# Question 2:

## Part (a):
| Working | Mark | Guidance |
|---------|------|----------|
| $|w| = \sqrt{(4\sqrt{3})^2 + (-4)^2} = 8$ | B1 | Correct modulus |
| $\arg w = \arctan\!\left(\dfrac{\pm 4}{\pm 4\sqrt{3}}\right) = \arctan\!\left(\pm\dfrac{1}{\sqrt{3}}\right)$ | M1 | Allow wrong quadrant for this mark |
| $= -\dfrac{\pi}{6}$ | A1 | Must be in fourth quadrant; accept $\dfrac{11\pi}{6}$ or other difference of $2\pi$ |
| $w = 8\!\left(\cos\!\left(-\dfrac{\pi}{6}\right) + \mathrm{i}\sin\!\left(-\dfrac{\pi}{6}\right)\right)$ | A1 | Must have positive $r=8$ and $\theta = -\dfrac{\pi}{6}$ |

**Note:** Using degrees scores B1 M1 A0 A0

## Part (b)(i) & (ii):
| Working | Mark | Guidance |
|---------|------|----------|
| $w$ plotted in 4th quadrant with $(4\sqrt{3}, -4)$ seen or $-\dfrac{\pi}{4} < \arg w < 0$ | B1 | Correct quadrant with coordinate or argument range |
| Half line with positive gradient emanating from imaginary axis | M1 | Need not be labelled |
| Half line passes between $O$ and $w$, starting from point on imaginary axis below $w$ | A1 | Assumed to start at $-10\mathrm{i}$ unless otherwise stated |

**Note:** Loci on separate diagrams: maximum B1 M1 A0

## Part (c):
| Working | Mark | Guidance |
|---------|------|----------|
| $\triangle OAX$ right-angled at $X$; $OX = 10\sin\dfrac{\pi}{6} = 5$ | M1 | Formulates correct strategy using right angle |
| $WX = OW - OX = 8 - 5 = \ldots$ | M1 | Full method to achieve shortest distance |
| Minimum distance $= 3$ | A1 | cao |

**Alternative 1:**
| Working | Mark | Guidance |
|---------|------|----------|
| Line from $O$ to $w$: $y = -\dfrac{1}{\sqrt{3}}x$; half line: $y = \sqrt{3}x - 10$; intersection $X\!\left(\dfrac{5\sqrt{3}}{2}, -\dfrac{5}{2}\right)$ | M1 | Correct method for both equations and intersection |
| $WX = \sqrt{\!\left(\dfrac{5\sqrt{3}}{2} - 4\sqrt{3}\right)^2 + \!\left(-\dfrac{5}{2} - (-4)\right)^2}$ | M1 | Condone sign slip in brackets |
| Minimum distance $= 3$ | A1 | cao |

**Alternative 2:**
| Working | Mark | Guidance |
|---------|------|----------|
| $AW = \sqrt{(4\sqrt{3})^2+(-4-(-10))^2} = \sqrt{84}$; angle between horizontal and $AW = \tan^{-1}\!\left(\dfrac{-4-(-10)}{4\sqrt{3}}\right) \approx 0.7137$ rad | M1 | Correct method for $AW$ and angle |
| $WX = \sqrt{84}\times\sin\!\left(\dfrac{\pi}{3} - 0.7137\right)$ | M1 | |
| Minimum distance $= 3$ | A1 | cao |

**Alternative 3:**
| Working | Mark | Guidance |
|---------|------|----------|
| Vector equation of half line $r = \begin{pmatrix}0\\-10\end{pmatrix} + \lambda\begin{pmatrix}1\\\sqrt{3}\end{pmatrix}$; sets $XW \cdot \begin{pmatrix}1\\\sqrt{3}\end{pmatrix} = 0$, solves $\lambda = \dfrac{5}{2}\sqrt{3}$ | M1 | Or minimises $|XW|^2$ by differentiation |
| $WX = \sqrt{\!\left(4\sqrt{3} - \dfrac{5\sqrt{3}}{2}\right)^2 + \!\left(-4 - \left(-\dfrac{5}{2}\right)\right)^2}$ | M1 | |
| Minimum distance $= 3$ | A1 | cao |

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