LEARN · EXPLAIN · REVISE

Vectors as displacements

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

TOPIC 01

Vectors as displacements

Translate, add and scale arrows; distinguish a journey from its net displacement.

What you will be able to explain

Represent a displacement independently of its starting point.
Use head-to-tail addition and absolute scaling of lengths.
Choose a velocity that meets a directional constraint.

Why an arrow can move

A route can contain many turns but its net displacement needs only the start and finish.

A displacement records change in position. Translating both endpoints preserves that change; reversing them negates it.

\overrightarrow{AB}=B-A,\quad|t u|=|t||u|

Intermediate positions cancel in a joined route.
$(B-A)+(C-B)=C-A$
Squaring a scaled length produces t², so its nonnegative square root is |t|.
$|tu|=\sqrt{t^2(u_x^2+u_y^2)}=|t||u|$

When it applies

Normalize only a nonzero vector.
Equal vectors have both the same direction and the same length.

Counterexample: Two opposite unit arrows have resultant zero, although the sum of their lengths is two.

Drawn arrow → ordered components → net displacement → magnitude.

PREDICT → EXPLORE → EXPLAIN → TRANSFER

Make a prediction, then explore the relationship.

Lesson question: Predict which features change when a vector is multiplied by a negative scalar.

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

Blue is u; red is tu from a translated start. Both arrows use the same adaptive scale so their complete lengths stay visible. Length uses |t|; a zero arrow has no unit direction.

Numerical data for the diagram; printable
Quantity	Value
u	3, 2
tu	3, 2
\|u\|	3.60555128
\|tu\|	3.60555128

Explain: The model plots u and tu from different starts; the table separates signed components from lengths.

Transfer: Choose controls producing zero displacement but a nonzero original vector; explain why unit direction then fails.

Try it. Leave your reasoning visible.

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

Examples 1–3: study a full demonstration; finish a solution from its first step; then choose a method independently before opening the solution. The remaining exercises may continue after class.

01 / Foundation#Worked example

Find the resultant displacement length after moving east 3 and north 4.

u=(3,0),\ v=(0,4)

Skills and prerequisite lessons

Scope: JM02

Working and explanation

BUILD THE REASONING

Hint 1

Join the arrows head to tail.

Hint 2

The resultant joins the first and last points.

Worked solution

Add horizontal and vertical displacements.
$u+v=(3,4)$
Use the right triangle, not the walked distance.
$|u+v|=\sqrt{3^2+4^2}=5$

The displacement length is 5.

Checks and common pitfalls: The distance walked is 7, but displacement measures the straight connection.

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

02 / Standard#Worked example

Find D so that CD represents the same vector as AB.

A=(1,2),B=(4,6),C=(-2,1)

First step provided — complete the reasoning:Compute the original displacement.

B-A=(3,4)

Skills and prerequisite lessons

Scope: JM02

Working and explanation

BUILD THE REASONING

Hint 1

Compare end minus start.

Hint 2

Translate the whole arrow without rotating it.

Worked solution

Compute the original displacement.
$B-A=(3,4)$
Add it to the new start.
$D=C+(3,4)=(1,5)$

D=(1,5).

Checks and common pitfalls: Equal free vectors need not share a starting point.

Reasoning checklist · self / teacher assessment

D=(1,5).
Equal free vectors need not share a starting point.

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

03 / Transfer#Worked example

Does |u+v|=|u|+|v| hold for every pair? Refute or prove.

u=(1,0),\ v=(-1,0)

Skills and prerequisite lessons

Scope: JM02

Working and explanation

BUILD THE REASONING

Hint 1

Use the displayed opposing arrows.

Hint 2

Calculate each side separately.

Worked solution

Opposite displacements cancel.
$u+v=(0,0)$
The two norms are positive.
$0=|u+v|<|u|+|v|=2$

False; these opposing unit vectors are a counterexample.

Checks and common pitfalls: A norm does not distribute over addition.

