Geometric Proofs with Vectors

Worked Example: Constructing a Vector Path

In a triangle $OAB$, where $\vec{OA} = \mathbf{a}$ and $\vec{OB} = \mathbf{b}$. Find the vector $\vec{AB}$ in its simplest form.

Step 1: Construct the vector journey. To get from $A$ to $B$, travel from $A$ to $O$, then $O$ to $B$.

$$\vec{AB} = \vec{AO} + \vec{OB}$$

Step 2: Substitute the known vectors, remembering to reverse the sign for $\vec{AO}$. Since $\vec{OA} = \mathbf{a}$, reversing the direction means $\vec{AO} = -\mathbf{a}$.

$$\vec{AB} = -\mathbf{a} + \mathbf{b}$$

Step 3: State the final resultant vector.

$$\vec{AB} = \mathbf{b} - \mathbf{a}$$

Ratios and Midpoints in Vectors

Dividing a piece of string into a $2:3$ ratio requires cutting it into $5$ equal segments first. In vector geometry, if a point $P$ divides a line segment $AB$ in the ratio $m:n$, you must divide by the total number of parts ($m+n$) to find the correct fraction of the journey.

The segment $\vec{AP}$ is calculated as $\frac{m}{m+n}\vec{AB}$. If $M$ is the midpoint of line $AB$, the ratio is exactly $1:1$, which splits the total journey into two equal parts. This means $\vec{AM} = \frac{1}{2}\vec{AB}$ and $\vec{MB} = \frac{1}{2}\vec{AB}$.

Worked Example: Finding a Vector via Ratio

Point $P$ lies on the line $AB$ such that $AP:PB = 2:3$. Given $\vec{OA} = \mathbf{a}$ and $\vec{OB} = \mathbf{b}$, find $\vec{OP}$.

Step 1: Find the full vector $\vec{AB}$.

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

Step 2: Calculate the fractional vector $\vec{AP}$ using the total ratio parts ($2+3=5$).

$$\vec{AP} = \frac{2}{5}\vec{AB} = \frac{2}{5}(\mathbf{b} - \mathbf{a})$$

Step 3: Construct the path for $\vec{OP}$ and simplify.

$$\vec{OP} = \vec{OA} + \vec{AP}$$ $$\vec{OP} = \mathbf{a} + \frac{2}{5}(\mathbf{b} - \mathbf{a}) = \frac{3}{5}\mathbf{a} + \frac{2}{5}\mathbf{b}$$

Scalar Multiples and Parallelism

You can instantly tell two rulers are parallel by looking at them, but proving lines are strictly parallel mathematically requires algebraic conditions. Two vectors are parallel if and only if one is a scalar multiple of the other, meaning one vector can be multiplied by a constant $k$ to equal the other ($\mathbf{v} = k\mathbf{u}$).

The scalar $k$ changes the magnitude (length) of the vector. If $k > 0$, the vectors point in the same direction, but if $k < 0$, they point in opposite directions while maintaining parallelism.

To prove parallelism algebraically, factorise both vector expressions to reveal a common vector bracket. For example, $4\mathbf{a} + 6\mathbf{b}$ and $6\mathbf{a} + 9\mathbf{b}$ are parallel because they can be factorised into $2(2\mathbf{a} + 3\mathbf{b})$ and $1.5(2\mathbf{a} + 3\mathbf{b})$, revealing the shared directional vector $(2\mathbf{a} + 3\mathbf{b})$.

Worked Example: Proving Parallelism

Given $\vec{MN} = 2\mathbf{a} + 3\mathbf{b}$ and $\vec{XY} = 4\mathbf{a} + 6\mathbf{b}$, prove that $MN$ is parallel to $XY$.

Step 1: Factorise the larger vector to find a scalar multiple of the smaller vector.

$$\vec{XY} = 2(2\mathbf{a} + 3\mathbf{b})$$

Step 2: Substitute the smaller vector into the equation.

$$\vec{XY} = 2\vec{MN}$$

Step 3: State the formal conclusion. Since $\vec{XY}$ is a scalar multiple of $\vec{MN}$, the two vectors are parallel.

Geometric Proofs: Collinearity

If three distant stars perfectly align in the night sky, how do we mathematically prove they lie on the exact same straight line? When three or more points lie on a single straight line, they are collinear.

To construct a causal geometric proof for collinearity, two distinct conditions must be met and explicitly written down. First, you must prove that two vector segments connecting these points are parallel (one is a scalar multiple of the other). Second, you must explicitly state that these two segments share a common point (a shared vertex).

Why are both conditions required? Because if two lines are mathematically parallel and physically pass through the exact same point, they cannot be separate parallel lines — they must be sections of the same straight line.

Worked Example: Proving Collinearity

Given points $O, X,$ and $Y$ where $\vec{OX} = 2\mathbf{a} + \mathbf{b}$ and $\vec{OY} = 6\mathbf{a} + 3\mathbf{b}$, prove that $O, X,$ and $Y$ are collinear.

Step 1: Factorise the larger vector to show parallelism.

$$\vec{OY} = 3(2\mathbf{a} + \mathbf{b})$$

Step 2: Identify the scalar multiple relationship.

$$\vec{OY} = 3\vec{OX}$$

Step 3: Write the final causal proof statement mentioning both conditions. Since $\vec{OY}$ is a scalar multiple of $\vec{OX}$, the vectors are parallel. Because they both share the common point $O$, the points $O, X,$ and $Y$ are collinear.