Kenneth ChengMathematical modeling
← Senior Mathematics Studio

LEARN · EXPLAIN · REVISE

The fundamental theorem of space vectors

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

高二選擇性必修 第一册(A版).pdf · 1.2 · PDF 16 / printed page 11

Revisit first: Space vectors and their operations

TOPIC 01

The fundamental theorem of space vectors

Recognise a basis and use unique vector coordinates and affine combinations.

What you will be able to explain

  • Recognise a basis and use unique vector coordinates and affine combinations.
  • Justify the method and check the conditions in a new situation.

Basis

Three noncoplanar vectors span space uniquely.

v=xe1+ye2+ze3v=xe_1+ye_2+ze_3

Independence and affine points

A basis must be independent. Weights representing a point in a plane sum to one.

PREDICT → EXPLORE → EXPLAIN → TRANSFER

Make a prediction, then explore the relationship.

Lesson question: Can three nonzero vectors still fail to describe all space?

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

Space vector v=(2,1,1); |v|=2.4495; v·(2,1,3)=8. The drawing shows the xy-projection, not its full spatial length.

Space vector v=(2,1,1); |v|=2.4495; v·(2,1,3)=8. The drawing shows the xy-projection, not its full spatial length.

Explain: Calculate two valid cases and explain the change using the defining relation.

Transfer: Construct a dependent triple and two different representations of the same vector.

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

In the standard basis, find the coefficient of e3.

v=2e1−e2+2e3v=2e_1-e_2+2e_3
  • A basis must be independent. Weights representing a point in a plane sum to one.
Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Use vector components and the distributive laws.
Hint 2
Use this intermediate relation.
v=(x,y,z)v=(x,y,z)
Worked solution

  1. Use vector components and the distributive laws.

  2. Apply the stated relation and retain its conditions.

    (x,y,z)=(2,−1,2)(x,y,z)=(2,-1,2)

  3. A basis coefficient is a coordinate relative to that basis.

The requested value is 2.

Checks and common pitfalls: A basis coefficient is a coordinate relative to that basis.

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

02 / Standard#Worked example

P=αA+βB+γC is affine, α=1/(t+1), β=1/2. Find γ.

t=3t=3
  • A basis must be independent. Weights representing a point in a plane sum to one.
Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Translate the geometric condition into a coordinate equation.
Hint 2
Use this intermediate relation.
α+β+γ=1α+β+γ=1
Worked solution

  1. Translate the geometric condition into a coordinate equation.

  2. Apply the stated relation and retain its conditions.

    γ=1−1/4−1/2\gamma=1-1/4-1/2

  3. Affine weights sum to one.

The requested value is 0.25.

Checks and common pitfalls: Affine weights sum to one.

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

03 / Transfer#Worked example

Explain why these three vectors do not give unique coefficients.

e1=(1,0,0),e2=(0,1,0),e3=(4,1,0)e_1=(1,0,0), e_2=(0,1,0), e_3=(4,1,0)
  • A basis must be independent. Weights representing a point in a plane sum to one.
Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Use vector components and the distributive laws.
Hint 2
Use this intermediate relation.
e3=te1+e2e_3=te_1+e_2
Worked solution

  1. Use vector components and the distributive laws.

  2. Apply the stated relation and retain its conditions.

    (4,1,0)=4(1,0,0)+(0,1,0)(4,1,0)=4(1,0,0)+(0,1,0)

  3. The dependent triple permits more than one representation.

The requested relation or conclusion is shown below.

e3=4e1+e2e_3=4e_1+e_2

Checks and common pitfalls: The dependent triple permits more than one representation.

Reasoning checklist · self / teacher assessment
  • State a valid definition or model and its assumptions.
  • Show the intermediate mathematical relations, not only the final claim.
  • Check exclusions, units or the interpretation of the result.

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

04 / Foundation#Your turn

In the standard basis, find the coefficient of e3.

v=2e1−e2+5e3v=2e_1-e_2+5e_3
  • A basis must be independent. Weights representing a point in a plane sum to one.
Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Use vector components and the distributive laws.
Hint 2
Use this intermediate relation.
v=(x,y,z)v=(x,y,z)
Worked solution

  1. Use vector components and the distributive laws.

  2. Apply the stated relation and retain its conditions.

    (x,y,z)=(2,−1,5)(x,y,z)=(2,-1,5)

  3. A basis coefficient is a coordinate relative to that basis.

The requested value is 5.

Checks and common pitfalls: A basis coefficient is a coordinate relative to that basis.

