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← Senior 1 · short mathematical inquiries

Teacher reference: contains answers, conditions and proofs.

08 Plane vectors · g10-08-01 · 5–10 min · TEACHER · ANSWERS

Move the arrow without changing the vector

Must equal vectors start at the same point?

Predict: Move an arrow of displacement(3,2) to a different starting point. Predict its length and components.

CompareComputed observation
Origin start
Start x = 0; Start y = 0; Displacement x = 3; Displacement y = 2		Position changed; components and magnitude determine the vector.
Translated start
Start x = 2; Start y = -1; Displacement x = 3; Displacement y = 2		Position changed; components and magnitude determine the vector.
Zero vector
Start x = 2; Start y = -1; Displacement x = 0; Displacement y = 0		The zero vector has no defined direction.

Conclusion: A free vector is determined by magnitude and direction, not by its position. Translation preserves its component differences.

Conditions: Use directed displacement. The zero vector has magnitude0 but no defined direction.

Justification: If both endpoints gain(t₁,t₂), their difference remains(B+t)−(A+t)=B−A.

Ask: Which subtraction removes the common starting position?

Watch for: Calling two translated arrows different vectors because their endpoints differ.

Extension: Explain why reversing the arrow usually changes the vector.

Exit: Are arrows from(0,0) to(2,−1) and(3,4) to(5,3) equal?

Yes, both have displacement(2,−1).

Before: Coordinates, displacement and Pythagoras. · Timing: predict 1, compare 3, explain 2, exit 1 minute. Finite observations do not prove a universal claim.

08 Plane vectors · g10-08-02 · 5–10 min · TEACHER · ANSWERS

A detour and its final displacement

Does changing the order of two displacements change the destination?

Predict: Compare a+b andb+a for a=(2,1),b=(−1,3).

CompareComputed observation
Add displacements
a₁ = 2; a₂ = 1; b₁ = -1; b₂ = 3; Operation = a+b		The resultant joins the original start to the final endpoint.
Subtract displacements
a₁ = 2; a₂ = 1; b₁ = -1; b₂ = 3; Operation = a−b		The resultant joins the original start to the final endpoint.
Return to the start
a₁ = 2; a₂ = 1; b₁ = -2; b₂ = -1; Operation = a+b		The resultant joins the original start to the final endpoint.

Conclusion: Vector addition joins arrows head-to-tail and adds components. Subtraction adds the opposite arrow; path length is a different quantity.

Conditions: All arrows use the same coordinate scale. Resultant magnitude need not equal the sum of magnitudes.

Justification: (a₁+b₁,a₂+b₂)=(b₁+a₁,b₂+a₂) by commutativity of real addition. Subtraction uses a+(−b).

Ask: Which arrow shows the trip, and which joins the first start to the final end?

Watch for: Adding lengths while ignoring direction.

Extension: Construct a trip with positive distance but zero displacement.

Exit: Find(2,1)−(−1,3).

(3,−2).

Before: Coordinates, displacement and Pythagoras. · Timing: predict 1, compare 3, explain 2, exit 1 minute. Finite observations do not prove a universal claim.

08 Plane vectors · g10-08-03 · 5–10 min · TEACHER · ANSWERS

A negative scalar reverses direction

What does multiplication by−2 change?

Predict: Compare a,2a,−2a and0a for a=(2,1).

CompareComputed observation
Enlarge
a₁ = 2; a₂ = 1; Scalar λ = 2		Magnitude uses |λ|; direction comparison excludes zero.
Reverse
a₁ = 2; a₂ = 1; Scalar λ = -2		Magnitude uses |λ|; direction comparison excludes zero.
Zero scalar
a₁ = 2; a₂ = 1; Scalar λ = 0		Magnitude uses |λ|; direction comparison excludes zero.

Conclusion: λa has magnitude |λ||a|; λ<0 reverses direction. A zero product has no defined direction.

Conditions: Direction comparisons require both vectors to be nonzero. Scalar multiples are collinear algebraically, including a zero product.

Justification: The components become(λa₁,λa₂), whose norm is√(λ²(a₁²+a₂²))=|λ||a|.

Ask: What changes sign: the components or the length?

Watch for: Saying a negative multiple has a negative length.

Extension: Explain why0a cannot be called opposite in direction to a.

Exit: Find−3(1,−2) and its magnitude.

(−3,6), magnitude3√5.

Before: Coordinates, displacement and Pythagoras. · Timing: predict 1, compare 3, explain 2, exit 1 minute. Finite observations do not prove a universal claim.

08 Plane vectors · g10-08-04 · 5–10 min · TEACHER · ANSWERS

Two directions need to be independent

Can any pair of vectors locate every point uniquely?

