Vectors and Matrices | Highers

A vector has both magnitude and direction. A scalar has only magnitude.

A vector in 2D can be written as $\mathbf{a} = \begin{pmatrix} a_1 \\ a_2 \end{pmatrix}$ or as the Column vector $(a_1, a_2)$. In 3D: $\mathbf{a} = \begin{pmatrix} a_1 \\ a_2 \\ a_3 \end{pmatrix}$.

Magnitude (Modulus):

In 3D: $|\mathbf{a}| = \sqrt{a_1^2 + a_2^2 + a_3^2}$.

Unit Vector:

A unit vector has magnitude 1. The standard unit vectors are $\mathbf{i} = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}$ $\mathbf{j} = \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}$ $\mathbf{k} = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}$.

Any vector in 3D can be written as $\mathbf{a} = a_1\mathbf{i} + a_2\mathbf{j} + a_3\mathbf{k}$.

Example: Find the unit vector in the direction of $\mathbf{a} = (3, -4)$.

$|\mathbf{a}| = \sqrt{9 + 16} = 5$

Addition:

Vector addition is commutative ($\mathbf{a} + \mathbf{b} = \mathbf{b} + \mathbf{a}$) and associative ($( \mathbf{a} + \mathbf{b}) + \mathbf{c} = \mathbf{a} + (\mathbf{b} + \mathbf{c})$).

Scalar Multiplication:

Scalar multiplication distributes over vector addition: $k(\mathbf{a} + \mathbf{b}) = k\mathbf{a} + k\mathbf{b}$.

Scalar Product (Dot Product):

Where $\theta$ is the angle between $\mathbf{a}$ and $\mathbf{b}$.

Proof of the dot product formula. By the cosine rule in the triangle formed by $\mathbf{a}$ $\mathbf{b}$ And $\mathbf{a} - \mathbf{b}$:

Expanding the left side: $(\mathbf{a} - \mathbf{b}) \cdot (\mathbf{a} - \mathbf{b}) = |\mathbf{a}|^2 - 2\mathbf{a} \cdot \mathbf{b} + |\mathbf{b}|^2$.

Comparing: $-2\mathbf{a} \cdot \mathbf{b} = -2|\mathbf{a}||\mathbf{b}|\cos\theta$Hence $\mathbf{a} \cdot \mathbf{b} = |\mathbf{a}||\mathbf{b}|\cos\theta$.

Example: Find the angle between $\mathbf{a} = (2, 1, -1)$ and $\mathbf{b} = (1, -3, 2)$.

The position vector of point $A$ relative to an origin $O$ is $\overrightarrow{OA} = \mathbf{a}$.

The displacement from $A$ to $B$ is $\overrightarrow{AB} = \mathbf{b} - \mathbf{a}$.

Triangle law of vector addition: $\overrightarrow{AB} + \overrightarrow{BC} = \overrightarrow{AC}$.

Example: Given $\overrightarrow{OA} = (3, 1, -2)$ and $\overrightarrow{OB} = (-1, 4, 3)$Find $\overrightarrow{AB}$and the distance $AB$.

• $\mathbf{a} \cdot \mathbf{a} = |\mathbf{a}|^2$
• $\mathbf{a} \cdot \mathbf{b} = 0$ if and only if $\mathbf{a}$ is perpendicular to $\mathbf{b}$ (for non-zero vectors)
• $\mathbf{a} \cdot \mathbf{b} = \mathbf{b} \cdot \mathbf{a}$ (commutative)
• $\mathbf{a} \cdot (\mathbf{b} + \mathbf{c}) = \mathbf{a} \cdot \mathbf{b} + \mathbf{a} \cdot \mathbf{c}$ (distributive)
• $\mathbf{a} \cdot (k\mathbf{b}) = k(\mathbf{a} \cdot \mathbf{b})$

Example: Determine the value of $k$ for which the vectors $\mathbf{a} = (k, 2, -1)$ and $\mathbf{b} = (3, k, 4)$ are perpendicular.

