Kenneth ChengMathematical modeling
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LEARN · EXPLAIN · REVISE

Concept of plane vectors

Read the idea, work independently, then explain what changed.

高一必修 第二册(A版).pdf · 6.1 · PDF 9 / printed page 2

Revisit first: Trigonometric identities and transformations

TOPIC 01

Concept of plane vectors

Build understanding of concept of plane vectors through definitions, contrasting cases and justified applications.

What you will be able to explain

  • Connect geometric and algebraic forms of concept of plane vectors.
  • Connect representations and justify the steps, including boundary cases.
  • Explain a solution and apply the idea to a changed situation.

Magnitude and direction

A nonzero vector has length and direction, independent of where it is drawn.

Zero and unit vectors

Zero has magnitude zero; a unit vector has magnitude one.

PREDICT → EXPLORE → EXPLAIN → TRANSFER

Make a prediction, then explore the relationship.

Lesson question: Before calculating, predict how the conclusion changes when one defining condition in concept of plane vectors changes. Record a reason.

Model exploration: predict which displayed result changes with the parameters, check the values, and compare the observation with the lesson question.

v=(2,1); |v|=2.2361; v·(2,1)=5.

v=(2,1); |v|=2.2361; v·(2,1)=5.

Explain: Compare two admissible cases and one boundary or invalid case. Explain the observed difference using the stated definition.

Transfer: Construct a new example and a tempting incorrect solution. Repair the solution by naming the missing condition.

Try it. Leave your reasoning visible.

Use one hint at a time. A correction explains what changed, not just the final answer.

01 / Foundation#Worked example

Find the vector magnitude.

a=(6,8)\mathbf a=(6,8)
  • Use the domain, units and sampling assumptions stated in the question.
Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Separate magnitude, direction and position; compare vectors by displacement.
Hint 2
Use the distance from the origin in coordinate space.
Worked solution

  1. Separate magnitude, direction and position; compare vectors by displacement.

  2. Calculate or simplify this relation.

    ∣a∣=(6)2+(8)2=10|\mathbf a|=\sqrt{(6)^2+(8)^2}=10

  3. The magnitude is a nonnegative scalar.

The requested value is 10.

Checks and common pitfalls: The magnitude is a nonnegative scalar.

Think first. Reveal a hint when the class is ready.

02 / Standard#Worked example

Find the displacement vector from A to B.

A=(2,1),B=(5,5)A=(2,1),\quad B=(5,5)
  • Use the domain, units and sampling assumptions stated in the question.
Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Separate magnitude, direction and position; compare vectors by displacement.
Hint 2
Subtract start coordinates from end coordinates.
Worked solution

  1. Separate magnitude, direction and position; compare vectors by displacement.

  2. Calculate or simplify this relation.

    AB→=(5−2,5−1)=(3,4)\overrightarrow{AB}=(5-2,5-1)=(3,4)

  3. Reversing the endpoints reverses the vector.

(3,4).

Checks and common pitfalls: Reversing the endpoints reverses the vector.

Reasoning checklist · self / teacher assessment
  • State the relevant definition, condition or model.
  • Show a valid calculation, proof or counterexample.
  • Interpret the conclusion with its restrictions.

Think first. Reveal a hint when the class is ready.

03 / Transfer#Worked example

A walker travels 2 m east then 2 m west. Find the magnitude of displacement.

  • Use the domain, units and sampling assumptions stated in the question.
Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Separate magnitude, direction and position; compare vectors by displacement.
Hint 2
Displacement depends only on start and finish.
Worked solution

  1. Separate magnitude, direction and position; compare vectors by displacement.

  2. Calculate or simplify this relation.

    (2,0)+(−2,0)=(0,0)(2,0)+(-2,0)=(0,0)

  3. The distance travelled is 4 m, although displacement is zero.

The requested value is 0.

Checks and common pitfalls: The distance travelled is 4 m, although displacement is zero.

Think first. Reveal a hint when the class is ready.

04 / Foundation#Your turn

Find the vector magnitude.

a=(9,12)\mathbf a=(9,12)
  • Use the domain, units and sampling assumptions stated in the question.
Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Separate magnitude, direction and position; compare vectors by displacement.
Hint 2
Use the distance from the origin in coordinate space.
Worked solution

  1. Separate magnitude, direction and position; compare vectors by displacement.

  2. Calculate or simplify this relation.

    ∣a∣=(9)2+(12)2=15|\mathbf a|=\sqrt{(9)^2+(12)^2}=15

  3. The magnitude is a nonnegative scalar.

The requested value is 15.