Predict: Keep e₁=(1,0). Compare e₂=(0,1) ande₂=(1,0) for target(3,2).

CompareComputed observation
Standard basis
e₁ x = 1; e₁ y = 0; e₂ x = 0; e₂ y = 1; Target x = 3; Target y = 2		The independent directions give one coefficient pair.
Oblique basis
e₁ x = 1; e₁ y = 1; e₂ x = -1; e₂ y = 1; Target x = 3; Target y = 2		The independent directions give one coefficient pair.
Dependent directions
e₁ x = 1; e₁ y = 0; e₂ x = 2; e₂ y = 0; Target x = 3; Target y = 2		Dependent directions do not form a plane basis.

Conclusion: Two noncollinear vectors form a basis and give unique coefficients. Collinear vectors cover only a line, and may give no solution or nonunique solutions.

Conditions: The determinant e₁x e₂y−e₁y e₂x must be nonzero for a plane basis.

Justification: The two component equations have a unique solution precisely when their2×2 determinant is nonzero. A zero determinant makes the directions dependent.

Ask: What region can be reached when the two arrows point along the same line?

Watch for: Assuming two arrows are a basis simply because there are two of them.

Extension: Give a target on the common line with more than one coefficient pair.

Exit: Do(1,0) and(2,0) form a plane basis?

No; they are collinear and cannot reach(0,1).

Before: Coordinates, displacement and Pythagoras. · Timing: predict 1, compare 3, explain 2, exit 1 minute. Finite observations do not prove a universal claim.

08 Plane vectors · g10-08-05 · 5–10 min · TEACHER · ANSWERS

What does the sign of a dot product tell us?

Does a dot product of0 always mean a right angle?

Predict: Compare two nonzero vectors at60°,90°,120°, then let one length become0.

CompareComputed observation
Acute angle
|a| = 3; |b| = 2; Angle (degrees) = 60		Positive dot product: acute angle.
Right angle
|a| = 3; |b| = 2; Angle (degrees) = 90		Both nonzero: zero dot product means perpendicular.
Zero vector
|a| = 0; |b| = 2; Angle (degrees) = 90		Zero dot product here does not establish a right angle.

Conclusion: For nonzero vectors, a·b=|a||b|cosθ is positive,zero or negative for acute,right or obtuse angles. With a zero vector the angle is undefined.

Conditions: Both vectors must be nonzero before concluding perpendicularity from a zero dot product.

Justification: The lengths are strictly positive for nonzero vectors, so the dot-product sign equals the sign of cosθ. At θ=90°,cosθ=0.

Ask: Which assumption lets us divide the dot product by both lengths?

Watch for: Declaring a zero vector perpendicular to every vector because its dot products vanish.

Extension: Produce a nonzero pair with negative dot product.

Exit: What extra condition is needed in “a·b=0 implies a⊥b”?

a andb must both be nonzero.

Before: Coordinates, displacement and Pythagoras. · Timing: predict 1, compare 3, explain 2, exit 1 minute. Finite observations do not prove a universal claim.

08 Plane vectors · g10-08-06 · 5–10 min · TEACHER · ANSWERS

Translate a midpoint and a distance

What changes when both endpoints receive the same translation?

Predict: Translate A=(−1,0),B=(3,2) by(2,1); predict the new midpoint and distance.

CompareComputed observation
Original endpoints
A x = -1; A y = 0; B x = 3; B y = 2; Translation x = 0; Translation y = 0		The midpoint translates once; the distance remains fixed.
Translate both
A x = -1; A y = 0; B x = 3; B y = 2; Translation x = 2; Translation y = 1		The midpoint translates once; the distance remains fixed.
Coincident endpoints
A x = 1; A y = 1; B x = 1; B y = 1; Translation x = 2; Translation y = 1		The midpoint translates once; the distance remains fixed.

Conclusion: A common translation preserves endpoint differences and distance. The midpoint receives the same translation as the endpoints.

Conditions: Distance uses the Euclidean metric with the same unit on both axes; coincident endpoints have distance0.

Justification: M=(A+B)/2, so M′=((A+t)+(B+t))/2=M+t; B′−A′=B−A gives unchanged distance.

Ask: Why is the translation added once after dividing the sum by2?

Watch for: Adding the translation twice when calculating the midpoint.

Extension: Check whether doubling all coordinates is a translation.

Exit: Find the midpoint of(−1,0),(3,2), then translate it by(2,1).

Midpoint(1,1); translated midpoint(3,2).

Before: Coordinates, displacement and Pythagoras. · Timing: predict 1, compare 3, explain 2, exit 1 minute. Finite observations do not prove a universal claim.