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2015 WAEC Further Mathematics Theory (a) Given that \(\overrightarrow{AB} = \begin{pmatrix} 4 \\ 5 \end{pmatrix}\) and \(\overrightarrow{BC} = \begin{pmatrix} -3...

Further Mathematics
WAEC 2015

(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 \(\overrightarrow{PM}\).

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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) \(\overrightarrow{PQ} = \begin{pmatrix} 2 \\ 8 \end{pmatrix} ; \overrightarrow{PR} = \begin{pmatrix} 11 \\ -12 \end{pmatrix}\)

\(\overrightarrow{QR} = \overrightarrow{QP} + \overrightarrow{PR} \)

= \(-\overrightarrow{PQ} + \overrightarrow{PR}\)

= \(\begin{pmatrix} -2 \\ -8 \end{pmatrix} + \begin{pmatrix} 11 \\ -12 \end{pmatrix}\)

= \(\begin{pmatrix} 9 \\ -20 \end{pmatrix}\)

\(\frac{QM}{MR} = \frac{QP + PM}{MP + PR} = \frac{-PQ + PM}{-PM + PR}\)

\(\frac{QM}{MR} = \frac{3}{7}\)

\(7QM = 3MR\)

\(7(-PQ + PM) = 3(-PM + PR)\)

\(7(-PQ) + 7PM = 3(-PM) + 3(PR)\)

\(7(\begin{pmatrix} -2 \\ -8 \end{pmatrix} - 3(\begin{pmatrix} 11 \\ -12 \end{pmatrix} = -10(PM)\)

\(\begin{pmatrix} -47 \\ -20 \end{pmatrix} = -10(PM)\)

\(PM = \begin{pmatrix} 4.7 \\ 2 \end{pmatrix}\)


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