A2 June 2025 Paper 1 Q5

OCR ACurrent spec8 marks3D Lines & Planes Matrices

5 A vector equation of the plane \(\Pi_1\) is \(\mathbf{r} = \begin{pmatrix} 1 \\ 4 \\ 3 \end{pmatrix} + \lambda\begin{pmatrix} 3 \\ 1 \\ -1 \end{pmatrix} + \mu\begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}\).

(a) Verify that a cartesian equation of \(\Pi_1\) is \(x - y + 2z = 3\). [1]

For some real constant \(a\), cartesian equations of planes \(\Pi_2\) and \(\Pi_3\) are

\(\begin{aligned} \Pi_2&: \quad x \phantom{{}-y} - 3z = 1 \\ \Pi_3&: \quad ax - y - z = 4 \end{aligned}\)

(b) By considering a suitable matrix, show that \(\Pi_1\), \(\Pi_2\) and \(\Pi_3\) intersect at a single point for all values of \(a\) except \(a = 2\). [3]

(c) Use the matrix from part (b) to find the coordinates of the point of intersection of \(\Pi_1\), \(\Pi_2\) and \(\Pi_3\) in the case where \(a = 3\). [2]

(d) In the case where \(a = 2\), determine the geometrical arrangement of \(\Pi_1\), \(\Pi_2\) and \(\Pi_3\). [2]

Mark scheme (a)

Scheme	Marks	AO
\((1 + 3\lambda + \mu) - (4 + \lambda + \mu) + 2(3 - \lambda)\) \(= 1 + 3\lambda + \mu - 4 - \lambda - \mu + 6 - 2\lambda = 3\)	B1	2.2a
	[1]

Notes

B1: AG substitutes into cartesian equation and obtains 3 or showing that 3 = 3 – must be at least one line of intermediate working from substitution to given answer – as a minimum allow correct expression without any brackets followed by 3. Alternative method would be to find at least three points from the vector equation and show that all three satisfy the cartesian equation

Alternative method

Scheme	Marks
\(\begin{pmatrix} 3 \\ 1 \\ -1 \end{pmatrix} \times \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix}\) \(\mathbf{r} \cdot \begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix} = \begin{pmatrix} 1 \\ 4 \\ 3 \end{pmatrix} \cdot \begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix} = 1 - 4 + 6\) \(x - y + 2z = 3\)	B1

By direct calculation of cartesian equation
Or \(\begin{pmatrix} 3 \\ 1 \\ -1 \end{pmatrix} \times \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix}\) so \(x - y + 2z = 1 - 4 + 2(3) = 3\)

Mark scheme (b)

Scheme	Marks	AO
\(\begin{vmatrix} 1 & -1 & 2 \\ 1 & 0 & -3 \\ a & -1 & -1 \end{vmatrix} \; (= 0)\)	M1*	1.1
\(= 1\begin{vmatrix} 0 & -3 \\ -1 & -1 \end{vmatrix} - (-1)\begin{vmatrix} 1 & -3 \\ a & -1 \end{vmatrix} + 2\begin{vmatrix} 1 & 0 \\ a & -1 \end{vmatrix}\) \(\big(= -3 + (-1 + 3a) + 2(-1) = 3a - 6\big)\)	M1dep*	1.1
Planes intersect at a single point if and only if \(\det\mathbf{M} \neq 0\). Therefore \(3a - 6 \neq 0 \Rightarrow a \neq 2\).	A1	2.2a
	[3]

Notes

M1*: For determinant of a relevant matrix (e.g. rows interchanged but not columns) – no MR of values in this part but allow a slip in no more than 2 (of the 9) values

M1dep*: Finds determinant of their matrix as far as calculating \(2 \times 2\) determinants. May expand by any row or column. Ignore sign errors in \(2 \times 2\) determinant calculations if shown, but cofactors must have correct signs.

A1: AG - correct determinant followed by 2 and either mention of that for a single point of intersection determinant \(\neq 0\) or for no single point of intersection determinant \(= 0\) - allow substitution of 2 into correct determinant to obtain 0

Mark scheme (c)

Scheme	Marks	AO
\(\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 1 & -1 & 2 \\ 1 & 0 & -3 \\ 3 & -1 & -1 \end{pmatrix}^{-1} \begin{pmatrix} 3 \\ 1 \\ 4 \end{pmatrix}\)	B1	1.1
\(\left(0, -\dfrac{11}{3}, -\dfrac{1}{3}\right)\)	B1	1.1
	[2]

Notes

B1: This mark is for a clear intention to consider \(\begin{pmatrix} 1 & -1 & 2 \\ 1 & 0 & -3 \\ 3 & -1 & -1 \end{pmatrix}^{-1} \begin{pmatrix} 3 \\ 1 \\ 4 \end{pmatrix}\) so \(\begin{pmatrix} 1 & -1 & 2 \\ 1 & 0 & -3 \\ 3 & -1 & -1 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 3 \\ 1 \\ 4 \end{pmatrix}\) only is B0. For reference the correct inverse is \(\dfrac{1}{3}\begin{pmatrix} -3 & -3 & 3 \\ -8 & -7 & 5 \\ -1 & -2 & 1 \end{pmatrix}\). Allow \(\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \mathbf{M}^{-1}\begin{pmatrix} 3 \\ 1 \\ 4 \end{pmatrix}\) providing \(\mathbf{M}\) defined (in this part or in part (b))

B1: BC - values must be exact but allow \(x = 0, y = -\frac{11}{3}, z = -\frac{1}{3}\) and condone position vector – this mark is independent of the previous B1 mark

Mark scheme (d)

Scheme	Marks	AO
\([1]: x - y + 2z = 3\) \([2]: x - 3z = 1\) \([3]: 2x - y - z = 4\) So [1] + [2] = [3] or [3] – [2] = [1]	B1*	1.1
So, equations are consistent (and as none of the planes are coincident so) geometrically, the planes meet in a straight line/they form a sheaf.	B1dep*	3.2a
	[2]

Notes

B1*: For showing consistency (must explicitly link all three equations together). Note that eliminating \(x\) gives \(5z - y = 2\) or eliminating \(z\) gives \(5x - 3y = 11\), but these must be derived twice for B1. No MR in this part (so must be using the correct equations). Stating the equation of the line where the three planes meet with no working scores zero marks in this part

B1dep*: For “consistent/infinite solutions” and either “line” or “sheaf”
If B0 B0 then SC B1 for implicitly showing consistency (e.g. [1] + [2] = \(2x - y - z = 4\) but not explicitly linking this to [3]) together with ‘consistent/ infinite solutions’ and either ‘line’ or ‘sheaf’