Notes

Note: Condone the use of column vectors throughout this question.
There may be working on the diagram that can be awarded marks.

M1: Attempts either \(\left(\overrightarrow{AD} = \overrightarrow{AB} - \overrightarrow{DB} =\right)\ 2\mathbf{a} + 3\mathbf{b} - \left(-4\mathbf{a} + k\mathbf{b}\right)\) or \(\left(\overrightarrow{DA} =\right)\ -4\mathbf{a} + k\mathbf{b} - \left(2\mathbf{a} + 3\mathbf{b}\right)\) or \(\left(\overrightarrow{DB} = \overrightarrow{DA} + \overrightarrow{AB} =\right)\ \alpha\left(15\mathbf{a} - 5\mathbf{b}\right) + 2\mathbf{a} + 3\mathbf{b}\)
Allow subtraction either way round and may be implied by one correct component or by e.g. \(2\mathbf{a} + 3\mathbf{b} = \overrightarrow{AD} - 4\mathbf{a} + k\mathbf{b}\)
For reference: \(\overrightarrow{AD} = 6\mathbf{a} + (3 - k)\mathbf{b}\) and \(\overrightarrow{DA} = -6\mathbf{a} + (k - 3)\mathbf{b}\)
Allow e.g. \(\left(\overrightarrow{AD} =\right) \begin{pmatrix} 6\mathbf{a} \\ (3 - k)\mathbf{b} \end{pmatrix}\) or \(\left(\overrightarrow{AD} =\right) \begin{pmatrix} 6 \\ 3 - k \end{pmatrix}\) including without the brackets \(\begin{matrix} 6 \\ 3 - k \end{matrix}\)
Condone the use of gradients or ratios for \(\overrightarrow{AD}\) e.g. \(\dfrac{6}{3 - k}\) or “\(6 : 3 - k\)” to imply this mark (either way round).
There are alternatives using e.g. \(\overrightarrow{DC}\) but, in these cases, we require two expressions for the same vector, of which one expression must use \(\overrightarrow{DB}\).

dM1: A full method to solve the problem. Some possible approaches:

• attempts to find a scale factor and uses it to find \(k\)
• sets up equivalent fractions e.g. \(\dfrac{15}{6} = -\dfrac{5}{3 - k}\) or e.g. \(\dfrac{2 - -4}{3 - k} = \dfrac{15}{-5}\) and solves for \(k\)
• sets up equivalent ratios e.g. \(6 : 15 = 3 - k : -5\) and solves for \(k\)
• sets up simultaneous equations and solves for \(k\) e.g. \(2\mathbf{a} + 3\mathbf{b} + 4\mathbf{a} - k\mathbf{b} = \alpha\left(15\mathbf{a} - 5\mathbf{b}\right)\) o.e. or \(\left(\overrightarrow{DB} = \overrightarrow{DA} + \overrightarrow{AB} =\right)\ \alpha\left(-15\mathbf{a} + 5\mathbf{b}\right) + 2\mathbf{a} + 3\mathbf{b} = -4\mathbf{a} + k\mathbf{b}\) leading to \(6 = 15\alpha\) \(\left(\alpha = \dfrac{2}{5}\right)\) and \(3 - k = -5\alpha\) hence \(k = \ldots\)
  Note that their coefficient might be the reciprocal, from e.g. \(\beta\left(2\mathbf{a} + 3\mathbf{b} + 4\mathbf{a} - k\mathbf{b}\right) = 15\mathbf{a} - 5\mathbf{b}\) \((\beta = 2.5)\) and directions might be reversed in which case \(\alpha = -0.4\) or \(\beta = -2.5\) can be used.

A1*: Arrives at \(k = 5\) via a correct method. Usually this will be following:

