Create a Vector Space
For a set to be a vector space, it must satisfy the following rules for all vectors u, v, and w, and all scalars a and b:[br][br][list=1][*]Closure under addition: u + v is always a vector in the set.[/*][*]Commutativity: u + v = v + u.[/*][*]Associativity of addition: (u + v) + w = u + (v + w).[/*][*]Additive identity: There exists a zero vector 0 such that u + 0 = u.[/*][*]Additive inverse: For every vector u, there exists a vector -u such that u + (-u) = 0.[/*][*]Closure under scalar multiplication: a * u is always a vector in the set.[/*][*]Distributivity of scalars over vectors: a * (u + v) = au + av.[/*][*]Distributivity of scalars over scalars: (a + b) * u = au + bu.[/*][*]Associativity of scalar multiplication: a * (b * u) = (a * b) * u.[/*][*]Scalar identity: 1 * u = u.[/*][/list][br]We will prove these 10 items using visual representations created through construction, and algebraic proof where the given vectors are non-specific, they could be any vectors.[br][br]Manipulate the given examples to ensure you feel the proof is visually sound.
It clearly shows that adding two vectors AB and BC results in another vector AC that is also a vector which is exactly what "closed under addition" means.
Vector AB = Vector EF[br]Vector BC = Vector DE[br]Vector AC = Vector DF[br][br]Vector AB + Vector BC = Vector AC[br]Since AB = EF and BC = DE, we can substitute:[br]Vector DE + Vector EF = Vector BC + Vector AB [br]Vector DF = Vector AC[br][br]Therefore Vector AB + Vector BC = Vector AC [br]and Vector BC + Vector AB = Vector DF = Vector AC[br][br]This shows that changing the order of addition still gives the same resultant vector. [br][br]As this is an algebraic proof these vectors could be any vectors, there for we've proven that vector addition is commutative.
In your own words, describe Figure 5.2 and specifically how it can be used to demonstrate that vector addition is commutative.
(1 mark) Correctly describes the figure: There are two different paths from a starting point to an ending point. One path goes AB then BC. The other path goes DE then EF (or BC then AB).[br](1 mark) Correctly states that both paths start at the same point and end at the same point (AC = DF).[br](1 mark) Correctly explains that AB + BC = AC and BC + AB = DF, and since AC = DF, the sums are equal.
Given (equal lengths):[br]E₀F₀ = E₃F₃ = E₄F₄[br]F₀G₀ = F₄G₄ and E₂H₂ = E₅H₅.[br]G₀H₀ = G₁H₁ = G₄H₄[br][br][br]The left side shows:[br](E₀F₀ + F₀G₀) + G₀H₀[br]= E₁G₁ + G₁H₁ (since E₁G₁ = E₀F₀ + F₀G₀)[br]= E₂H₂ (adding E₁G₁ + G₁H₁)[br][br]The right side shows:[br]= E₄F₄ + (F₄G₄ + G₄H₄)[br]= E₃F₃ + F₃H₃ (since F₄G₄ + G₄H₄ = F₃H₃)[br]= E₅H₅[br][br]where E₂H₂ = E₅H₅.[br][br]Therefore, (u + v) + w = u + (v + w), proving that vector addition is associative.
What does this figure prove about vector addition? Explain in your own words.
(1 mark) States associativity[br](2 mark) Explains that order of addition does not change the result
The figure shows:[br]Vector AB + Vector CC = Vector AB[br][br]Vector CC can be considered as just Point C, which has zero length and zero direction.[br]Therefore, any point can be used as the zero vector.
What do you notice about Vector CC?
starts and ends at the same point or CC has zero length (1 mark)
The figure shows [br]Vector AB and Vector CD [br][br]AB + CD = 0 [br]Therefore CD = -AB
What do you notice about the length and direction of Vector AB and Vector CD?
AB and CD are pointing in opposite directions. (0.5)[br] AB and CD are the same length (0.5)[br](1 mark) States equal length and opposite direction.
The figure shows Vector AB.[br]Point C is given on line CJ, which is parallel to AB.[br]Circle C has radius AB and intersects line CJ at point D.[br]Circle D has radius AB and intersects line CJ at point E.[br]Circle E has radius AB and intersects line CJ at point F.[br]This continues for n steps, ending at point J.[br][br]Therefore, CJ = n * AB, where n is the number of segments (CD, DE, EF, ...).