Reasoning checklist · self / teacher assessment

False; these opposing unit vectors are a counterexample.
A norm does not distribute over addition.

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

04 / Foundation#Your turn

Find the arrow length.

v=(-6,8)

Skills and prerequisite lessons

Scope: JM02

Working and explanation

BUILD THE REASONING

Hint 1

A negative component is a direction.

Hint 2

Square both components.

Worked solution

Use the Euclidean length.
$|v|^2=36+64$
Take the nonnegative root.
$|v|=10$

Length 10.

Checks and common pitfalls: Do not subtract squared lengths because a component is negative.

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

05 / Foundation#Your turn

Find the length after multiplication by −3.

|v|=2

Skills and prerequisite lessons

Scope: JM02

Working and explanation

BUILD THE REASONING

Hint 1

The sign reverses direction.

Hint 2

Length scales by the absolute multiplier.

Worked solution

Separate length from orientation.
$|-3v|=|-3||v|$
Apply the given length.
$3\cdot2=6$

Length 6.

Checks and common pitfalls: A vector length cannot be −6.

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

06 / Foundation#Your turn

Simplify the displacement around triangle ABC.

\overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CA}

Skills and prerequisite lessons

Scope: JM02

Working and explanation

BUILD THE REASONING

Hint 1

First combine the first two arrows.

Hint 2

Where is the final point?

Worked solution

The first two arrows join A to C.
$AB+BC=AC$
The final arrow returns to A.
$AC+CA=\mathbf0$

The zero vector.

Checks and common pitfalls: Zero displacement does not imply zero travelled distance.

Reasoning checklist · self / teacher assessment

The zero vector.
Zero displacement does not imply zero travelled distance.

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

07 / Foundation#Your turn

Give a unit vector pointing in the same direction.

v=(0,-5)

Skills and prerequisite lessons

Scope: JM02

Working and explanation

BUILD THE REASONING

Hint 1

Divide by the positive length.

Hint 2

Keep the downward direction.

Worked solution

Compute the nonzero norm.
$|v|=5$
Normalize both components.
$v/|v|=(0,-1)$

(0,−1).

Checks and common pitfalls: Dividing by −5 would reverse direction.

Reasoning checklist · self / teacher assessment

(0,−1).
Dividing by −5 would reverse direction.

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

08 / Standard#Your turn

In parallelogram ABCD, express AC and BD using u=AB and v=AD.

\overrightarrow{AB}=u,\ \overrightarrow{AD}=v

Skills and prerequisite lessons

Scope: JM02

Working and explanation

BUILD THE REASONING

Hint 1

Use A as the common origin.

Hint 2

For BD, subtract B from D.

Worked solution

The opposite vertex has position u+v.
$AC=u+v$
The other diagonal runs from u to v.
$BD=v-u$

AC=u+v and BD=v−u.

Checks and common pitfalls: Both diagonals are not u+v; their endpoints differ.

Reasoning checklist · self / teacher assessment

AC=u+v and BD=v−u.
Both diagonals are not u+v; their endpoints differ.

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

09 / Standard#Your turn

Walk east 3, north 4, west 3. Find displacement magnitude.

Skills and prerequisite lessons

Scope: JM02

Working and explanation

BUILD THE REASONING

Hint 1

Cancel the opposing horizontal movements.

Hint 2

Only one component survives.

Worked solution

Sum signed displacements.
$(3,0)+(0,4)+(-3,0)=(0,4)$
Measure the resultant.
$|(0,4)|=4$

Displacement 4; walked distance 10.

Checks and common pitfalls: Use 10 only when asked for distance travelled.

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

10 / Standard#Your turn

Two unit vectors satisfy u+v=0. Find their angle in degrees.

|u|=|v|=1

Skills and prerequisite lessons

Scope: JM02

Working and explanation

BUILD THE REASONING

Hint 1

Rearrange to v=−u.

Hint 2

The vectors are nonzero.

Worked solution

They point in opposite directions.
$v=-u$
The angle is a half turn.
$\theta=180^\circ$

180°.

