## Question 8:

### Part (a):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{\begin{pmatrix}2\\a\\0\end{pmatrix}\cdot\begin{pmatrix}0\\1\\-1\end{pmatrix}}{\sqrt{2^2+a^2}\sqrt{1^2+(-1)^2}} = \cos 120°$ | M1 | 3.1b – Complete method using scalar product of direction vectors with angle 120° to form equation in $a$ |
| $\frac{a}{\sqrt{4+a^2}\sqrt{2}} = -\frac{1}{2}$ | A1 | 1.1b – Correct simplified equation in $a$ |
| $2a = -\sqrt{4+a^2}\sqrt{2} \Rightarrow 4a^2 = 8+2a^2 \Rightarrow a^2 = 4 \Rightarrow a = \ldots$ | M1 | 1.1b – Solve quadratic by squaring, equation of form $a^2 = K$, $K>0$ |
| $a = -2$ | A1 | 2.2a – Deduces correct value from correct equation |

**Notes:** M1: Allow sign slip and cos 60. A1: cos 120 must be evaluated to $-\frac{1}{2}$. If candidate states $\left|\frac{a\cdot b}{|a||b|}\right| = \cos\theta$ then $\frac{a}{\sqrt{4+a^2}\sqrt{2}} = \frac{1}{2}$ award A1. If module of dot product not seen award A0. dM1: Solve quadratic by squaring. A1: Must use angle between the lines. **Alternative cross product method:** M1: $\begin{vmatrix}2 & a & 0\\0 & 1 & -1\end{vmatrix} = \sqrt{2^2+a^2}\sqrt{1^2+(-1)^2}\sin 120°$; A1: $\sqrt{a^2+8} = \sqrt{4+a^2}\sqrt{2}\cdot\frac{\sqrt{3}}{2}$. **Note: If they use the point of intersection to find a value for $a$ this scores no marks.**

---

### Part (b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Any two of: **i:** $-1+2\lambda=4$ (1); **j:** $5+\text{'their}-2'\lambda = -1+\mu$ (2); **k:** $2=3-\mu$ (3) | M1 | 3.4 – Uses model to write down any two correct equations |
| Solves to find $\lambda\left\{=\frac{5}{2}\right\}$ and $\mu\{=1\}$ | M1 | 1.1b – Solve two equations simultaneously |
| $r_1 = \begin{pmatrix}-1\\5\\2\end{pmatrix}+\frac{5}{2}\begin{pmatrix}2\\\text{'their}-2'\\0\end{pmatrix}$ or $r_2=\begin{pmatrix}4\\-1\\3\end{pmatrix}+1\begin{pmatrix}0\\1\\-1\end{pmatrix}$ | dM1 | 1.1b – Substitutes $\mu$ and $\lambda$ into relevant equation (dependent on previous M) |
| $(4,0,2)$ or $\begin{pmatrix}4\\0\\2\end{pmatrix}$ | A1 | 1.1b – Correct coordinates |
| Checks third equation e.g. $\lambda=\frac{5}{2}$: **L** $\text{HS}=5-2\lambda=5-5=0$; $\mu=1$: **R** $\text{HS}=-1+\mu=-1+1=0$, therefore **common point/intersect/consistent** | B1 | 2.1 – Shows values give same third coordinate and draws conclusion |

**Notes:** dM1: Dependent on previous M. If no method shown, two correct ordinates implies mark. B1: Note if incorrect $a$ found in (a) but in (b) they find $a=-2$ this scores B0 but all other marks available.

---

### Part (c):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Minimum distance $= \frac{|2\times'4'+(-3)\times'0'+1\times'2'-2|}{\sqrt{2^2+(-3)^2+1^2}}$ | M1 | 3.1b – Full attempt to find minimum distance from point of intersection to the plane |
| $\lambda = \ldots\left\{-\frac{4}{7}\right\}$ (via $\mathbf{r}=\begin{pmatrix}'4'\\{}'0'\\{}'2'\end{pmatrix}+\lambda\begin{pmatrix}2\\-3\\1\end{pmatrix}$) | A1ft | 3.4 – Following through on their point of intersection |
| $\frac{8}{\sqrt{14}}$ or $\frac{4\sqrt{14}}{7}$ or awrt 2.1 | A1 | 2.2b – Correct distance |

**Alternative:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| Find perpendicular distance from plane $2x-3y+z=2$ to origin: $|n|=\sqrt{2^2+(-3)^2+1^2}=\sqrt{14}$, shortest distance $=\frac{2}{\sqrt{14}}$. Find perpendicular distance from plane through point of intersection to origin: $2x-3y+z=\begin{pmatrix}4\\0\\2\end{pmatrix}\cdot\begin{pmatrix}2\\-3\\1\end{pmatrix}=10$, shortest distance $=\frac{10}{\sqrt{14}}$ | M1 | 3.1b |
| Minimum distance $= \frac{10}{\sqrt{14}}-\frac{2}{\sqrt{14}}$ | A1ft | 3.4 |
| $\frac{8}{\sqrt{14}}$ or $\frac{4\sqrt{14}}{7}$ or awrt 2.1 | A1 | 2.2b |

---

### Part (d):

| Answer/Working | Mark | Guidance |
|---|---|---|
| e.g. Not reliable as birds will not fly in a straight line / angle between flight paths will not always be 120° / ground will not be flat/smooth / bird's nest is not a point | B1 | 3.2b – Comments on one of the models, then states unreliable |

---