These Ten-Year-Series (TYS) worked solutions with video explanations for 2011 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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Referred to the origin $O$, the points $A$ and $B$ are such that $\overrightarrow{OA}=\mathbf{a}$ and $\overrightarrow{OB}=\mathbf{b}$. The point $P$ on $OA$ is such that $OP:PA=1:2$, and the point $Q$ on $OB$ is such that $OQ:QB=3:2$. The mid-point of $PQ$ is $M$ (see diagram).
(i)
Find $\overrightarrow{OM}$ in terms of $\mathbf{a}$ and $\mathbf{b}$ and show that the area of triangle $OPM$ can be written as $k\left| \mathbf{a}\times \mathbf{b} \right|$, where $k$ is a constant to be found.
[6]
(i) Find $\overrightarrow{OM}$ in terms of $\mathbf{a}$ and $\mathbf{b}$ and show that the area of triangle $OPM$ can be written as $k\left| \mathbf{a}\times \mathbf{b} \right|$, where $k$ is a constant to be found.
[6]
(ii)
The vectors $\mathbf{a}$ and $\mathbf{b}$ are now given by
$\mathbf{a}=2p\mathbf{i}-6p\mathbf{j}+3p\mathbf{k}$ and $\mathbf{b}=\mathbf{i}+\mathbf{j}-2\mathbf{k}$,
where $p$ is a positive constant. Given that $\mathbf{a}$ is a unit vector,
(ii) The vectors $\mathbf{a}$ and $\mathbf{b}$ are now given by
$\mathbf{a}=2p\mathbf{i}-6p\mathbf{j}+3p\mathbf{k}$ and $\mathbf{b}=\mathbf{i}+\mathbf{j}-2\mathbf{k}$,
where $p$ is a positive constant. Given that $\mathbf{a}$ is a unit vector,
(a) find the exact value of $p$,
[2]
(a) find the exact value of $p$,
[2]
(b) give a geometrical interpretation of $\left| \mathbf{a}\cdot \mathbf{b} \right|$,
[1]
(b) give a geometrical interpretation of $\left| \mathbf{a}\cdot \mathbf{b} \right|$,
[1]
(c) evaluate $\mathbf{a}\times \mathbf{b}$.
[2]
(c) evaluate $\mathbf{a}\times \mathbf{b}$.
[2]
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