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Mathematics · 11 / 23

Vectors: displacement, not location

High school foundations · Worked example and 4 exercises with solutions, at your pace.

UnderstandSee an exampleGuided exercise3 progressive challenges

The principle, simply

A point locates a position; a vector describes direction, orientation and length of a displacement. AB coordinates are B minus A. Vectors are collinear when one is a scalar multiple of the other; a two-dimensional determinant tests this.

x=5-2
THE PICTURE TO REMEMBERVectors: displacement, not location
  1. 01Departure A
  2. 02Arrival B
  3. 03B minus A
  4. 04Direction and orientation

WORKED EXAMPLE

From principle to reasoning

A(2, 1), B(5, 3) give AB = (3, 2). (6, 4) is twice AB and has zero determinant with it.

Solve a problem in this subject

  1. Define the objects

    State unknowns, domain and data. Real square roots need nonnegative arguments; denominators must be nonzero.

  2. Choose a transformation

    Name the rule: expansion, equivalent equations, factoring, differentiation. Check its conditions. Squaring an equation requires testing candidates in the original.

  3. Show the reasoning

    Each line should follow from the previous one. Keep exact values where possible and distinguish equality, approximation and implication.

  4. Test and conclude

    Substitute into the original expression. Check graph points and domain boundaries. State a solution set or the required unit.

YOUR TURN

01 · Try before checking

For A(−1, 2), B(3, −2), find AB.

One hint

Arrival minus departure, coordinate by coordinate.

Reveal and understand the solution

AB = (4, −4). Reversing direction gives BA = (−4, 4).

Compare method, conditions and answer; identify the exact part to revisit.

IMPROVE THROUGH PRACTICE

Three additional challenges with solutions

Start with the first challenge. Write your method before opening a hint; use the solution to find the first step that needs work.

Reflect after practising
02 · Recognise and understandLink to this exercise

For A(1,2), B(4,6), find vector AB.

A hint to get started

Subtract start from end in each coordinate.

Step-by-step solution
  1. AB = (4 − 1, 6 − 2).
  2. Thus AB = (3,4).
  3. BA would be (−3,−4): direction matters.

Which step did you miss? Retry without help, then explain why the method works.

03 · Apply the methodLink to this exercise

For u = (2,−1), v = (−3,4), find 2u + v.

A hint to get started

Multiply both coordinates of u by 2.

Step-by-step solution
  1. 2u = (4,−2).
  2. Add v coordinate by coordinate.
  3. 2u + v = (1,2).

Which step did you miss? Retry without help, then explain why the method works.

04 · Go furtherLink to this exercise

Are u = (2,3), v = (4,6) collinear and in the same direction?

A hint to get started

Find a scalar factor or a zero determinant.

Step-by-step solution
  1. v = 2u, so they are collinear.
  2. The positive factor means the same direction.
  3. Their determinant 2 × 6 − 3 × 4 is also zero.

Which step did you miss? Retry without help, then explain why the method works.

The trap to avoid

Do not reverse the subtraction in only one coordinate or confuse length with coordinates.

Without notes, can you…

  • Explain the principle in your own words?
  • Rebuild the diagram and its connections?
  • Solve the exercise and justify the method?

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