(a)

In a parallelogram PQRS with S diagonally opposite Q (vertices P, Q, R, S), opposite sides are equal: \( \mathbf{PQ} = \mathbf{SR} \).

Compute \( \mathbf{PQ} \):

\[ \mathbf{PQ} = \mathbf{Q} – \mathbf{P} = (-2 – (-1), 1 – 2, 0 – (-3)) = \begin{pmatrix} -1 \\ -1 \\ 3 \end{pmatrix} \]

Compute \( \mathbf{SR} \):

\[ \mathbf{SR} = \mathbf{R} – \mathbf{S} = (0 – x, 5 – y, 1 – z) = \begin{pmatrix} 0 – x \\ 5 – y \\ 1 – z \end{pmatrix} \] (M1)

Set \( \mathbf{PQ} = \mathbf{SR} \):

\[ \begin{pmatrix} -1 \\ -1 \\ 3 \end{pmatrix} = \begin{pmatrix} 0 – x \\ 5 – y \\ 1 – z \end{pmatrix} \]

Solve: \( -1 = 0 – x \implies x = 1 \), \( -1 = 5 – y \implies y = 6 \), \( 3 = 1 – z \implies z = -2 \).

Point S = (1, 6, -2) (A1)

[2 marks]

(b)

Using \( \mathbf{PQ} = \begin{pmatrix} -1 \\ -1 \\ 3 \end{pmatrix} \) and point S from part (a):

\[ \mathbf{PS} = \mathbf{S} – \mathbf{P} = (1 – (-1), 6 – 2, -2 – (-3)) = \begin{pmatrix} 2 \\ 4 \\ 1 \end{pmatrix} \] (A1)

Compute cross product \( \mathbf{PQ} \times \mathbf{PS} \):

\[ \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -1 & -1 & 3 \\ 2 & 4 & 1 \end{vmatrix} = \mathbf{i}((-1)(1) – (3)(4)) – \mathbf{j}((-1)(1) – (3)(2)) + \mathbf{k}((-1)(4) – (-1)(2)) \]

\[ = \mathbf{i}(-1 – 12) – \mathbf{j}(-1 – 6) + \mathbf{k}(-4 + 2) = \begin{pmatrix} -13 \\ 7 \\ -2 \end{pmatrix} \implies m = -2 \] (A1)

[2 marks]

(c)

The area of parallelogram PQRS is the magnitude of the cross product of adjacent sides:

\[ |\mathbf{PQ} \times \mathbf{PS}| = \sqrt{(-13)^2 + 7^2 + (-2)^2} = \sqrt{169 + 49 + 4} = \sqrt{222} \approx 14.9 \] (M1)(A1)

[2 marks]

(d)

The normal vector to plane \( \Pi_1 \) is \( \mathbf{PQ} \times \mathbf{PS} = \begin{pmatrix} -13 \\ 7 \\ -2 \end{pmatrix} \).

Plane equation: \[ -13x + 7y – 2z = d \] (M1)(A1)

Substitute P(-1, 2, -3): \[ -13(-1) + 7(2) – 2(-3) = 13 + 14 + 6 = 33 \] (A1)

\[ -13x + 7y – 2z = 33 \]

[3 marks]

(e)

The line perpendicular to \( \Pi_1 \) through (0, 0, 0) has direction vector equal to the normal \( \begin{pmatrix} -13 \\ 7 \\ -2 \end{pmatrix} \):

\[ \mathbf{r} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} + \lambda \begin{pmatrix} -13 \\ 7 \\ -2 \end{pmatrix} \] (A1)

[1 mark]

(f)

The closest point to the origin lies on the line from part (e): \( (-13\lambda, 7\lambda, -2\lambda) \).

Substitute into plane equation \( -13x + 7y – 2z = 33 \):

\[ -13(-13\lambda) + 7(7\lambda) – 2(-2\lambda) = 169\lambda + 49\lambda + 4\lambda = 222\lambda = 33 \] (M1)

\[ \lambda = \frac{33}{222} = \frac{11}{74} \] (A1)

Point: \[ \left( -13 \cdot \frac{11}{74}, 7 \cdot \frac{11}{74}, -2 \cdot \frac{11}{74} \right) = \left( -\frac{143}{74}, \frac{77}{74}, -\frac{11}{37} \right) \approx (-1.93, 1.04, -0.297) \] (A1)

[3 marks]

(g)

The angle between planes \( \Pi_1 \): \( -13x + 7y – 2z = 33 \) and \( \Pi_2 \): \( x – 2y + z = 3 \) is the angle between their normals.

Normals: \( \mathbf{n}_1 = \begin{pmatrix} -13 \\ 7 \\ -2 \end{pmatrix} \), \( \mathbf{n}_2 = \begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix} \). (R1)

Dot product: \[ (-13)(1) + 7(-2) + (-2)(1) = -13 – 14 – 2 = -29 \]

Magnitudes: \[ |\mathbf{n}_1| = \sqrt{(-13)^2 + 7^2 + (-2)^2} = \sqrt{222}, \quad |\mathbf{n}_2| = \sqrt{1^2 + (-2)^2 + 1^2} = \sqrt{6} \]

\[ \cos \theta = \frac{-29}{\sqrt{222} \cdot \sqrt{6}} \approx -0.795 \] (M1)(A1)

\[ \theta \approx \cos^{-1}(-0.795) \approx 143^\circ \text{ (or } 37^\circ\text{)} \] (A1)

[4 marks]