27 exam-style questions on Vectors: concept, components, algebra, magnitude — topic 3.12 of Geometry and Trigonometry in the IB Mathematics: Analysis and Approaches syllabus — with answers and explanations. Samples below; the full set is free with an account.
Determine the magnitude of the vector v = 2i − j + 3k.
Answer: √14
|v| = √(2² + (−1)² + 3²) = √(4 + 1 + 9) = √14; forgetting the square root gives the distractor 14.
The points A(1, 2, −1) and B(3, −2, 5) are given. Determine the vector AB.
Answer: (2, −4, 6)
AB = OB − OA = (3 − 1, −2 − 2, 5 − (−1)) = (2, −4, 6); subtracting in the wrong order gives (−2, 4, −6).
Which of the following vectors is parallel to 3i − 6j?
Answer: 6i − 12j
6i − 12j = 2(3i − 6j), so it is a scalar multiple and therefore parallel; the others are not multiples of 3i − 6j.
Determine the unit vector in the direction of 6i − 8j.
Answer: 0.6i − 0.8j
|6i − 8j| = √(36 + 64) = 10, so the unit vector is (6i − 8j)/10 = 0.6i − 0.8j; dividing by 6 + 8 = 14 is a classic error.
Given a = i + 2j − k and b = 3i − j + 2k, calculate 2a − b.
Answer: −i + 5j − 4k
2a = 2i + 4j − 2k, then subtracting b componentwise gives (2 − 3)i + (4 + 1)j + (−2 − 2)k = −i + 5j − 4k.
A vector a has magnitude 5. The vector ka has magnitude 15 and points in the opposite direction to a. Determine k.
Answer: −3
|ka| = |k||a| gives |k| = 3, and reversing the direction requires k < 0, so k = −3.
That was 6 of 27 Vectors: concept, components, algebra, magnitude questions.
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