Welcome to Further Vectors!
In your standard A Level Maths course, you’ve already mastered the basics of vectors—adding them, finding their magnitude, and using the scalar (dot) product. In Further Pure Mathematics 1 (FP1), we take these skills into the third dimension to solve much more interesting problems. We will learn how to find areas and volumes in 3D space and how to describe lines in a brand-new way. Think of this as the "GPS and 3D Graphics" chapter of your maths journey!
1. The Vector Product (\(\mathbf{a} \times \mathbf{b}\))
You already know the Scalar Product (which gives a number). Now, meet the Vector Product (also called the Cross Product). As the name suggests, the result is a vector.
What is it?
The vector product \(\mathbf{a} \times \mathbf{b}\) creates a new vector that is perpendicular (at 90 degrees) to both \(\mathbf{a}\) and \(\mathbf{b}\). If \(\mathbf{a}\) and \(\mathbf{b}\) lie on a table, \(\mathbf{a} \times \mathbf{b}\) points straight up or straight down.
How to calculate it
The easiest way to calculate \(\mathbf{a} \times \mathbf{b}\) for vectors \(\mathbf{a} = \begin{pmatrix} a_1 \\ a_2 \\ a_3 \end{pmatrix}\) and \(\mathbf{b} = \begin{pmatrix} b_1 \\ b_2 \\ b_3 \end{pmatrix}\) is to use a 3x3 determinant:
\(\mathbf{a} \times \mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix}\)
Step-by-step process:
1. Cover the \(\mathbf{i}\) column: \((a_2b_3 - a_3b_2)\mathbf{i}\)
2. Cover the \(\mathbf{j}\) column (don't forget to flip the sign!): \(-(a_1b_3 - a_3b_1)\mathbf{j}\)
3. Cover the \(\mathbf{k}\) column: \((a_1b_2 - a_2b_1)\mathbf{k}\)
Memory Aid: Think of the sign pattern for the columns as Plus, Minus, Plus.
Finding Area
One of the coolest features of the vector product is that its magnitude tells us about area:
Area of a parallelogram: \(Area = |\mathbf{a} \times \mathbf{b}|\)
Area of a triangle: \(Area = \frac{1}{2}|\mathbf{a} \times \mathbf{b}|\)
Analogy: If vectors \(\mathbf{a}\) and \(\mathbf{b}\) are two sides of a physical frame, the cross product's length literally measures how much "shadow" or surface that frame covers.
Common Mistake: Forgetting that the vector product is not commutative. \(\mathbf{a} \times \mathbf{b} = -(\mathbf{b} \times \mathbf{a})\). The direction flips!
Key Takeaway: The vector product gives you a perpendicular vector, and its magnitude gives you the area of the shapes formed by the two vectors.
2. The Scalar Triple Product \(\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})\)
Now we are combining the Dot Product and the Cross Product. This is used to find volumes in 3D space.
The Parallelepiped
A parallelepiped is basically a 3D box that has been squashed (all faces are parallelograms). If three edges meeting at a corner are represented by vectors \(\mathbf{a}\), \(\mathbf{b}\), and \(\mathbf{c}\):
Volume = \(|\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})|\)
The Tetrahedron
A tetrahedron is a 3D pyramid with a triangular base. It is much smaller than a parallelepiped. For the same three vectors:
Volume = \(\frac{1}{6}|\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})|\)
Did you know? If the scalar triple product equals zero, it means the three vectors are coplanar. This is because they have no "height," so the volume is zero. They all lie on the same flat sheet of paper!
Key Takeaway: Use the triple product for volumes. Use "1/6" for pyramids (tetrahedrons) and "1" for boxes (parallelepipeds).
3. Advanced Equations of Lines
You have seen the line equation \(\mathbf{r} = \mathbf{a} + \lambda\mathbf{b}\). In FP1, we use the vector product to write this differently.
The Vector Product Form
The equation of a line can be written as:
\((\mathbf{r} - \mathbf{a}) \times \mathbf{b} = \mathbf{0}\)
Why does this work?
Remember that \(\mathbf{r} - \mathbf{a}\) is a vector pointing from a fixed point on the line to any other point on the line. For this to be on the line, it must be parallel to the direction vector \(\mathbf{b}\). When two vectors are parallel, their vector product is zero (\(\mathbf{0}\)).
Don't worry if this seems tricky: It’s just a different "language" to say the same thing. If you see this form, \(\mathbf{a}\) is your position vector (a point) and \(\mathbf{b}\) is your direction vector.
Key Takeaway: If \((\text{vector}) \times \mathbf{b} = 0\), it means the vector is parallel to \(\mathbf{b}\). This identifies all points on that line.
4. Direction Ratios and Direction Cosines
These are ways to describe the "heading" or "tilt" of a line in 3D space.
Direction Ratios
If a line has a direction vector \(\mathbf{d} = \begin{pmatrix} l \\ m \\ n \end{pmatrix}\), then \(l, m, n\) are the direction ratios. You can multiply them all by any number \(k\), and they are still the same direction ratios. For example, \((1, 2, 2)\) is the same direction as \((10, 20, 20)\).
Direction Cosines
Direction cosines are more specific. They are the direction ratios of a unit vector (a vector with length 1). They are usually written as \((\cos \alpha, \cos \beta, \cos \gamma)\), where:
\(\alpha\) = angle with the x-axis
\(\beta\) = angle with the y-axis
\(\gamma\) = angle with the z-axis
The "Magic" Property:
For any line, the squares of the direction cosines always add up to 1:
\(\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1\)
Quick Review: How to find Direction Cosines
1. Take your direction vector \(\begin{pmatrix} a \\ b \\ c \end{pmatrix}\).
2. Find the magnitude: \(D = \sqrt{a^2 + b^2 + c^2}\).
3. Divide each component by \(D\). The results are your cosines: \(\frac{a}{D}, \frac{b}{D}, \frac{c}{D}\).
Key Takeaway: Direction ratios give you the general "slope." Direction cosines give you the precise angles with the axes using a unit vector.
Summary Checklist
Can you:
- Calculate \(\mathbf{a} \times \mathbf{b}\) using a determinant?
- Find the area of a triangle or parallelogram using vectors?
- Calculate the volume of a tetrahedron (\(\frac{1}{6}\)) or parallelepiped?
- Recognize a line in the form \((\mathbf{r} - \mathbf{a}) \times \mathbf{b} = \mathbf{0}\)?
- Convert a direction vector into direction cosines by making it a unit vector?
You've got this! Keep practicing the determinant calculations, as that is where most simple marks are gained or lost.