Note: Accept alternative notation for vectors (eg \(\langle a, b, c \rangle\) or \((a, b, c)\)).</pthorne

a. [4 marks]

\(\boldsymbol{n} = \begin{pmatrix} 1 \\ -2 \\ -3 \end{pmatrix}\) and \(\boldsymbol{m} = \begin{pmatrix} 2 \\ -1 \\ -1 \end{pmatrix}\) (A1).
\(\cos \theta = \frac{\boldsymbol{n} \cdot \boldsymbol{m}}{|\boldsymbol{n}| |\boldsymbol{m}|}\) (M1).
\(\cos \theta = \frac{2 + 2 + 3}{\sqrt{1 + 4 + 9} \sqrt{4 + 1 + 1}} = \frac{7}{\sqrt{14} \sqrt{6}}\) (A1).
\(\theta = 40.2^\circ (0.702 \text{ rad})\) (A1).

b. [5 marks]

METHOD 1
eliminate \(z\) from \(x – 2y – 3z = 2\) and \(2x – y – z = k\)
\(5x – y = 3k – 2 \Rightarrow x = \frac{y – (2 – 3k)}{5}\) (M1A1).
eliminate \(y\) from \(x – 2y – 3z = 2\) and \(2x – y – z = k\)
\(3x + z = 2k – 2 \Rightarrow x = \frac{z – (2k – 2)}{-3}\) (A1).
\(x = t, y = (2 – 3k) + 5t\) and \(z = (2k – 2) – 3t\) (A1A1).
\(r = \begin{pmatrix} 0 \\ 2 – 3k \\ 2k – 2 \end{pmatrix} + t \begin{pmatrix} 1 \\ 5 \\ -3 \end{pmatrix}\).
METHOD 2
\(\begin{pmatrix} 1 \\ -2 \\ -3 \end{pmatrix} \times \begin{pmatrix} 2 \\ -1 \\ -1 \end{pmatrix} = \begin{pmatrix} -1 \\ -5 \\ 3 \end{pmatrix} \Rightarrow \text{direction is } \begin{pmatrix} 1 \\ 5 \\ -3 \end{pmatrix}\) (M1A1).
Let \(x = 0\)
\(0 – 2y – 3z = 2\) and \(2 \times 0 – y – z = k\) (M1).
solve simultaneously (M1).
\(y = 2 – 3k\) and \(z = 2k – 2\) (A1).
therefore \(\mathbf{r} = \begin{pmatrix} 0 \\ 2 – 3k \\ 2k – 2 \end{pmatrix} + t \begin{pmatrix} 1 \\ 5 \\ -3 \end{pmatrix}\).
METHOD 3
substitute \(x = t, y = (2 – 3k) + 5t\) and \(z = (2k – 2) – 3t\) into \(\pi_1\) and \(\pi_2\) (M1).
for \(\pi_1: t – 2(2 – 3k + 5t) – 3(2k – 2 – 3t) = 2\) (A1).
for \(\pi_2: 2t – (2 – 3k + 5t) – (2k – 2 – 3t) = k\) (A1).
the planes have a unique line of intersection (R2).
therefore the line is \(r = \begin{pmatrix} 0 \\ 2 – 3k \\ 2k – 2 \end{pmatrix} + t \begin{pmatrix} 1 \\ 5 \\ -3 \end{pmatrix}\).

c. [5 marks]

\(5 – t = (2 – 3k + 5t) + 3 = 2 – 2(2k – 2 – 3t)\) (M1A1).
Note: Award M1A1 if candidates use vector or parametric equations of \(L_2\)
eg \(\begin{pmatrix} 0 \\ 2 – 3k \\ 2k – 2 \end{pmatrix} + t \begin{pmatrix} 1 \\ 5 \\ -3 \end{pmatrix} = \begin{pmatrix} 5 \\ -3 \\ 1 \end{pmatrix} + s \begin{pmatrix} -2 \\ 2 \\ -1 \end{pmatrix}\) or \(\begin{cases} t = 5 – 2s \\ 2 – 3k + 5t = -3 + 2s \\ 2k – 2 – 3t = 1 + s \end{cases}\)
solve simultaneously (M1).
\(k = 2, t = 1 (s = 2)\) (A1).
intersection point \((1, 1, -1)\) (A1).

d. [5 marks]

\(\overrightarrow{l_2} = \begin{pmatrix} 2 \\ -2 \\ 1 \end{pmatrix}\) (A1).
\(\overrightarrow{l_1} \times \overrightarrow{l_2} = \begin{vmatrix} \boldsymbol{i} & \boldsymbol{j} & \boldsymbol{k} \\ 1 & 5 & -3 \\ 2 & -2 & 1 \end{vmatrix} = \begin{pmatrix} -1 \\ -7 \\ -12 \end{pmatrix}\) (M1A1).
\(\boldsymbol{r} \cdot \begin{pmatrix} 1 \\ 7 \\ 12 \end{pmatrix} = \begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix} \cdot \begin{pmatrix} 1 \\ 7 \\ 12 \end{pmatrix}\) (M1).
\(x + 7y + 12z = -4\) (A1).

e. [5 marks]

Let \(\theta\) be the angle between the lines \(\overrightarrow{l_1} = \begin{pmatrix} 1 \\ 5 \\ -3 \end{pmatrix}\) and \(\overrightarrow{l_2} = \begin{pmatrix} 2 \\ -2 \\ 1 \end{pmatrix}\)
\(\cos \theta = \frac{|2 – 10 – 3|}{\sqrt{35} \sqrt{9}} \Rightarrow \theta = 0.902334… (51.699…^\circ)\) (M1).
as the triangle \(XYZ\) has a right angle at \(Y\),
\(XZ = a \Rightarrow YZ = a \sin \theta \text{ and } XY = a \cos \theta\) (M1).
area = 3 \(\Rightarrow \frac{a^2 \sin \theta \cos \theta}{2} = 3\) (M1).
\(a = 3.5122…\) (A1).
perimeter = \(a + a \sin \theta + a \cos \theta = 8.44537… = 8.45\) (A1).