(a) Vector equation of line \( (BC) \):

\[ \overrightarrow{\text{BC}} = (1, 3, 3) – (2, -1, 2) = \begin{pmatrix} -1 \\ 4 \\ 1 \end{pmatrix} \] (A1)

\[ \mathbf{r} = \begin{pmatrix} 2 \\ -1 \\ 2 \end{pmatrix} + \lambda \begin{pmatrix} -1 \\ 4 \\ 1 \end{pmatrix} \] (M1)(A1)

Or: \[ \mathbf{r} = \begin{pmatrix} 1 \\ 3 \\ 3 \end{pmatrix} + \lambda \begin{pmatrix} -1 \\ 4 \\ 1 \end{pmatrix} \]

Note: Do not award (A1) unless \( \mathbf{r} = \) or equivalent correct notation is seen.

[3 marks]

(b) Intersection of lines \( (OA) \) and \( (BC) \):

Parametric form of \( (OA) \): \[ \mathbf{r} = \mu \begin{pmatrix} 2 \\ 1 \\ -2 \end{pmatrix} \], and \( (BC) \): \[ \mathbf{r} = \begin{pmatrix} 2 \\ -1 \\ 2 \end{pmatrix} + \lambda \begin{pmatrix} -1 \\ 4 \\ 1 \end{pmatrix} \] (M1)

Equate: \[ \begin{cases} 2\mu = 2 – \lambda \\ \mu = -1 + 4\lambda \\ -2\mu = 2 + \lambda \end{cases} \] (A1)

Solving first two equations: \( \lambda = \frac{4}{9}, \mu = \frac{7}{9} \) (M1)(A1)

Substitute into third: \[ -2 \left( \frac{7}{9} \right) = 2 + \frac{4}{9} \implies -\frac{14}{9} \neq \frac{22}{9} \] (M1)

Since the values do not satisfy the third equation, the lines do not intersect (R1)

Note: Candidates may note that adding the first and third equations gives \( 0 = 4 \), a contradiction, deducing no intersection.

[6 marks]

(c) Cartesian equation of plane \( \Pi_1 \):

Method 1:

Plane form: \[ \mathbf{r} \cdot \begin{pmatrix} 2 \\ 1 \\ -2 \end{pmatrix} = d \] (A1)

Using point \( C(1, 3, 3) \): \[ d = \begin{pmatrix} 1 \\ 3 \\ 3 \end{pmatrix} \cdot \begin{pmatrix} 2 \\ 1 \\ -2 \end{pmatrix} = 2 + 3 – 6 = -1 \] (M1)

Hence: \[ 2x + y – 2z = -1 \] (A1)

Method 2:

Plane form: \[ 2x + y – 2z = d \] (A1)

Substitute \( C(1, 3, 3) \): \[ 2(1) + 3 – 2(3) = -1 \] (M1)

Hence: \[ 2x + y – 2z = -1 \] (A1)

[3 marks]

(d) Show line \( (BC) \) lies in \( \Pi_1 \):

Method 1:

Scalar product of direction vector \( \overrightarrow{\text{BC}} = (-1, 4, 1) \) and normal \( (2, 1, -2) \): \[ (-1)(2) + (4)(1) + (1)(-2) = -2 + 4 – 2 = 0 \] (M1)(A1)

Hence, \( (BC) \) lies in \( \Pi_1 \) (AG)

Method 2:

Substitute line \[ \mathbf{r} = \begin{pmatrix} 2 \\ -1 \\ 2 \end{pmatrix} + \lambda \begin{pmatrix} -1 \\ 4 \\ 1 \end{pmatrix} \] into plane \[ 2x + y – 2z = -1 \]: (M1)

\[ 2(2 – \lambda) + (-1 + 4\lambda) – 2(2 + \lambda) = 4 – 2\lambda – 1 + 4\lambda – 4 – 2\lambda = -1 \] (A1)

Hence, \( (BC) \) lies in \( \Pi_1 \) (AG)

Note: Candidates may substitute \( B(2, -1, 2) \), which satisfies \( 2(2) + (-1) – 2(2) = -1 \), since \( C \) is given to lie on \( \Pi_1 \).

[2 marks]

(e) Verify \( 2 \mathbf{j} + \mathbf{k} \) is perpendicular to \( \Pi_2 \):

Method 1:

Scalar products with vectors in \( \Pi_2 \):

\[ (2 \mathbf{j} + \mathbf{k}) \cdot (2 \mathbf{i} + \mathbf{j} – 2 \mathbf{k}) = 0 + 2 – 2 = 0 \] (M1)(A1)

\[ (2 \mathbf{j} + \mathbf{k}) \cdot (2 \mathbf{i} – \mathbf{j} + 2 \mathbf{k}) = 0 – 2 + 2 = 0 \] (A1)

Method 2:

Cross product: \[ \overrightarrow{\text{OA}} \times \overrightarrow{\text{OB}} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & 1 & -2 \\ 2 & -1 & 2 \end{vmatrix} = -8 \mathbf{j} – 4 \mathbf{k} \] (M1)(A1)

\[ -8 \mathbf{j} – 4 \mathbf{k} = -4 (2 \mathbf{j} + \mathbf{k}) \] (R1)

Hence, \( 2 \mathbf{j} + \mathbf{k} \) is perpendicular to \( \Pi_2 \).

[3 marks]

(f) Vector perpendicular to \( \Pi_3 \):

\[ \overrightarrow{\text{OA}} \times \overrightarrow{\text{OC}} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & 1 & -2 \\ 1 & 3 & 3 \end{vmatrix} = 9 \mathbf{i} – 8 \mathbf{j} + 5 \mathbf{k} \] (A1)

[1 mark]

(g) Acute angle between planes \( \Pi_2 \) and \( \Pi_3 \):

Normals: \( \Pi_2 \): \( (0, 2, 1) \), \( \Pi_3 \): \( (9, -8, 5) \) (M1)

\[ \cos \theta = \frac{(0, 2, 1) \cdot (9, -8, 5)}{\sqrt{0^2 + 2^2 + 1^2} \cdot \sqrt{9^2 + (-8)^2 + 5^2}} = \frac{0 – 16 + 5}{\sqrt{5} \cdot \sqrt{170}} = \frac{-11}{\sqrt{5} \sqrt{170}} \] (M1)(A1)

Acute angle: \[ \theta = \cos^{-1} \left( \frac{11}{\sqrt{5} \sqrt{170}} \right) \approx 67.8^\circ \] (A1)

[4 marks]