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Math 2030 Assignment 1 SOLUTIONS (1) Q14 from Poole: In the following figure A, B, C, D, E, and F are vertices of a regular hexagon centred at the origin O. C B b D b b b O b b b E F A −→ −−→ Express each of the following vectors in terms of a = OA, b = OB. − − → −→ −−→ −→ −−→ (a) AB = AO + OB = −OA + OB −→ −−→ −→ (b) BC = AO = −OA −−→ −→ (c) AD = −2OA − −→ − − → − − → −→ −−→ (d) CF = 2BA = −2AB = 2OA − 2OB −→ − − → −−→ −−→ −→ −→ −−→ −→ (e) AC = AB + BC = OB − OA − OA = OB − 2OA −−→ −→ −−→ −→ −−→ −−→ −→ (f) BC + DE + F A = −OA + (OA − OB) + OB = 0 (2) Express v as a linear combination of a, b and (where given) c. (a) v = (3, −4), a = (1, 0), b = (0, 1). v = 3a − 4b (b) v = (6, 3), a = (2, 4), b = (0, 1). Use a to give the first coordinate, then correct the second coordinate with b. v = 3a − 9b (c) v = (−3, 3, 5), a = (1, 0, 0), b = (1, 1, 1), c = (0, 0, 1). Use c to give the second coordinate, then correct the first coordinate with a and the third with c. v = −6a + 3b + 2c (d) v = (0, 15, 3), a = (4, −4, 5), b = (−6, 1, −2), c = (0, 5, −5). Note that 3a + 2b + 2c = (0, 0, 1); this may help. Write v using a linear combination of c and (0, 0, 1), then expand. v = 3c + 18(0, 0, 1) = 3c + 18(3a + 2b + 2c) = 54a + 36b + 39c (3) What is the angle between u = (1, −3, 5, 1) and v = (2, 6, 3, 1) in R4 ? Since u · v = 1 · 2 − 3 · 6 + 5 · 3 + 1 · 1 = 2 − 18 + 15 + 1 = 0, we have u·v cos θ = = 0. kukkvk As cos θ = 0, θ = π/2. 2 (4) Q42 and Q43 from Poole: Find all values of k for which the two vectors are orthogonal. (a) u = (2, 3), v = (k + 1, k − 1). We know that u and v are orthogonal if and only if u · v = 0. But u · v = 2k + 2 + 3k − 3 = 5k − 1. So, u and v are orthogonal when 5k − 1 = 0; k = 1/5. (b) u = (1, −1, 2), v = (k 2 , k, −3). Similarly u and v are orthogonal when 0 = u · v = k 2 − k − 6 = (k + 2)(k − 3) and hence when k = −2 or k = 3. (5) Q62b from Poole: Prove that if u is orthogonal to both v and w, then u is orthogonal to any linear combination of v and w (i.e. u is orthogonal to sv + tw for any scalars s, t ∈ R). We know that u · v = 0 and u · w = 0. Thus, u · (sv + tw) = u · sv + u · tw = s(u · v) + t(u · w) = s0 + t0 = 0 As u · (sv + tw) = 0, we know u and sv + tw are orthogonal.