M5 June 2015 Q1
1. A particle \(P\) moves from the point \(A\), with position vector \((2\mathbf{i} + 4\mathbf{j} + a\mathbf{k})\) m, where \(a\) is a positive constant, to the point \(B\), with position vector \((-\mathbf{i} + a\mathbf{j} - \mathbf{k})\) m, under the action of a constant force \(\mathbf{F} = (2\mathbf{i} + a\mathbf{j} - 3\mathbf{k})\) N. The work done by \(\mathbf{F}\), as it moves the particle \(P\) from \(A\) to \(B\), is 3 J. Find the value of \(a\). (6)
| Scheme | Marks |
|---|---|
| \(\mathbf{AB} = (-3\mathbf{i} + (a - 4)\mathbf{j} + (-1 - a)\mathbf{k})\) | B1 |
| Work done \(= 3 = (2\mathbf{i} + a\mathbf{j} - 3\mathbf{k}) \cdot (-3\mathbf{i} + (a - 4)\mathbf{j} + (-1 - a)\mathbf{k})\) | M1 |
| \(3 = -6 + a(a - 4) - 3(-1 - a)\) | A1 |
| \(0 = a^2 - a - 6\) | A1 |
| \(0 = (a - 3)(a + 2)\) | M1 |
| \(a = 3\) since \(a \gt 0\). | A1 |
| (6 marks) |
Notes
B1 for correct \(\mathbf{AB}\) in any form.
First M1 for \(3 = (2\mathbf{i} + a\mathbf{j} - 3\mathbf{k}) \cdot\) their \(\mathbf{AB}\) (allow \(\mathbf{BA}\)) Need an attempt.
First A1 for any correct equation
Second A1 for \(0 = a^2 - a - 6\)
Second M1 for solving a quadratic (2 solutions)
Third A1 for \(a = 3\)