Kenneth ChengMathematical modeling
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Coordinate representation 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 · 1.3 · PDF 21 / printed page 16

Revisit first: Space vectors and their operationsThe fundamental theorem of space vectors

TOPIC 01

Coordinate representation of space vectors

Calculate spatial lengths, angles, projections and perpendicular directions from coordinates.

What you will be able to explain

  • Calculate spatial lengths, angles, projections and perpendicular directions from coordinates.
  • Justify the method and check the conditions in a new situation.

Coordinates

Subtract point coordinates to obtain a displacement vector.

AB→=B−A\overrightarrow{AB}=B-A

Projection conditions

Projection requires a nonzero direction; signed scalar projection can be negative.

proj⁡uv=v⋅uu⋅uu\operatorname{proj}_{u}v=\frac{v\cdot u}{u\cdot u}u

PREDICT → EXPLORE → EXPLAIN → TRANSFER

Make a prediction, then explore the relationship.

Lesson question: Predict what happens to projection when the direction vector is reversed.

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: Compare the projection vector and signed component under u→−u.

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 AB².

A=(1,2,3), B=(3,5,7)A=(1,2,3),\ B=(3,5,7)
  • Projection requires a nonzero direction; signed scalar projection can be negative.
Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Use the dot product to connect coordinates with length or angle.
Hint 2
Use this intermediate relation.
AB2=(B−A)⋅(B−A)AB^2=(B-A)\cdot(B-A)
Worked solution

  1. Use the dot product to connect coordinates with length or angle.

  2. Apply the stated relation and retain its conditions.

    B−A=(2,3,4)B-A=(2,3,4)

  3. Apply the stated relation and retain its conditions.

    AB2=4+9+16=29AB^2=4+9+16=29

  4. Distance uses the displacement, not either point vector alone.

The requested value is 29.

Checks and common pitfalls: Distance uses the displacement, not either point vector alone.

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

02 / Standard#Worked example

Find the signed scalar projection of v onto u.

v=(3,2,0), u=(1,0,0)v=(3,2,0),\ u=(1,0,0)
  • Projection requires a nonzero direction; signed scalar projection can be negative.
Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Use the dot product to connect coordinates with length or angle.
Hint 2
Use this intermediate relation.
s=v⋅u/∣u∣s=v\cdot u/|u|
Worked solution

  1. Use the dot product to connect coordinates with length or angle.

  2. Apply the stated relation and retain its conditions.

    s=3/1=3s=3/1=3

  3. A unit coordinate direction picks out the matching component.

The requested value is 3.

Checks and common pitfalls: A unit coordinate direction picks out the matching component.

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

03 / Transfer#Worked example

Find a nonzero vector perpendicular to both a and b, with third component 1.

a=(1,0,4), b=(0,1,2)a=(1,0,4),\ b=(0,1,2)
  • Projection requires a nonzero direction; signed scalar projection can be negative.
Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Use the dot product to connect coordinates with length or angle.
Hint 2
Use this intermediate relation.
u⋅a=u⋅b=0u\cdot a=u\cdot b=0
Worked solution

  1. Use the dot product to connect coordinates with length or angle.

  2. Apply the stated relation and retain its conditions.

    x+4z=0,y+2z=0x+4z=0,\quad y+2z=0

  3. Apply the stated relation and retain its conditions.

    z=1⇒x=−4, y=−2z=1\Rightarrow x=-4,\ y=-2

  4. Two linear constraints determine a perpendicular direction up to scale.

The requested relation or conclusion is shown below.

u=(−4,−2,1)u=(-4,-2,1)

Checks and common pitfalls: Two linear constraints determine a perpendicular direction up to scale.

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

Find AB².

A=(1,2,3), B=(6,5,7)A=(1,2,3),\ B=(6,5,7)
  • Projection requires a nonzero direction; signed scalar projection can be negative.
Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Use the dot product to connect coordinates with length or angle.
Hint 2
Use this intermediate relation.
AB2=(B−A)⋅(B−A)AB^2=(B-A)\cdot(B-A)
Worked solution

  1. Use the dot product to connect coordinates with length or angle.

  2. Apply the stated relation and retain its conditions.

    B−A=(5,3,4)B-A=(5,3,4)

  3. Apply the stated relation and retain its conditions.

    AB2=25+9+16=50AB^2=25+9+16=50

  4. Distance uses the displacement, not either point vector alone.

The requested value is 50.

