27 exam-style questions on Vectors: concept, components, unit/base vectors, normalizing — topic 3.10 of Geometry and Trigonometry in the IB Mathematics: Applications and Interpretation syllabus — with answers and explanations. Samples below; the full set is free with an account.
Given the vectors a = (3, -4) and b = (1, 2), calculate a + 2b.
Answer: (5, 0)
a + 2b = (3, -4) + (2, 4) = (5, 0). Add corresponding components after scaling b by 2.
A vector in three dimensions is given by u = 2i - 3j + 6k. Determine its magnitude.
Answer: 7
|u| = √(2² + (-3)² + 6²) = √(4 + 9 + 36) = √49 = 7.
Determine the unit vector in the direction of v = (3, 4).
Answer: (0.6, 0.8)
|v| = √(9 + 16) = 5, so the unit vector is (3/5, 4/5) = (0.6, 0.8).
Given a = (5, -2) and b = (-1, 3), calculate a - b.
Answer: (6, -5)
a - b = (5 - (-1), -2 - 3) = (6, -5). Subtract corresponding components.
The base vectors i, j, k represent unit vectors along which respective axes?
Answer: x, y, z axes
By convention i = (1,0,0), j = (0,1,0), k = (0,0,1), the unit vectors along the x, y and z axes.
Determine the value of t for which the vectors a = (t, 2) and b = (6, -3) are perpendicular.
Answer: 1
Perpendicular means a·b = 0: 6t + 2(-3) = 0, so 6t = 6, giving t = 1.
That was 6 of 27 Vectors: concept, components, unit/base vectors, normalizing questions.
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27 Vectors: concept, components, unit/base vectors, normalizing questions — free with an account. Spaced repetition, streaks and full exam simulations included.
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