Checks and common pitfalls: The magnitude is a nonnegative scalar.

Think first. Reveal a hint when the class is ready.

05 / Foundation#Your turn

Find the displacement vector from A to B.

A=(3,1),B=(6,5)A=(3,1),\quad B=(6,5)
  • Use the domain, units and sampling assumptions stated in the question.
Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Separate magnitude, direction and position; compare vectors by displacement.
Hint 2
Subtract start coordinates from end coordinates.
Worked solution

  1. Separate magnitude, direction and position; compare vectors by displacement.

  2. Calculate or simplify this relation.

    AB→=(6−3,5−1)=(3,4)\overrightarrow{AB}=(6-3,5-1)=(3,4)

  3. Reversing the endpoints reverses the vector.

(3,4).

Checks and common pitfalls: Reversing the endpoints reverses the vector.

Reasoning checklist · self / teacher assessment
  • State the relevant definition, condition or model.
  • Show a valid calculation, proof or counterexample.
  • Interpret the conclusion with its restrictions.

Think first. Reveal a hint when the class is ready.

06 / Foundation#Your turn

Do equal magnitudes guarantee equal vectors? Give a counterexample.

∣a∣=∣b∣=3|\mathbf a|=|\mathbf b|=3
  • Use the domain, units and sampling assumptions stated in the question.
Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Separate magnitude, direction and position; compare vectors by displacement.
Hint 2
Equality requires both length and direction.
Worked solution

  1. Separate magnitude, direction and position; compare vectors by displacement.

  2. Calculate or simplify this relation.

    ∣(3,0)∣=∣(0,3)∣=3|(3,0)|=|(0,3)|=3

  3. The two perpendicular nonzero vectors cannot be equal.

No: (3,0) and (0,3) have different directions.

Checks and common pitfalls: The two perpendicular nonzero vectors cannot be equal.

Reasoning checklist · self / teacher assessment
  • State the relevant definition, condition or model.
  • Show a valid calculation, proof or counterexample.
  • Interpret the conclusion with its restrictions.

Think first. Reveal a hint when the class is ready.

07 / Foundation#Your turn

Give a unit vector in the same direction.

a=(9,12)\mathbf a=(9,12)
  • Use the domain, units and sampling assumptions stated in the question.
Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Separate magnitude, direction and position; compare vectors by displacement.
Hint 2
Divide by the positive magnitude.
Worked solution

  1. Separate magnitude, direction and position; compare vectors by displacement.

  2. Calculate or simplify this relation.

    a∣a∣=(3/5,4/5)\frac{\mathbf a}{|\mathbf a|}=(3/5,4/5)

  3. Normalizing a zero vector would be undefined.

(3/5,4/5).

Checks and common pitfalls: Normalizing a zero vector would be undefined.

Reasoning checklist · self / teacher assessment
  • State the relevant definition, condition or model.
  • Show a valid calculation, proof or counterexample.
  • Interpret the conclusion with its restrictions.

Think first. Reveal a hint when the class is ready.

08 / Standard#Your turn

Explain whether the zero vector has a unique direction.

0\mathbf0
  • Use the domain, units and sampling assumptions stated in the question.
Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Separate magnitude, direction and position; compare vectors by displacement.
Hint 2
Direction requires a nonzero displacement.
Worked solution

  1. Separate magnitude, direction and position; compare vectors by displacement.

  2. Calculate or simplify this relation.

    ∣0∣=0|\mathbf0|=0

  3. Conventions about parallelism do not create a physical direction for zero.

It has magnitude zero and no uniquely determined direction.

Checks and common pitfalls: Conventions about parallelism do not create a physical direction for zero.

Reasoning checklist · self / teacher assessment
  • State the relevant definition, condition or model.
  • Show a valid calculation, proof or counterexample.
  • Interpret the conclusion with its restrictions.

Think first. Reveal a hint when the class is ready.

09 / Standard#Your turn

Find the opposite vector.

a=(3,−5)\mathbf a=(3,-5)
  • Use the domain, units and sampling assumptions stated in the question.
Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Separate magnitude, direction and position; compare vectors by displacement.
Hint 2
Negate both coordinates.
Worked solution

  1. Separate magnitude, direction and position; compare vectors by displacement.

  2. Calculate or simplify this relation.

    −a=(−3,5);a+(−a)=0-\mathbf a=(-3,5);\quad\mathbf a+(-\mathbf a)=\mathbf0

  3. An opposite vector has the same magnitude and reversed direction.

(−3,5).

Checks and common pitfalls: An opposite vector has the same magnitude and reversed direction.