$\mathbf{a} \cdot \mathbf{b} = 0$

$3k + 2k - 4 = 0$

$5k = 4$

The position vector of a point that divides the line segment from $A$ to $B$ in the ratio $m : n$ Is:

The midpoint of $AB$ (when $m = n = 1$):

Example: Find the position vector of the point dividing the line from $A(1, 2, 3)$ to $B(7, -2, 5)$ in the ratio $2:1$.

Three points $A$, $B$, $C$ are collinear if and only if $\overrightarrow{AB}$is parallel to $\overrightarrow{AC}$I.e., $\overrightarrow{AB} = k\overrightarrow{AC}$for some scalar $k$.

Example: Determine whether $A(1, 2, -1)$, $B(3, 5, 1)$ And $C(5, 8, 3)$ are collinear.

$\overrightarrow{AB} = (2, 3, 2)$ $\overrightarrow{AC} = (4, 6, 4) = 2(2, 3, 2) = 2\overrightarrow{AB}$.

Since $\overrightarrow{AC} = 2\overrightarrow{AB}$The points are collinear. $B$ is the midpoint of $AC$.

A matrix is a rectangular array of numbers. An $m \times n$ matrix has $m$ rows and $n$ columns.

Addition: $C = A + B$ where $c_{ij} = a_{ij} + b_{ij}$ (matrices must have the same dimensions).

Scalar Multiplication: $kA$ has entries $ka_{ij}$.

Matrix Multiplication: If $A$ is $m \times n$ and $B$ is $n \times p$ Then $C = AB$ is $m \times p$ where:

Why the dimensions must match. The entry $c_{ij}$ is the dot product of row $i$ of $A$ with Column $j$ of $B$. For this dot product to be defined, row $i$ of $A$ and column $j$ of $B$ must Have the same length, which means the number of columns of $A$ equals the number of rows of $B$.

Matrix multiplication is associative ($(AB)C = A(BC)$) and distributive over addition ($A(B+C) = AB + AC$), but not commutative ($AB \neq BA$).

Example: Calculate $AB$ where $A = \begin{pmatrix} 2 & 1 \\ -1 & 3 \end{pmatrix}$ and $B = \begin{pmatrix} 1 & 4 \\ 0 & -2 \end{pmatrix}$.

Verify non-commutativity:

Determinant:

The determinant measures the area scaling factor of the linear transformation represented by $A$. If $\det A = 0$The transformation collapses the plane onto a line (or a point), and the matrix is not Invertible.

Inverse:

The inverse exists if and only if $\det A \neq 0$.

Verification: $AA^{-1} = \frac{1}{ad-bc}\begin{pmatrix} a & b \\ c & d \end{pmatrix}\begin{pmatrix} d & -b \\ -c & a \end{pmatrix} = \frac{1}{ad-bc}\begin{pmatrix} ad-bc & 0 \\ 0 & ad-bc \end{pmatrix} = I$.

Example: Find the inverse of $A = \begin{pmatrix} 3 & 5 \\ 1 & 2 \end{pmatrix}$.

$\det A = 3(2) - 5(1) = 6 - 5 = 1$

A system $A\mathbf{x} = \mathbf{b}$ has solution $\mathbf{x} = A^{-1}\mathbf{b}$ (provided $A$ is Invertible). If $\det A = 0$The system has either no solutions or infinitely many solutions.

Example: Solve the system:

$3x + 5y = 11$ $x + 2y = 5$

Solution: $x = -3$, $y = 4$.

2D Transformations:

The determinant of a transformation matrix gives the area scale factor. A negative determinant Indicates the transformation involves a reflection.

Example: Rotate the point $(3, 2)$ by $90^\circ$ anticlockwise about the origin.

The image is $(-2, 3)$.

Example: Reflect the point $(4, -1)$ in the $y$-axis.

