(a) Given that (overrightarrow{AB} = begin{pmatrix} 4 \ 5 end{pmatrix}) and (overrightarrow{BC} = begin{pmatrix} -3 \ 5 end{pmatrix}); find the :
(i) angle between the vectors AB and AC ; (ii) unit vector along (overrightarrow{AB} – overrightarrow{BC}).
(b) P, Q, R and M are points in the (O_{XY}) plane. If (overrightarrow{PQ} = 2i + 8j , overrightarrow{PR} = 11i – 12j) and M divides QR internally in the ratio 3 : 7, find (overrightarrow{PM}).
Explanation
(a) (overrightarrow{AB} = begin{pmatrix} 4 \ 5 end{pmatrix} = 4i + 5j)
(overrightarrow{AC} = overrightarrow{AB} + overrightarrow{BC})
= (begin{pmatrix} 4 \ 5 end{pmatrix} + begin{pmatrix} -3 \ 5 end{pmatrix})
= (begin{pmatrix} 1 \ 10 end{pmatrix})
(i) Let (alpha) be the angle between (overrightarrow{AB}) and (overrightarrow{AC}).
(overrightarrow{AB} cdot overrightarrow{AC} = |AB||AC| cos alpha)
((4i + 5j) cdot (i + 10j) = (sqrt{4^{2} + 5^{2}})(sqrt{1^{2} + 10^{2}}) cos alpha)
(4 + 50 = (sqrt{41})(sqrt{101}) cos alpha)
(cos alpha = frac{54}{(sqrt{41})(sqrt{101})})
(cos alpha = 0.8392)
(alpha = 32.94°)
(ii) (overrightarrow{AB} – overrightarrow{BC} = begin{pmatrix} 4 \ 5 end{pmatrix} – begin{pmatrix} -3 \ 5 end{pmatrix})
=(begin{pmatrix} 7 \ 0 end{pmatrix} = 7i)
(overrightarrow{AB} – overrightarrow{BC} = 7i)
Unit vector in the direction of 7i = (frac{7i}{7} = i)
(b) (overrightarrow{PQ} = begin{pmatrix} 2 \ 8 end{pmatrix} ; overrightarrow{PR} = begin{pmatrix} 11 \ -12 end{pmatrix})
(overrightarrow{QR} = overrightarrow{QP} + overrightarrow{PR} )
= (-overrightarrow{PQ} + overrightarrow{PR})
= (begin{pmatrix} -2 \ -8 end{pmatrix} + begin{pmatrix} 11 \ -12 end{pmatrix})
= (begin{pmatrix} 9 \ -20 end{pmatrix})
(frac{QM}{MR} = frac{QP + PM}{MP + PR} = frac{-PQ + PM}{-PM + PR})
(frac{QM}{MR} = frac{3}{7})
(7QM = 3MR)
(7(-PQ + PM) = 3(-PM + PR))
(7(-PQ) + 7PM = 3(-PM) + 3(PR))
(7(begin{pmatrix} -2 \ -8 end{pmatrix} – 3(begin{pmatrix} 11 \ -12 end{pmatrix} = -10(PM))
(begin{pmatrix} -47 \ -20 end{pmatrix} = -10(PM))
(PM = begin{pmatrix} 4.7 \ 2 end{pmatrix})