# Question 5:

## Part (a):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\overrightarrow{AB} = \begin{pmatrix}11\\-9\\0\end{pmatrix} - \begin{pmatrix}8\\-7\\-2\end{pmatrix} = \begin{pmatrix}3\\-2\\2\end{pmatrix}$ | M1 | Subtracting the position vectors of $A$ and $B$ (in either order). Calculation or answer is enough for M1. Allow M1 for a row vector i.e. $(3,-2,2)$ |
| $\mathbf{r} = \begin{pmatrix}8\\-7\\-2\end{pmatrix} + \lambda\begin{pmatrix}3\\-2\\2\end{pmatrix}$ | A1 | Must be "$\mathbf{r}=$" (or "$\begin{pmatrix}x\\y\\z\end{pmatrix}=$"). Allow $\mathbf{r}$ not to be underlined as a vector, but if $L_1$ used this must be indicated as a vector. $\mathbf{r}=\begin{pmatrix}8\\-7\\-2\end{pmatrix}$ or $\begin{pmatrix}11\\-9\\0\end{pmatrix} \pm \lambda\begin{pmatrix}3\\-2\\2\end{pmatrix}$. Note — could use other "starting points". Must be column vectors here. |

## Part (b):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $8+3\lambda=26$ or $-7-2\lambda=-19$ or $-2+2\lambda=-14$ | M1 | Writing down a correct equation for any component of their declared line. Note — values of $\lambda$ will depend on the "starting point" used in part (a) and different multiples of the direction vector. |
| ($x$ gives $\lambda=6$ or $y$ gives $\lambda=6$) [but] $z$ gives $\lambda=-6$ so $(26,-19,-14)$ does **not** lie on the line | A1FT | Finding a correct inconsistency and reaching correct conclusion. Not all values need be given but values given must be correct. FT their declared line. Could see e.g. $\lambda=6 \Rightarrow y=-19$ but $z=10$ which is not $-14$. Do not need to see the word "inconsistent" or explicit comparison. Finding two different values of $\lambda$ and then stating does not lie on line is enough. |
| **Alternate method:** $\begin{pmatrix}18\\-12\\-12\end{pmatrix} = \lambda\begin{pmatrix}3\\-2\\2\end{pmatrix}$ | M1 | Rearranging vector equation |
| But $\begin{pmatrix}3\\-2\\2\end{pmatrix}$ is not parallel to $\begin{pmatrix}18\\-12\\-12\end{pmatrix}$ so the point does not lie on the line | A1FT | Or finding $\lambda$ inconsistency as before. A0 for "there is no value of $\lambda$ that works" without justification |

## Part (c):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\overrightarrow{OC} = \begin{pmatrix}8\\-7\\-2\end{pmatrix} + \lambda\begin{pmatrix}3\\-2\\2\end{pmatrix}$ [for some particular value of $\lambda$] | B1FT | Can be embedded or implied. Can be same or different symbol as parameter in (a). |
| $[\overrightarrow{OC} = \mu\begin{pmatrix}a\\b\\c\end{pmatrix}]$ and $\begin{pmatrix}a\\b\\c\end{pmatrix}.\begin{pmatrix}3\\-2\\2\end{pmatrix}=0$ | B1FT | Condition for perpendicularity stated. Don't have to see an equation for $OC$ in terms of $\mu$ here. |
| $3a-2b+2c=0$; $3(8+3\lambda)-2(-7-2\lambda)+2(-2+2\lambda)=0$ | M1 | Forming a correct dot product and using it with the other definition of $\overrightarrow{OC}$ to form an equation in the parameter. $\mu$ may or may not be present here, but must be dealt with appropriately for the A mark. |
| $\Rightarrow \lambda=-2 \Rightarrow \overrightarrow{OC} = \begin{pmatrix}8\\-7\\-2\end{pmatrix} - 2\begin{pmatrix}3\\-2\\2\end{pmatrix} = \begin{pmatrix}2\\-3\\-6\end{pmatrix}$ so the equation of $L_2$ is $\mathbf{r} = \mu\begin{pmatrix}2\\-3\\-6\end{pmatrix}$ | A1 | Condone use of the same parameter as $L_1$. Could see e.g. $\mathbf{r}=\begin{pmatrix}2\\-3\\-6\end{pmatrix}+\mu\begin{pmatrix}2\\-3\\-6\end{pmatrix}$. BOD lack of "$\mathbf{r}=$" in this part if has already been penalised in part (a). If full marks awarded in part (a) then must have correct notation here for full marks. |

