# Question 2:

## Part (i)

| Answer | Marks | Guidance |
|--------|-------|----------|
| $AB = \sqrt{5^2 + (-2)^2} = \sqrt{29}$ | B1 | 5.39 or better (condone sign error in vector for B1) |
| $AC = \sqrt{3^2 + 4^2} = 5$ | B1 | Accept $\sqrt{25}$ (condone sign error in vector for B1) |
| $\cos\theta = \dfrac{\begin{pmatrix}5\\0\\-2\end{pmatrix}\cdot\begin{pmatrix}3\\4\\0\end{pmatrix}}{\sqrt{29}\cdot 5} = \dfrac{15+0+0}{5\sqrt{29}} = 0.5571$ | M1 | $\cos\theta = \dfrac{\text{scalar product of AB with AC}}{|AB||AC|}$ (accept BA/CA) **with substitution**; condone a single numerical error provided method is clearly understood. **[OR** Cosine Rule, as far as $\cos\theta =$ correct numerical expression**]** |
| | A1 | **www** $\pm 0.5571$, $0.557$, $\frac{15}{5\sqrt{29}}$, $\frac{15}{\sqrt{25}\sqrt{29}}$ oe or better soi ($\pm$ for method only) |
| $\Rightarrow \theta = 56.15°$ | A1 | www Accept answers that round to 56.1° or 56.2° or 0.98 radians (or better). **NB vector $5\mathbf{i}+0\mathbf{j}+2\mathbf{k}$ leads to apparently correct answer but loses all A marks in part(i)** |
| Area $= \frac{1}{2} \times 5 \times \sqrt{29} \times \sin\theta$ | M1 | Using their $AB$, $AC$, $\angle CAB$. Accept any valid method using trigonometry |
| $= 11.18$ | A1 | Accept $5\sqrt{5}$ and answers that round to 11.18 or 11.19 (2dp) **www** or SCA1 for accurate work soi rounded at the last stage to 11.2 (but not from an incorrect answer). We will not accept inaccurate work from over rounded answers for the final mark. |
| **[7]** | | |

## Part (ii)(A)

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\overrightarrow{AB}\cdot\begin{pmatrix}4\\-3\\10\end{pmatrix} = \begin{pmatrix}5\\0\\-2\end{pmatrix}\cdot\begin{pmatrix}4\\-3\\10\end{pmatrix} = 5\cdot4 + 0\cdot(-3) + (-2)\cdot10 = 0$ | B1 | Scalar product with one vector in the plane with numerical expansion shown |
| $\overrightarrow{AC}\cdot\begin{pmatrix}4\\-3\\10\end{pmatrix} = \begin{pmatrix}3\\4\\0\end{pmatrix}\cdot\begin{pmatrix}4\\-3\\10\end{pmatrix} = 3\times4 + 4\times(-3) + 0\times10 = 0$ | B1 | Scalar product, as above, with evaluation, with a second vector. **NB vectors are not unique**. SCB2 finding the equation of plane first by any valid method (or using vector product) and then clearly stating that the normal is proportional to the coefficients. SC For candidates who substitute all three points in the plane $4x-3y+10z=c$ and show that they give the same result, award M1. If they include a statement explaining why this means that $4\mathbf{i}-3\mathbf{j}+10\mathbf{k}$ is normal they can gain A1. |
| **[2]** | | |

## Part (ii)(B)

| Answer | Marks | Guidance |
|--------|-------|----------|
| $4x - 3y + 10z = c$ | M1 | Required form **and** substituting the co-ordinates of a point on the plane |
| $\Rightarrow 4x - 3y + 10z + 12 = 0$ | A1 | oe. If found in (A) it must be clearly referred to in (B) to gain the marks. Do not accept vector equation of the plane, as 'Hence'. $4\mathbf{i}-3\mathbf{j}+10\mathbf{k} = -12$ is M1A0 |
| **[2]** | | |

## Part (iii)

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\mathbf{r} = \begin{pmatrix}0\\4\\5\end{pmatrix}$ | B1 | Need $\mathbf{r} = \left(\text{or } \begin{pmatrix}x\\y\\z\end{pmatrix}\right)$ |
| $+ \lambda\begin{pmatrix}4\\-3\\10\end{pmatrix}$ | B1 | oe |
| Meets $4x - 3y + 10z + 12 = 0$ when $16\lambda - 3(4-3\lambda) + 10(5+10\lambda) + 12 = 0$ | M1 | Subst their $4\lambda$, $4-3\lambda$, $5+10\lambda$ in equation of their plane from (ii) |
| $\Rightarrow 125\lambda = -50$, $\lambda = -0.4$ | A1 | $\lambda = -0.4$ (NB not unique) |
| So meets plane ABC at $(-1.6, 5.2, 1)$ | A1 | cao www (condone vector) |
| **[5]** | | |

## Part (iv)

| Answer | Marks | Guidance |
|--------|-------|----------|
| height $= \sqrt{1.6^2 + (-1.2)^2 + 4^2} = \sqrt{20}$ | B1ft | ft their (iii) |
| volume $= 11.18 \times \sqrt{20} / 3 = 16.7$ | B1cao | $\frac{50}{3}$ or answers that round to 16.7 www and not from incorrect answers from (iii) ie not from say $(1.6, 2.8, 9)$ |
| **[2]** | | |

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