With respect to a fixed origin \(O\), the line \(l_1\) is given by the equation
$$\mathbf{r} = \mathbf{i} + 2\mathbf{j} + 5\mathbf{k} + \lambda(8\mathbf{i} - \mathbf{j} + 4\mathbf{k})$$
where \(\lambda\) is a scalar parameter.
The point \(A\) lies on \(l_1\)
Given that \(|\overrightarrow{OA}| = 5\sqrt{10}\)
- show that at \(A\) the parameter \(\lambda\) satisfies
$$81\lambda^2 + 52\lambda - 220 = 0$$
[3]
Hence
- show that one possible position vector for \(A\) is \(-15\mathbf{i} + 4\mathbf{j} - 3\mathbf{k}\)
- find the other possible position vector for \(A\).
[3]
The line \(l_2\) is parallel to \(l_1\) and passes through \(O\).
Given that
• \(\overrightarrow{OA} = -15\mathbf{i} + 4\mathbf{j} - 3\mathbf{k}\)
• point \(B\) lies on \(l_2\) where \(|\overrightarrow{OB}| = 4\sqrt{10}\)
- find the area of triangle \(OAB\), giving your answer to one decimal place.
[4]