2012 A-level H2 Mathematics (9740) Paper 1 Question 9 Suggested Solutions

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

\vec{AB} = \begin{pmatrix}-1\\-8\\1\end{pmatrix} - \begin{pmatrix}7\\8\\9\end{pmatrix} = \begin{pmatrix}--8\\-16\\-8\end{pmatrix}

l_{AB}: r = \begin{pmatrix}7\\8\\9\end{pmatrix} + \lambda \begin{pmatrix}1\\2\\1\end{pmatrix}, \lambda \in \mathbb{R}

\vec{ON} = \begin{pmatrix}{7+\lambda}\\{8+2\lambda}\\{9+\lambda}\end{pmatrix} for some \lambda.

\vec{CN} = \begin{pmatrix}{7+\lambda}\\{8+2\lambda}\\{9+\lambda}\end{pmatrix} - \begin{pmatrix}1\\8\\3\end{pmatrix} = \begin{pmatrix}{6+\lambda}\\{2\lambda}\\{6+\lambda}\end{pmatrix}

Since \vec{CN} is perpendicular to l_{AB}

\Rightarrow \begin{pmatrix}{6+\lambda}\\{2\lambda}\\{6+\lambda}\end{pmatrix} \bullet \begin{pmatrix}1\\2\\1\end{pmatrix} = 0

\lambda = -2

\therefore, \vec{ON} = \begin{pmatrix}5\\4\\7\end{pmatrix}

\vec{AN} = \begin{pmatrix}-2\\-4\\-2\end{pmatrix} = \frac{1}{4} \vec{AB}

\therefore, AN : AB = 1: 3

Let C’ be the point of reflection of C in line AB.

\vec{AC'} = \vec{AC} + 2\vec{CN} = \begin{pmatrix}2\\-8\\2\end{pmatrix}

l_{AB}^R: x - 7 = \frac{y-8}{-4} = z-9

KS Comments:

A few steps above are being skipped as it is really inconvenient to type the vectors in LaTeX. Please let me know if it caused problems for you.

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