Welcome to Vectors and 3D Geometry!
Hello! This chapter takes your understanding of vectors—those powerful tools for representing direction and magnitude—and launches them into the real world: three-dimensional space. If you struggled with 2D vectors, don't worry! 3D vectors follow the same core rules, but we add some exciting new operations like the Vector Product (which is unique to 3D) and apply them to describe lines and planes.
Mastering this topic is essential because it forms the foundation for many high-level physics and engineering applications. Let's make 3D geometry simple!
1. The Vector Product (Cross Product): \(\mathbf{a} \times \mathbf{b}\)
In standard A-Level Maths, you met the Scalar Product (Dot Product), which results in a number (a scalar). The Further Maths vector product is fundamentally different: it results in a vector.
1.1 Definition and Properties
The vector product of two vectors, \(\mathbf{a}\) and \(\mathbf{b}\), is denoted by \(\mathbf{a} \times \mathbf{b}\). The resulting vector, \(\mathbf{n}\), has two crucial properties:
- It is perpendicular (normal) to both \(\mathbf{a}\) and \(\mathbf{b}\).
- Its magnitude \(|\mathbf{a} \times \mathbf{b}|\) equals the area of the parallelogram spanned by \(\mathbf{a}\) and \(\mathbf{b}\).
The Right-Hand Rule (Memory Aid)
Imagine pointing the fingers of your right hand in the direction of the first vector (\(\mathbf{a}\)), and curling them towards the second vector (\(\mathbf{b}\)). Your thumb points in the direction of the resultant vector (\(\mathbf{a} \times \mathbf{b}\)).
- Important Property: The vector product is anti-commutative. This means the order matters!
\(\mathbf{b} \times \mathbf{a} = -(\mathbf{a} \times \mathbf{b})\). - If \(\mathbf{a}\) and \(\mathbf{b}\) are parallel, \(\mathbf{a} \times \mathbf{b} = \mathbf{0}\).
1.2 Calculating the Vector 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 easiest way to calculate the product is using a 3x3 determinant structure (like finding the normal to a plane):
$$\mathbf{a} \times \mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix}$$
Step-by-step Process:
- For the \(\mathbf{i}\) component: Cover the \(\mathbf{i}\) column and calculate \((a_2 b_3 - a_3 b_2)\).
- For the \(\mathbf{j}\) component: Cover the \(\mathbf{j}\) column and calculate \(-(a_1 b_3 - a_3 b_1)\). (Remember the minus sign for the middle term!)
- For the \(\mathbf{k}\) component: Cover the \(\mathbf{k}\) column and calculate \((a_1 b_2 - a_2 b_1)\).
1.3 Applications of the Vector Product (Area)
The magnitude of the vector product relates directly to geometric area:
Area of Parallelogram: If the sides are defined by vectors \(\mathbf{a}\) and \(\mathbf{b}\), the Area is:
$$A = |\mathbf{a} \times \mathbf{b}|$$
Area of Triangle: If the sides are defined by vectors \(\mathbf{a}\) and \(\mathbf{b}\) (sharing a vertex), the Area is:
$$A = \frac{1}{2} |\mathbf{a} \times \mathbf{b}|$$
Key Takeaway (Vector Product)
The vector product gives you a new vector perpendicular to the first two, and its magnitude measures the area of the shape they form.
2. The Scalar Triple Product: \(\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})\)
This operation combines the vector product and the scalar product. It involves three vectors and results in a scalar (a number, not a vector).
2.1 Definition and Calculation
The scalar triple product \(\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})\) is calculated by setting up a 3x3 determinant using the components of the three vectors:
$$\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) = \begin{vmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{vmatrix}$$
2.2 Applications (Volume and Coplanarity)
The scalar triple product has two major geometric uses:
1. Volume of a Parallelepiped:
The volume \(V\) of the parallelepiped (a 3D solid similar to a slanted cuboid) formed by the three adjacent vectors \(\mathbf{a}\), \(\mathbf{b}\), and \(\mathbf{c}\) is given by the modulus of the scalar triple product:
$$V = |\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})|$$
Analogy: Think of the scalar triple product as calculating length \(\times\) width \(\times\) height, where the vector product gives the area of the base (b and c), and the scalar product projects the height onto the normal vector (\(\mathbf{a}\)).
