## Question 10:

**Equation of plane:**

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k}\\ 1 & 1 & 1\\ 4 & 6 & 1\end{vmatrix} = -5\mathbf{i} + 3\mathbf{j} + 2\mathbf{k}$ | M1A1 | Uses vector product to find normal to plane |
| Equation of plane: $5x - 3y - 2z = \text{constant}$ | M1 | Uses $\mathbf{r}\cdot\mathbf{n} = \text{constant}$ |
| $30 - 15 - 8 = 7$; so $5x - 3y - 2z = 7$ | A1 | Obtains Cartesian equation of plane. **Part total: 4** |

**Alternatively:**

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\begin{pmatrix}x\\y\\z\end{pmatrix} = \begin{pmatrix}6\\5\\4\end{pmatrix} + \lambda\begin{pmatrix}1\\1\\1\end{pmatrix} + \mu\begin{pmatrix}4\\6\\1\end{pmatrix}$ | (M1) | |
| $x=6+\lambda+4\mu,\; y=5+\lambda+6\mu,\; z=4+\lambda+\mu$ | (A1) | |
| Eliminates $\lambda$ and $\mu$ | (M1) | |
| Obtains $5x-3y-2z=7$ | (A1) | |

**Foot of perpendicular:**

| Answer/Working | Marks | Guidance |
|---|---|---|
| Equation of perpendicular: $\mathbf{r} = \mathbf{i} + 10\mathbf{j} + 3\mathbf{k} + t(5\mathbf{i} - 3\mathbf{j} - 2\mathbf{k})$ | M1 | Finds equation of perpendicular to plane through given point |
| $5(1+5t) - 3(10-3t) - 2(3-2t) = 7$ | M1 | Finds value of parameter at point in plane |
| $\Rightarrow t = 1$ | A1 | |
| Foot of perpendicular is $6\mathbf{i} + 7\mathbf{j} + \mathbf{k}$ | A1 | Obtains foot of perpendicular. **Part total: 4** |

**Alternatively (foot of perpendicular):**

| Answer/Working | Marks | Guidance |
|---|---|---|
| Let foot be $a\mathbf{i}+b\mathbf{j}+c\mathbf{k}$; use orthogonality conditions: $\begin{pmatrix}a-1\\b-10\\c-3\end{pmatrix}\cdot\begin{pmatrix}1\\1\\1\end{pmatrix}=0 \Rightarrow a+b+c=14$ | | Form sufficient equations using orthogonality |
| $\begin{pmatrix}a-1\\b-10\\c-3\end{pmatrix}\cdot\begin{pmatrix}4\\6\\1\end{pmatrix}=0 \Rightarrow 4a+6b+c=67$ | | Two conditions suffice if foot expressed using parametric equation |
| $5a-3b-2c=7$ | (M1A1) | $a\mathbf{i}+b\mathbf{j}+c\mathbf{k}$ lies in plane of $l_1$ and $l_2$ |
| $\Rightarrow 6\mathbf{i}+7\mathbf{j}+\mathbf{k}$ | (M1A1) | |

**Shortest distance between lines:**

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k}\\ 1 & 1 & 1\\ 2 & -3 & 1\end{vmatrix} = 4\mathbf{i}+\mathbf{j}-5\mathbf{k}$ | M1A1 | Finds direction of common perpendicular |
| $\begin{pmatrix}6\\5\\4\end{pmatrix} - \begin{pmatrix}1\\10\\3\end{pmatrix} = \begin{pmatrix}5\\-5\\1\end{pmatrix}$ | | Forms vector between known points on $l_1$ and $l_3$ |
| $\frac{1}{\sqrt{16+1+25}}\left|\begin{pmatrix}5\\-5\\1\end{pmatrix}\cdot\begin{pmatrix}4\\1\\-5\end{pmatrix}\right| = \frac{10}{\sqrt{42}} \approx 1.54$ | M1A1 A1 | Finds shortest distance by projection. **Part total: 5** |
| | | **Total: [13]** |

**Alternative for last part:**

| Answer/Working | Marks | Guidance |
|---|---|---|
| Let P be on $l_1$, Q be on $l_3$: $\mathbf{p}=\begin{pmatrix}6+\lambda\\5+\lambda\\4+\lambda\end{pmatrix}$, $\mathbf{q}=\begin{pmatrix}1+2v\\10-3v\\3+v\end{pmatrix}$ | | |
| $\Rightarrow \overrightarrow{PQ}=\begin{pmatrix}-5-\lambda+2v\\5-\lambda-3v\\-1-\lambda-3v\end{pmatrix}$ | (M1) | |
| $\overrightarrow{PQ}\cdot\begin{pmatrix}1\\1\\1\end{pmatrix}=0 \Rightarrow -1-3\lambda=0 \Rightarrow \lambda=-\frac{1}{3}$ | (M1) | Uses orthogonality conditions |
| $\overrightarrow{PQ}\cdot\begin{pmatrix}2\\-3\\1\end{pmatrix}=0 \Rightarrow -26+14v=0 \Rightarrow v=\frac{13}{7}$ | (A1) | |
| $\Rightarrow \overrightarrow{PQ}=\frac{1}{21}\begin{pmatrix}-20\\-5\\25\end{pmatrix}$ | (A1) | |
| $\Rightarrow |\overrightarrow{PQ}|=\frac{5}{21}\sqrt{4^2+1^2+5^2}=\frac{5}{21}\sqrt{42}$ | (A1) | **Part total: (5)** |