Pure Mathematics | Vectors

Welcome to Vectors in FP3!

Hello future mathematician! This chapter, Vectors, is one of the most powerful and beautiful areas of Further Pure Mathematics. In FP3, we leave the familiar 2D world behind and dive fully into 3D geometry. You will learn the advanced tools needed to describe planes, calculate volumes, and find distances and angles in space.

Don't worry if vectors seemed abstract before. We will break down the key new concept—the Vector Product (Cross Product)—into simple, manageable steps. Mastering this will unlock the geometry of the plane!

Section 1: Foundation Review (The Scalar Product)

Before tackling the new concepts, let’s quickly remind ourselves of the tool you already know: the Scalar Product.

1.1 The Scalar Product (Dot Product) Review

The scalar product takes two vectors and produces a scalar (just a number). If \(\mathbf{a} = a_1 \mathbf{i} + a_2 \mathbf{j} + a_3 \mathbf{k}\) and \(\mathbf{b} = b_1 \mathbf{i} + b_2 \mathbf{j} + b_3 \mathbf{k}\):

$ \mathbf{a} \cdot \mathbf{b} = a_1 b_1 + a_2 b_2 + a_3 b_3 $

• Angle calculation: $\mathbf{a} \cdot \mathbf{b} = |\mathbf{a}||\mathbf{b}| \cos \theta$.
• Perpendicularity Test: If $\mathbf{a} \cdot \mathbf{b} = 0$, the vectors are perpendicular (orthogonal). This is a vital test!

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Section 2: The Vector Product (The Cross Product)

This is the centerpiece of the FP3 Vectors chapter. The key difference here is that the result is not a scalar, but another vector!

2.1 Definition and Direction

The vector product of two vectors, \(\mathbf{a}\) and \(\mathbf{b}\), is written as $\mathbf{a} \times \mathbf{b}$.

The most important property: The resulting vector, $\mathbf{c} = \mathbf{a} \times \mathbf{b}$, is perpendicular (orthogonal) to both \(\mathbf{a}\) and \(\mathbf{b}\). This makes the cross product the essential tool for finding the equation of a plane.

Magnitude of the Cross Product

The magnitude of the resulting vector is given by: $$ |\mathbf{a} \times \mathbf{b}| = |\mathbf{a}||\mathbf{b}| \sin \theta $$ where \(\theta\) is the angle between \(\mathbf{a}\) and \(\mathbf{b}\).

Direction: The Right-Hand Rule

How do we know which way the resulting vector points? We use the Right-Hand Rule:
Imagine turning vector \(\mathbf{a}\) towards vector \(\mathbf{b}\) through the angle \(\theta\). Curl the fingers of your right hand in this direction. Your extended thumb points in the direction of $\mathbf{a} \times \mathbf{b}$.

Memory Aid: Because of the right-hand rule, the order matters! $$ \mathbf{b} \times \mathbf{a} = -(\mathbf{a} \times \mathbf{b}) $$ (The cross product is anti-commutative.)

2.2 Calculating the Vector Product

To find the components of the resulting vector, we use the method of determinants.

Let $\mathbf{a} = a_1 \mathbf{i} + a_2 \mathbf{j} + a_3 \mathbf{k}$ and $\mathbf{b} = b_1 \mathbf{i} + b_2 \mathbf{j} + b_3 \mathbf{k}$.

We set up a 3x3 determinant matrix: $$ \mathbf{a} \times \mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix} $$

Step-by-Step Calculation

1. For the \(\mathbf{i}\) component: Cover the first column. Calculate the determinant of the remaining 2x2 matrix: $$ (a_2 b_3) - (a_3 b_2) $$
2. For the \(\mathbf{j}\) component: Cover the second column. Calculate the determinant (remember to subtract this term, or multiply by \(-1\)): $$ - [(a_1 b_3) - (a_3 b_1)] $$
3. For the \(\mathbf{k}\) component: Cover the third column. Calculate the determinant: $$ (a_1 b_2) - (a_2 b_1) $$

The resulting vector is: $$ \mathbf{a} \times \mathbf{b} = [(a_2 b_3 - a_3 b_2)]\mathbf{i} - [(a_1 b_3 - a_3 b_1)]\mathbf{j} + [(a_1 b_2 - a_2 b_1)]\mathbf{k} $$

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Section 3: Applications of the Vector Product

The cross product provides elegant shortcuts for finding areas and volumes in 3D space.

