(i) If we can write c as a linear combination of a and b, then there exist scalars k₁ and k₂ such that
[tex]c = k_1a + k_2b[/tex]
Solve for k₁ and k₂ :
[tex]3u + 4v = k_1 (3u - v) + k_2 (-u + 2v)[/tex]
[tex]3u + 4v = (3k_1 - k_2) u + (-k_1 + 2k_2) v[/tex]
Then we must have
[tex]\begin{cases}3k_1 - k_2 = 3 \\ -k_1 + 2k_2 = 4\end{cases} \implies k_1 = 2, k_2 = 3[/tex]
and so
[tex]\boxed{c = 2a + 3b}[/tex]
(ii) Given that u and v form an angle of 60° between them, we have by the dot product identity
[tex]u \cdot v = \|u\| \|v\| \cos(60^\circ)[/tex]
where ||u|| denotes the magnitude of the vector u. Both u and v are unit vectors, so ||u|| = ||v|| = 1, and
[tex]u \cdot v = \dfrac12[/tex]
A special case of the identity with u = v tells us that for any vector u,
[tex]u \cdot u = \|u\|^2 \cos(0^\circ) = \|u\|^2[/tex]
which means
[tex]u \cdot u = v \cdot v = 1[/tex]
This in turn means we have
[tex]a \cdot a = (3u - v) \cdot (3u - v) = 9 (u \cdot u) + (v \cdot v) - 6 (u \cdot v) = 7[/tex]
[tex]a \cdot b = (3u - v) \cdot (-u + 2v) = -3 (u\cdot u) - 2 (v \cdot v) + 7 (u \cdot v) = -\dfrac32[/tex]
[tex]a \cdot c = (3u - v) \cdot (3u + 4v) = 9 (u\cdot u) - 4 (v\cdot v) + 9 (u \cdot v) = \dfrac{19}2[/tex]
[tex]b\cdot b = (-u + 2v) \cdot (-u + 2v) = (u \cdot u) + 4 (v\cdot v) - 4 (u \cdot v) = 3[/tex]
[tex]b\cdot c = (-u + 2v) \cdot (3u + 4v) = -3 (u\cdot u) + 8 (v\cdot v) + 2 (u\cdot v) = 6[/tex]
[tex]c \cdot c = (3u + 4v) \cdot (3u + 4v) = 9 (u\cdot u) + 16 (v\cdot v) + 24 (u\cdot v) = 37[/tex]
and hence
[tex]d \cdot d = (3a + 2b + c) \cdot (3a + 2b + c)[/tex]
[tex]d \cdot d = 9 (a\cdot a) + 4 (b\cdot b) + (c \cdot c) + 12 (a\cdot b) + 6 (a\cdot c) + 4 (b\cdot c)[/tex]
[tex]d\cdot d = 175[/tex]
and so ||d|| = √175 = 5√7.
