(a) Given that \(m = (6i + 8j)\) and \(n = (-8i + \frac{7}{3}j)\), find the : (i) magnitudes and direction of m and n ; (ii) angle between m and n. (b) The ...

Answer Details

(a)(i) Magnitudes and directions.

\(m=6i+8j\): \(|m|=\sqrt{6^{2}+8^{2}}=10\). Direction \(=\tan^{-1}\dfrac{8}{6}=53.1^{\circ}\) above the positive \(x\)-axis.

\(n=-8i+\tfrac{7}{3}j\): \(|n|=\sqrt{(-8)^{2}+\left(\tfrac{7}{3}\right)^{2}}=\sqrt{\dfrac{576+49}{9}}=\sqrt{\dfrac{625}{9}}=\dfrac{25}{3}\approx8.33\). It is in the second quadrant, direction \(=180^{\circ}-\tan^{-1}\dfrac{7/3}{8}=180^{\circ}-16.3^{\circ}=163.7^{\circ}\).

(ii) Angle between m and n.

\[m\cdot n=6(-8)+8\left(\tfrac{7}{3}\right)=-48+\tfrac{56}{3}=-\tfrac{88}{3}\]

\[\cos\theta=\frac{m\cdot n}{|m|\,|n|}=\frac{-88/3}{10\times25/3}=\frac{-88}{250}=-0.352\]

\[\theta=\cos^{-1}(-0.352)\approx110.6^{\circ}\]

(b) \(\overrightarrow{PQ}=Q-P=\binom{10}{4}-\binom{-2}{3}=\binom{12}{1}\); \(\overrightarrow{RS}=S-R=\binom{4}{0}-\binom{3}{12}=\binom{1}{-12}\).

\[\overrightarrow{PQ}\cdot\overrightarrow{RS}=(12)(1)+(1)(-12)=0\]

Since the scalar product is zero, \(\overrightarrow{PQ}\perp\overrightarrow{RS}\).