• A correct expression for \(\overrightarrow{AD}\) (e.g. \(2\mathbf{a} + 3\mathbf{b} - \left(-4\mathbf{a} + k\mathbf{b}\right)\)) or \(\overrightarrow{DA}\) or \(\overrightarrow{BD}\) or \(\overrightarrow{DB}\) which may be mislabelled.
• Correct scale factor stated (\(\pm 2.5\) or \(\pm 0.4\)) or implied (e.g., by \(\dfrac{-5}{3 - k} = \dfrac{15}{6}\) or \(3(3 - k) = -6\))
• A correct intermediate equation that leads to \(k = 5\)
  An example minimal response might look like:
  e.g. \(\alpha\left(15\mathbf{a} - 5\mathbf{b}\right) = 6\mathbf{a} + (3 - k)\mathbf{b} \rightarrow 6 = 15\alpha \rightarrow \alpha = \dfrac{2}{5} \rightarrow -5\left(\dfrac{2}{5}\right) = 3 - k \rightarrow k = 5\)
  or e.g. \(6\mathbf{a} + (3 - k)\mathbf{b} \rightarrow \dfrac{15}{6} = -\dfrac{5}{3 - k} \rightarrow k = 5\)
  Condone missing/invisible brackets if recovered.

Alternative by verification

M1: Sets \(k = 5\), substitutes into \(\overrightarrow{DB}\) and attempts \(\left(\overrightarrow{AD} = \overrightarrow{AB} - \overrightarrow{DB} =\right)\ 2\mathbf{a} + 3\mathbf{b} - \left(-4\mathbf{a} + 5\mathbf{b}\right)\) o.e.

dM1: Attempts to compare \(\overrightarrow{AD}\) o.e. with \(\overrightarrow{BC}\) (usually \(6\mathbf{a} - 2\mathbf{b} = \alpha\left(15\mathbf{a} - 5\mathbf{b}\right)\)) and finds \(\alpha\)

A1*: Requires:

• Correct \(\overrightarrow{AD}\) o.e. e.g. \(\overrightarrow{DA}\)
• Correct scale factor \(\alpha = \pm\dfrac{2}{5}\) or \(\beta = \pm 2.5\) (sign dependent on their approach)
• Conclusion referencing the lines being parallel e.g. “hence \(\overrightarrow{AD}\) and \(\overrightarrow{BC}\) are parallel.”

Notes

Note: Condone the use of column vectors throughout this question.
There may be working on the diagram that can be awarded marks.

B1: Deduces a correct interpretation of the ratio \(BN : NC = 1 : 4\) that enables a start to a solution.
i.e., progresses to a correct statement that is at least as far as \(\left(\overrightarrow{BN} =\right)\ \dfrac{1}{5}\overrightarrow{BC}\) (or e.g. \(3\mathbf{a} - \mathbf{b}\)) or \(\left(\overrightarrow{CN} =\right)\ \dfrac{4}{5}\overrightarrow{CB}\) (or e.g. \(4\mathbf{b} - 12\mathbf{a}\)). May be embedded in e.g. \(\left(\overrightarrow{AN} =\right)\ 2\mathbf{a} + 3\mathbf{b} + \dfrac{1}{5}\overrightarrow{BC}\)
Allow e.g. \(\dfrac{1}{5}\begin{pmatrix} 15 \\ -5 \end{pmatrix}\) or \(\begin{pmatrix} 3 \\ -1 \end{pmatrix}\) or \(\begin{pmatrix} 3a \\ -b \end{pmatrix}\) for this mark.

M1: For the key step in attempting a valid expression for \(\overrightarrow{BX}\) (or \(\overrightarrow{AX}\) or \(\overrightarrow{DX}\) or \(\overrightarrow{CX}\) or \(\overrightarrow{NX}\)) in terms of \(\mathbf{a}\) and \(\mathbf{b}\). See diagram/notes below for helpful vectors.
Condone slips provided their intention is clear.
Note that they might be using \(\lambda\) and \(\mu\) the other way round or alternative variables.

Note: May be seen as a single expression such as \(\overrightarrow{AB} = \overrightarrow{AX} + \overrightarrow{XB}\) \(\left(\text{i.e. } \overrightarrow{AB} = p\overrightarrow{AN} + q\overrightarrow{DB}\right)\) or \(\overrightarrow{BN} = \overrightarrow{BX} + \overrightarrow{XN}\) \(\left(\text{i.e. } \overrightarrow{BN} = p\overrightarrow{BD} + q\overrightarrow{AN}\right)\) either of which scores M1dM1 simultaneously.

