Vector Arithmetic and Geometric Proofs for the ESAT

A vector is a method for describing the displacement required to move from one point to another. Unlike scalars, vectors describe both a specific size and a specific direction. In printed text, vectors are usually denoted by bold letters, such as $\mathbf{b}$. In handwriting, vectors are indicated by underlining the letter, for example $\underline{b}$, because underlining is the standard printing proof symbol for bold text. Vectors are typically represented either as column vectors or diagrammatically using line segments.

Operations with Column Vectors

Column vectors allow for precise algebraic manipulation. To add or subtract column vectors, you simply perform the operation across the corresponding rows of the vectors.

$\binom{a}{b} + \binom{d}{c} - \binom{f}{e} = \binom{a + d - f}{b + c - e}$

Multiplying by a Scalar

A scalar is a physical quantity that possesses magnitude only and no direction. For example, velocity is a vector because it has a direction, whereas speed is a scalar. When a column vector is multiplied by a scalar, every component within the vector is multiplied by that value.

If $\mathbf{x} = \binom{a}{b}$ then $6\mathbf{x} = \binom{6a}{6b}$ and $-\frac{1}{2}\mathbf{x} = \binom{-\frac{1}{2}a}{-\frac{1}{2}b}$

Parallel Vectors

Identifying parallel vectors is a common requirement in geometric problems. If two vectors $\mathbf{x}$ and $\mathbf{y}$ are parallel, then $\mathbf{x} = c\mathbf{y}$ where $c$ is a constant scalar. This means one vector is simply a scaled version of the other, pointing in either the same or the exact opposite direction.

Diagrammatic Representation

When a vector $\mathbf{a}$ is shown diagrammatically by a line segment $OA$ going from point $O$ to point $A$, any other line segment that is parallel to $OA$, has the same length, and points in the same direction also represents the vector $\mathbf{a}$.

If a line segment is parallel to $OA$ and points in the same direction but is twice as long, it represents the vector $2\mathbf{a}$. If it is parallel but points in the opposite direction and is twice as long, it represents $-2\mathbf{a}$.

Triangular Addition and Subtraction

Vectors can be combined geometrically using the triangle or parallelogram rules. Consider a parallelogram $OACB$ where $OA = \mathbf{a}$ and $OB = \mathbf{b}$.

The vector $OC$ represents the journey from $O$ to $C$. An alternative route is to travel from $O$ to $B$ (vector $\mathbf{b}$) and then from $B$ to $C$. Since $BC$ is parallel to $OA$, equal in length, and in the same direction, $BC = \mathbf{a}$. Therefore, the total journey is:

$OC = OB + BC = \mathbf{b} + \mathbf{a} = \mathbf{a} + \mathbf{b}$

Similarly, we can find vectors between other points:

$AB = AO + OB = -\mathbf{a} + \mathbf{b} = \mathbf{b} - \mathbf{a}$

$BA = BO + OA = -\mathbf{b} + \mathbf{a} = \mathbf{a} - \mathbf{b}$

To add vectors $\mathbf{a}$ and $\mathbf{b}$, you can either use the parallelogram construction or draw a triangle where the start of $\mathbf{b}$ is placed at the end of $\mathbf{a}$. The resultant vector is the line joining the start of the first vector to the end of the second.

Worked Examples: Vector Calculations

Example 1: Column Vector Arithmetic

Given $\mathbf{a} = \binom{2}{-3}$, $\mathbf{b} = \binom{-3}{5}$, and $\mathbf{c} = \binom{1}{6}$, find $\mathbf{a} - \mathbf{b} + \mathbf{c}$.

$\mathbf{a} - \mathbf{b} + \mathbf{c} = \binom{2}{-3} - \binom{-3}{5} + \binom{1}{6} = \binom{2 - (-3) + 1}{-3 - 5 + 6} = \binom{6}{-2}$

Example 2: Scalar Multiples

Using the same vectors, find $2\mathbf{a} - 3\mathbf{b}$.

First, calculate the individual multiples: $2\mathbf{a} = \binom{4}{-6}$ $3\mathbf{b} = \binom{-9}{15}$

Then subtract them: $2\mathbf{a} - 3\mathbf{b} = \binom{4}{-6} - \binom{-9}{15} = \binom{4 - (-9)}{-6 - 15} = \binom{13}{-21}$

Example 3: Identifying Parallel Vectors

If $\mathbf{x} = \binom{2}{-3}$ and $\mathbf{y} = \binom{-6}{-1}$, which of the following are parallel to $\mathbf{x} - \mathbf{y}$?

1. $\binom{16}{-4}$
2. $\binom{-4}{1}$
3. $\binom{8}{2}$

First, find $\mathbf{x} - \mathbf{y}$: $\mathbf{x} - \mathbf{y} = \binom{2}{-3} - \binom{-6}{-1} = \binom{8}{-2}$

Now compare: $\binom{16}{-4} = 2 \times \binom{8}{-2}$, so this is parallel. $\binom{-4}{1} = -\frac{1}{2} \times \binom{8}{-2}$, so this is parallel. $\binom{8}{2}$ is not a multiple of $\binom{8}{-2}$, so it is not parallel.

Worked Examples: Diagrammatic Representation

Example 4: Labelling Vectors

Label the lines in the following diagram using $\mathbf{a}$, $\mathbf{b}$, and $\mathbf{c}$.

1. Line (i) is parallel to $\mathbf{a}$ and in the same direction but half the length, so it is $\frac{1}{2}\mathbf{a}$.
2. Line (ii) is parallel to $\mathbf{b}$ in the opposite direction and 3 times as long, so it is $-3\mathbf{b}$.
3. Line (iii) is parallel to $\mathbf{c}$ and in the same direction but $\frac{2}{5}$ of the length, so it is $\frac{2}{5}\mathbf{c}$.
4. Line (iv) is parallel to $\mathbf{a}$ in the opposite direction and 3 times as long, so it is $-3\mathbf{a}$.

Example 5: Constructing Resultants

Draw a vector equal to $2\mathbf{a} + 3\mathbf{b}$ based on the following diagram.

Draw the vector $2\mathbf{a}$ (twice the length of $\mathbf{a}$, same direction). Then draw $3\mathbf{b}$ starting from the end of $2\mathbf{a}$. The resulting path represents the vector sum.

Geometric Proofs Using Vectors

Vectors are powerful tools for proving geometric theorems. Consider the following problem: in triangle $ABC$, $D$ is the midpoint of $AB$ and $E$ is the midpoint of $AC$. Prove that $DE$ is parallel to $BC$ and half its length.

Let $AB = \mathbf{x}$ and $AC = \mathbf{y}$.

1. Express $BC$ in terms of $\mathbf{x}$ and $\mathbf{y}$: $BC = BA + AC = -\mathbf{x} + \mathbf{y} = \mathbf{y} - \mathbf{x}$

2. Express $AD$ and $AE$ as fractions of the full sides: $AD = \frac{1}{2}\mathbf{x}$ and $AE = \frac{1}{2}\mathbf{y}$

3. Express $DE$ in terms of the known vectors: $DE = DA + AE = -\frac{1}{2}\mathbf{x} + \frac{1}{2}\mathbf{y} = \frac{1}{2}(\mathbf{y} - \mathbf{x})$

4. Conclusion: Since $DE = \frac{1}{2}BC$, the vector $DE$ is a scalar multiple of $BC$. Therefore, $DE$ must be parallel to $BC$ and exactly half its length.