Mathematics - Further (9231) | Vectors

Further Pure Mathematics 1: Chapter 1.6 Vectors

Hello and welcome to the Vectors chapter! You’ve met vectors before in your AS Level studies, focusing mainly on lines and the dot product. In Further Maths, we take vectors into the big leagues: 3D space, dealing with planes, calculating areas, and finding the shortest distance between two non-intersecting lines (called skew lines).

This chapter requires strong visualization skills. Don’t worry if it feels abstract—we’ll break down every concept, use analogies, and focus on the formulas you need to master for Paper 1. Mastering these techniques is vital for solving complex geometry problems!

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1. Review: The Scalar Product (Dot Product)

Before jumping into the new material, let's quickly recall the fundamental tool for measuring angles and perpendicularity, the Scalar Product.

If \(\mathbf{a} = a_1\mathbf{i} + a_2\mathbf{j} + a_3\mathbf{k}\) and \(\mathbf{b} = b_1\mathbf{i} + b_2\mathbf{j} + b_3\mathbf{k}\), the scalar product is: \[\mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3\]

It is also defined geometrically by: \[\mathbf{a} \cdot \mathbf{b} = |\mathbf{a}||\mathbf{b}|\cos \theta\] Where \(\theta\) is the angle between the two vectors.

• Key Use: If two non-zero vectors are perpendicular (at 90°), then \(\cos 90^\circ = 0\), so \(\mathbf{a} \cdot \mathbf{b} = 0\). This is your go-to test for perpendicularity!

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2. The Vector Product (Cross Product) \(\mathbf{a} \times \mathbf{b}\)

The vector product is the most powerful new tool in this chapter. Unlike the scalar product (which yields a number), the vector product yields a vector.

2.1 Geometric Definition and Direction

The vector product \(\mathbf{a} \times \mathbf{b}\) is a vector whose magnitude and direction are defined as follows:

• Magnitude: \(|\mathbf{a} \times \mathbf{b}| = |\mathbf{a}||\mathbf{b}|\sin \theta\). This magnitude represents the area of the parallelogram formed by vectors \(\mathbf{a}\) and \(\mathbf{b}\).
• Direction: The resulting vector is perpendicular to both \(\mathbf{a}\) and \(\mathbf{b}\). We denote this direction by the unit vector \(\mathbf{n}\).

Thus, the syllabus defines the vector product as: \[\mathbf{a} \times \mathbf{b} = |\mathbf{a}||\mathbf{b}|\sin \theta \mathbf{n}\]

Did you know? The direction of \(\mathbf{n}\) is determined by the Right-Hand Rule. If you curl your fingers from \(\mathbf{a}\) to \(\mathbf{b}\), your thumb points in the direction of \(\mathbf{a} \times \mathbf{b}\). This means that \(\mathbf{b} \times \mathbf{a} = - (\mathbf{a} \times \mathbf{b})\). Order matters!

2.2 Component Form Calculation

Calculating \(\mathbf{a} \times \mathbf{b}\) is usually done using components. This process looks complex but is mechanical once you learn the pattern.

If \(\mathbf{a} = a_1\mathbf{i} + a_2\mathbf{j} + a_3\mathbf{k}\) and \(\mathbf{b} = b_1\mathbf{i} + b_2\mathbf{j} + b_3\mathbf{k}\), then: \[\mathbf{a} \times \mathbf{b} = (a_2b_3-a_3b_2)\mathbf{i} + (a_3b_1-a_1b_3)\mathbf{j} + (a_1b_2-a_2b_1)\mathbf{k}\]

Struggling Student Tip: If this formula seems daunting, use the determinant method (even if you don't write it out as a determinant in the exam). Write out the components of \(\mathbf{a}\) and \(\mathbf{b}\) twice, and cross-multiply, covering the column corresponding to the unit vector you are calculating:

For \(\mathbf{i}\): Ignore the first column, calculate \((a_2b_3 - a_3b_2)\).
For \(\mathbf{j}\): Ignore the middle column, calculate \((a_3b_1 - a_1b_3)\). (Note the swapped order to account for the sign change in the determinant calculation.)
For \(\mathbf{k}\): Ignore the last column, calculate \((a_1b_2 - a_2b_1)\).

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3. Equations of Planes

A plane is an infinite 2D surface in 3D space. It can be defined in several equivalent ways. The syllabus requires you to understand and convert between three main forms.

3.1 Cartesian Form: \(ax + by + cz = d\)

This is the standard algebraic form.

• The coefficients \((a, b, c)\) form the normal vector, \(\mathbf{n}\).
• The constant \(d\) determines the position of the plane relative to the origin.

3.2 Vector Normal Form: \(\mathbf{r} \cdot \mathbf{n} = p\)

This is often the most useful form for calculations.

