A2 June 2023 Paper 1 Q14

OCR MEICurrent spec13 marks3D Lines & Planes Matrices

14 Three planes have equations

\[\begin{aligned} kx \phantom{{}+ky} - z &= 2, \\ -x + ky + 2z &= 1, \\ 2kx + 2y + 3z &= 0, \end{aligned}\]

where \(k\) is a constant.

(a) By considering a suitable determinant, show that the three planes meet at a point for all values of \(k\). [5]

(b) Using a matrix method, find, in terms of \(k\), the coordinates of the point of intersection of the planes. [8]

Mark scheme (a)

Scheme	Marks	AO
\(\begin{vmatrix} k & 0 & -1 \\ -1 & k & 2 \\ 2k & 2 & 3 \end{vmatrix} = k(3k - 4) - 1(-2 - 2k^2)\)	M1 M1	1.1 1.1
\(= 5k^2 - 4k + 2\)	A1	1.1
discriminant \(= 16 - 40 = -24 \lt 0\)	M1	3.1a
so determinant is never zero \(\Rightarrow\) planes always meet at a point	A1	2.2a
	[5]

M1: (1st) considering correct determinant

M1: (2nd) finding determinant using any row or column

M1: (3rd) oe, e.g. completing the square, considering quadratic formula or showing no real roots

A1: (2nd) www. Both statements required.

Mark scheme (b)

Scheme	Marks	AO
\(\mathbf{M}^{-1} = \dfrac{1}{5k^2 - 4k + 2}\begin{pmatrix} 3k - 4 & -2 & k \\ 4k + 3 & 5k & -2k + 1 \\ -2 - 2k^2 & -2k & k^2 \end{pmatrix}\)	M1 A1 M1 M1 A1	1.1 1.1 1.1 3.1a 1.1
\(\mathbf{M}^{-1}\begin{pmatrix} 2 \\ 1 \\ 0 \end{pmatrix} = \dfrac{1}{5k^2 - 4k + 2}\begin{pmatrix} 6k - 10 \\ 13k + 6 \\ -4k^2 - 2k - 4 \end{pmatrix}\)	M1	1.1
\(\left(\dfrac{6k - 10}{5k^2 - 4k + 2}, \dfrac{13k + 6}{5k^2 - 4k + 2}, \dfrac{-4k^2 - 2k - 4}{5k^2 - 4k + 2}\right)\)	A2,1,0	1.1, 1.1
	[8]

M1: (1st) at least 5 cofactors correct (need not be in matrix)

A1: (1st) all cofactors correct

M1: (2nd) cofactor matrix transposed

M1: (3rd) Multiplying by \(\frac{1}{\text{their determinant}}\)

A1: (2nd) correct inverse

M1: (4th) finding \(\mathbf{M}^{-1}\begin{pmatrix} 2 \\ 1 \\ 0 \end{pmatrix}\)

A2,1,0: Award A1 for one of \(x\), \(y\) or \(z\) coordinates correct or all three correct FT their determinant