(a) Given that \(m = (6i + 8j)\) and \(n = (-8i + \frac{7}{3}j)\), find

Explanation

(a) \(m = (6i + 8j) ; n = (-8i + \frac{7}{3}j)\)

(i) \(|m| = \sqrt{6^{2} + 8^{2}} = \sqrt{36 + 64}\)

= 10 units.

Direction of m: \(\tan \alpha = \frac{8}{6} = 1.333\)

\(\alpha = 53.13°\)

\(|n| = \sqrt{(-8)^{2} + (\frac{7}{3})^{2}} = \sqrt{64 + \frac{49}{9}}\)

= \(\sqrt{\frac{625}{9}} = \frac{25}{3} units\)

Direction of n :

\(\tan \beta = \frac{\frac{7}{3}}{-8}\)

= \(-\frac{7}{24} = -0.2917\)

\(\beta = \tan^{-1} (-0.2917) = -16.26°\)

(ii) Let the angle between m and n be \(\theta\).

\(m \cdot n = |m||n| \cos \theta\)

\(\cos \theta = \frac{m \cdot n}{|m||n|}\)

\(m \cdot n = (6i + 8j) \cdot (-8i + \frac{7}{3}j)\)

= \(-48 + \frac{56}{3}\)

= \(\frac{-88}{3}\)

\(\cos \theta = \frac{-\frac{88}{3}}{10 \times \frac{25}{3}}\)

= \(\frac{-88}{250}\)

\(\cos \theta = -0.352\)

\(\theta = \cos^{-1} (-0.352) = 110.61°\)

(b) For PQ to be perpendicular to RS, the angle between them should be 90°. That is, \(\cos \theta = 0\).

\(\overrightarrow{PQ} = \begin{pmatrix} 10 \\ 4 \end{pmatrix} - \begin{pmatrix} -2 \\ 3 \end{pmatrix} = \begin{pmatrix} 12 \\ 1 \end{pmatrix}\)

\(\overrightarrow{RS} = \begin{pmatrix} 4 \\ 0 \end{pmatrix} - \begin{pmatrix} 3 \\ 12 \end{pmatrix} = \begin{pmatrix} 1 \\ -12 \end{pmatrix}\)

\(|PQ| = \sqrt{12^{2} + 1^{2}} = \sqrt{144 + 1} = \sqrt{145}\)

\(|RS| = \sqrt{1^{2} + (-12)^{2}} = \sqrt{1 + 144} = \sqrt{145}\)

\(\overrightarrow{PQ} \cdot \overrightarrow{RS} = (12i + j) \cdot (i - 12j)\)

= \(0\)

\(\overrightarrow{PQ} \cdot \overrightarrow{RS} = |PQ||RS| \cos \theta\)

\(\cos \theta = \frac{0}{(\sqrt{145})(\sqrt{145})} = 0\)