dM1: For the key step in attempting a second valid expression for their \(\overrightarrow{BX}\) (or \(\overrightarrow{AX}\) or \(\overrightarrow{DX}\) or \(\overrightarrow{CX}\) or \(\overrightarrow{NX}\)) in terms of \(\mathbf{a}\) and \(\mathbf{b}\) which enables the problem to be solved, i.e., it must not be parallel in approach to the first. See diagram/notes below for helpful vectors.
One expression should involve \(\text{``}\lambda\text{''}\left(4\mathbf{a} - 5\mathbf{b}\right)\) and the other should involve \(\text{``}\mu\text{''}\left(\text{``}5\mathbf{a} + 2\mathbf{b}\text{''}\right)\)
Condone slips provided their intention is clear.
They must use different parameters in their approaches, e.g., \(\lambda\) and \(\mu\).
If using e.g. \(\overrightarrow{DX} = -6\mathbf{a} + 2\mathbf{b} + \mu\left(5\mathbf{a} + 2\mathbf{b}\right)\) and \(\overrightarrow{XB} = \lambda\left(-4\mathbf{a} + 5\mathbf{b}\right)\) this mark is not scored until they set \(\overrightarrow{DX} + \overrightarrow{XB} = \overrightarrow{DB}\) as \(-6\mathbf{a} + 2\mathbf{b} + \mu\left(5\mathbf{a} + 2\mathbf{b}\right) + \lambda\left(-4\mathbf{a} + 5\mathbf{b}\right) = -4\mathbf{a} + 5\mathbf{b}\)
Dependent on the previous method mark.

ddM1: Compares coefficients of \(\mathbf{a}\) and \(\mathbf{b}\) to create simultaneous equations in their parameters and attempts to solve (which may be by calculator) leading to a value for one of their parameters.
Condone slips provided the intention is clear.
This mark may be implied by a correct value for e.g. \(\lambda\) following their two correct expressions for e.g. \(\overrightarrow{BX}\)
Dependent on both previous method marks.

A1: \(1 : 2\) o.e. Must follow a correct value for their parameter.
The correct ratio seen does not imply full marks. Candidates must show detailed reasoning.
Allow equivalent ratios e.g. \(\dfrac{1}{3} : \dfrac{2}{3}\) and ISW (e.g. \(1 : 3\)) but they must be the correct way round.
There may be attempts using similar triangles. Send to review.

Helpful Diagram:

Note: Some examples of valid expressions for the M and dM marks for part (b) are:
In each expression they may use different parameters and e.g. \(1 - \lambda\) might just be e.g. \(\phi\).

• \(\overrightarrow{BX} = \lambda\overrightarrow{BD} = \lambda\left(4\mathbf{a} - 5\mathbf{b}\right)\)
• \(\overrightarrow{BX} = \overrightarrow{BA} + \mu\overrightarrow{AN} = \left(-2\mathbf{a} - 3\mathbf{b}\right) + \mu\left(2\mathbf{a} + 3\mathbf{b} + \text{``}3\mathbf{a} - \mathbf{b}\text{''}\right)\)
• \(\overrightarrow{BX} = \overrightarrow{BN} + (1 - \mu)\overrightarrow{NA} = \left(\text{``}3\mathbf{a} - \mathbf{b}\text{''}\right) + (1 - \mu)\left(-2\mathbf{a} - 3\mathbf{b} + \text{``}{-}3\mathbf{a} + \mathbf{b}\text{''}\right)\)
• \(\overrightarrow{DX} = (1 - \lambda)\overrightarrow{DB} = (1 - \lambda)\left(-4\mathbf{a} + 5\mathbf{b}\right)\)
• \(\overrightarrow{DX} = \overrightarrow{DA} + \mu\overrightarrow{AN} = \left(\text{``}{-}6\mathbf{a} + 2\mathbf{b}\text{''}\right) + \mu\left(2\mathbf{a} + 3\mathbf{b} + \text{``}3\mathbf{a} - \mathbf{b}\text{''}\right)\)
• \(\overrightarrow{DX} = \overrightarrow{DN} + (1 - \mu)\overrightarrow{NA} = \left(\text{``}{-}\mathbf{a} + 4\mathbf{b}\text{''}\right) + (1 - \mu)\left(-2\mathbf{a} - 3\mathbf{b} + \text{``}{-}3\mathbf{a} + \mathbf{b}\text{''}\right)\)
• \(\overrightarrow{AX} = \mu\overrightarrow{AN} = \mu\left(2\mathbf{a} + 3\mathbf{b} + \text{``}3\mathbf{a} - \mathbf{b}\text{''}\right)\)
• \(\overrightarrow{AX} = \overrightarrow{AB} + \lambda\overrightarrow{BD} = \left(2\mathbf{a} + 3\mathbf{b}\right) + \lambda\left(4\mathbf{a} - 5\mathbf{b}\right)\)
• \(\overrightarrow{AX} = \overrightarrow{AD} + (1 - \lambda)\overrightarrow{DB} = \left(\text{``}6\mathbf{a} - 2\mathbf{b}\text{''}\right) + (1 - \lambda)\left(-4\mathbf{a} + 5\mathbf{b}\right)\)
• \(\overrightarrow{XN} = \mu\overrightarrow{AN} = \mu\left(2\mathbf{a} + 3\mathbf{b} + \text{``}3\mathbf{a} - \mathbf{b}\text{''}\right)\)
• \(\overrightarrow{XN} = \lambda\overrightarrow{DB} + \overrightarrow{BN} = \lambda\left(-4\mathbf{a} + 5\mathbf{b}\right) + \left(\text{``}3\mathbf{a} - \mathbf{b}\text{''}\right)\)
• \(\overrightarrow{XN} = (1 - \lambda)\overrightarrow{BD} + \overrightarrow{DN} = (1 - \lambda)\left(4\mathbf{a} - 5\mathbf{b}\right) + \left(\text{``}{-}\mathbf{a} + 4\mathbf{b}\text{''}\right)\)