• \(\mathbf{r}\) is the position vector of any point \((x, y, z)\) on the plane.
• \(\mathbf{n}\) is the normal vector (perpendicular to the plane).
• \(p\) is a scalar constant. If \(\mathbf{n}\) is a unit vector, \(p\) is the shortest distance from the origin to the plane. In general, \(p = \mathbf{a} \cdot \mathbf{n}\), where \(\mathbf{a}\) is any position vector lying in the plane.

Conversion Tip: If you have \(\mathbf{r} \cdot (a\mathbf{i} + b\mathbf{j} + c\mathbf{k}) = p\), expanding \(\mathbf{r}\) as \(x\mathbf{i} + y\mathbf{j} + z\mathbf{k}\) immediately gives you the Cartesian form: \(ax + by + cz = p\).

3.3 Parametric Vector Form: \(\mathbf{r} = \mathbf{a} + \lambda \mathbf{b} + \mu \mathbf{c}\)

This form defines the plane using two direction vectors.

• \(\mathbf{a}\) is the position vector of a fixed point on the plane.
• \(\mathbf{b}\) and \(\mathbf{c}\) are non-parallel direction vectors that lie in the plane.
• \(\lambda\) and \(\mu\) are independent scalar parameters.
• Analogy: If a line is defined by moving in one direction (\(\mathbf{d}\)), a plane is defined by moving in two independent directions (\(\mathbf{b}\) and \(\mathbf{c}\)).

Conversion from Parametric to Normal Form:

The key linkage is the Vector Product! Since \(\mathbf{b}\) and \(\mathbf{c}\) lie in the plane, their cross product must give the normal vector \(\mathbf{n}\): \[\mathbf{n} = \mathbf{b} \times \mathbf{c}\] Once you have \(\mathbf{n}\), calculate \(p = \mathbf{a} \cdot \mathbf{n}\), and you have the equation \(\mathbf{r} \cdot \mathbf{n} = p\).

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4. Solving Problems Involving Lines, Planes, and Angles

This is where you combine the Scalar Product and the Vector Product to solve complex geometrical scenarios.

4.1 Relationships Between a Line and a Plane

Given a line \(\mathbf{L}: \mathbf{r} = \mathbf{a} + \lambda \mathbf{d}\) (direction \(\mathbf{d}\)) and a plane \(\mathbf{\Pi}: \mathbf{r} \cdot \mathbf{n} = p\) (normal \(\mathbf{n}\)), three cases can occur:

1. Intersection: The line pierces the plane at one point.
2. Parallel (No Intersection): The line is parallel to the plane but does not lie in it.
3. Line Lies in the Plane: The line is parallel to the plane and every point lies on the plane.

How to determine the case:

• Step 1: Check for Parallelism. If the line is parallel to the plane, its direction vector \(\mathbf{d}\) must be perpendicular to the normal vector \(\mathbf{n}\).
  Check: \(\mathbf{d} \cdot \mathbf{n} = 0\)?
  If \(\mathbf{d} \cdot \mathbf{n} \neq 0\), the line intersects the plane (Case 1).


• Step 2: If Parallel, Check if the Line Lies in the Plane. If \(\mathbf{d} \cdot \mathbf{n} = 0\), the line is either parallel or lies within. Take a known point \(\mathbf{a}\) on the line and check if it satisfies the plane equation.
  Check: Does \(\mathbf{a} \cdot \mathbf{n} = p\)?
  If Yes, the line lies in the plane (Case 3). If No, the line is strictly parallel (Case 2).

4.2 Finding the Point of Intersection of a Line and a Plane

Substitute the general point of the line (\(\mathbf{a} + \lambda \mathbf{d}\)) into the plane equation (\(\mathbf{r} \cdot \mathbf{n} = p\)):
\[(\mathbf{a} + \lambda \mathbf{d}) \cdot \mathbf{n} = p\]
Rearrange this to solve for the parameter \(\lambda\). Once \(\lambda\) is found, substitute it back into the line equation to get the intersection coordinates.

4.3 Foot of the Perpendicular from a Point to a Plane

Imagine shining a laser straight from a point P onto the plane. The 'foot' is where the beam hits.

Process:

1. Identify the given Point \(P\) (position vector \(\mathbf{p}\)) and the Plane (\(\mathbf{r} \cdot \mathbf{n} = p\)).
2. Since the perpendicular line must be parallel to the plane's normal, write the equation of the line L passing through P with direction \(\mathbf{n}\): \[\mathbf{r} = \mathbf{p} + \lambda \mathbf{n}\]
3. Find the intersection point of this new line L and the plane \(\Pi\) (using the method in 4.2).
4. This intersection point is the foot of the perpendicular.

4.4 Angles Involving Planes

When dealing with angles, remember this golden rule: The normal vector \(\mathbf{n}\) is your best friend.