## Question 5(c) [Alternative Method 1]:

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\overrightarrow{OC} = \begin{pmatrix} 8 \\ -7 \\ -2 \end{pmatrix} + \lambda\begin{pmatrix} 3 \\ -2 \\ 2 \end{pmatrix}$ for some particular value of $\lambda$ | B1FT | Can be embedded or implied. Can be same or different symbol as parameter in (a). |
| $\overrightarrow{OC} \cdot \mathbf{b}_{L_1} = 0$ | B1FT | Condition for perpendicularity stated. |
| $\left(\begin{pmatrix} 8 \\ -7 \\ -2 \end{pmatrix} + \lambda\begin{pmatrix} 3 \\ -2 \\ 2 \end{pmatrix}\right) \cdot \begin{pmatrix} 3 \\ -2 \\ 2 \end{pmatrix}$ forming dot product equation | M1 | Forming a correct dot product and using it to form an equation in the parameter. |
| $= 24+14-4+(9+4+4)\lambda = 0 \Rightarrow \lambda = -2$ | | |
| $\Rightarrow \overrightarrow{OC} = \begin{pmatrix} 2 \\ -3 \\ -6 \end{pmatrix}$, equation of $L_2$ is $\mathbf{r} = \mu\begin{pmatrix} 2 \\ -3 \\ -6 \end{pmatrix}$ | A1 | Condone use of same parameter as $L_1$. BOD lack of "$\mathbf{r} =$". Could see e.g. $\mathbf{r} = \begin{pmatrix} 2 \\ -3 \\ -6 \end{pmatrix} + \mu\begin{pmatrix} 2 \\ -3 \\ -6 \end{pmatrix}$ |

## Question 5(c) [Alternative Method 2]:

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\overrightarrow{OC} = \begin{pmatrix} 8 \\ -7 \\ -2 \end{pmatrix} + \lambda\begin{pmatrix} 3 \\ -2 \\ 2 \end{pmatrix}$ for some particular value of $\lambda$ | B1FT | Can be embedded or implied. |
| $|\overrightarrow{OC}|^2 = (8+3\lambda)^2+(-7-2\lambda)^2+(-2+2\lambda)^2$ $= 17\lambda^2 + 68\lambda + 117$ | M1 | Using the fact that minimum distance is the perpendicular distance. |
| Equating derivative to 0: $34\lambda + 68 = 0 \Rightarrow \lambda = -2$ | M1 | |
| $\Rightarrow \overrightarrow{OC} = \begin{pmatrix} 2 \\ -3 \\ -6 \end{pmatrix}$, equation of $L_2$ is $\mathbf{r} = \mu\begin{pmatrix} 2 \\ -3 \\ -6 \end{pmatrix}$ | A1 | Condone use of same parameter as $L_1$. Condone presence of zero vector as first point. |

## Question 5(d):

| Answer | Marks | Guidance |
|--------|-------|----------|
| Because $C$ lies on $L_1$ and $L_2$ and $OC$ is perpendicular to $L_1$, $OC$ must be the shortest route from $O$ to $L_1$ | M1 | Can be implied by sight of $|\overrightarrow{OC}|$. Or attempt at evaluating $|OC|$ |
| $|\overrightarrow{OC}| = \sqrt{2^2+(-3)^2+(-6)^2} = \sqrt{49} = 7$, so shortest distance from $O$ to $L_1$ is 7 units | A1 | Follow through sign errors on coordinates of $C$ |
| **[2]** | | If just an evaluation seen, need to see the vector identified as $C$ in previous part. |