(a) Three vectors a, b and c are \(\begin{pmatrix} 8 \\ 3 \end{pmatrix}, \begin{pmatrix} 6 \\ -5 \end{pmatrix}\) and \(\begin{pmatrix} 2 \\ -3 \end{pmatrix}...

Answer Details

(a) First compute the direction vector \( \mathbf{a} + \mathbf{b} - 2\mathbf{c} \):

\[ \begin{pmatrix} 8 \\ 3 \end{pmatrix} + \begin{pmatrix} 6 \\ -5 \end{pmatrix} - 2\begin{pmatrix} 2 \\ -3 \end{pmatrix} = \begin{pmatrix} 14 - 4 \\ -2 + 6 \end{pmatrix} = \begin{pmatrix} 10 \\ 4 \end{pmatrix}. \]

Its magnitude is \( \sqrt{10^2 + 4^2} = \sqrt{116} = 2\sqrt{29} \). The required vector \( \mathbf{d} \) has magnitude \( \sqrt{41} \) in this direction:

\[ \mathbf{d} = \sqrt{41}\cdot\frac{1}{\sqrt{116}}\begin{pmatrix} 10 \\ 4 \end{pmatrix} = \sqrt{\frac{41}{116}}\begin{pmatrix} 10 \\ 4 \end{pmatrix} \approx \begin{pmatrix} 5.94 \\ 2.38 \end{pmatrix}. \]

So \( \mathbf{d} \approx 5.94\mathbf{i} + 2.38\mathbf{j} \) (check: \( 5.94^2 + 2.38^2 \approx 41 \)).

(b) With \( O \) the origin, \( \vec{OA} = (3,4) \) so \( |OA| = 5 \), and \( \vec{OB} = (3,n) \) with \( |OB| = \sqrt{9+n^2} \). The angle \( AOB = 30^\circ \):

\[ \cos 30^\circ = \frac{\vec{OA}\cdot\vec{OB}}{|OA||OB|} = \frac{9 + 4n}{5\sqrt{9+n^2}} = \frac{\sqrt3}{2}. \]

\[ 2(9 + 4n) = 5\sqrt3\,\sqrt{9+n^2} \Rightarrow (18 + 8n)^2 = 75(9 + n^2). \]

\[ 324 + 288n + 64n^2 = 675 + 75n^2 \Rightarrow 11n^2 - 288n + 351 = 0. \]

\[ n = \frac{288 \pm \sqrt{288^2 - 4(11)(351)}}{22} = \frac{288 \pm \sqrt{67500}}{22}. \]

\[ n \approx \frac{288 \pm 259.81}{22} \Rightarrow n \approx 24.90 \text{ or } n \approx 1.28. \]