Angle between Two Planes \(\Pi_1\) and \(\Pi_2\)

The angle between two planes is equal to the angle between their normal vectors, \(\mathbf{n}_1\) and \(\mathbf{n}_2\).
Use the scalar product formula: \[\cos \theta = \frac{|\mathbf{n}_1 \cdot \mathbf{n}_2|}{|\mathbf{n}_1||\mathbf{n}_2|}\]

Note: We use the absolute value \(| \mathbf{n}_1 \cdot \mathbf{n}_2 |\) to ensure the angle \(\theta\) calculated is always acute (between 0° and 90°), which is standard convention for the angle between planes.

Angle between a Line L (direction \(\mathbf{d}\)) and a Plane \(\Pi\) (normal \(\mathbf{n}\))

This is a common trap! The angle calculated using \(\mathbf{d} \cdot \mathbf{n}\) is NOT the required angle.

If you use the formula \(\cos \alpha = \frac{|\mathbf{d} \cdot \mathbf{n}|}{|\mathbf{d}||\mathbf{n}|}\), you find the angle \(\alpha\) between the line and the normal vector. Since the normal is perpendicular to the plane, the angle you actually need, \(\theta\), is: \[\theta = 90^\circ - \alpha\]

Alternatively, you can just calculate \(\sin \theta\): \[\sin \theta = \frac{|\mathbf{d} \cdot \mathbf{n}|}{|\mathbf{d}||\mathbf{n}|}\] (Since \(\sin(90^\circ - \alpha) = \cos \alpha\).)

4.5 Intersection of Two Planes (Finding the Line)

When two non-parallel planes intersect, they form a straight line, L.

The line of intersection must be perpendicular to both plane normals, \(\mathbf{n}_1\) and \(\mathbf{n}_2\).

• Step 1: Find the direction vector, \(\mathbf{d}\). Use the Vector Product: \[\mathbf{d} = \mathbf{n}_1 \times \mathbf{n}_2\]
• Step 2: Find a common point, \(\mathbf{a}\). Choose a value for one coordinate (e.g., \(z=0\) or \(y=0\)) and solve the resulting pair of simultaneous Cartesian equations for the other two coordinates.
• Step 3: Write the line equation. Use the common point \(\mathbf{a}\) and the direction \(\mathbf{d}\): \(\mathbf{r} = \mathbf{a} + \lambda \mathbf{d}\).

5. Shortest Distance Between Two Skew Lines

Two lines are skew if they are not parallel and do not intersect (think of non-adjacent edges on a cube). Finding the shortest distance between them is often considered the trickiest part of the Vectors chapter.

Let the lines be: \[\mathbf{L}_1: \mathbf{r} = \mathbf{a}_1 + \lambda \mathbf{d}_1\] \[\mathbf{L}_2: \mathbf{r} = \mathbf{a}_2 + \mu \mathbf{d}_2\]

5.1 Finding the Direction of the Common Perpendicular

The line representing the shortest distance must be perpendicular to both \(\mathbf{d}_1\) and \(\mathbf{d}_2\). We find the direction of this common perpendicular, \(\mathbf{N}\), using the vector product: \[\mathbf{N} = \mathbf{d}_1 \times \mathbf{d}_2\]

5.2 Calculating the Shortest Distance (D)

The shortest distance D is the projection of the vector connecting the two fixed points \((\mathbf{a}_2 - \mathbf{a}_1)\) onto the common perpendicular direction \(\mathbf{N}\).

The formula is: \[D = \left| \frac{(\mathbf{a}_2 - \mathbf{a}_1) \cdot \mathbf{N}}{|\mathbf{N}|} \right|\]

Analogy: Imagine Line 1 and Line 2 are two pipes. \(\mathbf{a}_2 - \mathbf{a}_1\) is a vector connecting any point on Pipe 1 to any point on Pipe 2. The shortest distance D is found by projecting this connecting vector onto the shortest, most perpendicular path, \(\mathbf{N}\).

5.3 Finding the Equation of the Common Perpendicular

This is an extension, requiring you to find the specific points P on \(\mathbf{L}_1\) and Q on \(\mathbf{L}_2\) that define the shortest distance.

1. Let \(P\) and \(Q\) be general points on \(\mathbf{L}_1\) and \(\mathbf{L}_2\) respectively, using parameters \(\lambda\) and \(\mu\).
2. The vector \(\vec{PQ}\) must be parallel to the common perpendicular \(\mathbf{N}\) (i.e., \(\vec{PQ} = k\mathbf{N}\) or \(\vec{PQ} \cdot \mathbf{d}_1 = 0\) and \(\vec{PQ} \cdot \mathbf{d}_2 = 0\)).
3. Set up the two equations: \(\vec{PQ} \cdot \mathbf{d}_1 = 0\) and \(\vec{PQ} \cdot \mathbf{d}_2 = 0\).
4. Solve the simultaneous equations for \(\lambda\) and \(\mu\).
5. Substitute \(\lambda\) and \(\mu\) back into the line equations to find the coordinates of P and Q.
6. The equation of the common perpendicular is \(\mathbf{r} = \mathbf{p} + \tau (\mathbf{q} - \mathbf{p})\).