The vector product, areas and the scalar triple product
Answers are for parents and teachers.
1
Find A × B for A = (−3, 2, −2) and B = (1, 2, 2), and the unit vectors perpendicular to both.
i part: 2 × 2 − (−2) × 2 = 8. j part: −((−3) × 2 − (−2) × 1) = −(−6 + 2) = 4. k part: (−3) × 2 − 2 × 1 = −8. So A × B = (8, 4, −8), size 12, and the unit vectors are ±(2/3, 1/3, −2/3).
2
Find the area of the triangle with vertices D(2, 0, 1), E(3, 2, 3) and F(0, 1, 5).
DE = (1, 2, 2) and DF = (−2, 1, 4). DE × DF = (2 × 4 − 2 × 1, −(1 × 4 − 2 × (−2)), 1 × 1 − 2 × (−2)) = (6, −8, 5), size √125 = 5√5. The area is 5√5/2 ≈ 5.59.
3
Find the volume of the parallelepiped with edges (2, 1, 0), (1, 3, 1) and (0, 1, 4).
The determinant is 2 × (3 × 4 − 1 × 1) − 1 × (1 × 4 − 1 × 0) + 0 = 22 − 4 = 18. The volume is 18 cubic units.
4
For which a and b is (a, −6, 9) parallel to (2, b, −3)?
(a, −6, 9) = λ(2, b, −3): 9 = −3λ gives λ = −3, so a = −6 and −6 = −3b gives b = 2.
5
If ‖A‖ = 4, ‖B‖ = 5 and A · B = −12, find ‖A × B‖.
cos θ = −12/20 = −3/5, so sin θ = 4/5 (θ is obtuse, and sin is positive). Then ‖A × B‖ = 20 × 4/5 = 16.