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IB Maths AA

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Vectors practice questions — IB Maths AA

12 free multiple-choice problems on vectors, ordered to match IB Maths AA difficulty, each with a full worked solution. No account needed — and there's a fresh problem set every day.

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Problem #0429 International
Advancedvectors
Let and be two non-collinear vectors. Find so that and are collinear.

Problems & worked solutions

Problem #0429 International

Problem 1 — Vectors

Let a⃗ and b⃗ be two non-collinear vectors. Find m∈R so that u⃗=3a⃗−(m+1)b⃗ and v⃗=(m−1)a⃗−5b⃗ are collinear.

Show answer & worked solution
  1. A. {−4}
  2. B. {4}
  3. C. {−4,4}✓ correct
  4. D. {−2,8}

3⋅(−5)=(m−1) (−(m+1)), i.e. −15=−(m−1)(m+1)=1−m2, so m2=16 and m∈{−4,4}.

Problem #0430 International

Problem 2 — Vectors

For u⃗=i⃗+j⃗ and v⃗=ai⃗−2j⃗, find a∈R so that ∣u⃗+v⃗∣2=∣u⃗∣2+∣v⃗∣2.

Show answer & worked solution
  1. A. −2
  2. B. −1
  3. C. 2✓ correct
  4. D. 4

u⃗⋅v⃗=a+(−2)=a−2=0⇒a=2.

Problem #0425 International

Problem 3 — Vectors

Are the vectors u⃗=i⃗+2j⃗ and v⃗=2i⃗+4j⃗ collinear?

Show answer & worked solution
  1. A. Yes, v⃗=2u⃗✓ correct
  2. B. No, the dot product is non-zero
  3. C. Yes, but only because they are perpendicular
  4. D. No, they have different magnitudes

v⃗=2u⃗, so the two vectors are collinear (parallel).

Problem #0427 International

Problem 4 — Vectors

Find m∈R so that u⃗=mi⃗+3j⃗ and v⃗=4i⃗+(m+1)j⃗ are perpendicular.

Show answer & worked solution
  1. A. −1
  2. B. −37✓ correct
  3. C. 37
  4. D. 1

u⃗⋅v⃗=4m+3(m+1)=7m+3=0⇒m=−37.

Problem #0426 International

Problem 5 — Vectors

For u⃗=i⃗+j⃗ and v⃗=ai⃗−2j⃗, find a∈R so that u⃗ and v⃗ are collinear.

Show answer & worked solution
  1. A. −2✓ correct
  2. B. −12
  3. C. 12
  4. D. 2

v⃗=ku⃗ for some k ⇒ a=k and −2=k. So a=−2.

Problem #0428 International

Problem 6 — Vectors

The vector AB→ from A(1,2) to B(4,6) equals:

Show answer & worked solution
  1. A. 5i⃗+8j⃗
  2. B. 3i⃗+4j⃗✓ correct
  3. C. −3i⃗−4j⃗
  4. D. i⃗+2j⃗

AB→=(4−1)i⃗+(6−2)j⃗=3i⃗+4j⃗.

Problem #1640 International

Problem 7 — Vectors

The vectors u⃗=(m, m−2) and v⃗=(m+3, 1) are perpendicular. The sum of all possible values of m is:

Show answer & worked solution
  1. A. 2
  2. B. −4✓ correct
  3. C. 4
  4. D. −2

The perpendicularity condition: u⃗⋅v⃗=0: m(m+3)+(m−2)⋅1=0  ⟹  m2+3m+m−2=0  ⟹  m2+4m−2=0.

This is a quadratic equation with two real roots (discriminant 16+8=24>0). By Vieta's formulas, the sum of the roots is −41=−4 (without needing to compute them explicitly, which are m=−2±6).

Check: (−2+6)+(−2−6)=−4. ✓ And, by direct computation: the product of the roots =−21=−2, consistent with (−2+6)(−2−6)=4−6=−2. ✓

Problem #1636 International

Problem 8 — Vectors

The points A(2,−1), B(5,3) and C(m,11) are collinear for m=

Show answer & worked solution
  1. A. 8
  2. B. 11✓ correct
  3. C. 5
  4. D. 443

AB→=(5−2, 3−(−1))=(3,4), AC→=(m−2, 11−(−1))=(m−2,12).

Collinearity: 3⋅12−4⋅(m−2)=0  ⟹  36−4m+8=0  ⟹  4m=44  ⟹  m=11.

Check: slope of AB=43; slope of AC=12m−2=129=43. ✓ Equal, so collinear.

Problem #0424 International

Problem 9 — Vectors

The dot product u⃗⋅v⃗ for u⃗=2i⃗+3j⃗, v⃗=i⃗−4j⃗ equals:

Show answer & worked solution
  1. A. −10✓ correct
  2. B. −6
  3. C. 5
  4. D. 14

u⃗⋅v⃗=2⋅1+3⋅(−4)=2−12=−10.

Problem #0422 International

Problem 10 — Vectors

For u⃗=i⃗+2j⃗ and v⃗=3i⃗−j⃗, the sum u⃗+v⃗ equals:

Show answer & worked solution
  1. A. 4i⃗+2j⃗
  2. B. 4i⃗−3j⃗
  3. C. 4i⃗+j⃗✓ correct
  4. D. −2i⃗+3j⃗

Add components: (1+3)i⃗+(2−1)j⃗=4i⃗+j⃗.

2 more Vectors questions in the app

Also covered in Vectors practice across every exam.

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