Welcome to Vectors in 3D Space!
Welcome to one of the most visual and rewarding topics in AS 1: Pure Mathematics! In standard A-Level Mathematics, you met vectors in 2D and basic 3D, using the scalar product (dot product) to find angles. In Further Mathematics, we take things to the next dimension.
Here, you will discover powerful tools that engineers, game developers, and physicists use every single day:
• The Vector Product (Cross Product) to find vectors perpendicular to surfaces and calculate areas.
• The Scalar Triple Product to find 3D volumes.
• Vector and Cartesian Equations of Lines and Planes in three dimensions.
• Methods to find intersections, angles, and shortest distances between objects in 3D space.
Don't worry if working in 3D feels a bit mind-bending at first! We will break down every single concept into small, easy-to-follow steps with clear examples.
---1. The Vector Product (Cross Product)
What is the Vector Product?
You already know the scalar product (\(\mathbf{a} \cdot \mathbf{b}\)), which multiplies two vectors to give a single number (a scalar). The vector product (written with a cross: \(\mathbf{a} \times \mathbf{b}\)) takes two 3D vectors and produces a new vector that is strictly perpendicular (at \(90^\circ\)) to both original vectors!
Analogy: Imagine opening a door. One vector points along the door from the hinge to the handle. Another vector is the direction you push. The vector product points straight along the hinge axis (up or down) — perpendicular to both the door and your push!
How to Calculate \(\mathbf{a} \times \mathbf{b}\)
Let \(\mathbf{a} = \begin{pmatrix} a_1 \\ a_2 \\ a_3 \end{pmatrix}\) and \(\mathbf{b} = \begin{pmatrix} b_1 \\ b_2 \\ b_3 \end{pmatrix}\).
We calculate \(\mathbf{a} \times \mathbf{b}\) using a \(3 \times 3\) determinant:
\(\mathbf{a} \times \mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix} = \mathbf{i}(a_2 b_3 - a_3 b_2) - \mathbf{j}(a_1 b_3 - a_3 b_1) + \mathbf{k}(a_1 b_2 - a_2 b_1)\)
Step-by-Step Rule:
1. For the \(\mathbf{i}\)-component: cover up the \(\mathbf{i}\) column, evaluate the \(2 \times 2\) determinant: \(a_2 b_3 - a_3 b_2\).
2. For the \(\mathbf{j}\)-component: cover up the \(\mathbf{j}\) column, evaluate \(a_1 b_3 - a_3 b_1\), and remember the minus sign: \(-(a_1 b_3 - a_3 b_1)\).
3. For the \(\mathbf{k}\)-component: cover up the \(\mathbf{k}\) column, evaluate \(a_1 b_2 - a_2 b_1\).
Quick Example:
Let \(\mathbf{a} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}\) and \(\mathbf{b} = \begin{pmatrix} 4 \\ 0 \\ -1 \end{pmatrix}\).
\(\mathbf{a} \times \mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 2 & 3 \\ 4 & 0 & -1 \end{vmatrix}\)
• \(\mathbf{i}\)-component: \((2)(-1) - (3)(0) = -2\)
• \(\mathbf{j}\)-component: \(-[(1)(-1) - (3)(4)] = -[-1 - 12] = +13\)
• \(\mathbf{k}\)-component: \((1)(0) - (2)(4) = -8\)
So, \(\mathbf{a} \times \mathbf{b} = \begin{pmatrix} -2 \\ 13 \\ -8 \end{pmatrix}\).
Check your answer: Take the dot product of your result with \(\mathbf{a}\):
\((-2)(1) + (13)(2) + (-8)(3) = -2 + 26 - 24 = 0\). It is indeed perpendicular!
Key Properties of the Vector Product
• Anti-commutative: \(\mathbf{b} \times \mathbf{a} = -(\mathbf{a} \times \mathbf{b})\). Swapping the order reverses the direction!
• Self-cross: \(\mathbf{a} \times \mathbf{a} = \mathbf{0}\).
• Parallel Vectors: If \(\mathbf{a}\) and \(\mathbf{b}\) are parallel, \(\mathbf{a} \times \mathbf{b} = \mathbf{0}\).