Checks and common pitfalls: Distance uses the displacement, not either point vector alone.

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

05 / Foundation#Your turn

Find the z-coordinate of the midpoint of A and B.

A=(1,2,6), B=(3,4,12)A=(1,2,6),\ B=(3,4,12)
  • Projection requires a nonzero direction; signed scalar projection can be negative.
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.
M=(A+B)/2M=(A+B)/2
Worked solution

  1. Translate the geometric condition into a coordinate equation.

  2. Apply the stated relation and retain its conditions.

    Mz=(6+12)/2=9M_z=(6+12)/2=9

  3. Midpoint averaging applies separately to all three coordinates.

The requested value is 9.

Checks and common pitfalls: Midpoint averaging applies separately to all three coordinates.

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

06 / Foundation#Your turn

Find λ so that a and b are perpendicular.

a=(1,2,3), b=(7,1,λ)a=(1,2,3),\ b=(7,1,\lambda)
  • Projection requires a nonzero direction; signed scalar projection can be negative.
Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Use the dot product to connect coordinates with length or angle.
Hint 2
Use this intermediate relation.
a⋅b=0a\cdot b=0
Worked solution

  1. Use the dot product to connect coordinates with length or angle.

  2. Apply the stated relation and retain its conditions.

    7+2+3λ=07+2+3\lambda=0

  3. Apply the stated relation and retain its conditions.

    λ=−9/3\lambda=-9/3

  4. Use a zero dot product, not componentwise multiplication equal to zero.

The requested value is -3.

Checks and common pitfalls: Use a zero dot product, not componentwise multiplication equal to zero.

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

07 / Foundation#Your turn

Find cos θ between a and b.

a=(1,1,0),b=(0,8,1)a=(1,1,0), b=(0,8,1)
  • Projection requires a nonzero direction; signed scalar projection can be negative.
Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Use the dot product to connect coordinates with length or angle.
Hint 2
Use this intermediate relation.
cosθ=(a⋅b)/(∣a∣∣b∣)cosθ=(a·b)/(|a||b|)
Worked solution

  1. Use the dot product to connect coordinates with length or angle.

  2. Apply the stated relation and retain its conditions.

    a⋅b=8,∣a∣=2,∣b∣=65a\cdot b=8, |a|=\sqrt2, |b|=\sqrt{65}

  3. Apply the stated relation and retain its conditions.

    cos⁡θ=8/130\cos\theta=8/\sqrt{130}

  4. Both vector lengths enter the denominator.

The requested value is 0.701646415446.

Checks and common pitfalls: Both vector lengths enter the denominator.

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

08 / Standard#Your turn

Find cos θ between a and b.

a=(1,1,0),b=(0,9,1)a=(1,1,0), b=(0,9,1)
  • Projection requires a nonzero direction; signed scalar projection can be negative.
Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Use the dot product to connect coordinates with length or angle.
Hint 2
Use this intermediate relation.
cosθ=(a⋅b)/(∣a∣∣b∣)cosθ=(a·b)/(|a||b|)
Worked solution

  1. Use the dot product to connect coordinates with length or angle.

  2. Apply the stated relation and retain its conditions.

    a⋅b=9,∣a∣=2,∣b∣=82a\cdot b=9, |a|=\sqrt2, |b|=\sqrt{82}

  3. Apply the stated relation and retain its conditions.

    cos⁡θ=9/164\cos\theta=9/\sqrt{164}

  4. Both vector lengths enter the denominator.

The requested value is 0.702781928499.

Checks and common pitfalls: Both vector lengths enter the denominator.

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

09 / Standard#Your turn

Find the signed scalar projection of v onto u.

v=(10,2,0), u=(1,0,0)v=(10,2,0),\ u=(1,0,0)
  • Projection requires a nonzero direction; signed scalar projection can be negative.
Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Use the dot product to connect coordinates with length or angle.
Hint 2
Use this intermediate relation.
s=v⋅u/∣u∣s=v\cdot u/|u|
Worked solution

  1. Use the dot product to connect coordinates with length or angle.

  2. Apply the stated relation and retain its conditions.

    s=10/1=10s=10/1=10

  3. A unit coordinate direction picks out the matching component.

The requested value is 10.

Checks and common pitfalls: A unit coordinate direction picks out the matching component.

