Scale a vector and handle sign and zero
Multiply every component by one scalar · Use |c| for the magnitude factor · Distinguish same, opposite and undefined directions
Exam scope and assessment notes
Extended content; not a Core requirement. Historical study of Cambridge IGCSE Mathematics (US) 0444, 2025 Version 2. The final assessment was June 2025; the syllabus was available only to schools in Arizona, USA. This is not a current exam-entry option. Original self-check teaching and explained practice are not Cambridge papers, official marks or predicted grades. Linked teaching foundations do not establish full syllabus mastery. Calibrated cumulative assessment, reliable drawn-work review and qualified mathematics-teacher review remain unfinished.
Before you begin
- Vector magnitude and subtraction
- Signed multiplication
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What should you establish before applying the method in this lesson?
What you will learn
- Multiply every component by one scalar
- Use |c| for the magnitude factor
- Distinguish same, opposite and undefined directions
Scale both components
For v = [a; b], cv = [ca; cb]. Magnitude changes by |c|: |cv| = |c||v|. If c > 0 and v is non-zero, the direction is unchanged; if c < 0, it is opposite. The magnitude is never made negative by a negative scalar.
Zero is a separate case
If c = 0 or v = [0; 0], the result is the zero vector and has no direction. The along/against rule requires c|v| ≠ 0. For non-zero v, multiplying by 1/|v| gives length 1 in the same direction, but that operation cannot be used for a zero vector.
Sources for this lesson
Pause and explain
Does a negative scalar make a vector's magnitude negative?
Worked example
For v = [3; 4], find −2v and its magnitude and direction.
Show the worked solution
- −2v = [−6; −8].
- |v| = 5, so |−2v| = 2×5 = 10; direction is opposite.
Answer [−6; −8]; magnitude 10; opposite direction.
Try a topic test
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Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Find 3[−4; 3] and its magnitude.
Give me a hint
Scale both components; the original length is 5.
Compare my reasoning
- The result is [−12; 9].
- Its magnitude is 3×5 = 15.
[−12; 9]; magnitude 15.
Look for these in your work
- Scale both components
- Use a non-negative length
Find (−1/2)[6; −8] and state its direction relative to the original.
Give me a hint
The negative factor reverses direction.
Compare my reasoning
- The result is [−3; 4].
- Its length is 5, half the original length 10, in the opposite direction.
[−3; 4]; magnitude 5; opposite direction.
Look for these in your work
- Handle sign and size separately
- Check the length ratio
Compare 0[2; 7] with 4[0; 0]. May either be described as pointing along its original vector?
Give me a hint
The direction rule needs a non-zero result.
Compare my reasoning
- Both products are [0; 0].
- Neither zero result has a direction; the along/against condition does not apply.
Both are [0; 0] with undefined direction.
Look for these in your work
- Check the zero condition
- Do not invent a direction
Common mistakes
- Multiply every component.
- Use the scalar's absolute value for length.
What can you explain now?
Does a negative scalar make a vector's magnitude negative?
Compare with the explanation
No; it reverses direction and multiplies length by its absolute value.
Magnitude is a non-negative scalar.
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Make it stick
Does a negative scalar make a vector's magnitude negative?
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.