Term 1 review: binomial expansions, complex numbers, and vectors, lines and planes
Answers are for parents and teachers.
1
Find the coefficient of x⁵ in (x² − 2/x)⁷, and show that x⁶ does not occur.
The power is 2(7 − r) − r = 14 − 3r. For x⁵: r = 3 and the coefficient is 7C3 × (−2)³ = 35 × (−8) = −280. For x⁶: 14 − 3r = 6 gives r = 8/3, not a whole number, so there is no x⁶ term.
2
Write z = −√2 + √2 i in trigonometric form and find z⁶.
|z| = 2 and the amplitude is 3π/4, so z = 2(cos 3π/4 + i sin 3π/4). z⁶ = 64(cos 9π/2 + i sin 9π/2) = 64(cos π/2 + i sin π/2) = 64i.
3
Solve Z³ = −27, and say what shape the three roots form on the Argand plane.
−27 = 27(cos 180° + i sin 180°), so Z = 3(cos((180° + 360°k)/3) + i sin(…)): amplitudes 60°, 180°, 300°. The roots are 3/2 + (3√3/2)i, −3 and 3/2 − (3√3/2)i. They are the corners of an equilateral triangle on the circle of radius 3.
4
For A(2, 1, −1), B(4, 3, 0) and C(1, 2, 2) find the angle at A, the area of triangle ABC and the equation of its plane.
AB = (2, 2, 1), AC = (−1, 1, 3). AB · AC = 3, norms 3 and √11, so cos A = 3/(3√11) = 1/√11 and A ≈ 72.5°. AB × AC = (5, −7, 4), of size √90 = 3√10, so the area is 3√10/2 ≈ 4.74. The plane is 5x − 7y + 4z = 10 − 7 − 4 = −1, that is 5x − 7y + 4z + 1 = 0 (check B: 20 − 21 + 0 = −1, C: 5 − 14 + 8 = −1).
5
Find where the line r = (0, 1, 1) + t(1, 2, 2) meets the plane 2x + 2y − z = 9, and the distance between that plane and the plane 2x + 2y − z = −3.
The point is (t, 1 + 2t, 1 + 2t): 2t + 2 + 4t − 1 − 2t = 4t + 1 = 9, so t = 2 and the point is (2, 5, 5). Check: 4 + 10 − 5 = 9. The distance between the planes is |9 − (−3)|/3 = 4.