• Magnitude Formula: \(|\mathbf{a} \times \mathbf{b}| = |\mathbf{a}| |\mathbf{b}| \sin\theta\), where \(\theta\) is the angle between the vectors.
Geometric Applications of \(\mathbf{a} \times \mathbf{b}\)
• Area of a Parallelogram: The area of a parallelogram with adjacent sides \(\mathbf{a}\) and \(\mathbf{b}\) is:
\(\text{Area} = |\mathbf{a} \times \mathbf{b}|\)
• Area of a Triangle: A triangle formed by vectors \(\mathbf{a}\) and \(\mathbf{b}\) is half of the parallelogram:
\(\text{Area} = \frac{1}{2} |\mathbf{a} \times \mathbf{b}|\)
• Triangle formed by three points \(A, B, C\):
\(\text{Area} = \frac{1}{2} |\vec{AB} \times \vec{AC}|\)
Key Takeaway: The cross product produces a vector perpendicular to two given vectors. Its magnitude gives the area of the parallelogram spanned by those vectors.
---2. The Scalar Triple Product
Definition and Determinant Form
The scalar triple product combines the dot product and the cross product of three vectors \(\mathbf{a}\), \(\mathbf{b}\), and \(\mathbf{c}\):
\(\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})\)
Instead of calculating the cross product first and then the dot product, you can evaluate it in one go using the determinant of a \(3 \times 3\) matrix containing the components of all three vectors:
\(\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) = \begin{vmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{vmatrix}\)
Geometric Meaning & Applications
• Volume of a Parallelepiped: A parallelepiped is a 3D slanted box with 6 parallelogram faces. If three adjacent edges are given by vectors \(\mathbf{a}\), \(\mathbf{b}\), and \(\mathbf{c}\), then:
\(\text{Volume} = |\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})|\)
Note: We take the absolute value because volume must always be positive!
• Volume of a Tetrahedron (Pyramid with triangular base):
\(\text{Volume} = \frac{1}{6} |\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})|\)
• Coplanar Vectors: If three vectors lie in the same flat 2D plane (they are coplanar), the 3D volume of the box they form is zero:
\(\mathbf{a}, \mathbf{b}, \mathbf{c} \text{ are coplanar} \iff \mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) = 0\)
Key Takeaway: The scalar triple product gives a scalar value. Its absolute value represents the volume of a parallelepiped, and a result of zero means the three vectors lie in the same plane.
---3. Straight Lines in 3D Space
Vector Equation of a Line
To define a straight line in 3D, you need two things:
1. A fixed starting point \(A\) with position vector \(\mathbf{a}\).
2. A direction vector \(\mathbf{d}\).
The vector equation of a line is:
\(\mathbf{r} = \mathbf{a} + \lambda \mathbf{d}\)
where \(\lambda \in \mathbb{R}\) is a scalar parameter, and \(\mathbf{r} = \begin{pmatrix} x \\ y \\ z \end{pmatrix}\) represents the position vector of any general point on the line.
Cartesian Form of a Line
Writing out the components:
\(x = a_1 + \lambda d_1 \implies \lambda = \frac{x - a_1}{d_1}\)
\(y = a_2 + \lambda d_2 \implies \lambda = \frac{y - a_2}{d_2}\)
\(z = a_3 + \lambda d_3 \implies \lambda = \frac{z - a_3}{d_3}\)
Equating these gives the Cartesian equation of a line:
\(\frac{x - a_1}{d_1} = \frac{y - a_2}{d_2} = \frac{z - a_3}{d_3} \quad (= \lambda)\)
Special Case: If a direction component is zero (e.g., \(d_1 = 0\)), write: \(x = a_1, \; \frac{y - a_2}{d_2} = \frac{z - a_3}{d_3}\).
Relationships Between Two Lines in 3D
In 2D, two lines are either parallel or they intersect. In 3D, there is a third exciting possibility: skew lines (lines that are not parallel, but never meet because they pass each other on different levels, like planes flying at different altitudes!).
