Answer: (a) \(\displaystyle \frac1{17}\begin{pmatrix}-15\\8\end{pmatrix}\); (b) \(a=2\), \(b=-\frac54\).

Answer: (a) \(\displaystyle \frac1{17}\begin{pmatrix}-15\\8\end{pmatrix}\); (b) \(a=2\), \(b=-\frac54\).

(a) The magnitude of \(\begin{pmatrix}-15\\8\end{pmatrix}\) is

\(\sqrt{(-15)^2+8^2}=\sqrt{225+64}=17.\)

So the unit vector is

(b) Add the vectors on the left:

\(\begin{pmatrix}2a+4b-12\\-2\end{pmatrix}=\begin{pmatrix}4b-4a\\4a+8b\end{pmatrix}.\)

Equating the first components gives

\(2a+4b-12=4b-4a,\)

so \(6a=12\), hence \(a=2\).

Equating the second components gives

\(-2=4a+8b.\)

Substitute \(a=2\):

\(-2=8+8b,\)

so \(8b=-10\), and therefore

\(b=-\frac54.\)