Welcome to the World of Geometric Proofs!

In your previous studies, you’ve learned how to add vectors and find their magnitudes. Now, we are going to use those skills to solve "geometric puzzles." In this chapter, we use vectors to prove whether lines are parallel, whether points lie on a straight line, and whether multiple lines meet at the same point. These are powerful tools used in engineering, navigation, and even computer game design!

1. Parallel Lines

Two vectors are parallel if one is simply a scaled-up or scaled-down version of the other. In vector language, we say they are scalar multiples.

If you have two vectors, \( \mathbf{a} \) and \( \mathbf{b} \), they are parallel if:
\( \mathbf{a} = k\mathbf{b} \)
...where \( k \) is any constant (a number) other than zero.

How to spot them:
  • If \( \vec{PQ} = 3\mathbf{a} + 2\mathbf{b} \) and \( \vec{RS} = 6\mathbf{a} + 4\mathbf{b} \), notice that \( \vec{RS} = 2(3\mathbf{a} + 2\mathbf{b}) \). Since \( \vec{RS} = 2\vec{PQ} \), the lines \( PQ \) and \( RS \) are parallel.
  • If \( k \) is negative, the lines are still parallel, but the vectors point in opposite directions!

Quick Tip: If two vectors are parallel and they also share a common point, they aren't just parallel—they are the same line!

2. Collinearity (Points on a Straight Line)

The word collinear sounds fancy, but it just means "lying on the same straight line." To prove that three points \( A \), \( B \), and \( C \) are collinear, you need to show two things:

  1. The vector \( \vec{AB} \) is parallel to the vector \( \vec{BC} \) (or \( \vec{AC} \)).
  2. The vectors share a common point (like \( B \)).
Example:

If \( \vec{AB} = 2\mathbf{a} + \mathbf{b} \) and \( \vec{AC} = 6\mathbf{a} + 3\mathbf{b} \):
1. We see that \( \vec{AC} = 3(2\mathbf{a} + \mathbf{b}) \), so \( \vec{AC} = 3\vec{AB} \). This means they are parallel.
2. Since both vectors start at point \( A \), the points \( A, B, \) and \( C \) must lie on the same straight line.

Key Takeaway: Parallel + Common Point = Collinear.

3. The Power of Non-Parallel Vectors

This is a "secret weapon" for solving exam questions. The syllabus states that if \( \mathbf{a} \) and \( \mathbf{b} \) are non-parallel vectors, and we have an equation like this:

\( \alpha_1 \mathbf{a} + \beta_1 \mathbf{b} = \alpha_2 \mathbf{a} + \beta_2 \mathbf{b} \)

Then the coefficients must be equal! This means:
\( \alpha_1 = \alpha_2 \) and \( \beta_1 = \beta_2 \)

Why is this useful?

Don't worry if this seems abstract. In exams, you will often find two different "paths" to the same point. By setting the two vector paths equal to each other, you can create two simple algebraic equations to find unknown values. It’s like saying "If 2 apples and 3 bananas cost the same as \( x \) apples and \( y \) bananas, then \( x \) must be 2 and \( y \) must be 3."

4. Concurrency

Lines are concurrent if they all intersect at the same single point. In Further Pure Maths, we usually prove this by showing that the point where two lines meet is the exact same point where the third line passes through.

Common Strategy for Exam Questions:

Often, you are asked to find the position vector of an intersection point \( X \). To do this, you might:

  1. Find the position vector of \( X \) using one line (e.g., \( \vec{OX} = \vec{OA} + \lambda\vec{AB} \)).
  2. Find the position vector of \( X \) using a second line (e.g., \( \vec{OX} = \vec{OC} + \mu\vec{CD} \)).
  3. Equate the two expressions and solve for \( \lambda \) and \( \mu \).

Note: For more on dividing lines into ratios (which is often the first step here), check the chapter "Dividing a line in a given ratio with vectors".

5. Summary and Common Pitfalls

Quick Review:

  • Parallel: \( \vec{U} = k\vec{V} \).
  • Collinear: \( \vec{AB} = k\vec{BC} \) (Parallel + Common Point).
  • Independence: If \( \mathbf{a} \) and \( \mathbf{b} \) are not parallel, you can "compare coefficients" on both sides of an equals sign.
Common Mistakes to Avoid:
  • Forgetting the common point: Just showing two vectors are parallel doesn't prove points are collinear. You must mention the common point.
  • Mixing up scalars and vectors: Remember that \( \lambda \) and \( \mu \) are just numbers (scalars), while \( \mathbf{a} \) and \( \mathbf{b} \) are directions (vectors). Keep your notation clear!
  • Misreading the ratio: If a point divides a line \( AB \) in the ratio \( 2:3 \), the vector \( \vec{AP} \) is actually \( \frac{2}{5} \) of the way along the line, not \( \frac{2}{3} \). Always add the ratio parts together (\( 2+3=5 \)) for the denominator!

Did you know? The concept of concurrency is why the three medians of any triangle always meet perfectly at a single point called the centroid! You can prove this using the very vector methods you are learning right now.