Multiple choice · 16
Q1Which of these is a vector quantity?
- AMass
- BTemperature
- CVelocity
- DTime
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Correct answer: C — Velocity
Velocity is speed together with a direction. Speed on its own would be a scalar, which is precisely the distinction being tested.
Q2If u = ⟨2, 3⟩ and v = ⟨5, −1⟩, then u + v is:
- A⟨7, 2⟩
- B⟨3, 4⟩
- C⟨10, −3⟩
- D⟨7, 4⟩
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Correct answer: A — ⟨7, 2⟩
Add componentwise: 2 + 5 = 7 and 3 + (−1) = 2.
Q3The magnitude of ⟨6, 8⟩ is:
- A14
- B10
- C48
- D√14
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Correct answer: B — 10
√(36 + 64) = √100 = 10. This is the 3–4–5 triple scaled by 2, which examiners use frequently.
Q4If A = (2, 7) and B = (5, 3), then AB is:
- A⟨7, 10⟩
- B⟨3, −4⟩
- C⟨−3, 4⟩
- D⟨3, 4⟩
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Correct answer: B — ⟨3, −4⟩
AB = b − a = ⟨5 − 2, 3 − 7⟩ = ⟨3, −4⟩. Option C is BA, the vector in the opposite direction.
Q5A unit vector in the direction of ⟨0, 4⟩ is:
- A⟨0, 4⟩
- B⟨0, 1⟩
- C⟨1, 0⟩
- D⟨0, 1/4⟩
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Correct answer: B — ⟨0, 1⟩
The magnitude is 4, so dividing gives ⟨0, 1⟩, which points the same way and has length 1.
Q6⟨6, −9⟩ is parallel to:
- A⟨2, −3⟩
- B⟨9, 6⟩
- C⟨−3, −2⟩
- D⟨3, 9⟩
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Correct answer: A — ⟨2, −3⟩
⟨6, −9⟩ = 3⟨2, −3⟩, a scalar multiple, so they are parallel. Option B is perpendicular to it, not parallel.
Q7If AB = DC in quadrilateral ABCD, then ABCD is:
- Aa rectangle
- Ba parallelogram
- Ca rhombus
- Da trapezium only
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Correct answer: B — a parallelogram
Equal vectors mean the sides are the same length and parallel, which defines a parallelogram. It might additionally be a rectangle or rhombus, but nothing here establishes that.
Q83⟨2, −1⟩ − 2⟨1, 4⟩ equals:
- A⟨4, −11⟩
- B⟨8, 5⟩
- C⟨4, 5⟩
- D⟨8, −11⟩
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Correct answer: A — ⟨4, −11⟩
3⟨2, −1⟩ = ⟨6, −3⟩ and 2⟨1, 4⟩ = ⟨2, 8⟩. Subtracting gives ⟨6 − 2, −3 − 8⟩ = ⟨4, −11⟩.
Q9The magnitude of 2i − 3j + 6k is:
- A5
- B7
- C11
- D49
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Correct answer: B — 7
Square each component, add, then take the root: √(2² + (−3)² + 6²) = √(4 + 9 + 36) = √49 = 7. The minus sign on the j component disappears when it is squared, so a negative component never reduces the magnitude.
Q10The dot product of two perpendicular vectors is:
- A1
- B0
- Ctheir magnitudes multiplied
- Dundefined
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Correct answer: B — 0
a · b = |a||b| cos 90° = 0. This is the standard test for perpendicularity.
Q11a × b is:
- Aa scalar
- Ba vector perpendicular to both
- Ca vector parallel to a
- Dalways zero
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Correct answer: B — a vector perpendicular to both
The cross product returns a vector at right angles to the plane containing a and b, with direction given by the right-hand rule.
Q12If a × b = 0 for non-zero vectors, then a and b are:
- Aperpendicular
- Bparallel
- Cequal
- Dunit vectors
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Correct answer: B — parallel
|a × b| = |a||b| sin θ, which vanishes only when sin θ = 0, so θ is 0° or 180° — the vectors are parallel. Perpendicular is the condition for the dot product.
Q13i · j equals:
- A0
- B1
- Ck
- Di
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Correct answer: A — 0
i and j are perpendicular unit vectors, so their dot product is 1 × 1 × cos 90° = 0. Note that i × j = k, which is a different product entirely.
Q14The area of the triangle formed by vectors a and b is:
- A|a × b|
- B½|a × b|
- Ca · b
- D½(a · b)
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Correct answer: B — ½|a × b|
The cross product magnitude gives the parallelogram area, and a triangle is half of that parallelogram.
Q15If a · (b × c) = 0, the three vectors are:
- Amutually perpendicular
- Bcoplanar
- Call zero
- Dunit vectors
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Correct answer: B — coplanar
The triple product is the volume of the solid they span. Zero volume means they all lie in one plane.
Q16Which is true?
- Aa · b = b · a and a × b = b × a
- Ba · b = b · a but a × b = −(b × a)
- Cboth anticommute
- Dboth are vectors
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Correct answer: B — a · b = b · a but a × b = −(b × a)
The dot product is commutative because multiplication of components is. The cross product reverses direction when the order is swapped, which is why the order must be preserved exactly as the question gives it.
Exam-style questions · 12
Q1[2 marks]
Distinguish between a scalar and a vector, giving one example of each.
Answer
A scalar has magnitude only — for example a mass of 5 kg. A vector has both magnitude and direction — for example a displacement of 5 m due east.
Q2[2 marks]
If u = ⟨3, −4⟩, find |u| and a unit vector in the direction of u.
