Ratio Theorem: While the midpoint is included, the formula for a general point dividing a line segment in the ratio \(\lambda:\mu\) is not required knowledge.
  • Shortest Distance between Skew Lines: You only need to *determine* if lines are skew, not calculate the shortest distance between them.
  • Equation of a Plane: Plane geometry is beyond the scope of P3 vectors.

Focus on magnitudes, line equations, checking intersection/skewness, and the scalar product applications!

You've got this! Vectors are a powerful skill that link algebra and geometry beautifully. Keep practising those steps!