The image is $(-4, -1)$.

Example: Use the matrix $\begin{pmatrix} 3 & 0 \\ 0 & 3 \end{pmatrix}$ to describe the Transformation.

This is an enlargement by scale factor 3 centred at the origin. The determinant is $9 > 0$ Confirming it is a pure enlargement (no reflection). Every point $(x, y)$ maps to $(3x, 3y)$.

The relationship with the identity matrix $\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$ is that This matrix represents an enlargement by scale factor 1 (i.e., the identity/no transformation).

The vector product of $\mathbf{a} = (a_1, a_2, a_3)$ and $\mathbf{b} = (b_1, b_2, b_3)$ is:

Properties:

• $\mathbf{a} \times \mathbf{b} = -(\mathbf{b} \times \mathbf{a})$ (anti-commutative)
• $|\mathbf{a} \times \mathbf{b}| = |\mathbf{a}||\mathbf{b}|\sin\theta$
• $\mathbf{a} \times \mathbf{b}$ is perpendicular to both $\mathbf{a}$ and $\mathbf{b}$
• $\mathbf{a} \times \mathbf{b} = \mathbf{0}$ if $\mathbf{a}$ and $\mathbf{b}$ are parallel

Geometric interpretation: $|\mathbf{a} \times \mathbf{b}|$ is the area of the parallelogram with Sides $\mathbf{a}$ and $\mathbf{b}$. The area of the triangle is $\frac{1}{2}|\mathbf{a} \times \mathbf{b}|$.

Example: Find $\mathbf{a} \times \mathbf{b}$ where $\mathbf{a} = (1, 2, 3)$ and $\mathbf{b} = (4, -1, 2)$.

Verification: $\mathbf{a} \cdot (\mathbf{a} \times \mathbf{b}) = 1(7) + 2(10) + 3(-9) = 7 + 20 - 27 = 0$. Confirmed perpendicular.

This equals the volume of the parallelepiped formed by vectors $\mathbf{a}$, $\mathbf{b}$ And $\mathbf{c}$.

The three vectors are coplanar if and only if $[\mathbf{a}, \mathbf{b}, \mathbf{c}] = 0$.

Properties:

• $[\mathbf{a}, \mathbf{b}, \mathbf{c}] = [\mathbf{b}, \mathbf{c}, \mathbf{a}] = [\mathbf{c}, \mathbf{a}, \mathbf{b}]$ (cyclic permutation)
• $[\mathbf{a}, \mathbf{b}, \mathbf{c}] = -[\mathbf{b}, \mathbf{a}, \mathbf{c}]$ (swapping two vectors negates)

Example: Show that the vectors $(1, 2, -1)$, $(3, 1, 2)$ And $(0, 5, -5)$ are coplanar.

Wait, that is not zero. Let me recompute. $\mathbf{b} = (3, 1, 2)$, $\mathbf{c} = (0, 5, -5)$.

Wait, let me be careful. The cross product formula gives:

$\mathbf{b} \times \mathbf{c} = (b_2 c_3 - b_3 c_2, b_3 c_1 - b_1 c_3, b_1 c_2 - b_2 c_1)$ $= (1 \cdot (-5) - 2 \cdot 5, 2 \cdot 0 - 3 \cdot (-5), 3 \cdot 5 - 1 \cdot 0)$ $= (-15, 15, 15)$.

$\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) = 1(-15) + 2(15) + (-1)(15) = -15 + 30 - 15 = 0$.

Confirmed coplanar. $\blacksquare$

Equation of a plane: $\mathbf{r} \cdot \mathbf{n} = d$Where $\mathbf{n}$ is the normal vector And $d$ is a constant.

In Cartesian form: $ax + by + cz = d$.

The normal vector $\mathbf{n} = (a, b, c)$ is perpendicular to every vector in the plane.

Angle between two planes: The angle between their normal vectors.