The Figure shows vector AB.[br]Vector A₀B₀ is a copy of vector AB.[br][br]Line P₁B₀ starts at B₀ and extends outward.[br]Points P₁, P₂, P₃, P₄, P₅ are equally spaced using the method from 5.6.1, [br]Therefore P₁B₀ = P₁P₂ = P₂P₃ = P₃P₄ = P₄P₅.[br][br]Parallel lines are drawn to meet A₀B₀ at points B₁, B₂, B₃, B₄, and A₀ [br]Parallel lines are drawn from P₁, P₂, P₃, P₄, and P₅ respectively.[br][br]This forms similar triangles: B₀P₁B₁, B₀P₂B₂, B₀P₃B₃, B₀P₄B₄, and B₀P₅A₀.[br][br]Because the points on line P₁B₀ are equally spaced, A₀B₀ is cut into 5 equal pieces.[br][br]A point Pₙ is animated to travel along P₁B₀. As it moves, it shows that any rational scalar multiple of AB can be obtained in this way.[br]
Why are the triangles B₀P₁B₁, B₀P₂B₂, B₀P₃B₃, B₀P₄B₄, and B₀P₅A₀ similar to each other?
(1 mark) States they share angle at B₀.[br](1 mark) States parallel lines give equal corresponding angles [br](2marks) Triangles are similar due to AAA.
The animation shows point Pₙ moving along line P₁B₀. What does this demonstrate about scalar multiplication of a vector?
(1 mark) States any scalar multiple can be constructed.[br](2 mark) States this proves closure under scalar multiplication.[br][br]
The figure shows[br]Slider "a" where 0 ≤ a ≤ 5.[br]The construction Vector u + Vector v = Vector w is given.[br]As the slider is adjusted, Vector aw scales correspondingly.[br]The construction au + av = aw is shown.[br]As the slider is adjusted, all vectors scale accordingly.
Move the slider from a = 1 to a = 3. Compare what happens to vectors u, v, and w versus vectors au, av, and aw.
(1 mark) States that u, v, and w do not change when the slider moves they remain fixed.[br](1 mark) States that au, av, and aw scale by the same factor they become 3 times longer.[br](1 mark) States that the directions of au, av, and aw remain the same as u, v, and w.
The figure shows[br]Slider a and slider b, where 0 ≤ a ≤ 5 and 0 ≤ b ≤ 5.[br]Vector u is given.[br]Vector cu is shown, where c = a + b.[br]Vector au and vector bu are shown head-to-tail.[br][br]Move the sliders and observe the vectors. You will see that vector au + vector bu is always equal to vector cu.[br][br]By inspection, vector au + bu = cu.
The figure shows slider a and slider b, where 0 ≤ a ≤ 5, and vector u can be manipulated to be any vector. The construction demonstrates that (a + b)u = au + bu for all values of a and b between 0 and 5, and for any vector u you choose.[br][br]Does the fact that a and b are limited to values between 0 and 5 truly limit the proof? Or is it easy to assume that the property would also hold for a = 500 or a = 5,000,000? Explain your reasoning.
(1 mark) Correctly states that the slider range (0 to 5) does not truly limit the proof if u can be any vector.[br][br](1 mark) Explains that scaling a vector by 500 or 5 million is the same operation as scaling by 5, just repeated or extended. The algebraic property (a + b)u = au + bu does not depend on the size of a or b.[br][br](1 mark) States that since u can be any vector, the property is demonstrated for all directions and magnitudes of u.
The figure shows[br]Slider a and slider b, where 0 ≤ a ≤ 5 and 0 ≤ b ≤ 5.[br]Vector u is given.[br][br]Vector b[i]u is shown.[br]Vector a(b[/i]u) is shown, this is vector u scaled by b, then by a.[br]Vector (a[i]b)[/i]u is shown, this is vector u scaled by the product a*b.[br][br]Move the sliders and observe that a(b[i]u) and (a[/i]b)*u are always equal in both length and direction.[br][br]The value of a*b is also displayed to help you verify.
Since vector u can be any vector and you can move the sliders, does this prove that the property holds for all values a and b and all vectors u? Explain.
(1 mark) Explains that only a finite range.[br](1 mark) Explains that u can be any vector, so direction/magnitude is covered.[br](1 mark) States that a formal algebraic proof is needed.
5.10 - Scalar identity: 1 * u = u.[br][br]The figure shows[br]Slider a, where 0 ≤ a ≤ 5[br]Vector AB labelled [br]Vector CD = a*AB
For what value of slider a are vector AB and vector CD equal?
If and only if a = 1, vector CD = 1 * AB = AB , so AB and CD are identical. This demonstrates the scalar identity property: 1 * u = u.