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June 2008 p3 q10
2167
The points A and B have position vectors, relative to the origin O, given by \(\overrightarrow{OA} = \mathbf{i} + 2\mathbf{j} + 3\mathbf{k}\) and \(\overrightarrow{OB} = 2\mathbf{i} + \mathbf{j} + 3\mathbf{k}\).
The line \(l\) has vector equation \(\mathbf{r} = (1 - 2t)\mathbf{i} + (5 + t)\mathbf{j} + (2 - t)\mathbf{k}\).
(i) Show that \(l\) does not intersect the line passing through A and B.
(ii) The point P lies on \(l\) and is such that angle \(PAB\) is equal to 60°. Given that the position vector of P is \((1 - 2t)\mathbf{i} + (5 + t)\mathbf{j} + (2 - t)\mathbf{k}\), show that \(3t^2 + 7t + 2 = 0\). Hence find the only possible position vector of P.
Solution
(i) The vector equation for the line through A and B is \(\mathbf{r} = \mathbf{i} + 2\mathbf{j} + 3\mathbf{k} + s(\mathbf{i} - \mathbf{j})\). Equating components with the line \(l\), we have:
\(1 + s = 1 - 2t\)
\(2 - s = 5 + t\)
\(3 = 2 - t\)
Solving these gives \(s = -6, t = 3\) for the first equation, \(s = 2, t = -1\) for the second, and \(t = -1\) for the third. The values do not satisfy all equations simultaneously, so \(l\) does not intersect the line through A and B.
(ii) The direction vector for \(AP\) is \((-2t, 3 + t, -1 - t)\). The cosine of angle \(PAB\) is given by:
Expanding the scalar product and expressing the product of the moduli in terms of \(t\), we obtain the equation \(3t^2 + 7t + 2 = 0\).
Solving the quadratic equation, we find roots \(t = -2\) and \(t = -\frac{1}{3}\). The position vector of P for \(t = -2\) is \(5\mathbf{i} + 3\mathbf{j} + 4\mathbf{k}\), which is valid as it satisfies the angle condition.
