## Question 3:

### Part (a)(i):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\begin{pmatrix}1\\2\\-1\end{pmatrix} \times \begin{pmatrix}3\\5\\-2\end{pmatrix} = \begin{pmatrix}1\\-1\\-1\end{pmatrix}$ | B1 | |

### Part (a)(ii):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\begin{pmatrix}1\\-1\\-1\end{pmatrix}$ is perpendicular to both $\begin{pmatrix}1\\2\\-1\end{pmatrix}$ and $\begin{pmatrix}3\\5\\-2\end{pmatrix}$ [also $\begin{pmatrix}3\\5\\-2\end{pmatrix}$] | B1FT | This can be stated as a generality (e.g. $\mathbf{a} \times \mathbf{b}$ is perpendicular to both $\mathbf{a}$ and $\mathbf{b}$). FT their answer to (a)(i). Accept "perpendicular to both vectors" or "they are perpendicular". If vectors stated must be correct vectors (or their vectors if MR in part (i)). If cross product vector stated must be their cross product vector. |

### Part (a)(iii):

| Answer | Marks | Guidance |
|--------|-------|----------|
| If the dot product is zero then the vectors are perpendicular | M1 | M1 can be implied by attempt to find a relevant dot product and show it is 0 |
| $\begin{pmatrix}1\\2\\-1\end{pmatrix}\cdot\begin{pmatrix}1\\-1\\-1\end{pmatrix} = 1-2+1 = 0$; $\begin{pmatrix}3\\5\\-2\end{pmatrix}\cdot\begin{pmatrix}1\\-1\\-1\end{pmatrix} = 3-5+2 = 0$ (so answer to (a)(i) is perpendicular to both as claimed) | A1 | Correct calculation of $\mathbf{a}.(\mathbf{a}\times\mathbf{b})$ and $\mathbf{b}.(\mathbf{a}\times\mathbf{b})$, showing some details of calculation (not simply stating "$= 0$" without justification). NB both marks still available to candidates whose answer to (a)(i) is $\begin{pmatrix}-1\\1\\1\end{pmatrix}$ |

# Question (b) [Vectors dot product]:

| Answer | Marks | Guidance |
|--------|-------|----------|
| $(2\mathbf{i} - 2\mathbf{j} + \mathbf{k}).(4\mathbf{i} - \mathbf{j} + 8\mathbf{k}) = 8 + 2 + 8 = 18$ | B1 | Can be implied by correct answer |
| $\cos\theta = \frac{\mathbf{a}.\mathbf{b}}{|\mathbf{a}||\mathbf{b}|} = \frac{18}{\sqrt{2^2+(-2)^2+1^2}\sqrt{4^2+(-1)^2+(-8)^2}}$ | M1 | Correct method for evaluation of cosine of required angle, including correct form for both moduli. M1 can be awarded after correct rearrangement of $\mathbf{a.b} = |\mathbf{a}||\mathbf{b}|\cos\theta$ once correct form for moduli seen even if subsequent (calculation) error. |
| $= \frac{18}{\sqrt{9}\sqrt{81}} = \frac{2}{3}$, $\therefore \theta = \cos^{-1}\left(\frac{2}{3}\right) = 48.2°$ (1 dp) | A1 | awrt $48.2°$ or $0.841$ rads. (48.1896851… or 0.8410686706) |

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