LEARN · EXPLAIN · REVISE

Dot products and constrained projection

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

高一必修第二册(A版).pdf · 6.2 · PDF 14 / printed page 7

Revisit first: Coordinates, bases and division

TOPIC 01

Dot products and constrained projection

Connect signed projection, angle and squared distance; distinguish a line projection from a nearest point on a segment.

What you will be able to explain

Interpret the sign of a dot product.
Expand squared vector norms correctly.
Check zero vectors and segment endpoints before interpreting an angle or minimizer.

Projection is an orthogonal decomposition

Closest-point questions become manageable by splitting motion into parallel and perpendicular components.

Subtract a multiple of u from v so that the remainder has zero dot product with u. The resulting multiple is a projection coefficient, not a projected length.

t=\frac{u\cdot v}{u\cdot u},\quad |v-tu|^2=|v|^2-2t(u\cdot v)+t^2|u|^2

Force the residual to have zero dot product.
$u\cdot(v-tu)=0$
Solve only when the direction vector is nonzero.
$t=(u\cdot v)/|u|^2$

When it applies

The angle formula requires two nonzero vectors.
For a segment A+t(B−A), restrict 0≤t≤1 and compare endpoints when needed.

Counterexample: A perpendicular foot at (3,0) is outside the segment from (0,0) to (2,0); the nearest segment point is (2,0).

Geometric shadow ↔ scalar coefficient ↔ completed square in t.

PREDICT → EXPLORE → EXPLAIN → TRANSFER

Make a prediction, then explore the relationship.

Lesson question: Predict the sign of u·v as v crosses the perpendicular direction.

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

The green vector is the projection tu. Its residual has dot product zero with u; the angle is undefined at v=0.

Numerical data for the diagram; printable
Quantity	Value
u·v	10
t	2
u·(v−tu)	0
angle / degrees / 角度	26.56505118

Explain: The residual v−tu has zero dot product with u; when v=0, projection is zero but its angle is undefined.

Transfer: Choose v whose projection has negative coefficient and explain the direction of the projected vector.

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

Calculate u·v and interpret its sign.

u=(2,1),\ v=(3,-4)

Skills and prerequisite lessons

Scope: JM02

Working and explanation

BUILD THE REASONING

Hint 1

Multiply corresponding components.

Hint 2

For nonzero vectors, the sign matches cosθ.

Worked solution

Add the two products.
$u\cdot v=6-4=2$
A positive cosine means an acute angle.

Dot product 2; the angle is acute.

Checks and common pitfalls: Dot product is a scalar, not a pair of coordinates.

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

02 / Standard#Worked example

Find the angle in degrees.

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

First step provided — complete the reasoning:The normalized dot product is 1/√2.

\cos\theta=1/(1\cdot\sqrt2)

Skills and prerequisite lessons

Scope: JM02

Working and explanation

BUILD THE REASONING

Hint 1

Compute dot product and both lengths.

Hint 2

Divide by the product of nonzero lengths.

Worked solution

The normalized dot product is 1/√2.
$\cos\theta=1/(1\cdot\sqrt2)$
Choose the angle in [0°,180°].
$\theta=45^\circ$

45°.

Checks and common pitfalls: A vector angle uses the specified principal interval.

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

03 / Transfer#Worked example

Find t minimizing |v−tu| and explain the geometry.

u=(2,0),\ v=(3,4)

Skills and prerequisite lessons

Scope: JM02

Working and explanation

BUILD THE REASONING

Hint 1

The residual should be perpendicular to u.

Hint 2

Alternatively complete a square.

Worked solution

Write the squared distance.
$|v-tu|^2=(3-2t)^2+16$
Its square term vanishes at the projection.
$t=3/2,\ v-tu=(0,4)$

t=3/2; minimum distance 4.

Checks and common pitfalls: The coefficient t is not itself the projection length when |u|≠1.