Checks and common pitfalls: Opposite vectors are not perpendicular.

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

11 / Standard#Your turn

A student normalizes the zero vector by dividing by its length. Explain the failure.

v=(0,0)

Skills and prerequisite lessons

Scope: JM02

Working and explanation

BUILD THE REASONING

Hint 1

Find the denominator first.

Hint 2

A unit vector must have length one.

Worked solution

Zero has length zero.
$|v|=0$
Division by zero gives no direction or unit vector.
$v/|v|\text{ is undefined}$

Normalization is undefined.

Checks and common pitfalls: Assigning an arbitrary direction changes the question rather than normalizing zero.

Reasoning checklist · self / teacher assessment

Normalization is undefined.
Assigning an arbitrary direction changes the question rather than normalizing zero.

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

12 / Transfer#Your turn

A robot moves by (2,1) then (1,3). Find the shortest possible return distance to its start.

Skills and prerequisite lessons

Scope: JM02

Working and explanation

BUILD THE REASONING

Hint 1

Locate the robot relative to its start.

Hint 2

Any bent return path is at least the straight distance.

Worked solution

The return displacement is the negative sum.
$r=-(3,4)=(-3,-4)$
A straight return attains the lower bound.
$d_{\min}=|r|=5$

Minimum 5, attained by a straight return.

Checks and common pitfalls: Reversing both original moves is valid but is not the shortest route.

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

13 / Transfer#Your turn

A current moves east at 3 m/s. A boat moves at 5 m/s relative to water. Choose its velocity to travel due north across a 20 m river, and find the time.

\text{east}=x,\ \text{north}=y

Skills and prerequisite lessons

Scope: JM02

Working and explanation

BUILD THE REASONING

Hint 1

Cancel the current horizontally.

Hint 2

Use the fixed speed to obtain the north component.

Worked solution

Due north requires the boat to supply −3 horizontally.
$v_x=-3$
Choose the positive vertical root.
$v_y=\sqrt{25-9}=4$
Ground velocity is (0,4), so time is distance over speed.
$t=20/4=5\text{ s}$

Boat velocity (−3,4) m/s; crossing time 5 s.

Checks and common pitfalls: Pointing the boat due north does not produce a due-north ground path.

Reasoning checklist · self / teacher assessment

Boat velocity (−3,4) m/s; crossing time 5 s.
Pointing the boat due north does not produce a due-north ground path.

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

End-of-lesson check and correction

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Teacher preparation and assessment

40-minute lesson

Minutes	Focus	Teaching and student action
5	Prediction and diagnosis	Compare two translated arrows and vote on equality.
8	Concept derivation	Derive head-to-tail addition and negative scaling using a closed route.
10	Guided application	Work the translated-arrow example; pairs explain the length counterexample.
12	Transfer and explanation	Solve the constrained river-crossing task and discuss when it fails.
5	Exit evidence	Write and check the exit statement about displacement.

Question sequence

If I slide this arrow to another desk, what changes?
Can travelling farther leave you closer to your start?

Board development

Draw east 3, north 4 and the single diagonal; label distance 7 and displacement 5.
Display boat + current = ground velocity, then impose the zero horizontal component.

Expected correct and incorrect responses

A student gives −6 as a scaled length. Ask them to locate a negative distance on a ruler.
Correct: “The length is 6; the negative multiplier reverses the arrow.”
Incorrect: “The length is −6 because the multiplier is negative.”

Follow-up questions

If the current is faster than the boat, can a due-north path still exist? Explain with a squared component.

Remediation

Use two opposite paper arrows, then separately measure the two lengths and the zero resultant.

Exit check

State one quantity preserved by translating an arrow and one route property it does not encode.

Assessment checklist and sample responses

Credit the chosen orientation, signed components and a nonnegative magnitude separately. Sample: (−3,4)+(3,0)=(0,4), so crossing time is 20/4.
Correct: “The length is 6; the negative multiplier reverses the arrow.”