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

10 / Standard#Your turn

A unit direction has cos²α=1/t², cos²β=1/9. Find cos²γ.

t=11t=11
  • Projection requires a nonzero direction; signed scalar projection can be negative.
Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Use the dot product to connect coordinates with length or angle.
Hint 2
Use this intermediate relation.
cos2α+cos2β+cos2γ=1cos²α+cos²β+cos²γ=1
Worked solution

  1. Use the dot product to connect coordinates with length or angle.

  2. Apply the stated relation and retain its conditions.

    cos⁡2γ=1−1/121−1/9\cos^2\gamma=1-1/121-1/9

  3. The squared direction cosines sum to one.

The requested value is 0.880624426079.

Checks and common pitfalls: The squared direction cosines sum to one.

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

11 / Standard#Your turn

Find a nonzero vector perpendicular to both a and b, with third component 1.

a=(1,0,12), b=(0,1,2)a=(1,0,12),\ b=(0,1,2)
  • Projection requires a nonzero direction; signed scalar projection can be negative.
Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Use the dot product to connect coordinates with length or angle.
Hint 2
Use this intermediate relation.
u⋅a=u⋅b=0u\cdot a=u\cdot b=0
Worked solution

  1. Use the dot product to connect coordinates with length or angle.

  2. Apply the stated relation and retain its conditions.

    x+12z=0,y+2z=0x+12z=0,\quad y+2z=0

  3. Apply the stated relation and retain its conditions.

    z=1⇒x=−12, y=−2z=1\Rightarrow x=-12,\ y=-2

  4. Two linear constraints determine a perpendicular direction up to scale.

The requested relation or conclusion is shown below.

u=(−12,−2,1)u=(-12,-2,1)

Checks and common pitfalls: Two linear constraints determine a perpendicular direction up to scale.

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

A unit direction has cos²α=1/t², cos²β=1/9. Find cos²γ.

t=13t=13
  • Projection requires a nonzero direction; signed scalar projection can be negative.
Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Use the dot product to connect coordinates with length or angle.
Hint 2
Use this intermediate relation.
cos2α+cos2β+cos2γ=1cos²α+cos²β+cos²γ=1
Worked solution

  1. Use the dot product to connect coordinates with length or angle.

  2. Apply the stated relation and retain its conditions.

    cos⁡2γ=1−1/169−1/9\cos^2\gamma=1-1/169-1/9

  3. The squared direction cosines sum to one.

The requested value is 0.882971729126.

Checks and common pitfalls: The squared direction cosines sum to one.

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

13 / Transfer#Your turn

Find a nonzero vector perpendicular to both a and b, with third component 1.

a=(1,0,14), b=(0,1,2)a=(1,0,14),\ b=(0,1,2)
  • Projection requires a nonzero direction; signed scalar projection can be negative.
Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Use the dot product to connect coordinates with length or angle.
Hint 2
Use this intermediate relation.
u⋅a=u⋅b=0u\cdot a=u\cdot b=0
Worked solution

  1. Use the dot product to connect coordinates with length or angle.

  2. Apply the stated relation and retain its conditions.

    x+14z=0,y+2z=0x+14z=0,\quad y+2z=0

  3. Apply the stated relation and retain its conditions.

    z=1⇒x=−14, y=−2z=1\Rightarrow x=-14,\ y=-2

  4. Two linear constraints determine a perpendicular direction up to scale.

The requested relation or conclusion is shown below.

u=(−14,−2,1)u=(-14,-2,1)

Checks and common pitfalls: Two linear constraints determine a perpendicular direction up to scale.

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

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    Choose a foundation skill to revisit ↗

    Teacher preparation and assessment

    Question sequence

    • Calculate spatial lengths, angles, projections and perpendicular directions from coordinates.
    • Which condition is essential in coordinate representation of space vectors?
    • Predict what happens to projection when the direction vector is reversed.

    Board plan

    • Coordinates: Subtract point coordinates to obtain a displacement vector.
      AB→=B−A\overrightarrow{AB}=B-A
    • Projection conditions: Projection requires a nonzero direction; signed scalar projection can be negative.
      proj⁡uv=v⋅uu⋅uu\operatorname{proj}_{u}v=\frac{v\cdot u}{u\cdot u}u

    Anticipated thinking

    • A projection vector and its signed scalar component are different objects.

    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.

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    Curriculum and source notes ↗