9 The position vector of point \(A\) relative to the origin \(O\) is \(\overrightarrow { O A } = 8 \mathbf { i } - 5 \mathbf { j } + 6 \mathbf { k }\).
The line \(l\) passes through \(A\) and is parallel to the vector \(2 \mathbf { i } + \mathbf { j } + 4 \mathbf { k }\).
- State a vector equation for \(l\).
- The position vector of point \(B\) relative to the origin \(O\) is \(\overrightarrow { O B } = - t \mathbf { i } + 4 t \mathbf { j } + 3 t \mathbf { k }\), where \(t\) is a constant. The line \(l\) also passes through \(B\).
Find the value of \(t\).
- The line \(m\) has vector equation \(\mathbf { r } = 5 \mathbf { i } - \mathbf { j } + 2 \mathbf { k } + \mu ( a \mathbf { i } - \mathbf { j } + 3 \mathbf { k } )\). The acute angle between the directions of \(l\) and \(m\) is \(\theta\), where \(\cos \theta = \frac { 1 } { \sqrt { 6 } }\).
Find the possible values of \(a\).
\includegraphics[max width=\textwidth, alt={}, center]{656df2a8-fc4d-49f3-a649-746103b4576e-18_542_559_251_753}
A large cylindrical tank is used to store water. The base of the tank is a circle of radius 4 metres. At time \(t\) minutes, the depth of the water in the tank is \(h\) metres. There is a tap at the bottom of the tank. When the tap is open, water flows out of the tank at a rate proportional to the square root of the volume of water in the tank. - Show that \(\frac { \mathrm { d } h } { \mathrm {~d} t } = - \lambda \sqrt { h }\), where \(\lambda\) is a positive constant.
\includegraphics[max width=\textwidth, alt={}, center]{656df2a8-fc4d-49f3-a649-746103b4576e-18_2718_42_107_2007} - At time \(t = 0\) the tap is opened. It is given that \(h = 4\) when \(t = 0\) and that \(h = 2.25\) when \(t = 20\).
Solve the differential equation to obtain an expression for \(t\) in terms of \(h\), and hence find the time taken to empty the tank.
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