Reasoning checklist · self / teacher assessment
  • State the relevant definition, condition or model.
  • Show a valid calculation, proof or counterexample.
  • Interpret the conclusion with its restrictions.

Think first. Reveal a hint when the class is ready.

10 / Standard#Your turn

A walker travels 3 m east then 3 m west. Find the magnitude of displacement.

  • Use the domain, units and sampling assumptions stated in the question.
Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Separate magnitude, direction and position; compare vectors by displacement.
Hint 2
Displacement depends only on start and finish.
Worked solution

  1. Separate magnitude, direction and position; compare vectors by displacement.

  2. Calculate or simplify this relation.

    (3,0)+(−3,0)=(0,0)(3,0)+(-3,0)=(0,0)

  3. The distance travelled is 6 m, although displacement is zero.

The requested value is 0.

Checks and common pitfalls: The distance travelled is 6 m, although displacement is zero.

Think first. Reveal a hint when the class is ready.

11 / Standard#Your turn

Find the vector magnitude.

a=(12,16)\mathbf a=(12,16)
  • Use the domain, units and sampling assumptions stated in the question.
Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Separate magnitude, direction and position; compare vectors by displacement.
Hint 2
Use the distance from the origin in coordinate space.
Worked solution

  1. Separate magnitude, direction and position; compare vectors by displacement.

  2. Calculate or simplify this relation.

    ∣a∣=(12)2+(16)2=20|\mathbf a|=\sqrt{(12)^2+(16)^2}=20

  3. The magnitude is a nonnegative scalar.

The requested value is 20.

Checks and common pitfalls: The magnitude is a nonnegative scalar.

Think first. Reveal a hint when the class is ready.

12 / Transfer#Your turn

Find the displacement vector from A to B.

A=(4,1),B=(7,5)A=(4,1),\quad B=(7,5)
  • Use the domain, units and sampling assumptions stated in the question.
Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Separate magnitude, direction and position; compare vectors by displacement.
Hint 2
Subtract start coordinates from end coordinates.
Worked solution

  1. Separate magnitude, direction and position; compare vectors by displacement.

  2. Calculate or simplify this relation.

    AB→=(7−4,5−1)=(3,4)\overrightarrow{AB}=(7-4,5-1)=(3,4)

  3. Reversing the endpoints reverses the vector.

(3,4).

Checks and common pitfalls: Reversing the endpoints reverses the vector.

Reasoning checklist · self / teacher assessment
  • State the relevant definition, condition or model.
  • Show a valid calculation, proof or counterexample.
  • Interpret the conclusion with its restrictions.

Think first. Reveal a hint when the class is ready.

13 / Transfer#Your turn

Do equal magnitudes guarantee equal vectors? Give a counterexample.

∣a∣=∣b∣=4|\mathbf a|=|\mathbf b|=4
  • Use the domain, units and sampling assumptions stated in the question.
Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Separate magnitude, direction and position; compare vectors by displacement.
Hint 2
Equality requires both length and direction.
Worked solution

  1. Separate magnitude, direction and position; compare vectors by displacement.

  2. Calculate or simplify this relation.

    ∣(4,0)∣=∣(0,4)∣=4|(4,0)|=|(0,4)|=4

  3. The two perpendicular nonzero vectors cannot be equal.

No: (4,0) and (0,4) have different directions.

Checks and common pitfalls: The two perpendicular nonzero vectors cannot be equal.

Reasoning checklist · self / teacher assessment
  • State the relevant definition, condition or model.
  • Show a valid calculation, proof or counterexample.
  • Interpret the conclusion with its restrictions.

Think first. Reveal a hint when the class is ready.

Focus on one question

End-of-lesson check and correction

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    Teacher preparation and assessment

    Question sequence

    • What must be true before using the main rule for concept of plane vectors?
    • Which representation makes this task easier, and why?
    • Change one assumption. Does the conclusion survive?

    Board plan

    • Connect geometric and algebraic forms of concept of plane vectors.
    • Magnitude and direction: A nonzero vector has length and direction, independent of where it is drawn.
    • Zero and unit vectors: Zero has magnitude zero; a unit vector has magnitude one.
    • Close with: conditions → representation → reasoning → check.

    Anticipated thinking

    • Expected reasoning: A nonzero vector has length and direction, independent of where it is drawn.
    • Expected reasoning: Zero has magnitude zero; a unit vector has magnitude one.
    • Expected correction: Equal lengths do not imply equal vectors.

    Assessment checklist

    • 1: identify the givens and required quantity.
    • 1: choose a valid definition, representation or method.
    • 1: present connected, correct reasoning.
    • 1: check conditions and explain the result.

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