Angle between a line and a plane: If the line has direction $\mathbf{d}$ and the plane has Normal $\mathbf{n}$The angle $\phi$ between the line and the plane satisfies:

Distance from a point to a plane:

Where $\mathbf{r}_0$ is the position vector of the point and $d_0$ is the constant in the plane Equation.

Example: Find the equation of the plane through $(1, 2, -1)$, $(3, 0, 2)$ And $(0, 1, 4)$.

Using point $(1, 2, -1)$: $-7x - 13y - 4z = -7(1) - 13(2) - 4(-1) = -7 - 26 + 4 = -29$.

Two non-parallel planes intersect in a line. To find the line of intersection:

1. The direction vector is $\mathbf{d} = \mathbf{n}_1 \times \mathbf{n}_2$
2. Find a point on both planes by setting one variable (e.g., $z = 0$) and solving the resulting $2 \times 2$ system

Example: Find the line of intersection of $x + y + z = 6$ and $2x - y + z = 3$.

Direction: $\mathbf{d} = (1, 1, 1) \times (2, -1, 1) = (1+1, 2-1, -1-2) = (2, 1, -3)$.

Set $z = 0$: $x + y = 6$ and $2x - y = 3$. Adding: $3x = 9$ So $x = 3$, $y = 3$.

Line: $\mathbf{r} = (3, 3, 0) + t(2, 1, -3)$.

Example: Find the shortest distance from the point $(2, 1, 3)$ to the plane $2x - y + 2z = 5$.

$\mathbf{n} = (2, -1, 2)$, $|\mathbf{n}| = \sqrt{4 + 1 + 4} = 3$.

Example: Find the angle between the planes $2x - y + 2z = 5$ and $x + 3y - z = 2$.

$\mathbf{n}_1 = (2, -1, 2)$, $\mathbf{n}_2 = (1, 3, -1)$.

$|\mathbf{n}_1| = 3$, $|\mathbf{n}_2| = \sqrt{11}$.

$\mathbf{n}_1 \cdot \mathbf{n}_2 = 2 - 3 - 2 = -3$.

This is the cofactor expansion along the first row. You can expand along any row or column; choose The one with the most zeros for efficiency.

Properties of determinants:

• $\det(AB) = \det(A)\det(B)$
• $\det(A^{-1}) = \frac{1}{\det(A)}$
• $\det(A^T) = \det(A)$
• Swapping two rows negates the determinant
• A row of zeros gives determinant zero
• If two rows are equal, the determinant is zero

Where \mathrm{adj(A) is the adjugate (transpose of the cofactor matrix).

The cofactor $C_{ij}$ is $(-1)^{i+j}$ times the determinant of the $2 \times 2$ matrix obtained by Deleting row $i$ and column $j$.

Example: Find the inverse of $A = \begin{pmatrix} 1 & 0 & 2 \\ -1 & 3 & 1 \\ 2 & 1 & 0 \end{pmatrix}$.

$\det A = 1(0 - 1) - 0(0 - 2) + 2(-1 - 6) = -1 + 0 - 14 = -15$.

Cofactor matrix:

\mathrm{adj(A) = C^T = \begin{pmatrix} -1 & 2 & -6 \\ 2 & -4 & -3 \\ -7 & -1 & 3 \end{pmatrix}.

A scalar $\lambda$ is an eigenvalue of $A$ if there exists a non-zero vector $\mathbf{v}$ such That:

The vector $\mathbf{v}$ is called an eigenvector corresponding to $\lambda$.

Geometric interpretation: When $A$ acts on $\mathbf{v}$It only stretches or compresses $\mathbf{v}$ (by factor $\lambda$) without changing its direction.

Finding Eigenvalues: Solve the characteristic equation $\det(A - \lambda I) = 0$.

Example: Find the eigenvalues and eigenvectors of $A = \begin{pmatrix} 4 & 1 \\ 2 & 3 \end{pmatrix}$.

