(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. ...
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(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.
Answer Details
(a) Equating components of \(m\begin{pmatrix}2\\1\end{pmatrix}+n\begin{pmatrix}-1\\2\end{pmatrix}=\begin{pmatrix}5\\-4\end{pmatrix}\):
\[2m-n=5,\qquad m+2n=-4.\]From the first, \(n=2m-5\). Substituting: \(m+2(2m-5)=-4\Rightarrow 5m=6\Rightarrow m=\tfrac{6}{5}\), and \(n=2(\tfrac65)-5=-\tfrac{13}{5}\).
\[m+n=\frac{6}{5}-\frac{13}{5}=-\frac{7}{5}=-1.4.\](b) \(A(-1,3),B(2,-1),C(5,3)\). \(\vec{AB}=\begin{pmatrix}3\\-4\end{pmatrix},\ |\vec{AB}|=5\); \(\vec{AC}=\begin{pmatrix}6\\0\end{pmatrix},\ |\vec{AC}|=6\).
(i) Unit vectors: \(\hat{AB}=\begin{pmatrix}3/5\\-4/5\end{pmatrix},\quad \hat{AC}=\begin{pmatrix}1\\0\end{pmatrix}.\)
(ii) \(\vec{AB}\cdot\vec{AC}=3(6)+(-4)(0)=18\), so
\[\cos(\widehat{BAC})=\frac{18}{5\times 6}=\frac{18}{30}=0.6\Rightarrow \widehat{BAC}=53^\circ.\]Download The App On Google Playstore
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