Given two lines \(L_1: \mathbf{r} = \mathbf{a} + \lambda \mathbf{d}_1\) and \(L_2: \mathbf{r} = \mathbf{b} + \mu \mathbf{d}_2\):
• Parallel: Direction vectors are scalar multiples: \(\mathbf{d}_1 = k \mathbf{d}_2\).
• Intersecting: Set the expressions for \(\mathbf{r}\) equal to each other. Solve two equations for \(\lambda\) and \(\mu\). If these values satisfy the third equation, the lines intersect.
• Skew: Not parallel AND do not intersect.
Key Takeaway: A 3D line needs a point and a direction. Lines in 3D can be parallel, intersecting, or skew.
---4. Planes in 3D Space
What is a Plane?
A plane is a flat, infinite 2D surface extending in 3D space. To fix a plane in space, you need:
1. A known point on the plane with position vector \(\mathbf{a}\).
2. A normal vector \(\mathbf{n} = \begin{pmatrix} n_1 \\ n_2 \\ n_3 \end{pmatrix}\), which is perpendicular to every single line lying in the plane.
Equations of a Plane
1. Vector Form (Scalar Product Form):
If \(\mathbf{r}\) is any point on the plane, the vector \((\mathbf{r} - \mathbf{a})\) lies in the plane, so \((\mathbf{r} - \mathbf{a}) \cdot \mathbf{n} = 0\). This simplifies to:
\(\mathbf{r} \cdot \mathbf{n} = d\)
where \(d = \mathbf{a} \cdot \mathbf{n}\) is a constant scalar.
2. Cartesian Form:
Expanding \(\begin{pmatrix} x \\ y \\ z \end{pmatrix} \cdot \begin{pmatrix} n_1 \\ n_2 \\ n_3 \end{pmatrix} = d\) gives:
\(n_1 x + n_2 y + n_3 z = d\)
Tip: The coefficients of \(x, y, z\) in the Cartesian equation are simply the components of the normal vector \(\mathbf{n}\)!
3. Parametric / 2-Vector Form:
A plane can also be defined by a point \(\mathbf{a}\) and two non-parallel direction vectors \(\mathbf{u}\) and \(\mathbf{v}\) lying within the plane:
\(\mathbf{r} = \mathbf{a} + \lambda \mathbf{u} + \mu \mathbf{v}\)
To convert this to \(\mathbf{r} \cdot \mathbf{n} = d\) form, find the normal vector using the cross product: \(\mathbf{n} = \mathbf{u} \times \mathbf{v}\)!
Finding the Equation of a Plane Through Three Points \(A, B, C\)
Step-by-Step Method:
1. Find two direction vectors in the plane: \(\vec{AB} = \mathbf{b} - \mathbf{a}\) and \(\vec{AC} = \mathbf{c} - \mathbf{a}\).
2. Find the normal vector: \(\mathbf{n} = \vec{AB} \times \vec{AC}\).
3. Calculate \(d = \mathbf{a} \cdot \mathbf{n}\).
4. Write the equation: \(\mathbf{r} \cdot \mathbf{n} = d\) or \(n_1 x + n_2 y + n_3 z = d\).
Key Takeaway: The normal vector \(\mathbf{n}\) is the key to working with planes. The Cartesian equation \(n_1 x + n_2 y + n_3 z = d\) directly reveals the normal vector \(\mathbf{n} = \begin{pmatrix} n_1 \\ n_2 \\ n_3 \end{pmatrix}\).
---5. Angles in 3D Space
Whenever calculating angles in 3D, we always use the scalar product formula:
\(\cos\theta = \frac{|\mathbf{u} \cdot \mathbf{v}|}{|\mathbf{u}| |\mathbf{v}|}\)
1. Angle Between Two Lines
Use the direction vectors \(\mathbf{d}_1\) and \(\mathbf{d}_2\) of the two lines:
\(\cos\theta = \frac{|\mathbf{d}_1 \cdot \mathbf{d}_2|}{|\mathbf{d}_1| |\mathbf{d}_2|}\)
2. Angle Between Two Planes
The angle between two planes is equal to the angle between their normal vectors \(\mathbf{n}_1\) and \(\mathbf{n}_2\):
\(\cos\theta = \frac{|\mathbf{n}_1 \cdot \mathbf{n}_2|}{|\mathbf{n}_1| |\mathbf{n}_2|}\)
3. Angle Between a Line and a Plane (Watch out!)
Common Exam Trap: The line has direction vector \(\mathbf{d}\), but the plane has a normal vector \(\mathbf{n}\) (which is perpendicular to the plane).
