PART I — Content Questions (Q1–Q12)
Q1 — Multiple Choice (1 correct answer) · 1 pts
An adventurer sets out from an unknown starting point
- First, she travels by the vector from
to . - Then, she travels by twice the vector from
to .
She finds herself at
What is
- (a)
- (b) ★
- (c)
- (d)
Solution.
. , so . Total displacement: . Since , we get .
Q2 — Multiple Choice (1 correct answer) · 1 pt
On the surface of the Earth, if you walk 500 km south and then 500 km east, you arrive at a different location than if you walk 500 km east and then 500 km south. This shows that displacement addition on the sphere is not commutative.
In
- (a) ★ On a flat surface, direction vectors like “south” and “east” mean the same thing regardless of where you are. On a sphere, moving east shifts the meaning of “south” — by the time you’ve moved, the direction you call “south” has rotated relative to what it was — so the two moves you are composing are genuinely different depending on which you do first.
- (b) On a flat surface, the force of gravity is uniform, ensuring paths always coincide.
- (c) On a sphere, the finite size means your speed changes along the path, adding extra distance in one direction.
- (d) On a flat surface, any two paths between two points have the same total length.
Solution. The core point is that on a flat plane, “south” is a fixed direction vector
— it is the same wherever you stand. On a sphere, once you’ve moved east, the local notion of “south” is no longer “parallel” to what it was. You are composing different displacement vectors depending on the order. Options (b) and (c) are physically false; (d) confuses commutativity with arc-length equality.
Q3 — Numerical Answer · 1 pts
The vector
Find
Answer:
Solution. Substituting
into each equation:
- Equation 1:
, i.e. . - Equation 2:
, i.e. . Adding the two equations:
. Back-substituting into equation 2: . Thus .
Q4 — Multiple Answers (select all that apply) · 2 pts
Consider a general linear system
and its associated homogeneous system
Suppose
- (a) ★
- (b)
- (c)
- (d) ★
Solution. For each equation
, contributes and contributes .
- (a)
- (b)
in general - (c)
, so satisfies , not - (d)
This is the “particular solution + homogeneous” structure: you can freely add any homogeneous solution to a particular solution and stay in the solution set of
.
Q5 — Multiple Choice (1 correct answer) · 1 pt
Consider the equation
Two students each write the solution set in parametric form:
- Student A:
, for . - Student B:
, for .
Which of the following is correct?
- (a) Only Student A is right — the particular solution must be a unique, canonical point.
- (b) ★ Both are correct — they describe the same line using different particular solutions, and any point on the solution line is a valid choice for the “anchor.”
- (c) Neither is correct — the direction vector should be
, not . - (d) The two descriptions define parallel but distinct lines.
Solution. Both
and satisfy (check: and ). The direction satisfies the homogeneous equation . Both parametrizations trace the same geometric line. Option (d) is wrong because the direction vectors are identical — two lines with the same direction through points on the same line are the same line. Option (c) confuses the direction with the coefficient vector.
Q6 — Numerical Answer · 1 pts
A robot is placed at the origin of a factory floor. Its drive system allows it to move in exactly two directions:
The robot must reach the target position
Set up and solve the linear system. Enter the value of
Answer:
Solution. The system is
, i.e. From the first equation:
. Substituting: , then . So .
Q7 — Multiple Choice (1 correct answer) · 1 pt
In Lecture 3, we established that
The step
- (a) (A1):
- (b) (S2):
- (c) ★ (D2):
- (d) (D1):
Solution. The step factors a scalar sum out of two scalar multiples of the same vector — that is precisely (D2), distributivity over scalar addition. (D1) distributes a scalar over a vector sum. (S2) handles products of scalars. (A1) is about vector commutativity.
Q8 — Multiple Answers (select all that apply) · 2 pts
Consider the following four sets built from the empty set:
Select all statements below that are TRUE.
- (a) ★
- (b)
- (c) ★
- (d) ★
- (e) ★
- (f)
Solution.
- (a) ★ TRUE:
is literally listed as an element of . - (b) FALSE: The only element of
is , not . - (c) ★ TRUE:
is listed as the element of . - (d) ★ TRUE: The empty set is a subset of every set (it has no elements that could fail to belong). This is easy to mistake for (b).
- (e) ★ TRUE: The only element of
is , which is indeed in . - (f) FALSE:
contains ; contains . These are different elements, so the sets are different.
Q9 — Multiple Choice (1 correct answer) · 1 pt
In the row picture of a linear system (as introduced in Lecture 3), each homogeneous equation
In
- (a) A single point (the origin only)
- (b) A line through the origin
- (c) ★ A plane through the origin
- (d) All of
Solution. In
, fixing one linear constraint (the dot product with ) reduces the degrees of freedom from 3 to 2 — the solution set is 2-dimensional, i.e., a plane. Geometrically, it is the set of all vectors perpendicular to . A line would result from two independent constraints; the origin alone from three; all of from zero constraints.
Q10 — Multiple Answers (select all that apply) · 2 pts
Which of the following sets, with the standard operations (usual polynomial addition and scalar multiplication, or componentwise operations), are vector spaces? (Select all that apply.)
- (a) ★ All polynomials of degree at most 3
- (b) All polynomials of degree exactly 3
- (c) ★ All real-valued sequences
, with termwise addition and scalar multiplication - (d) The unit circle
, with the standard operations inherited from - (e) ★
for any fixed , with component-wise operations
Solution.
- (a) ★ Adding two polynomials of degree
gives degree ; scalar multiple also stays . Vector space axioms hold. - (b) ✗ Not closed under addition:
has degree 0, not 3. - (c) ★ Termwise operations are well-defined; all axioms hold.
- (d) ✗ Not closed under addition:
, and . - (e) ★ The canonical example from the course.
Q11 — Multiple Choice (1 correct answer) · 1 pts
The homogeneous equation
The student wants to use the homogeneous lemma (and scalar multiples of solutions) to generate a solution that is not a scalar multiple of
Can the student achieve this using only
- (a) ★ No. Since
, every linear combination reduces to — always a scalar multiple of . No new direction can be produced. - (b) Yes. The lemma guarantees that
is a solution, and since , their sum must point in a new direction. - (c) Yes. Multiplying
by a non-integer scalar will produce a solution outside the line through . - (d) No. The lemma does not apply here because
has more than one solution.
Solution.
, so — a line through the origin. The lemma guarantees that sums of solutions are solutions, but says nothing about obtaining independent solutions. To find a solution outside the line through , the student would need a starting vector not proportional to — which requires actually solving the equation more carefully.
Answer Summary
| # | Type | Correct Answer | Points |
|---|---|---|---|
| Q1 | MCQ | (b) | 1 |
| Q2 | MCQ | (a) Directions are location-dependent on a sphere | 1 |
| Q3 | Numeric | 1 | |
| Q4 | Multi-select | (a) and (d) | 2 |
| Q5 | MCQ | (b) Both parametrizations are correct | 1 |
| Q6 | Numeric | 1 | |
| Q7 | MCQ | (c) Axiom (D2) | 1 |
| Q8 | Multi-select | (a), (c), (d), (e) | 2 |
| Q9 | MCQ | (c) A plane through the origin | 1 |
| Q10 | Numeric | 1 | |
| Q11 | Multi-select | (a), (c), (e) | 2 |
| Q12 | MCQ | (a) No: | 1 |