# Question 9(a):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\mathbf{n} = \begin{pmatrix}3\\0\\1\end{pmatrix}\times\begin{pmatrix}1\\-2\\2\end{pmatrix} = \begin{pmatrix}2\\-5\\-6\end{pmatrix}$ | **M1** | Calculates vector product of two vectors in $\Pi_1$ (two components correct) |
| $\begin{pmatrix}5\\3\\0\end{pmatrix}\cdot\begin{pmatrix}2\\-5\\-6\end{pmatrix} = \ldots \quad \{-5\}$ | **M1** | Calculates scalar product of a point in the plane and their normal. Must follow attempt at vector product. Value must be correct if no indication of correct method to evaluate scalar product |
| $\mathbf{r}\cdot\begin{pmatrix}2\\-5\\-6\end{pmatrix} = \begin{pmatrix}5\\3\\0\end{pmatrix}\cdot\begin{pmatrix}2\\-5\\-6\end{pmatrix} \Rightarrow 2x - 5y - 6z = -5$ | **A1** | Any correct Cartesian equation e.g. $-2x+5y+6z=5$, $2x-5y-6z+5=0$ |

**Total: (3)**

**Alt (Sim eqns):**

| Answer/Working | Mark | Guidance |
|---|---|---|
| $x=5+3s+t,\; y=3-2t,\; z=s+2t \Rightarrow$ e.g. $y+z=3+s$ | **M1** | Forms simultaneous equations in $x,y,z,s,t$ and obtains equation eliminating at least one of $s$ and $t$ |
| $x = 5+3(y+z-3)+\frac{1}{2}z - \frac{1}{2}(y+z-3) \Rightarrow x = \frac{5}{2}y + 3z - \frac{5}{2}$ | **M1** | Proceeds to equation in $x,y,z$ only |
| (collected form) | **A1** | Any correct equation with terms collected |

**Total: (3)**

---

# Question 9(b):

## Way 1:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $2x-5y-6z=-5,\quad 5x-2y+3z=1 \Rightarrow$ e.g. $12x-9y=-3$ | **M1** | Uses both plane equations to eliminate one variable. May see $21y+36z=27$, $21x+27z=15$ |
| e.g. $4x-3y=-1 \Rightarrow x=\dfrac{3y-1}{4} \Rightarrow y=\dfrac{4x+1}{3}$ and $3z=1-\dfrac{5(3y-1)}{4}+2y=\dfrac{4-15y+5+8y}{4} \Rightarrow z=\dfrac{9-7y}{12} \Rightarrow y=\dfrac{12z-9}{-7}$ | **dM1** | Expresses one variable in terms of other two, or two variables in terms of the other one. May set variable equal to parameter. Requires previous M mark |
| e.g. $\dfrac{4x+1}{3} = y = \dfrac{12z-9}{-7}$ or $\mathbf{r}=\begin{pmatrix}-\frac{1}{4}\\0\\\frac{3}{4}\end{pmatrix}+\lambda\begin{pmatrix}\frac{3}{4}\\1\\-\frac{7}{12}\end{pmatrix}$ | **ddM1 A1** | ddM1: Correct method to form RHS of vector equation (allow copying slips, must not be clearly incorrect method). A1: Any correct **equation** (with any parameter). Do not condone e.g. $l=\ldots$ |

**Total: (4)**

## Way 2 (Finds point and takes vector product of normals):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $2x-5y-6z=-5,\quad 5x-2y+3z=1$; Let $y=0 \Rightarrow 2x-6z=-5,\; 5x+3z=1 \Rightarrow 12x=-3 \Rightarrow x=-\frac{1}{4},\; y=0,\; z=\frac{3}{4}$ | **M1** | Assigns value to one variable to obtain two equations in other variables, or eliminates one variable |
| May see $\left(0,\frac{1}{3},\frac{5}{9}\right)$ or $\left(\frac{5}{7},\frac{9}{7},0\right)$ | **dM1** | Solves or assigns value to find values for other variables. Requires previous M mark |
| $\begin{pmatrix}2\\-5\\-6\end{pmatrix}\times\begin{pmatrix}5\\-2\\3\end{pmatrix}=\begin{pmatrix}-27\\-36\\21\end{pmatrix} \Rightarrow \mathbf{r}=\begin{pmatrix}-\frac{1}{4}\\0\\\frac{3}{4}\end{pmatrix}+\lambda\begin{pmatrix}-27\\-36\\21\end{pmatrix}$ | **ddM1 A1** | ddM1: Calculates vector product of normals (two components correct) and forms RHS of vector equation. A1: Any correct equation (with any parameter). Correct points have form $\left(\frac{3\alpha-1}{4},\alpha,\frac{9-7\alpha}{12}\right)$ |

**Total: (4)**

## Way 3 (2 points):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Finding 2 points on the line and **subtract** for direction e.g. finds $\left(-\frac{1}{4},0,\frac{3}{4}\right)$ | **M1 dM1** | As Way 2 |
| Then finds $\left(0,\frac{1}{3},\frac{5}{9}\right) \Rightarrow$ direction $=\left(\frac{1}{4},\frac{1}{3},-\frac{7}{36}\right) \Rightarrow$ forms RHS of vector equation | **ddM1** | |
| Then A1 for correct equation | **A1** | |

**Total: (4)**

## Question 9(c):

| Answer/Working | Mark | Guidance Notes |
|---|---|---|
| Substitutes parametric form of line into $\Pi_3$: $\mathbf{r} = \begin{pmatrix} -\frac{1}{4} \\ 0 \\ \frac{3}{4} \end{pmatrix} + \lambda \begin{pmatrix} 9 \\ 12 \\ -7 \end{pmatrix} = \begin{pmatrix} -\frac{1}{4}+9\lambda \\ 12\lambda \\ \frac{3}{4}-7\lambda \end{pmatrix}$ and solves for $\lambda$ | M1 | Substitutes the parametric form of their line (allow slips but must not clearly confuse position and direction) from (b) into $\Pi_3$ and solves for $\lambda$. The "$=0$" could be implied by a solution. |
| $4\!\left(-\tfrac{1}{4}+9\lambda\right) - 3(12\lambda) - \left(\tfrac{3}{4}-7\lambda\right) = 0 \Rightarrow 7\lambda = \tfrac{7}{4} \Rightarrow \lambda = \tfrac{1}{4}$ | | |
| $\Rightarrow \left(9\!\left(\tfrac{1}{4}\right)-\tfrac{1}{4},\ 12\!\left(\tfrac{1}{4}\right),\ -7\!\left(\tfrac{1}{4}\right)+\tfrac{3}{4}\right) = \ldots$ | dM1 | Substitutes their $\lambda$ into their line and obtains a point/position vector with values for all coordinates/components. If no working shown, at least two coordinates/components should be consistent with their equation or parametric form. Isw if the point/position is altered by multiplication. Requires previous M mark. |
| $(2,\ 3,\ -1)$ | A1 | Correct point. No others. Allow $x = \ldots,\ y = \ldots,\ z = \ldots$ and condone as a position vector. Do not isw. |

**Total for 9(c): (3) marks**

**Total for Question 9: 10 marks**

**PAPER TOTAL: 75 marks**