(a) Given that \(p = (4i - 3j)\) and \(q = (-i + 5j)\), find r such that \(|r| = 15\) and is in the direction \((2p + 3q)\). (b) Forces of magnitude 8N, 6N ...

(a) With \(p=4i-3j\) and \(q=-i+5j\):

\[2p+3q=2(4i-3j)+3(-i+5j)=(8i-6j)+(-3i+15j)=5i+9j.\]

Its magnitude is

\[|2p+3q|=\sqrt{5^2+9^2}=\sqrt{25+81}=\sqrt{106}.\]

The unit vector in the direction of \(2p+3q\) is \(\dfrac{5i+9j}{\sqrt{106}}\). Since \(|r|=15\) and \(r\) is in this direction,

\[r=15\cdot\frac{5i+9j}{\sqrt{106}}=\frac{75}{\sqrt{106}}i+\frac{135}{\sqrt{106}}j\approx 7.28i+13.11j.\]

(b) From the diagram, taking angles anticlockwise from the positive \(x\)-axis (horizontal): the \(6\ \text{N}\) is vertical (\(90^\circ\)); the \(8\ \text{N}\) is \(30^\circ\) to the left of the \(6\ \text{N}\), so at \(120^\circ\); the \(4\ \text{N}\) is \(60^\circ\) to the right of the \(6\ \text{N}\), so at \(30^\circ\).

Resolve horizontally:

\[\sum F_x=8\cos120^\circ+6\cos90^\circ+4\cos30^\circ=-4+0+3.464=-0.536\ \text{N}.\]

Resolve vertically:

\[\sum F_y=8\sin120^\circ+6\sin90^\circ+4\sin30^\circ=6.928+6+2=14.928\ \text{N}.\]

(i) Magnitude:

\[R=\sqrt{(-0.536)^2+14.928^2}=\sqrt{0.287+222.85}=\sqrt{223.14}\approx \mathbf{14.94\ \text{N}}.\]

(ii) Direction. \(\sum F_x\) is negative and \(\sum F_y\) positive, so the resultant lies in the second quadrant (just left of the upward vertical). The acute angle to the vertical is

\[\tan^{-1}\!\left(\frac{|\sum F_x|}{\sum F_y}\right)=\tan^{-1}\!\left(\frac{0.536}{14.928}\right)\approx 2.1^\circ.\]

Measured anticlockwise from the positive \(x\)-axis, the direction is \(180^\circ-\tan^{-1}(14.928/0.536)\approx \mathbf{92.1^\circ}\); that is, the resultant of about \(14.94\ \text{N}\) acts almost vertically, inclined \(2.1^\circ\) to the left of the \(6\ \text{N}\) force.