2. Identifying Coplanar Vectors:
If three vectors \(\mathbf{a}\), \(\mathbf{b}\), and \(\mathbf{c}\) lie on the same plane, they cannot form a 3D solid, meaning the parallelepiped has zero volume.
Therefore, vectors \(\mathbf{a}\), \(\mathbf{b}\), and \(\mathbf{c}\) are coplanar if and only if:
$$\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) = 0$$
Key Takeaway (Scalar Triple Product)
The scalar triple product measures volume. If the result is zero, the vectors are flat (coplanar).
3. Equations of Lines in Three Dimensions
You are already familiar with the basic vector equation of a line, but here we cover the various forms required, including the one involving the vector product.
3.1 The Standard Vector Equation
A line passing through a point with position vector \(\mathbf{a}\) and moving in the direction of vector \(\mathbf{b}\) is given by:
$$\mathbf{r} = \mathbf{a} + \lambda \mathbf{b}$$
Where \(\mathbf{r}\) is the position vector of any point on the line, and \(\lambda\) is a scalar parameter.
3.2 The Vector Product Form of a Line
The syllabus specifically requires understanding this alternative form:
$$(\mathbf{r} - \mathbf{a}) \times \mathbf{b} = \mathbf{0}$$
Why does this work?
The vector \((\mathbf{r} - \mathbf{a})\) is the vector running from the fixed point \(\mathbf{a}\) to the arbitrary point \(\mathbf{r}\). This vector must lie along the line, meaning it must be parallel to the direction vector \(\mathbf{b}\).
We know that if two vectors are parallel, their vector product is \(\mathbf{0}\). Hence, this equation states mathematically that the line segment from \(\mathbf{a}\) to \(\mathbf{r}\) is parallel to \(\mathbf{b}\).
Quick Review: Line Equations
- Standard (Parametric): \(\mathbf{r} = \mathbf{a} + \lambda \mathbf{b}\)
- Vector Product Form (Parallelism): \((\mathbf{r} - \mathbf{a}) \times \mathbf{b} = \mathbf{0}\)
4. Equations of Planes in Three Dimensions
Unlike a line, a plane requires two independent direction vectors, or more commonly, a single normal vector (a vector perpendicular to the entire plane).
4.1 The Parametric Form of a Plane
A plane passing through a point \(\mathbf{a}\) and spanned by two non-parallel direction vectors \(\mathbf{b}\) and \(\mathbf{c}\) is given by:
$$\mathbf{r} = \mathbf{a} + \lambda \mathbf{b} + \mu \mathbf{c}$$
Where \(\lambda\) and \(\mu\) are independent scalar parameters.
Analogy: Think of \(\mathbf{a}\) as where you start, and \(\lambda \mathbf{b} + \mu \mathbf{c}\) as how you move across the surface. Since you have two directions (\(\mathbf{b}\) and \(\mathbf{c}\)), you can reach any point on that surface.
4.2 The Normal Vector Form (Scalar Product Form)
This is the most useful form for calculating distances and angles. It requires a normal vector, \(\mathbf{n}\), which is perpendicular to every vector in the plane.
$$\mathbf{r} \cdot \mathbf{n} = d$$
Where \(\mathbf{r}\) is any point on the plane, \(\mathbf{n}\) is the normal vector, and \(d\) is a scalar constant representing the perpendicular distance from the origin to the plane, multiplied by the magnitude of \(\mathbf{n}\).
- Finding \(d\): If the plane passes through point \(\mathbf{a}\), then \(d = \mathbf{a} \cdot \mathbf{n}\).
- Finding \(\mathbf{n}\) from Parametric Form: If you have \(\mathbf{r} = \mathbf{a} + \lambda \mathbf{b} + \mu \mathbf{c}\), the normal vector is simply the vector product of the two direction vectors: \(\mathbf{n} = \mathbf{b} \times \mathbf{c}\).
5. Intersection and Angles
We use the scalar and vector products to solve problems involving intersections and angles in 3D.
5.1 Intersection of a Line and a Plane
To find the point where a line \(\mathbf{r}_L = \mathbf{a} + \lambda \mathbf{b}\) meets a plane \(\mathbf{r} \cdot \mathbf{n} = d\):
- Substitute the line equation into the plane equation:
\((\mathbf{a} + \lambda \mathbf{b}) \cdot \mathbf{n} = d\) - Expand the dot product:
\(\mathbf{a} \cdot \mathbf{n} + \lambda (\mathbf{b} \cdot \mathbf{n}) = d\) - Solve the resulting linear equation for the parameter \(\lambda\).
