Further Mathematics | Vectors

Welcome to FP3 Vectors: Navigating 3D Space!

Hello future mathematician! This chapter, Vectors, is where Further Pure Mathematics really opens up the three-dimensional world. If you found standard A-Level vectors straightforward, great! If you found them challenging, don't worry—we're going to build on those basics step-by-step.

In FP3, we introduce powerful new tools like the Vector Product and the Equation of a Plane. Mastering these concepts is essential for solving complex geometric problems involving volumes, areas, and relative positions of objects in space.

Let's dive in and make 3D geometry understandable!

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1. The Vector Product (The Cross Product)

You are already familiar with the Scalar Product (Dot Product), which results in a number (a scalar) and tells us about the angle between vectors. The Vector Product (or Cross Product) is different: it results in a new vector!

1.1 Definition and Calculation

If we have two vectors, \(\mathbf{a}\) and \(\mathbf{b}\), their vector product, \(\mathbf{a} \times \mathbf{b}\), is calculated using a determinant based on the unit vectors \(\mathbf{i}, \mathbf{j}, \mathbf{k}\).

Let \(\mathbf{a} = a_1\mathbf{i} + a_2\mathbf{j} + a_3\mathbf{k}\) and \(\mathbf{b} = b_1\mathbf{i} + b_2\mathbf{j} + b_3\mathbf{k}\).

The calculation is:

$$ \mathbf{a} \times \mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix} $$

Step-by-Step Calculation Trick:

To find the resulting vector components:

1. For the \(\mathbf{i}\) component: Cover the first column. Calculate \((a_2b_3 - a_3b_2)\).
2. For the \(\mathbf{j}\) component: Cover the second column. Calculate \((a_1b_3 - a_3b_1)\). CRITICAL STEP: Because the determinant calculation alternates sign, you must multiply this result by \(-1\), giving \(-(a_1b_3 - a_3b_1)\) or \((a_3b_1 - a_1b_3)\).
3. For the \(\mathbf{k}\) component: Cover the third column. Calculate \((a_1b_2 - a_2b_1)\).

Example: If \(\mathbf{a} = \mathbf{i} + 2\mathbf{j} + 0\mathbf{k}\) and \(\mathbf{b} = 3\mathbf{i} + 0\mathbf{j} + 4\mathbf{k}\).

$$\mathbf{a} \times \mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 2 & 0 \\ 3 & 0 & 4 \end{vmatrix}$$

i-comp: \((2)(4) - (0)(0) = 8\)
j-comp: \(-[(1)(4) - (0)(3)] = -4\)
k-comp: \((1)(0) - (2)(3) = -6\)

Therefore, \(\mathbf{a} \times \mathbf{b} = 8\mathbf{i} - 4\mathbf{j} - 6\mathbf{k}\).

1.2 Key Properties and Geometric Meaning

The Perpendicular Vector

The single most important fact about the cross product is that the resulting vector, \(\mathbf{a} \times \mathbf{b}\), is always perpendicular (orthogonal) to both \(\mathbf{a}\) and \(\mathbf{b}\).

Analogy: Imagine a floor (the plane containing \(\mathbf{a}\) and \(\mathbf{b}\)). The vector product is like a pole sticking straight up out of that floor.

Direction: The direction 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 \(\mathbf{a} \times \mathbf{b}\).

Order Matters! (Non-Commutative):

$$\mathbf{a} \times \mathbf{b} = -(\mathbf{b} \times \mathbf{a})$$

If you reverse the order, the resulting vector has the exact opposite direction (it points down instead of up).

Area of a Parallelogram

The magnitude of the vector product gives the area of the parallelogram formed by the two vectors \(\mathbf{a}\) and \(\mathbf{b}\) when placed tail-to-tail:

$$\text{Area} = |\mathbf{a} \times \mathbf{b}|$$

Tip for finding the Area of a Triangle: Since a triangle is half a parallelogram, the area of the triangle formed by \(\mathbf{a}\) and \(\mathbf{b}\) is:

$$\text{Area} = \frac{1}{2}|\mathbf{a} \times \mathbf{b}|$$

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2. The Equation of a Plane

A plane is a flat, infinite surface in 3D space. To uniquely define a plane, we don't need three points (though that helps); mathematically, we need a point on the plane and a vector that is perpendicular to the plane.

