Welcome to Further Vectors!

In A Level Mathematics, you learned how to work with vectors in 2D and 3D. In this chapter of Further Mathematics (8FM0), we take those skills to the next level! We will explore how to describe straight lines and flat surfaces (called planes) in three-dimensional space. These concepts are the backbone of modern technology—from how GPS satellites track your location to how video game engines render 3D worlds.

Don't worry if this seems tricky at first! We will break everything down into small, manageable steps. Let's dive in!


1. Straight Lines in 3D

Think of a line in 3D as a path through the air. To describe this path exactly, you only need two things: a starting point and the direction you are traveling.

The Vector Form

The vector equation of a line is written as: \( \mathbf{r} = \mathbf{a} + \lambda \mathbf{b} \)

  • \( \mathbf{r} \): Any point on the line (represented by the coordinates \(x, y, z\)).
  • \( \mathbf{a} \): A position vector of a known point that the line passes through (the "starting point").
  • \( \mathbf{b} \): The direction vector (which way the line is pointing).
  • \( \lambda \) (lambda): A scalar parameter. As you change \(\lambda\), you "slide" along the line to different points.

The Cartesian Form

Sometimes, we prefer to write the line using \(x, y,\) and \(z\) instead of vectors. This is the Cartesian form:

\( \frac{x - a_1}{b_1} = \frac{y - a_2}{b_2} = \frac{z - a_3}{b_3} \)

Quick Tip: To switch from Vector to Cartesian, rearrange the vector equation to make \(\lambda\) the subject for each component (\(x, y,\) and \(z\)) and equate them.

Parallel, Intersecting, and Skew Lines

In 2D, lines are either parallel or they intersect. In 3D, there is a third option:

  • Parallel lines: Have the same (or a scalar multiple of the same) direction vector \(\mathbf{b}\).
  • Intersecting lines: They meet at exactly one point.
  • Skew lines: Lines that are not parallel and never intersect.

Intersection of Two Lines

To determine if two lines intersect and find the point of intersection:

  1. Use distinct scalar parameters for each line: \(\mathbf{r}_1 = \mathbf{a}_1 + \lambda \mathbf{b}_1\) and \(\mathbf{r}_2 = \mathbf{a}_2 + \mu \mathbf{b}_2\).
  2. Equate the \(x, y,\) and \(z\) components to obtain three equations in \(\lambda\) and \(\mu\).
  3. Solve two of the equations simultaneously to find values for \(\lambda\) and \(\mu\).
  4. Substitute these values into the third equation. If it is satisfied, the lines intersect (substitute \(\lambda\) into \(\mathbf{r}_1\) to get the position vector of the point). If it is not satisfied, the lines do not intersect (they are skew, provided they are not parallel).
Key Takeaway: A line needs a point (\(\mathbf{a}\)) and a direction (\(\mathbf{b}\)). Always use different parameters (e.g. \(\lambda\) and \(\mu\)) when working with two lines.

2. The Equation of a Plane

Imagine a flat sheet of paper extending forever in all directions—that is a plane.

Vector Form (Two Directions)

If you have a point on the plane and two non-parallel direction vectors that lie in the plane: \( \mathbf{r} = \mathbf{a} + \lambda \mathbf{b} + \mu \mathbf{c} \)

The Normal Form (The Scalar Product Form)

Every plane has a normal vector (\(\mathbf{n}\)), which is perpendicular to the plane surface:

\( \mathbf{r} \cdot \mathbf{n} = \mathbf{a} \cdot \mathbf{n} = d \)

where \(d\) is a constant scalar found by taking the scalar product of a known point \(\mathbf{a}\) on the plane with \(\mathbf{n}\).

Cartesian Form of a Plane

If \(\mathbf{n} = \begin{pmatrix} a \\ b \\ c \end{pmatrix}\), the Cartesian equation is: \( ax + by + cz = d \)

Did you know? The coefficients \(a, b,\) and \(c\) in the Cartesian equation are exactly the components of the normal vector \(\mathbf{n}\)!

Key Takeaway: The most direct way to define a plane is using its normal vector (the vector perpendicular to its surface).

