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.