2013 A-level H2 Mathematics (9740) Paper 2 Question 4 Suggested Solutions

All solutions here are SUGGESTED. Mr. Teng will hold no liability for any errors. Comments are entirely personal opinions.

(i)
cos{\theta}=\frac{|\begin{pmatrix}2\\-2\\1\end{pmatrix}{\bullet}\begin{pmatrix}-6\\3\\2\end{pmatrix}|}{\sqrt{9} \sqrt{49}}=\frac{16}{21}

\theta = {40.4}^{\circ}

(ii) Using Graphing Calculator, equation of line,

r = \begin{pmatrix}{-\frac{1}{6}}\\{-\frac{2}{3}}\\0\end{pmatrix} + \lambda \begin{pmatrix}7\\10\\6\end{pmatrix}, \lambda \in \mathbb{R}

(iii)
Distance A to P_1 = \frac{|\begin{pmatrix}2\\-2\\1\end{pmatrix}{\bullet}\begin{pmatrix}4\\3\\c\end{pmatrix} - 1|}{\sqrt{9}} = \frac{|1+c|}{3}

Distance A to P_1 = \frac{|\begin{pmatrix}-6\\3\\2\end{pmatrix}{\bullet}\begin{pmatrix}4\\3\\c\end{pmatrix} + 1|}{\sqrt{49}} = \frac{|2c-14|}{7}

\frac{|1+c|}{3} = \frac{|2c-14|}{7}
49(1+c)^{2} = 9(2c-14)^{2}
c = -49 \mathrm{~or~} \frac{35}{13}

KS Comments:

Some students have slight confusion about how to evaluate the modulus, mainly cos they confuse over are the applying modulus to a scalar or vector quantity.

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