3.1 Area of a Parallelogram and Triangle

Remember that the magnitude of the cross product is $|\mathbf{a} \times \mathbf{b}| = |\mathbf{a}||\mathbf{b}| \sin \theta$. Geometrically, this is exactly the area of the parallelogram formed by the two vectors \(\mathbf{a}\) and \(\mathbf{b}\).

• Area of Parallelogram: If $\mathbf{a}$ and $\mathbf{b}$ are adjacent sides, then: $$ Area = |\mathbf{a} \times \mathbf{b}| $$
• Area of Triangle: Since a triangle is half a parallelogram: $$ Area = \frac{1}{2} |\mathbf{a} \times \mathbf{b}| $$

3.2 Volume of a Parallelepiped (The Scalar Triple Product)

A parallelepiped is a 3D figure like a stretched or squashed cube (six faces that are parallelograms). Its volume is found using the scalar triple product.

If the parallelepiped is defined by three vectors, \(\mathbf{a}\), \(\mathbf{b}\), and \(\mathbf{c}\), meeting at a corner, the volume (V) is: $$ V = |\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})| $$

Did you know? Since the dot product of two vectors is commutative, we often write the scalar triple product as a single determinant: $$ V = \left| \begin{vmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{vmatrix} \right| $$

Volume of a Tetrahedron

A tetrahedron is a pyramid with four triangular faces. It has 1/6th the volume of the parallelepiped defined by the same three edge vectors. $$ V_{tetrahedron} = \frac{1}{6} |\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})| $$

Key Takeaway: The cross product finds perpendicular vectors and areas; the scalar triple product finds volume.

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Section 4: Equations of a Plane

A plane in 3D space is defined by its orientation (tilt) and its position. The orientation is governed entirely by a single vector: the normal vector.

4.1 The Normal Vector (\(\mathbf{n}\))

The normal vector, \(\mathbf{n}\), is a vector that is perpendicular to the plane itself. It is perpendicular to every single line and vector that lies in the plane.

Finding the Normal Vector

If you are given three non-collinear points A, B, and C in the plane:

1. Form two vectors lying in the plane, e.g., $\mathbf{a} = \vec{AB}$ and $\mathbf{b} = \vec{AC}$.
2. Calculate the cross product: $\mathbf{n} = \mathbf{a} \times \mathbf{b}$. This result is the normal vector.

4.2 The Vector Equation of a Plane

A plane can be defined by a known position vector \(\mathbf{a}\) (a point in the plane) and its normal vector \(\mathbf{n}\).

Let $\mathbf{r}$ be the position vector of any arbitrary point P(x, y, z) on the plane. The vector $\mathbf{r} - \mathbf{a}$ lies entirely within the plane. Since $\mathbf{n}$ is perpendicular to the plane, it must be perpendicular to $\mathbf{r} - \mathbf{a}$.

This gives the standard vector equation: $$ (\mathbf{r} - \mathbf{a}) \cdot \mathbf{n} = 0 $$

Rearranging this gives the common form: $$ \mathbf{r} \cdot \mathbf{n} = \mathbf{a} \cdot \mathbf{n} $$ The quantity $\mathbf{a} \cdot \mathbf{n}$ is a fixed scalar value, often denoted as $D$. $$ \mathbf{r} \cdot \mathbf{n} = D $$

4.3 The Cartesian Equation of a Plane

If $\mathbf{r} = x\mathbf{i} + y\mathbf{j} + z\mathbf{k}$ and $\mathbf{n} = n_1\mathbf{i} + n_2\mathbf{j} + n_3\mathbf{k}$, the vector equation $\mathbf{r} \cdot \mathbf{n} = D$ expands directly to the Cartesian form:

$$ n_1 x + n_2 y + n_3 z = D $$

Crucial Insight: The coefficients of \(x, y, z\) in the Cartesian equation are immediately the components of the normal vector \(\mathbf{n}\).

Example: If the plane is $2x - 3y + z = 10$, then the normal vector is $\mathbf{n} = 2\mathbf{i} - 3\mathbf{j} + \mathbf{k}$.

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Section 5: Angles and Distances in 3D Space

The final applications of vectors involve calculating the relationship between lines and planes.

5.1 Angle Between Two Planes

When two planes intersect, the angle between them, \(\theta\), is defined as the angle between their respective normal vectors, \(\mathbf{n}_1\) and \(\mathbf{n}_2\).

We use the dot product formula, ensuring we find the acute angle (usually between $0^\circ$ and $90^\circ$): $$ \cos \theta = \frac{|\mathbf{n}_1 \cdot \mathbf{n}_2|}{|\mathbf{n}_1||\mathbf{n}_2|} $$ (We use the absolute value of the dot product to guarantee we get the acute angle.)