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

04 / Foundation#Your turn

Find k making u perpendicular to v.

u=(1,2),\ v=(k,3)

Skills and prerequisite lessons

Scope: JM02

Working and explanation

BUILD THE REASONING

Hint 1

Set the dot product to zero.

Hint 2

Both vectors are nonzero.

Worked solution

Form the scalar equation.
$k+2\cdot3=0$
Solve the parameter.
$k=-6$

k=−6.

Checks and common pitfalls: Parallelism uses a determinant, not a zero dot product.

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

05 / Foundation#Your turn

Find |u+v|².

|u|=3,\ |v|=4,\ u\cdot v=2

Skills and prerequisite lessons

Scope: JM02

Working and explanation

BUILD THE REASONING

Hint 1

Expand with the dot product.

Hint 2

There are two cross terms.

Worked solution

Use the norm identity.
$|u+v|^2=|u|^2+2u\cdot v+|v|^2$
Substitute all three quantities.
$9+4+16=29$

29.

Checks and common pitfalls: Do not replace 2u·v by u·v.

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

06 / Foundation#Your turn

Find the work in joules.

F=(3,4)\text{ N},\ s=(2,-1)\text{ m}

Skills and prerequisite lessons

Scope: JM02

Working and explanation

BUILD THE REASONING

Hint 1

Work is force dotted with displacement.

Hint 2

Opposed components contribute negatively.

Worked solution

Calculate the directional contributions.
$W=3\cdot2+4(-1)$
Combine with joule units.
$W=2\text{ J}$

2 J.

Checks and common pitfalls: Multiplying the two magnitudes would ignore the angle.

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

07 / Foundation#Your turn

Find the signed scalar projection of v onto the positive x-axis.

v=(-3,4)

Skills and prerequisite lessons

Scope: JM02

Working and explanation

BUILD THE REASONING

Hint 1

Use the unit direction (1,0).

Hint 2

A negative projection points opposite that direction.

Worked solution

Dot with the axis unit vector.
$v\cdot(1,0)=-3$
Keep the sign for a signed projection.

−3.

Checks and common pitfalls: Projection length would be 3; the signed projection is −3.

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

08 / Standard#Your turn

Prove that equal-length vectors have perpendicular sum and difference, with a zero-vector caveat.

|u|=|v|

Skills and prerequisite lessons

Scope: JM02

Working and explanation

BUILD THE REASONING

Hint 1

Expand (u+v)·(u−v).

Hint 2

The mixed terms cancel.

Worked solution

Use symmetry of dot product.
$(u+v)\cdot(u-v)=|u|^2-|v|^2=0$
If both resulting vectors are nonzero their angle is 90°; otherwise only the dot-product statement applies.

The dot product is zero; an angle requires both vectors nonzero.

Checks and common pitfalls: u=v makes the difference zero and its angle undefined.

Reasoning checklist · self / teacher assessment

The dot product is zero; an angle requires both vectors nonzero.
u=v makes the difference zero and its angle undefined.

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

09 / Standard#Your turn

Find u·v from the three norms.

|u|=2,\ |v|=3,\ |u-v|=4

Skills and prerequisite lessons

Scope: JM02

Working and explanation

BUILD THE REASONING

Hint 1

Expand the squared difference.

Hint 2

The cross term has a minus sign.

Worked solution

Relate the squares.
$16=4+9-2u\cdot v$
Solve the scalar equation.
$u\cdot v=-3/2$

−3/2.

Checks and common pitfalls: The negative result is consistent with an obtuse angle.

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

10 / Standard#Your turn

A student inserts u=0 into the angle formula and concludes θ=90°. Identify the invalid step.

\cos\theta=\frac{u\cdot v}{|u||v|}

Skills and prerequisite lessons

Scope: JM02

Working and explanation

BUILD THE REASONING

Hint 1

Check the denominator.

Hint 2

Zero divided by zero is not zero.

Worked solution

The denominator vanishes.
$|u||v|=0$
The ratio and vector angle are undefined.

