These Ten-Year-Series (TYS) worked solutions with video explanations for 1994 A Level H2 Mathematics are suggested by Mr Gan. For any comments or suggestions please contact us at support@timganmath.edu.sg.
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In the diagram, $O$ is centre of the square base $ABCD$ of a right pyramid, vertex $V$. Perpendicular unit vectors $\mathbf{i}$, $\mathbf{j}$, $\mathbf{k}$ are parallel to $AB$, $AD$, $OV$ respectively. The length of $AB$ is 4 units and the length of $OV$ is $2h$ units. $P$, $Q$, $M$ and $N$ are mid-points of $AB$, $BC$, $CV$ and $VA$ respectively. The point $O$ is taken as the origin for position vectors.
(i)
Show that the equation of the line $PM$ may be expressed as $\mathbf{r}=\left( \begin{matrix} 0 \\ -2 \\ 0 \\
\end{matrix} \right)+t\left( \begin{matrix} 1 \\ 3 \\ h \\
\end{matrix} \right)$, where $t$ is a parameter.
[2]
(i) Show that the equation of the line $PM$ may be expressed as $\mathbf{r}=\left( \begin{matrix} 0 \\ -2 \\ 0 \\
\end{matrix} \right)+t\left( \begin{matrix} 1 \\ 3 \\ h \\
\end{matrix} \right)$, where $t$ is a parameter.
[2]
(ii)
Find the equation for the line $QN$.
[2]
(ii) Find the equation for the line $QN$.
[2]
(iii)
Show that the lines $PM$ and $QN$ intersect, and that the position vector $\overrightarrow{OX}$ of their point of intersection is $\left( \begin{matrix}
\frac{1}{2} \\
-\frac{1}{2} \\
\frac{1}{2}h \\
\end{matrix} \right)$.
[3]
(iii) Show that the lines $PM$ and $QN$ intersect, and that the position vector $\overrightarrow{OX}$ of their point of intersection is $\left( \begin{matrix}
\frac{1}{2} \\
-\frac{1}{2} \\
\frac{1}{2}h \\
\end{matrix} \right)$.
[3]
(iv)
Given that $OX$ is perpendicular to $VB$, find the value of $h$ and calculate the acute angle between $PM$ and $QN$, giving your answer correct to the nearest $0.1{}^\circ $.
[4]
(iv) Given that $OX$ is perpendicular to $VB$, find the value of $h$ and calculate the acute angle between $PM$ and $QN$, giving your answer correct to the nearest $0.1{}^\circ $.
[4]
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