June 2025 Paper 1 Q4
4 The position vector of the point \(A\) is \(0.5\mathbf{i} - 0.5\mathbf{j}\).
The position vector of the point \(B\) is \(3.5\mathbf{i} + c\mathbf{j}\), where \(c\) is a constant.
| Scheme | Marks | AO |
|---|---|---|
| magnitude is \(\sqrt{0.5^2 + 0.5^2} = \sqrt{0.5}\) | M1 | 1.1 |
| \(\sqrt{0.5}\) (or 0.707) \(\neq 1\) OR \(0.5 \neq 1^2\) OR would need to be \(\frac{1}{2}\sqrt{2}\,\mathbf{i} - \frac{1}{2}\sqrt{2}\,\mathbf{j}\) hence not unit vector | A1 | 2.1 |
| [2] |
Notes
M1: Attempt to find magnitude (or magnitude \({}^2\)) of given vector. Pythagoras or trigonometry
A1: Make comparison using definition of a unit vector, and conclude that it is not a unit vector. Any values seen must be correct
Accept \(0.5 \neq 1\) as evidence
Geometric arguments can gain full credit, as long as fully reasoned and convincing eg to have a length of 1 unit the two components would have to be collinear, but this is not possible as they have to be perpendicular
| Scheme | Marks | AO |
|---|---|---|
| \((\tan^{-1} 1 =)\ 45^\circ\) | M1 | 1.1 |
| hence direction is \(315^\circ\) | A1 | 1.1 |
| [2] |
Notes
M1: Attempt to find a useful angle soi. Could be using their magnitude from (a) with sin or cos
Implied by \(-45^\circ\)
BOD for any incorrect diagrams seen
Allow 0.785 (radians)
A1: Obtain correct direction
A0 for \(-45^\circ\)
Condone both \(315^\circ\) and \(-45^\circ\) given
A0 if in radians
| Scheme | Marks | AO |
|---|---|---|
| \(3^2 + (c + 0.5)^2 = 5^2\) | M1 | 3.1a |
| \(c^2 + c - 15.75\ (= 0)\) | A1 | 1.1 |
| \(c = 3.5\), \(c = -4.5\) oe | A1 | 1.1 |
| [3] |
Notes
M1: Attempt distance equation involving \(c\), and equate to \(5^2\) oe. Allow one sign error eg \(3^2 + (c - 0.5)^2 = 5^2\)
If spotting that the \(y\) difference is 4 then must still link to \(c\) for M1
A1: Correct quadratic equation. Brackets expanded, but not necessarily simplified
A1: Obtain both correct values
SC if no method shown then B1 for each correct value of \(c\), as ‘determine’
Alt method for final two marks
| Scheme | Marks | AO |
|---|---|---|
| \((c + 0.5)^2 = 16\) \(c + 0.5 = 4\) \(c = 3.5\) oe | A1 | |
| \(c + 0.5 = -4\) \(c = -4.5\) oe | A1 |
A1: Solve to obtain one correct value for \(c\)
A1: Obtain second correct value, and no others