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

05 / Foundation#Your turn

Find the coefficient of e1 when e1=(1,0,0), e2=(1,1,0), e3=(0,0,1).

v=(8,2,3)v=(8,2,3)
  • A basis must be independent. Weights representing a point in a plane sum to one.
Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Use vector components and the distributive laws.
Hint 2
Use this intermediate relation.
(x+y,y,z)=v(x+y,y,z)=v
Worked solution

  1. Use vector components and the distributive laws.

  2. Apply the stated relation and retain its conditions.

    y=2,z=3,x=8−2=6y=2,\quad z=3,\quad x=8-2=6

  3. Changing the basis changes coefficients even when the geometric vector is unchanged.

The requested value is 6.

Checks and common pitfalls: Changing the basis changes coefficients even when the geometric vector is unchanged.

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

06 / Foundation#Your turn

Decide whether e1,e2,e3 form a basis.

e1=(1,0,0), e2=(0,1,0), e3=(7,2,0)e_1=(1,0,0),\ e_2=(0,1,0),\ e_3=(7,2,0)
  • A basis must be independent. Weights representing a point in a plane sum to one.
Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Use vector components and the distributive laws.
Hint 2
All third components are zero.
Worked solution

  1. Use vector components and the distributive laws.

  2. Apply the stated relation and retain its conditions.

    e3=7e1+2e2e_3=7e_1+2e_2

  3. They span only a plane and cannot represent (0,0,1).

The requested relation or conclusion is shown below.

e3=7e1+2e2e_3=7e_1+2e_2

Checks and common pitfalls: They span only a plane and cannot represent (0,0,1).

Reasoning checklist · self / teacher assessment
  • State a valid definition or model and its assumptions.
  • Show the intermediate mathematical relations, not only the final claim.
  • Check exclusions, units or the interpretation of the result.

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

07 / Foundation#Your turn

Find λ so that the third vector lies in the plane spanned by the first two.

a=(1,0,0), b=(0,1,0), c=(8,2,λ)a=(1,0,0),\ b=(0,1,0),\ c=(8,2,\lambda)
  • A basis must be independent. Weights representing a point in a plane sum to one.
Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Use vector components and the distributive laws.
Hint 2
Use this intermediate relation.
c=xa+ybc=xa+yb
Worked solution

  1. Use vector components and the distributive laws.

  2. Apply the stated relation and retain its conditions.

    c=(x,y,0)⇒λ=0c=(x,y,0)\Rightarrow\lambda=0

  3. The plane spanned by a and b has zero third coordinate.

The requested value is 0.

Checks and common pitfalls: The plane spanned by a and b has zero third coordinate.

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

08 / Standard#Your turn

Find λ so that the third vector lies in the plane spanned by the first two.

a=(1,0,0), b=(0,1,0), c=(9,2,λ)a=(1,0,0),\ b=(0,1,0),\ c=(9,2,\lambda)
  • A basis must be independent. Weights representing a point in a plane sum to one.
Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Use vector components and the distributive laws.
Hint 2
Use this intermediate relation.
c=xa+ybc=xa+yb
Worked solution

  1. Use vector components and the distributive laws.

  2. Apply the stated relation and retain its conditions.

    c=(x,y,0)⇒λ=0c=(x,y,0)\Rightarrow\lambda=0

  3. The plane spanned by a and b has zero third coordinate.

The requested value is 0.

Checks and common pitfalls: The plane spanned by a and b has zero third coordinate.

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

09 / Standard#Your turn

P=αA+βB+γC is affine, α=1/(t+1), β=1/2. Find γ.

t=10t=10
  • A basis must be independent. Weights representing a point in a plane sum to one.
Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Translate the geometric condition into a coordinate equation.
Hint 2
Use this intermediate relation.
α+β+γ=1α+β+γ=1
Worked solution

  1. Translate the geometric condition into a coordinate equation.

  2. Apply the stated relation and retain its conditions.

    γ=1−1/11−1/2\gamma=1-1/11-1/2

  3. Affine weights sum to one.

The requested value is 0.409090909091.

Checks and common pitfalls: Affine weights sum to one.

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

10 / Standard#Your turn

Find the representation of v in the given basis.

e1=(1,1,0), e2=(1,−1,0), e3=(0,0,1), v=(22,2,3)e_1=(1,1,0),\ e_2=(1,-1,0),\ e_3=(0,0,1),\ v=(22,2,3)
  • A basis must be independent. Weights representing a point in a plane sum to one.
Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Use vector components and the distributive laws.
Hint 2
Use this intermediate relation.
x+y=2t,x−y=2x+y=2t,\quad x-y=2
Worked solution

  1. Use vector components and the distributive laws.

  2. Apply the stated relation and retain its conditions.

    2x=24⇒x=122x=24\Rightarrow x=12

  3. Apply the stated relation and retain its conditions.

    y=10,z=3y=10,\quad z=3

  4. Solve the simultaneous component equations; uniqueness follows from independence.

The requested relation or conclusion is shown below.

v=12e1+10e2+3e3v=12e_1+10e_2+3e_3

Checks and common pitfalls: Solve the simultaneous component equations; uniqueness follows from independence.

