AS June 2018 Paper 1 Q10

OCR MEICurrent spec8 marks3D Lines & Planes Matrices

10 Three planes have equations

\[\begin{aligned} -x + 2y + z &= 0 \\ 2x - y - z &= 0 \\ x + y \phantom{{}- z} &= a \end{aligned}\]

where \(a\) is a constant.

(i) Investigate the arrangement of the planes:

when \(a = 0\);
when \(a \neq 0\). [6]

(ii) Chris claims that the position vectors \(-\mathbf{i} + 2\mathbf{j} + \mathbf{k}\), \(2\mathbf{i} - \mathbf{j} - \mathbf{k}\) and \(\mathbf{i} + \mathbf{j}\) lie in a plane. Determine whether or not Chris is correct. [2]

Mark scheme (i)

Scheme	Marks	AO
\(\det\mathbf{M} = 0\) [so no unique solution]	B1	3.1a
no planes parallel [so prism or sheaf]	B1	1.1
when \(a = 0\), they form a sheaf	B1	2.2a
as the system has solutions	B1	2.2a
when \(a \neq 0\), they form a prismatic intersection	B1	2.2a
as there are no solutions	B1	3.1a
	[6]

Notes

B1: (1st) or \(\mathbf{M}\) is singular
or state no unique solution by direct solution of equations

B1: (3rd) allow ‘intersect in a line’

B1: (4th) o.e. e.g. finding solutions

B1: (5th) allow ‘prism’

Mark scheme (ii)

Scheme	Marks	AO
These are the normals to the three planes	M1	1.1a
In either of the above cases, they must lie in the same plane	A1dep	2.3
	[2]

Notes

A1dep: dep previous part correct

Alternative

Scheme	Marks
(e.g) \(\mathbf{i} + \mathbf{j} = -\mathbf{i} + 2\mathbf{j} + \mathbf{k} + 2\mathbf{i} - \mathbf{j} - \mathbf{k}\)	M1
\(\Rightarrow\) coplanar	A1

M1: showing linear dependence