• Parallel Planes: If $\mathbf{n}_1$ is parallel to $\mathbf{n}_2$ (i.e., $\mathbf{n}_1 = k \mathbf{n}_2$), the angle $\theta = 0^\circ$.
• Perpendicular Planes: If $\mathbf{n}_1 \cdot \mathbf{n}_2 = 0$, the angle $\theta = 90^\circ$.

5.2 Angle Between a Line and a Plane

This is often a tricky spot! If the line L has direction vector \(\mathbf{d}\) and the plane $\Pi$ has normal vector $\mathbf{n}$.

The angle we usually find using the dot product, $\alpha$, is the angle between the line direction $\mathbf{d}$ and the normal $\mathbf{n}$.

$$ \cos \alpha = \frac{|\mathbf{d} \cdot \mathbf{n}|}{|\mathbf{d}||\mathbf{n}|} $$

However, the question usually asks for the angle $\theta$ between the line L and the plane $\Pi$.

Since the plane and the normal are perpendicular: $$ \theta = 90^\circ - \alpha \quad \text{OR} \quad \sin \theta = \cos \alpha $$

Crucial Tip: Always calculate the cosine of the angle between the line direction and the normal first ($\cos \alpha$), then use trigonometry to find the sine of the required angle ($\sin \theta$).

5.3 Shortest Distance from a Point to a Plane

We want the perpendicular distance from a point $P(x_0, y_0, z_0)$ to a plane $\Pi: n_1 x + n_2 y + n_3 z = D$.

First, rewrite the Cartesian equation so it equals zero: $n_1 x + n_2 y + n_3 z - D = 0$.

The shortest distance (S) is given by substituting the coordinates of the point into the plane equation and dividing by the magnitude of the normal vector:

$$ S = \frac{|n_1 x_0 + n_2 y_0 + n_3 z_0 - D|}{\sqrt{n_1^2 + n_2^2 + n_3^2}} $$

Encouragement: This looks complicated, but it is just a formula substitution! Identify your $x_0, y_0, z_0$ and the normal vector components $n_1, n_2, n_3$ from the equation and plug them in. The absolute value ensures the distance is positive.

5.4 Finding the Point of Intersection (Line and Plane)

If you have a line L and a plane \(\Pi\), they will either be parallel, or they will intersect at a unique point.

Process:

1. Start with the vector equation of the line L: $\mathbf{r} = \mathbf{a} + \lambda \mathbf{d}$.
2. Start with the vector equation of the plane $\Pi$: $\mathbf{r} \cdot \mathbf{n} = D$.
3. Substitute the expression for $\mathbf{r}$ from the line into the plane equation: $$ (\mathbf{a} + \lambda \mathbf{d}) \cdot \mathbf{n} = D $$
4. Expand this dot product: $\mathbf{a} \cdot \mathbf{n} + \lambda (\mathbf{d} \cdot \mathbf{n}) = D$.
5. This equation has only one unknown, $\lambda$. Solve for $\lambda$.
6. Substitute the value of $\lambda$ back into the line equation $\mathbf{r} = \mathbf{a} + \lambda \mathbf{d}$ to find the position vector of the intersection point.

What if $\mathbf{d} \cdot \mathbf{n} = 0$? If the dot product of the line direction and the plane normal is zero, the line is perpendicular to the normal. This means the line is parallel to the plane! In this case, there is either no intersection (if $\mathbf{a} \cdot \mathbf{n} \neq D$) or the line lies entirely within the plane (if $\mathbf{a} \cdot \mathbf{n} = D$).

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Chapter Summary: Key Takeaways

• The Vector Product $\mathbf{a} \times \mathbf{b}$ produces a vector perpendicular to both $\mathbf{a}$ and $\mathbf{b}$. This resulting vector is the foundation for finding plane normals and calculating area.
• The Normal Vector $\mathbf{n}$ is the defining feature of a plane.
• The two main forms of a plane equation are $\mathbf{r} \cdot \mathbf{n} = D$ (Vector) and $n_1 x + n_2 y + n_3 z = D$ (Cartesian).
• When finding the angle between a line and a plane, remember to calculate the angle to the normal ($\alpha$) and then convert it ($\theta = 90^\circ - \alpha$).

You have successfully navigated the deep waters of 3D geometry! Practice these formulas and techniques, especially the cross product calculation, and you will find FP3 Vectors highly rewarding. Good luck!