The formula requires two nonzero vectors.

Checks and common pitfalls: A zero dot product alone cannot assign an angle to zero.

Reasoning checklist · self / teacher assessment

The formula requires two nonzero vectors.
A zero dot product alone cannot assign an angle to zero.

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

11 / Standard#Your turn

Find t so v−tu is perpendicular to u.

u=(1,2),\ v=(4,3)

Skills and prerequisite lessons

Scope: JM02

Working and explanation

BUILD THE REASONING

Hint 1

Set (v−tu)·u=0.

Hint 2

Divide by u·u.

Worked solution

Calculate the two dot products.
$v\cdot u=10,\ u\cdot u=5$
Solve for the coefficient.
$t=10/5=2$

t=2.

Checks and common pitfalls: Use the squared norm in the denominator.

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

12 / Transfer#Your turn

Find the point of segment AB closest to P; explain why unrestricted projection is insufficient.

A=(0,0),B=(2,0),P=(3,4)

Skills and prerequisite lessons

Scope: JM02

Working and explanation

BUILD THE REASONING

Hint 1

First project onto the whole line.

Hint 2

Then impose the segment parameter interval.

Worked solution

The line projection has x=3, outside [0,2].
$|(x,0)-P|^2=(x-3)^2+16$
On [0,2] the square decreases, so the nearest point is the endpoint.
$Q=B=(2,0),\ PQ=\sqrt{17}$

B is nearest; distance √17.

Checks and common pitfalls: A foot outside the segment is not a feasible minimizer.

Reasoning checklist · self / teacher assessment

B is nearest; distance √17.
A foot outside the segment is not a feasible minimizer.

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

13 / Transfer#Your turn

Describe all points X satisfying the displayed equation, including A and B.

A=(-2,0),B=(2,0),\ (X-A)\cdot(X-B)=0

Skills and prerequisite lessons

Scope: JM02

Working and explanation

BUILD THE REASONING

Hint 1

Write X=(x,y).

Hint 2

Do not require nonzero vectors to use the algebraic equation.

Worked solution

Expand the dot product.
$(x+2)(x-2)+y^2=0$
Recognize the circle; endpoints also give zero.
$x^2+y^2=4$

The complete circle with diameter AB, including its endpoints.

Checks and common pitfalls: The right-angle interpretation excludes endpoints; the algebraic locus does not.

Reasoning checklist · self / teacher assessment

The complete circle with diameter AB, including its endpoints.
The right-angle interpretation excludes endpoints; the algebraic locus does not.

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 positive, zero and negative work.
8	Concept derivation	Derive projection by making the residual dot product zero.
10	Guided application	Compute and distinguish projection coefficient and length.
12	Transfer and explanation	Solve the nearest-segment problem and the endpoint-inclusive circle locus.
5	Exit evidence	Explain the zero-vector angle exception.

Question sequence

Why can work be negative even though distance is nonnegative?
Does a perpendicular foot always lie on the segment?

Board development

Keep coefficient t, signed projected length t|u| and vector tu in separate columns.
Sketch the external foot and highlight the admissible segment interval.

Expected correct and incorrect responses

Students divide by |u| instead of |u|² when finding t. Ask for the vector tu and check its residual dot product.
Correct: “The coefficient is 1.5, but the signed projection length is 3.”
Incorrect: “The projection coefficient and projection length are always the same.”

Follow-up questions

How does changing the length of u without changing its direction alter t and tu?

Remediation

Project (3,4) onto (2,0): label coefficient 1.5, projected length 3 and vector (3,0).

Exit check

Name the two assumptions needed before interpreting a dot-product quotient as a cosine.

Assessment checklist and sample responses

Credit expansion, admissible parameter range and geometric interpretation independently. Sample: t=3/2 is outside [0,1], so the nearest segment point occurs at t=1.
Correct: “The coefficient is 1.5, but the signed projection length is 3.”