27 exam-style questions on Vector (cross) product; area applications — topic 3.16 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.
Calculate a × b for a = (1, 0, 2) and b = (0, 3, 1).
Answer: (-6, -1, 3)
a × b = (a₂b₃ - a₃b₂, a₃b₁ - a₁b₃, a₁b₂ - a₂b₁) = (0 - 6, 0 - 1, 3 - 0) = (-6, -1, 3); forgetting the sign pattern in the middle component is the classic error.
Let θ be the angle between non-zero vectors v and w. State the correct expression for the magnitude of the vector product |v × w|.
Answer: |v||w| sin θ
|v × w| = |v||w| sin θ; the expression with cos θ is the scalar (dot) product v·w, a frequent mix-up.
For the standard unit vectors, determine i × j.
Answer: k
Following the right-hand rule and the cyclic order i → j → k, we get i × j = k (whereas j × i = -k).
A triangle has vertices A(1, 1, 0), B(3, 2, 0) and C(1, 4, 0). Using the vector product, calculate the area of triangle ABC.
Answer: 3
AB = (2, 1, 0) and AC = (0, 3, 0) give AB × AC = (0, 0, 6), so the area is (1/2)|AB × AC| = (1/2)(6) = 3; answer 6 comes from forgetting the factor 1/2.
For any vector a, determine a × a.
Answer: the zero vector
The angle between a and itself is 0, so |a × a| = |a||a| sin 0 = 0; the cross product of parallel vectors is always the zero vector.
Given that a × b = (2, -1, 3), determine b × a.
Answer: (-2, 1, -3)
The vector product is anticommutative: b × a = -(a × b) = (-2, 1, -3).
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