H2 JC Maths Tuition Foot of Perpendicular 2007 Paper 1 Q8

One of my students asked me how to solve 2007 Paper 1 Q8 (iii) using Foot of Perpendicular method.

The answer given in the TYS uses a sine method, which is actually shorter in this case, since we have found the angle in part (ii).

Nevertheless, here is how we solve the question using Foot of Perpendicular method.

(Due to copyright issues, I cannot post the whole question here, so please refer to your Ten Year Series.)

Firstly, let F be the foot of the perpendicular.

Then, \vec{AF}=k\begin{pmatrix}3\\-1\\2\end{pmatrix} ——– Eqn (1)

\vec{OF}\cdot\begin{pmatrix}3\\-1\\2\end{pmatrix}=17 ——– Eqn (2)

From Eqn (1), \vec{OF}-\vec{OA}=\begin{pmatrix}3k\\-k\\2k\end{pmatrix}

\begin{array}{rcl}\vec{OF}&=&\vec{OA}+\begin{pmatrix}3k\\-k\\2k\end{pmatrix}\\  &=&\begin{pmatrix}1\\2\\4\end{pmatrix}+\begin{pmatrix}3k\\-k\\2k\end{pmatrix}\\  &=&\begin{pmatrix}1+3k\\2-k\\4+2k\end{pmatrix}\end{array}

Substituting into Eqn (2),

\begin{pmatrix}1+3k\\2-k\\4+2k\end{pmatrix}\cdot\begin{pmatrix}3\\-1\\2\end{pmatrix}=17

14k+9=17

k=4/7

Substituting back into Eqn (1),

\displaystyle\vec{AF}=\frac{4}{7}\begin{pmatrix}3\\-1\\2\end{pmatrix}

\displaystyle|\vec{AF}|=\frac{4}{7}\sqrt{14}

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