This perpendicular vector is called the normal vector, denoted \(\mathbf{n}\).

2.1 The Vector Form of a Plane

Let \(\mathbf{a}\) be the position vector of a known fixed point \(A\) on the plane, and let \(\mathbf{r}\) be the position vector of any arbitrary point \(P\) on the plane.

The vector \(\vec{AP} = \mathbf{r} - \mathbf{a}\) must lie completely within the plane.

Since \(\mathbf{n}\) is perpendicular to the plane, it must be perpendicular to every vector lying in the plane, including \(\vec{AP}\).

Therefore, their scalar product must be zero:

$$(\mathbf{r} - \mathbf{a}) \cdot \mathbf{n} = 0$$

Rearranging gives the standard vector equation of a plane:

$$\mathbf{r} \cdot \mathbf{n} = \mathbf{a} \cdot \mathbf{n}$$

Since \(\mathbf{a} \cdot \mathbf{n}\) is just a constant number, we usually write:

$$\mathbf{r} \cdot \mathbf{n} = d$$

where \(d = \mathbf{a} \cdot \mathbf{n}\).

2.2 The Cartesian Form of a Plane

If we let \(\mathbf{r} = x\mathbf{i} + y\mathbf{j} + z\mathbf{k}\) and the normal vector \(\mathbf{n} = a\mathbf{i} + b\mathbf{j} + c\mathbf{k}\), the scalar product \(\mathbf{r} \cdot \mathbf{n} = d\) expands to the Cartesian equation of the plane:

$$ax + by + cz = d$$

Important Connection: The coefficients of \(x, y, z\) in the Cartesian equation are simply the components of the normal vector \(\mathbf{n}\).

2.3 Finding the Equation from Three Points

Often, you are given three non-collinear points \(A, B, C\) and asked to find the equation of the plane containing them.

Step-by-Step Process:

1. Find two vectors lying in the plane: For example, \(\mathbf{u} = \vec{AB}\) and \(\mathbf{v} = \vec{AC}\).
2. Calculate the Normal Vector (\(\mathbf{n}\)): Use the vector product: \(\mathbf{n} = \mathbf{u} \times \mathbf{v}\). (Remember, the cross product is perpendicular to both vectors, so it must be perpendicular to the plane).
3. Find the constant \(d\): Use the equation \(\mathbf{r} \cdot \mathbf{n} = d\). Substitute the coordinates of ANY of the three known points (\(\mathbf{a}\) or \(\mathbf{b}\) or \(\mathbf{c}\)) for \(\mathbf{r}\).
4. Write the final equation: Substitute \(\mathbf{n}\) and \(d\) into \(ax + by + cz = d\).

Did you know? If three points were collinear (on the same line), they wouldn't define a unique plane. Since \(\mathbf{u}\) and \(\mathbf{v}\) would be parallel, their cross product \(\mathbf{u} \times \mathbf{v}\) would be the zero vector, and you couldn't find a normal!

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3. The Scalar Triple Product

The scalar triple product involves three vectors, \(\mathbf{a}, \mathbf{b},\) and \(\mathbf{c}\). It results in a scalar (a number).

3.1 Definition and Calculation

The scalar triple product is defined as:

$$\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})$$

To calculate this, you calculate the vector product \(\mathbf{b} \times \mathbf{c}\) first, and then take the dot product of that resulting vector with \(\mathbf{a}\).

The easiest way to calculate it is using the 3x3 determinant where the rows are the components of the vectors \(\mathbf{a}, \mathbf{b},\) and \(\mathbf{c}\) in that order:

$$ \mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) = \begin{vmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{vmatrix} $$

Memory Aid: It's a scalar (number) result, so there are no \(\mathbf{i}, \mathbf{j}, \mathbf{k}\) columns in the determinant.

3.2 Geometric Meaning: Volume of a Parallelepiped

The absolute value (magnitude) of the scalar triple product is equal to the volume of the parallelepiped (a skewed cuboid, like a stack of playing cards) defined by the three vectors \(\mathbf{a}, \mathbf{b},\) and \(\mathbf{c}\) sharing the same vertex.

$$\text{Volume} = |\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})|$$

Why? The magnitude of \(\mathbf{b} \times \mathbf{c}\) gives the area of the base, and the dot product with \(\mathbf{a}\) calculates the projection of \(\mathbf{a}\) onto the normal vector, which gives the perpendicular height. Volume = Base Area \(\times\) Perpendicular Height.