Answer
|u| = √(9 + 16) = 5, and the unit vector is ⟨3/5, −4/5⟩.
Q3[2 marks]
Given AB = ⟨4, −2⟩, write down BA and explain the relationship.
Answer
BA = ⟨−4, 2⟩. It is the same length but in the exactly opposite direction, so BA = −AB.
Q4[4 marks]
P is (1, 3), Q is (5, 6) and R is (9, 9). Show that P, Q and R are collinear.
Mark scheme
PQ = ⟨5 − 1, 6 − 3⟩ = ⟨4, 3⟩head minus tail[1]QR = ⟨9 − 5, 9 − 6⟩ = ⟨4, 3⟩[1]PQ = QR, so PQ is a scalar multiple of QR with k = 1, hence they are parallel[1]- They share the point Q, so P, Q and R lie on a single straight line and are collinearthe shared point is essential — parallel alone is not enough[1]
PQ = QR = ⟨4, 3⟩ and they share Q, so the three points are collinear (with Q the midpoint of PR).
Q5[4 marks]
Given a = ⟨2, 5⟩ and b = ⟨−3, 1⟩, find 2a − 3b and its magnitude.
Mark scheme
2a = ⟨4, 10⟩both components multiplied[1]3b = ⟨−9, 3⟩[1]2a − 3b = ⟨4 − (−9), 10 − 3⟩ = ⟨13, 7⟩subtracting a negative is where marks are lost[1]|2a − 3b| = √(169 + 49) = √218 ≈ 14.8exact surd or 3 s.f. both accepted[1]
⟨13, 7⟩, magnitude √218 ≈ 14.8
Q6[6 marks]
A quadrilateral has vertices A(1, 1), B(5, 2), C(6, 6) and D(2, 5).
- Find the vectors AB and DC.
- Show that ABCD is a parallelogram.
- Find the position vector of the point where the diagonals meet.
Mark scheme
AB = ⟨5 − 1, 2 − 1⟩ = ⟨4, 1⟩[1]DC = ⟨6 − 2, 6 − 5⟩ = ⟨4, 1⟩take the vertices in the order that makes DC correspond to AB[1]AB = DC, so AB and DC are equal in length and parallel[1]- One pair of opposite sides equal and parallel is sufficient for a parallelogramthe reason must be stated, not just the equality[1]
- The diagonals of a parallelogram bisect each other, so they meet at the midpoint of ACor the midpoint of BD, which gives the same point[1]
- Midpoint
= (a + c)/2 = (⟨1,1⟩ + ⟨6,6⟩)/2 = ⟨3.5, 3.5⟩checking with BD: (⟨5,2⟩ + ⟨2,5⟩)/2 = ⟨3.5, 3.5⟩ ✓[1]
(a) AB = DC = ⟨4, 1⟩ (b) equal and parallel, so a parallelogram (c) ⟨3.5, 3.5⟩
Q7[2 marks]
Find the magnitude of v = 3i − 4j + 12k.
Answer
|v| = √(9 + 16 + 144) = √169 = 13.
Q8[2 marks]
State two differences between the dot product and the cross product.
Answer
The dot product gives a scalar and is commutative; the cross product gives a vector perpendicular to both, and reversing the order reverses its direction. The dot product is zero for perpendicular vectors, the cross product for parallel ones.
Q9[2 marks]
If a · b = 0 and neither vector is zero, what can you conclude?
Answer
Since a · b = |a||b| cos θ and neither magnitude is zero, cos θ = 0, so θ = 90° — the vectors are perpendicular.
Q10[4 marks]
Given a = i + 2j − k and b = 3i − j + 2k, find a · b and the angle between them.
Mark scheme
a · b = (1)(3) + (2)(−1) + (−1)(2) = 3 − 2 − 2 = −1a negative dot product means an obtuse angle[1]|a| = √(1 + 4 + 1) = √6[1]|b| = √(9 + 1 + 4) = √14[1]cos θ = −1/(√6 √14) = −1/9.165 = −0.109, so θ = 96.3°obtuse, as the negative dot product predicted[1]
a · b = −1; θ ≈ 96.3°
Q11[4 marks]
Find a × b for a = 2i + j + k and b = i − j + 3k, and hence the area of the triangle they form.
Mark scheme
- i component:
(1)(3) − (1)(−1) = 3 + 1 = 4[1] - j component:
−[(2)(3) − (1)(1)] = −5the minus sign on j is where marks are lost[1] - k component:
(2)(−1) − (1)(1) = −3, so a × b = 4i − 5j − 3k[1] - Area
= ½|a × b| = ½√(16 + 25 + 9) = ½√50 = 3.54accept (5√2)/2[1]
a × b = 4i − 5j − 3k; area ≈ 3.54 square units
Q12[6 marks]
Three vectors are given: a = i + j, b = j + k and c = i + k.
- Find
b × c. - Hence find the scalar triple product
a · (b × c). - State what your answer tells you about the three vectors, and give the volume of the parallelepiped they span.
Mark scheme
- b = ⟨0, 1, 1⟩ and c = ⟨1, 0, 1⟩; i component
= (1)(1) − (1)(0) = 1[1] - j component
= −[(0)(1) − (1)(1)] = 1; k component = (0)(0) − (1)(1) = −1so b × c = i + j − k[1] a · (b × c) = (1)(1) + (1)(1) + (0)(−1)a = ⟨1, 1, 0⟩[1]= 2[1]- The triple product is non-zero, so the three vectors are NOT coplanara zero value would have meant they all lay in one plane[1]
- Volume
= |2| = 2 cubic units[1]
(a) i + j − k (b) 2 (c) not coplanar; volume 2 cubic units