Reasoning checklist · self / teacher assessment
  • State a valid definition or model and its assumptions.
  • Show the intermediate mathematical relations, not only the final claim.
  • Check exclusions, units or the interpretation of the result.

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

11 / Standard#Your turn

Explain why these three vectors do not give unique coefficients.

e1=(1,0,0),e2=(0,1,0),e3=(12,1,0)e_1=(1,0,0), e_2=(0,1,0), e_3=(12,1,0)
  • A basis must be independent. Weights representing a point in a plane sum to one.
Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Use vector components and the distributive laws.
Hint 2
Use this intermediate relation.
e3=te1+e2e_3=te_1+e_2
Worked solution

  1. Use vector components and the distributive laws.

  2. Apply the stated relation and retain its conditions.

    (12,1,0)=12(1,0,0)+(0,1,0)(12,1,0)=12(1,0,0)+(0,1,0)

  3. The dependent triple permits more than one representation.

The requested relation or conclusion is shown below.

e3=12e1+e2e_3=12e_1+e_2

Checks and common pitfalls: The dependent triple permits more than one representation.

Reasoning checklist · self / teacher assessment
  • State a valid definition or model and its assumptions.
  • Show the intermediate mathematical relations, not only the final claim.
  • Check exclusions, units or the interpretation of the result.

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

12 / Transfer#Your turn

Find the representation of v in the given basis.

e1=(1,1,0), e2=(1,−1,0), e3=(0,0,1), v=(26,2,3)e_1=(1,1,0),\ e_2=(1,-1,0),\ e_3=(0,0,1),\ v=(26,2,3)
  • A basis must be independent. Weights representing a point in a plane sum to one.
Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Use vector components and the distributive laws.
Hint 2
Use this intermediate relation.
x+y=2t,x−y=2x+y=2t,\quad x-y=2
Worked solution

  1. Use vector components and the distributive laws.

  2. Apply the stated relation and retain its conditions.

    2x=28⇒x=142x=28\Rightarrow x=14

  3. Apply the stated relation and retain its conditions.

    y=12,z=3y=12,\quad z=3

  4. Solve the simultaneous component equations; uniqueness follows from independence.

The requested relation or conclusion is shown below.

v=14e1+12e2+3e3v=14e_1+12e_2+3e_3

Checks and common pitfalls: Solve the simultaneous component equations; uniqueness follows from independence.

Reasoning checklist · self / teacher assessment
  • State a valid definition or model and its assumptions.
  • Show the intermediate mathematical relations, not only the final claim.
  • Check exclusions, units or the interpretation of the result.

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

13 / Transfer#Your turn

Explain why these three vectors do not give unique coefficients.

e1=(1,0,0),e2=(0,1,0),e3=(14,1,0)e_1=(1,0,0), e_2=(0,1,0), e_3=(14,1,0)
  • A basis must be independent. Weights representing a point in a plane sum to one.
Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Use vector components and the distributive laws.
Hint 2
Use this intermediate relation.
e3=te1+e2e_3=te_1+e_2
Worked solution

  1. Use vector components and the distributive laws.

  2. Apply the stated relation and retain its conditions.

    (14,1,0)=14(1,0,0)+(0,1,0)(14,1,0)=14(1,0,0)+(0,1,0)

  3. The dependent triple permits more than one representation.

The requested relation or conclusion is shown below.

e3=14e1+e2e_3=14e_1+e_2

Checks and common pitfalls: The dependent triple permits more than one representation.

Reasoning checklist · self / teacher assessment
  • State a valid definition or model and its assumptions.
  • Show the intermediate mathematical relations, not only the final claim.
  • Check exclusions, units or the interpretation of the result.

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

Focus on one question

End-of-lesson check and correction

Review your latest checked answers and explanations. A draft change requires a fresh check. Written work needs your self-assessment or a teacher’s review.

Enable JavaScript for a summary of your local work.

    Choose a foundation skill to revisit ↗

    Teacher preparation and assessment

    Question sequence

    • Recognise a basis and use unique vector coordinates and affine combinations.
    • Which condition is essential in the fundamental theorem of space vectors?
    • Can three nonzero vectors still fail to describe all space?

    Board plan

    • Basis: Three noncoplanar vectors span space uniquely.
      v=xe1+ye2+ze3v=xe_1+ye_2+ze_3
    • Independence and affine points: A basis must be independent. Weights representing a point in a plane sum to one.

    Anticipated thinking

    • Three vectors are not automatically a basis merely because there are three of them.

    Assessment checklist

    • 1 mark: choose the correct representation and conditions.
    • 1 mark: establish the intermediate relation.
    • 1 mark: complete a connected calculation or proof.
    • 1 mark: interpret and check the conclusion.

    No sign-in. Work stays in this browser. Export before clearing browser data. Written reasoning is assessed with a checklist.

    Curriculum and source notes ↗