3.3 Coplanarity Test

If the three vectors \(\mathbf{a}, \mathbf{b},\) and \(\mathbf{c}\) all lie in the same plane (they are coplanar), they cannot form a volume. The parallelepiped collapses!

This provides a powerful test:

If \(\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) = 0\), then the vectors \(\mathbf{a}, \mathbf{b},\) and \(\mathbf{c}\) are coplanar.

This is a very common exam application. If you are asked to prove that four points lie in the same plane, first form three vectors between them (e.g., \(\vec{AB}, \vec{AC}, \vec{AD}\)), and then show the scalar triple product of those three vectors is zero.

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4. Working with Angles and Distances in 3D

Once you have mastered the equation of a plane, you can solve problems involving intersections, angles, and distances.

4.1 Angle Between Two Planes (\(\Pi_1\) and \(\Pi_2\))

Don't worry about the planes themselves! The angle between two planes is the angle between their normal vectors, \(\mathbf{n}_1\) and \(\mathbf{n}_2\).

We use the dot product formula to find the angle \(\theta\):

$$\mathbf{n}_1 \cdot \mathbf{n}_2 = |\mathbf{n}_1| |\mathbf{n}_2| \cos \theta$$

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

We use the absolute value \(|\mathbf{n}_1 \cdot \mathbf{n}_2|\) to ensure we always find the acute angle between the planes (which is generally the required answer unless specified otherwise).

4.2 Angle Between a Line (\(L\)) and a Plane (\(\Pi\))

This is tricky! The angle the line makes with the plane is NOT the angle between the direction vector of the line (\(\mathbf{d}\)) and the normal vector of the plane (\(\mathbf{n}\)).

Step-by-Step Process:

1. Find the angle \(\alpha\) between the line's direction vector \(\mathbf{d}\) and the plane's normal vector \(\mathbf{n}\) using the dot product: $$\cos \alpha = \frac{|\mathbf{d} \cdot \mathbf{n}|}{|\mathbf{d}| |\mathbf{n}|}$$
2. Since \(\mathbf{n}\) is perpendicular to the plane, the angle \(\theta\) the line makes with the plane is the complement of \(\alpha\). $$\theta = 90^\circ - \alpha \quad \text{or} \quad \sin \theta = \cos \alpha$$

Avoid this Common Mistake: Using the dot product calculation directly gives the angle between the line and the *normal*. You must convert it to the angle with the *plane* itself using \(90^\circ - \alpha\) or the sine relationship.

4.3 Distance from a Point to a Plane

The shortest distance from a point \(P(x_0, y_0, z_0)\) to a plane \(\Pi: ax + by + cz = d\) is given by a simple formula derived from vector projection.

If you write the plane equation as \(ax + by + cz - d = 0\), the distance \(D\) is:

$$ D = \frac{|ax_0 + by_0 + cz_0 - d|}{\sqrt{a^2 + b^2 + c^2}} $$

Explanation: The numerator takes the coordinates of the point \(P\) and substitutes them into the plane equation. The denominator is simply the magnitude of the normal vector \(\mathbf{n} = (a, b, c)\).

Tip: If the point is on the plane, the numerator will be zero, and the distance is zero—which makes perfect sense!

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Summary and Next Steps

You have now explored the advanced geometry of FP3 vectors!

Key Takeaways:

• The Vector Product gives you a vector perpendicular to two others, and its magnitude gives the area of the parallelogram. It is crucial for finding the normal vector (\(\mathbf{n}\)) of a plane.
• The Equation of a Plane relies entirely on its normal vector: \(\mathbf{r} \cdot \mathbf{n} = d\).
• The Scalar Triple Product helps find the volume of a parallelepiped and determines if three vectors are coplanar (if the volume is zero).

The biggest challenge in this chapter is knowing *which* tool to use. If a question involves area or finding a perpendicular, think Cross Product. If it involves angles or distance, think Dot Product and the relevant formula. Keep practicing those determinant calculations, and you will master 3D space!