The angle \(\phi\) between \(\mathbf{d}\) and \(\mathbf{n}\) is \(90^\circ - \theta\), where \(\theta\) is the angle between the line and the plane surface.
Since \(\cos(90^\circ - \theta) = \sin\theta\), we use \(\sin\theta\):
\(\sin\theta = \frac{|\mathbf{d} \cdot \mathbf{n}|}{|\mathbf{d}| |\mathbf{n}|}\)
Memory Trick:
• Same types (Line & Line, or Plane & Plane) \(\implies \mathbf{\cos\theta}\)
• Different types (Line & Plane) \(\implies \mathbf{\sin\theta}\)
Key Takeaway: Use \(\cos\theta\) for line-line and plane-plane angles, but switch to \(\sin\theta\) for line-plane angles.
---6. Intersections and Distances
Intersection of a Line and a Plane
To find the point where a line \(\mathbf{r} = \mathbf{a} + \lambda \mathbf{d}\) meets a plane \(\mathbf{r} \cdot \mathbf{n} = d\):
1. Substitute \(\mathbf{r} = \begin{pmatrix} a_1 + \lambda d_1 \\ a_2 + \lambda d_2 \\ a_3 + \lambda d_3 \end{pmatrix}\) into the Cartesian equation of the plane.
2. Solve the linear equation for \(\lambda\).
3. Substitute \(\lambda\) back into the line equation to get the coordinates of the intersection point.
Perpendicular Distance from a Point \(P(x_1, y_1, z_1)\) to a Plane \(ax + by + cz = d\)
The shortest (perpendicular) distance \(D\) from point \(P\) to the plane is given by the formula:
\(D = \frac{|a x_1 + b y_1 + c z_1 - d|}{\sqrt{a^2 + b^2 + c^2}}\)
Did you know? This is the exact 3D analogue of the 2D distance formula from a point to a line: \(\frac{|ax_1 + by_1 - c|}{\sqrt{a^2 + b^2}}\)!
Distance Between Two Parallel Planes
If two planes are parallel, their equations can be written as \(ax + by + cz = d_1\) and \(ax + by + cz = d_2\).
The perpendicular distance between them is:
\(D = \frac{|d_1 - d_2|}{\sqrt{a^2 + b^2 + c^2}}\)
Key Takeaway: To find intersections, substitute the line's parametric form into the plane equation. For perpendicular distances, use the standard modulus formula.
---Quick Review & Common Mistakes to Avoid
Top Exam Mistakes:
• Sign errors in the cross product: Forgetting the negative sign on the middle \(\mathbf{j}\)-component.
• Confusing \(\sin\) and \(\cos\): Using \(\cos\theta\) instead of \(\sin\theta\) when finding the angle between a line and a plane.
• Forgetting absolute value bars: Volumes and distances can never be negative; always apply modulus signs where appropriate.
• Using points instead of direction vectors: When finding line equations or angles, make sure you use the direction vector \(\mathbf{d}\), not a position vector \(\mathbf{a}\).
Formula Quick Reference:
• \(\text{Area of Triangle} = \frac{1}{2} |\mathbf{a} \times \mathbf{b}|\)
• \(\text{Volume of Parallelepiped} = |\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})|\)
• \(\text{Volume of Tetrahedron} = \frac{1}{6} |\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})|\)
• \(\text{Coplanar Vectors} \iff \mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) = 0\)
• \(\text{Line-Plane Angle} \implies \sin\theta = \frac{|\mathbf{d} \cdot \mathbf{n}|}{|\mathbf{d}| |\mathbf{n}|}\)