10
- 3 \end{array} \right) + \lambda \left( \begin{array} { r } 2
- 5
4 \end{array} \right)
& L _ { 2 } : \mathbf { r } = \left( \begin{array} { r } - 1
2
3 \end{array} \right) + \mu \left( \begin{array} { l } 1
2
2 \end{array} \right) \end{aligned}$$ （a）Show that \(L _ { 1 }\) and \(L _ { 2 }\) are perpendicular．
（b）Show that \(L _ { 1 }\) and \(L _ { 2 }\) are skew lines． The point \(A\) with position vector \(- \mathbf { i } + 2 \mathbf { j } + 3 \mathbf { k }\) lies on \(L _ { 2 }\) and the point \(X\) lies on \(L _ { 1 }\) such that \(\overrightarrow { A X }\) is perpendicular to \(L _ { 1 }\)
（c）Find the position vector of \(X\) ．
（d）Find \(| \overrightarrow { A X } |\) The point \(B\)（distinct from \(A\) ）also lies on \(L _ { 2 }\) and \(| \overrightarrow { B X } | = | \overrightarrow { A X } |\)
（e）Find the position vector of \(B\) ．
（f）Find the cosine of angle \(A X B\) ． 7．（a）Use the substitution \(x = \sec \theta\) to show that $$\int _ { \sqrt { 2 } } ^ { 2 } \frac { 1 } { \left( x ^ { 2 } - 1 \right) ^ { \frac { 3 } { 2 } } } \mathrm {~d} x = \frac { \sqrt { 6 } - 2 } { \sqrt { 3 } }$$ （b）Use integration by parts to show that $$\int \operatorname { cosec } \theta \cot ^ { 2 } \theta \mathrm {~d} \theta = \frac { 1 } { 2 } [ \ln | \operatorname { cosec } \theta + \cot \theta | - \operatorname { cosec } \theta \cot \theta ] + c$$ （6） \begin{figure}[h] \end{figure} Figure 2 shows a sketch of part of the curve with equation \(y = \frac { 1 } { \left( x ^ { 2 } - 1 \right) ^ { \frac { 3 } { 2 } } }\) for \(x > 1\)
The region \(R\) ，shown shaded in Figure 2，is bounded by the curve，the \(x\)－axis and the lines \(x = \sqrt { 2 }\) and \(x = 2\)
The region \(R\) is rotated through \(2 \pi\) radians about the \(x\)－axis．
（c）Show that the volume of the solid formed is $$\pi \left[ \frac { 3 } { 8 } \ln \left( \frac { 1 + \sqrt { 2 } } { \sqrt { 3 } } \right) + \frac { 7 } { 36 } - \frac { \sqrt { 2 } } { 8 } \right]$$