# Question 7(a):

**Show that two lines lie in the same plane**

**Way 1:**

| $1 + 2\lambda = 1 + t$, $\quad -1 - \lambda = -t$, $\quad 4 + 3\lambda = 3 + 2t$ $\Rightarrow t = \ldots$ or $\lambda = \ldots$ | M1 | 3.1a | Forms system of equations from lines |
| Checks third equation with $t=2$ and $\lambda=1$, or shows $(3,-2,7)$ lies on both lines | A1 | 1.1b | |
| As the lines intersect at a point the lines lie in the same plane | A1 | 2.4 | |

**Way 2:**

| Sets up equations; $t=2, \lambda=1$ or $t=-2, \lambda=-1$ depending on parameterisation | M1, A1 | 3.1a, 1.1b | |
| Second coordinates lie on the plane; therefore lines lie on same plane | A1 | 2.4 | |

**Way 3:**

| $x=1+t,\ y=-t,\ z=3+2t$; $\frac{1+t-1}{2} = \frac{-t+1}{-1} = \frac{3+2t-4}{3}$; solves pair of equations | M1 | 3.1a | |
| Solve two pairs to find $t=2$ | A1 | 1.1b | |
| As lines intersect they lie in same plane | A1 | 2.4 | |

**Way 4 (Further Pure):**

| $\begin{pmatrix}2\\-1\\3\end{pmatrix}\cdot\begin{pmatrix}x\\y\\z\end{pmatrix} \Rightarrow 2x-y+3z=0$ and $\begin{pmatrix}1\\-1\\2\end{pmatrix}\cdot\begin{pmatrix}x\\y\\z\end{pmatrix} \Rightarrow x-y+2z=0$; attempts cross product $\begin{pmatrix}2\\-1\\3\end{pmatrix}\times\begin{pmatrix}1\\-1\\2\end{pmatrix}$; finds equation of one plane OR dot product between normal and coordinate | M1 | 3.1a | |

# Question 7 (Vectors):

## Part (a):

| Working | Mark | Guidance |
|---------|------|----------|
| Sets up dot product: $\mathbf{r} \cdot \begin{pmatrix}1\\-1\\-1\end{pmatrix} = \begin{pmatrix}1\\-1\\4\end{pmatrix} \cdot \begin{pmatrix}1\\-1\\-1\end{pmatrix} = \ldots$ or equivalent with other position vector | M1 | Attempts to find planes containing each line; allow sign slips |
| $\mathbf{r} \cdot \begin{pmatrix}1\\-1\\-1\end{pmatrix} = -2$ or $x - y - z = -2$ | A1 | Achieves correct plane equation for each line |
| Both planes are the same, therefore the lines lie in the same plane | A1 | Conclusion required |

## Part (b):

| Working | Mark | Guidance |
|---------|------|----------|
| e.g. $\mathbf{r} = \begin{pmatrix}1\\0\\3\end{pmatrix} + p\begin{pmatrix}2\\-1\\3\end{pmatrix} + q\begin{pmatrix}1\\-1\\2\end{pmatrix}$ or $\mathbf{r} \cdot k\begin{pmatrix}1\\-1\\-1\end{pmatrix} = -2k$ | B1 | Correct vector equation; allow any letter for scalars; must use two of the direction vectors $\pm\begin{pmatrix}1\\-1\\2\end{pmatrix}$, $\pm\begin{pmatrix}2\\-1\\3\end{pmatrix}$, $\pm\begin{pmatrix}0\\-1\\1\end{pmatrix}$ |

## Part (c) — Way 1:

| Working | Mark | Guidance |
|---------|------|----------|
| $\begin{pmatrix}2\\-1\\3\end{pmatrix} \cdot \begin{pmatrix}1\\-1\\2\end{pmatrix} = 2+1+6$ | M1 | Calculates scalar product between direction vectors; allow one slip |
| $\sqrt{2^2+(-1)^2+3^2}\sqrt{1^2+(-1)^2+2^2}\cos\theta = 9 \Rightarrow \cos\theta = \dfrac{9}{\sqrt{2^2+(-1)^2+3^2}\sqrt{1^2+(-1)^2+2^2}}$ | dM1 | Dependent on M1; applies scalar product formula to find $\cos\theta$ |
| $\theta = 11°$ cao | A1 | Correct answer only |

## Part (c) — Way 2 (Further Pure 2):

| Working | Mark | Guidance |
|---------|------|----------|
| $\begin{pmatrix}2\\-1\\3\end{pmatrix} \times \begin{pmatrix}1\\-1\\2\end{pmatrix} = \begin{pmatrix}1\\-1\\-1\end{pmatrix}$ | M1 | Calculates vector product between direction vectors; allow one slip |
| $\sqrt{2^2+(-1)^2+3^2}\sqrt{1^2+(-1)^2+2^2}\sin\theta = \sqrt{1^2+(-1)^2+(-1)^2}$ $\Rightarrow \sin\theta = \dfrac{\sqrt{1^2+(-1)^2+(-1)^2}}{\sqrt{2^2+(-1)^2+3^2}\sqrt{1^2+(-1)^2+2^2}}$ | dM1 | Dependent on M1; applies vector product formula to find $\sin\theta$ |
| $\theta = 11°$ cao | A1 | Correct answer only |

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