- Substitute the value of \(\lambda\) back into the line equation \(\mathbf{r} = \mathbf{a} + \lambda \mathbf{b}\) to find the position vector of the intersection point.
Did you know? If the line is parallel to the plane, the term \((\mathbf{b} \cdot \mathbf{n})\) will be zero, as the direction vector \(\mathbf{b}\) is perpendicular to the normal \(\mathbf{n}\). In this case, either there is no solution (parallel and distinct) or infinite solutions (the line lies within the plane).
5.2 Angles Between Objects
When calculating angles, remember this golden rule: the scalar product always involves the directions that are perpendicular to the surface.
5.2.1 Angle Between Two Planes (\(\phi\))
This is the angle between their respective normal vectors, \(\mathbf{n}_1\) and \(\mathbf{n}_2\).
$$\cos \phi = \frac{|\mathbf{n}_1 \cdot \mathbf{n}_2|}{|\mathbf{n}_1| |\mathbf{n}_2|}$$
We usually take the modulus of the dot product to ensure we find the acute angle between the planes.
5.2.2 Angle Between a Line and a Plane (\(\theta\))
Let \(\mathbf{b}\) be the direction vector of the line and \(\mathbf{n}\) be the normal vector of the plane.
If we use the scalar product formula, we actually find the angle, \(\alpha\), between the line and the normal:
$$\cos \alpha = \frac{|\mathbf{b} \cdot \mathbf{n}|}{|\mathbf{b}| |\mathbf{n}|}$$
Since the normal is perpendicular to the plane, the actual angle we want, \(\theta\), is \(90^\circ - \alpha\) (or \(\frac{\pi}{2} - \alpha\)).
Common Mistake Alert! Do not forget the final step: \(\theta = 90^\circ - \alpha\). If the question asks for the angle between the line and the plane, finding \(\alpha\) is only the first part!
5.3 Line of Intersection of Two Planes
Two non-parallel planes intersect to form a straight line. To find the equation of this line, you need its direction vector and one point on it.
- Find the direction vector (\(\mathbf{b}\)): The direction of the line of intersection must be perpendicular to both plane normals, \(\mathbf{n}_1\) and \(\mathbf{n}_2\). Therefore, the direction vector is:
$$\mathbf{b} = \mathbf{n}_1 \times \mathbf{n}_2$$ - Find a point on the line (\(\mathbf{a}\)): You can usually do this by setting one coordinate (e.g., \(z=0\)) in both plane equations and solving the resulting simultaneous equations for the other two coordinates (e.g., \(x\) and \(y\)).
- Form the line equation: \(\mathbf{r} = \mathbf{a} + \lambda \mathbf{b}\).
6. Direction Ratios and Direction Cosines
These concepts provide a way to formally describe the orientation of a vector or line in space.
6.1 Direction Ratios \((l, m, n)\)
These are simply the components of any direction vector \(\mathbf{b}\) for the line. If \(\mathbf{b} = a\mathbf{i} + b\mathbf{j} + c\mathbf{k}\), then the direction ratios are \((a, b, c)\). Any scalar multiple \((ka, kb, kc)\) are also direction ratios for the same line.
6.2 Direction Cosines
The direction cosines are the cosines of the angles (\(\alpha, \beta, \gamma\)) that the line makes with the positive \(x, y,\) and \(z\) axes, respectively. We label them as \(l\), \(m\), and \(n\).
- $$l = \cos \alpha = \frac{a}{|\mathbf{b}|}$$
- $$m = \cos \beta = \frac{b}{|\mathbf{b}|}$$
- $$n = \cos \gamma = \frac{c}{|\mathbf{b}|}$$
The direction cosines are effectively the components of the unit vector in the direction of the line.
The Fundamental Identity
Since direction cosines are components of a unit vector, their squares must sum to 1. This is a crucial identity you must know:
$$l^2 + m^2 + n^2 = 1$$
Key Takeaway (Direction Cosines)
Direction cosines are just normalized direction ratios. They tell you the exact angles a line makes with the axes, and they always satisfy \(l^2 + m^2 + n^2 = 1\).
You’ve navigated the complexities of 3D space! Remember, every problem boils down to finding key vectors (position, direction, or normal) and applying the appropriate product (scalar or vector). Keep practicing those calculations!