(a) (m begin{pmatrix} 2 \ 1 end{pmatrix} + n begin{pmatrix} -1 \ 2 end{pmatrix} = begin{pmatrix} 5 \ -4 end{pmatrix}) where m and n are scalars. Find the value of (m + n).
(b) A(-1, 3), B(2, -1) and C(5, 3) are the vertices of (Delta) ABC.
(i) Express in column notation, the unit vectors parallel to AB and AC.
(ii) Use a dot product to calculate (stackrel frown{BAC}), correct to the nearest degree.
Explanation
(a) (m begin{pmatrix} 2 \ 1 end{pmatrix} + n begin{pmatrix} -1 \ 2 end{pmatrix} = begin{pmatrix} 5 \ -4 end{pmatrix})
(2m – n = 5 …(1))
(m + 2n = -4 … (2))
((1) times 2 : 4m – 2n = 10 … (3))
((2) + (3) : 5m = 6 implies m = frac{6}{5})
(2(frac{6}{5}) – n = 5 implies n = frac{12}{5} – 5 = frac{-13}{5})
(therefore (m + n) = (frac{6}{5} + frac{-13}{5}) = -frac{7}{5}).
(b) A(-1, 3) = (-i + 3j)
B(2, -1) = (2i – j)
(overrightarrow{AB} = overrightarrow{OB} – overrightarrow{OA})
(overrightarrow{AB} = begin{pmatrix} 2 \ -1 end{pmatrix} – begin{pmatrix} -1 \ 3 end{pmatrix})
= (overrightarrow{AB} = begin{pmatrix} 3 \ -4 end{pmatrix})
(|overrightarrow{AB}| = sqrt{3^{2} + (-4)^{2}} = 5)
Unit vector parallel to AB = (frac{1}{5} begin{pmatrix} 3 \ -4 end{pmatrix})
(overrightarrow{OA} + overrightarrow{AC} = overrightarrow{OC})
(overrightarrow{AC} = overrightarrow{OC} – overrightarrow{OA})
= (begin{pmatrix} 5 \3 end{pmatrix} – begin{pmatrix} -1 \ 3 end{pmatrix})
= (begin{pmatrix} 6 \ 0 end{pmatrix})
(|overrightarrow{AC}| = 6)
Unit vector parallel to AC = (frac{1}{6} begin{pmatrix} 6 \ 0 end{pmatrix})
(ii) Let the angle BAC = (theta)
(overrightarrow{AB} cdot overrightarrow{AC} = |overrightarrow{AB}||overrightarrow{AC}| cos theta)
((3i – 4j) cdot (6i + 0j) = |3i + 4j||6i + 0j| cos theta)
(18 = 5 times 6 times cos theta implies cos theta = frac{18}{30} = 0.6)
(theta = cos^{-1} (0.6) = 53.1°)