$\lambda = 5$ or $\lambda = 2$.

For $\lambda = 5$: $(A - 5I)\mathbf{v} = \mathbf{0}$:

$-v_1 + v_2 = 0$ So $v_1 = v_2$. Eigenvector: $\begin{pmatrix} 1 \\ 1 \end{pmatrix}$.

For $\lambda = 2$: $(A - 2I)\mathbf{v} = \mathbf{0}$:

$2v_1 + v_2 = 0$ So $v_2 = -2v_1$. Eigenvector: $\begin{pmatrix} 1 \\ -2 \end{pmatrix}$.

Example: Find the eigenvalues and eigenvectors of $\begin{pmatrix} 5 & 2 \\ 1 & 4 \end{pmatrix}$.

$\lambda = 6$ or $\lambda = 3$.

For $\lambda = 6$: $\begin{pmatrix} -1 & 2 \\ 1 & -2 \end{pmatrix}\mathbf{v} = \mathbf{0}$Giving $v_1 = 2v_2$. Eigenvector: $\begin{pmatrix} 2 \\ 1 \end{pmatrix}$.

For $\lambda = 3$: $\begin{pmatrix} 2 & 2 \\ 1 & 1 \end{pmatrix}\mathbf{v} = \mathbf{0}$Giving $v_1 = -v_2$. Eigenvector: $\begin{pmatrix} 1 \\ -1 \end{pmatrix}$.

If an $n \times n$ matrix $A$ has $n$ linearly independent eigenvectors, it can be diagonalised: $A = PDP^{-1}$Where $D$ is a diagonal matrix containing the eigenvalues and $P$ has the Eigenvectors as columns.

Applications:

• Computing $A^k = PD^kP^{-1}$ is efficient because $D^k$ is trivial (just raise diagonal entries to power $k$).
• Solving systems of differential equations.

See the examples integrated throughout the sections above.

Vectors are arrows in space: A vector is like an arrow — it has a direction and a length. The direction tells you where it points, and the length (magnitude) tells you how far. Vectors are essential for describing anything that has both direction and size: force, velocity, acceleration, displacement.

Why it matters: Vectors are the language of physics and engineering. They describe forces on structures, motion of objects, fields in space, and transformations in computer graphics. Understanding vector operations (addition, dot product, cross product) is fundamental to these fields.

The key insight: The dot product measures how aligned two vectors are (parallel = maximum, perpendicular = zero), while the cross product measures how perpendicular they are (parallel = zero, perpendicular = maximum).

1. Confusing scalar and vector products: The scalar product gives a scalar (number); the vector product gives a vector perpendicular to both inputs.

2. Matrix multiplication is not commutative: , $AB \neq BA$. Always multiply in the specified order.

3. Dimension mismatch: You can only multiply an $m \times n$ matrix by an $n \times p$ matrix. The “inner dimensions” must match.

4. Forgetting the determinant in the inverse: The inverse is \dfrac{1}{\det A} \mathrm{adj(A) not just \mathrm{adj(A). Forgetting the $1/\det A$ factor gives a wrong answer unless $\det A = 1$.

5. Normal vector direction: The normal to a plane can point in either direction; check consistency when computing angles. The angle between planes should be between $0$ and $\pi/2$.

6. Eigenvector scaling: Eigenvectors are not unique — any non-zero scalar multiple is also an eigenvector. Always state the direction, not a specific magnitude.

7. Sign errors in the cross product: The cross product is anti-commutative: $\mathbf{a} \times \mathbf{b} = -(\mathbf{b} \times \mathbf{a})$. Getting the order wrong negates the result.

8. Cofactor sign errors: The cofactor $C_{ij}$ includes a factor of $(-1)^{i+j}$. For position $(2, 3)$This is $(-1)^5 = -1$. Getting the sign wrong invalidates the entire inverse.

