(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 position vectors of points P, Q, R and S are (begin{pmatrix} -2 \ 3 end{pmatrix}, begin{pmatrix} 10 \ 4 end{pmatrix}, begin{pmatrix} 3 \ 12 end{pmatrix}) and (begin{pmatrix} 4 \ 0 end{pmatrix}) respectively. Show that (overrightarrow{PQ}) is perpendicular to (overrightarrow{RS}).
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)