A2 October 2021 Q4

EdexcelCurrent spec7 marksVectors

Diagram of a horizontal plane representing the field, with the aircraft approach vector vA meeting the plane, the normal n drawn dashed upwards from that point, and the landing direction vL along the field, all in a shaded vertical plane — Figure 2

A small aircraft is landing in a field.

In a model for the landing the aircraft travels in different straight lines before and after it lands, as shown in Figure 2.

The vector \(\mathbf{v}_{\mathbf{A}}\) is in the direction of travel of the aircraft as it approaches the field.

The vector \(\mathbf{v}_{\mathbf{L}}\) is in the direction of travel of the aircraft after it lands.

With respect to a fixed origin, the field is modelled as the plane with equation

\[x - 2y + 25z = 0\]

and

\[\mathbf{v}_{\mathbf{A}} = \begin{pmatrix}3\\ -2\\ -1\end{pmatrix}\]

(a) Write down a vector \(\mathbf{n}\) that is a normal vector to the field. (1)

(b) Show that \(\mathbf{n} \times \mathbf{v}_{\mathbf{A}} = \lambda\begin{pmatrix}13\\ 19\\ 1\end{pmatrix}\), where \(\lambda\) is a constant to be determined. (2)

When the aircraft lands it remains in contact with the field and travels in the direction \(\mathbf{v}_{\mathbf{L}}\)

The vector \(\mathbf{v}_{\mathbf{L}}\) is in the same plane as both \(\mathbf{v}_{\mathbf{A}}\) and \(\mathbf{n}\) as shown in Figure 2.

(d) State a limitation of the model. (1)

Mark scheme (a)

Scheme	Marks	AO
\(\mathbf{n} = \begin{pmatrix}1\\ -2\\ 25\end{pmatrix}\) or any non-zero scalar multiple thereof	B1	1.2
	(1)

Notes

B1: Correct normal vector (any non-zero scalar multiple thereof is fine)

Mark scheme (b)

Scheme	Marks	AO
\(\begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k}\\ 1 & -2 & 25\\ 3 & -2 & -1\end{vmatrix} = \begin{pmatrix}(-2)(-1) - (-2)(25)\\ -\left((1)(-1) - (3)(25)\right)\\ (1)(-2) - (3)(-2)\end{pmatrix} = \ldots\)	M1	1.1b
\(= \begin{pmatrix}52\\ 76\\ 4\end{pmatrix} = 4\begin{pmatrix}13\\ 19\\ 1\end{pmatrix}\) (or correct multiple for their normal vector used)	A1	2.1
	(2)

Notes

M1: Uses their \(\mathbf{n}\) and \(\mathbf{v}_{\mathbf{A}}\) in cross product formula. Allow slips in coordinates as long as the intent is clear.

A1: Correct work leading to a multiple of the required vector – may be a different multiple to the one shown if their \(\mathbf{n}\) was different.

Mark scheme (c)

Scheme	Marks	AO
Landing direction is perpendicular to \(\mathbf{n} \times \mathbf{v}_{\mathbf{A}}\) and \(\mathbf{n}\) so required direction is given by \(\begin{pmatrix}13\\ 19\\ 1\end{pmatrix} \times \begin{pmatrix}1\\ -2\\ 25\end{pmatrix} = \ldots\) Alternatively recognises the recognises the landing direction is the line of intersection of the plane containing \(\mathbf{n}\) and \(\mathbf{v}_{\mathbf{A}}\) and the plane representing the field. Finds the equation of the plane containing \(\mathbf{n}\) and \(\mathbf{v}_{\mathbf{A}}\) \(\begin{pmatrix}x\\ y\\ z\end{pmatrix}.\begin{pmatrix}13\\ 19\\ 1\end{pmatrix} = \begin{pmatrix}3\\ -2\\ -1\end{pmatrix}.\begin{pmatrix}13\\ 19\\ 1\end{pmatrix}\) \(x - 2y + 25z = 0\) and \(13x + 19y + z = 0\)	M1	3.1b
\(= \begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k}\\ 13 & 19 & 1\\ 1 & -2 & 25\end{vmatrix} = \begin{pmatrix}(19)(25) - (-2)(1)\\ -\left((13)(25) - (1)(1)\right)\\ (13)(-2) - (1)(19)\end{pmatrix} = \ldots\) Alternative Selects a value for either \(x\), \(y\) or \(z\) and solves simultaneously e.g \(z = -5\) leading to \(x - 2y = 125\) and \(13x + 19y = 5 \Rightarrow x = \ldots,\ y = \ldots\)	M1	3.4
\(= \begin{pmatrix}477\\ -324\\ -45\end{pmatrix}\) or any positive multiple thereof, e.g \(\begin{pmatrix}53\\ -36\\ -5\end{pmatrix}\) or \(\begin{pmatrix}1908\\ -1296\\ -180\end{pmatrix}\)	A1	1.1b
	(3)

Notes

M1: Uses a correct strategy to find the direction, ie realises \(\mathbf{v}_{\mathbf{L}}\) must be perpendicular to both the vector from (b) and \(\mathbf{n}\) (corrected from the printed mark scheme, which has \(\mathbf{v}_{\mathbf{A}}\) here). Allow for vectors used either way round.
Other methods may be possible – e.g finds the plane containing \(\mathbf{n}\) and \(\mathbf{v}_{\mathbf{A}}\) and solves with the plane representing the field.

M1: Uses their answer to (b) with the normal vector to find a vector in the direction required. Allow for vectors either way round.
Alternative find the line of intersection of the plane containing \(\mathbf{n}\) and \(\mathbf{v}_{\mathbf{A}}\) and the field.

A1: A correct direction vector, as shown or any positive multiple. For this mark the direction should be correct – so order of vectors must have been correct, or adapted to correct direction if initially incorrect.

(The equation of the plane containing \(\mathbf{n}\) and \(\mathbf{v}_{\mathbf{A}}\) is garbled in the printed mark scheme; it is shown here as \(\begin{pmatrix}x\\ y\\ z\end{pmatrix}.\begin{pmatrix}13\\ 19\\ 1\end{pmatrix} = \begin{pmatrix}3\\ -2\\ -1\end{pmatrix}.\begin{pmatrix}13\\ 19\\ 1\end{pmatrix}\), which gives the equation \(13x + 19y + z = 0\) that follows it.)

Mark scheme (d)

Scheme	Marks	AO
Any acceptable reason e.g Paths would not be linear May have some lateral movement Could be affected by cross winds Field might not be flat	B1	3.5b
	(1)
	(7 marks)

Notes

B1: Any correct limitation given. See scheme for examples.