9. Confusing rotation direction: A positive angle in the rotation matrix represents anticlockwise rotation. For clockwise rotation by $\theta$Use $-\theta$ or swap the signs of the off-diagonal entries.

1. Given $\mathbf{a} = (2, -1, 3)$ and $\mathbf{b} = (4, 2, -1)$Find \mathbf{a} \cdot \mathbf{b}$$|\mathbf{a}|$$|\mathbf{b}| And the angle between them.

2. Find the equation of the plane containing the points (1, 0, 2)$$(3, 1, -1) And $(0, 2, 4)$.

3. Calculate $\mathbf{a} \times \mathbf{b}$ for $\mathbf{a} = (1, 3, -2)$ and $\mathbf{b} = (4, -1, 5)$. Verify that $\mathbf{a} \times \mathbf{b}$ is perpendicular to both $\mathbf{a}$ and $\mathbf{b}$.

4. Find the eigenvalues and eigenvectors of $\begin{pmatrix} 5 & 2 \\ 1 & 4 \end{pmatrix}$.

5. Compute the determinant and inverse of $\begin{pmatrix} 2 & 0 & 1 \\ -1 & 3 & 2 \\ 1 & 1 & -1 \end{pmatrix}$.

6. Show that the vectors (1, 2, -1)$$(3, 1, 2) And $(0, 5, -5)$ are coplanar.

7. Find the shortest distance from the point $(2, 1, 3)$ to the plane $2x - y + 2z = 5$.

8. Solve the system of equations using matrices:

$2x + y - z = 8$ $x - y + 3z = 1$ $3x + 2y + z = 11$

1. Find the line of intersection of the planes $x + 2y - z = 4$ and $3x - y + 2z = 1$.

2. Find the angle between the planes $2x - y + 2z = 5$ and $x + 3y - z = 2$.

3. Use the matrix $\begin{pmatrix} 3 & 0 \\ 0 & 3 \end{pmatrix}$ to describe the transformation. What is the relationship between this matrix and $\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$?

4. Find the area of the triangle with vertices $A(1, 0, 2)$, $B(3, -1, 4)$ And $C(0, 2, -1)$.

5. Given $\overrightarrow{OA} = (1, -1, 3)$ and $\overrightarrow{OB} = (4, 2, -1)$Find the position vector of the point $P$ on $AB$ such that $AP : PB = 3 : 1$.

6. Find the angle between the lines $\mathbf{r} = (0, 0, 0) + s(1, 2, -1)$ and $\mathbf{r} = (1, 1, 0) + t(2, -1, 3)$.

7. A $2 \times 2$ matrix $A$ has eigenvalues $\lambda_1 = 2$ and $\lambda_2 = 5$ with corresponding eigenvectors $\begin{pmatrix} 1 \\ 1 \end{pmatrix}$ and $\begin{pmatrix} 1 \\ -2 \end{pmatrix}$. Find $A$ and use it to compute $A^3$.

8. Reflect the point $(2, 5)$ in the line $y = x$ using a matrix method. Verify your answer geometrically.

flowchart TD
    A[4_Vectors] --> B[Key Concepts]
    A --> C[Core Principles]
    A --> D[Practical Applications]
    B --> E[Fundamental definitions]
    C --> F[Design patterns]
    D --> G[Real-world usage]

This topic covers the mathematical techniques and concepts related to vectors and matrices, including key theorems, methods, and problem-solving approaches.

Key concepts include:

• fundamental definitions and theorems
• algebraic and graphical methods
• proof and logical reasoning
• problem-solving strategies
• applications and modelling

Regular practice with a variety of question types is essential to build fluency and confidence in applying these mathematical techniques.

• Algebra and Functions — Vector operations use algebraic techniques including dot product and cross product.
• Trigonometry — Angles between vectors are calculated using trigonometric relationships and the dot product formula.
• Mechanics — Forces, displacement, and velocity are vector quantities analysed using vector algebra.