3. The Scalar Product (Dot Product)

The scalar product is defined as:

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

Key Applications:

  1. Finding the angle (\(\theta\)): \( \cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}||\mathbf{b}|} \).
  2. Perpendicularity: Two non-zero vectors are perpendicular if and only if \( \mathbf{a} \cdot \mathbf{b} = 0 \).

4. Finding Angles Between Lines and Planes

When calculating angles, identify the relevant direction or normal vectors:

Angle between two lines

Use the direction vectors \(\mathbf{b}_1\) and \(\mathbf{b}_2\):

\( \cos \theta = \frac{|\mathbf{b}_1 \cdot \mathbf{b}_2|}{|\mathbf{b}_1||\mathbf{b}_2|} \)

Angle between two planes

Use the normal vectors \(\mathbf{n}_1\) and \(\mathbf{n}_2\):

\( \cos \theta = \frac{|\mathbf{n}_1 \cdot \mathbf{n}_2|}{|\mathbf{n}_1||\mathbf{n}_2|} \)

Angle between a line and a plane

Because the normal vector \(\mathbf{n}\) is perpendicular to the plane, the angle \(\theta\) between the line direction \(\mathbf{b}\) and the plane satisfies:

\( \sin \theta = \frac{|\mathbf{b} \cdot \mathbf{n}|}{|\mathbf{b}||\mathbf{n}|} \)

Memory Aid: Use \(\cos\) for line-line and plane-plane; use \(\sin\) for line-plane.


5. Intersections and Perpendicular Distances

Intersection of a Line and a Plane

To find where a line \(\mathbf{r} = \mathbf{a} + \lambda \mathbf{b}\) meets a plane \(ax + by + cz = d\):

  1. Express \(x, y, z\) in terms of \(\lambda\) using the line equation: \(x = a_1 + \lambda b_1\), \(y = a_2 + \lambda b_2\), \(z = a_3 + \lambda b_3\).
  2. Substitute these into the plane equation and solve the linear equation for \(\lambda\).
  3. Substitute \(\lambda\) back into the line equation to find the coordinates of intersection.

Perpendicular Distance from a Point to a Plane

The perpendicular distance from a point \(P(\alpha, \beta, \gamma)\) to the plane \(ax + by + cz = d\) is given by:

\( \text{Distance} = \frac{|a\alpha + b\beta + c\gamma - d|}{\sqrt{a^2 + b^2 + c^2}} \)

Perpendicular Distance from a Point to a Line

To find the perpendicular distance from a point \(P\) with position vector \(\mathbf{p}\) to a line \(\mathbf{r} = \mathbf{a} + \lambda \mathbf{b}\):

  1. Let \(F\) be the foot of the perpendicular on the line, so \(\vec{OF} = \mathbf{a} + \lambda \mathbf{b}\).
  2. Find the vector \(\vec{PF} = (\mathbf{a} + \lambda \mathbf{b}) - \mathbf{p}\).
  3. Since \(\vec{PF}\) is perpendicular to the line, set \(\vec{PF} \cdot \mathbf{b} = 0\) and solve for \(\lambda\).
  4. Substitute \(\lambda\) back to get \(\vec{PF}\), then calculate its magnitude \(|\vec{PF}|\).

Perpendicular Distance Between Two Lines

  • Parallel lines: Choose any point on the first line and use the point-to-line method above to calculate its perpendicular distance to the second line.
  • Skew lines: Let \(P\) be a general point on line 1 (\(\mathbf{a}_1 + \lambda \mathbf{b}_1\)) and \(Q\) be a general point on line 2 (\(\mathbf{a}_2 + \mu \mathbf{b}_2\)). The vector \(\vec{PQ}\) gives the shortest distance when it is perpendicular to both direction vectors: \(\vec{PQ} \cdot \mathbf{b}_1 = 0\) and \(\vec{PQ} \cdot \mathbf{b}_2 = 0\). Solve these two equations simultaneously for \(\lambda\) and \(\mu\), then evaluate \(|\vec{PQ}|\).

You've reached the end of the Further Vectors notes! Practice visualizing lines and planes in 3D—sketching quick diagrams will help you master every distance and angle calculation.