or alternatives using \(C\) or \(N\) as starting points, but these are unlikely.

Notes

If part (a) is not attempted and the correct \(\overrightarrow{PR}\) is seen in part (b) then M1A1 can be awarded

M1: Attempts to add \(\overrightarrow{PQ} = 2\mathbf{i} + 8\mathbf{j} - 2\mathbf{k}\) and \(\overrightarrow{QR} = 6\mathbf{i} + 6\mathbf{k}\) with at least one correct component of \(8\mathbf{i} + 8\mathbf{j} + 4\mathbf{k}\). A typical misread of \(\overrightarrow{QR}\) as \(6\mathbf{i} + 6\mathbf{j}\) can score for at least one correct component of \(8\mathbf{i} + 14\mathbf{j} - 2\mathbf{k}\)

A1: Correct vector. Allow \(8\mathbf{i} + 8\mathbf{j} + 4\mathbf{k}\) or \(\begin{pmatrix}8\\8\\4\end{pmatrix}\) but not \(\begin{pmatrix}8\mathbf{i}\\8\mathbf{j}\\4\mathbf{k}\end{pmatrix}\) and not (8, 8, 4)

Condone \(\begin{matrix}8\\8\\4\end{matrix}\) for \(\begin{pmatrix}8\\8\\4\end{pmatrix}\) Do not apply isw here but award for e.g. \(8\mathbf{i} + 8\mathbf{j} + 4\mathbf{k} = \begin{pmatrix}8\mathbf{i}\\8\mathbf{j}\\4\mathbf{k}\end{pmatrix}\)

E.g. if they obtain \(\overrightarrow{PR} = 8\mathbf{i} + 8\mathbf{j} + 4\mathbf{k}\) and then say \(\overrightarrow{PR} = 2\mathbf{i} + 2\mathbf{j} + \mathbf{k}\) then award A0

Notes

Note that M1A0M1A1 is not possible. If they have an incorrect vector in part (a) then the maximum score is M1A1M1A0. A misread of \(\overrightarrow{QR}\) as \(6\mathbf{i} + 6\mathbf{j}\) in (b) can only score a maximum M1A1M1A0

They may attempt to show the triangle is either isosceles or right-angled in either order. You will need to look through their solution and award the order which scores most marks. Usually it will be isosceles first. Condone slips to be recovered.

To show the triangle is isosceles they only need to show two sides (or two angles) are the same. They do not need to consider the other side to show it is isosceles.

M1: Attempts to show that the triangle is either isosceles (or right-angled). See table below.

A1: Fully shows that the triangle is isosceles (or right-angled). See table below. Allow slips in their method if recovered as long as they proceed to correct lengths or values. A conclusion that the triangle is isosceles or right-angled is not required for this mark.

M1: Attempts to show that the triangle is both isosceles and right-angled. Usually they will have shown the triangle is isosceles and attempt to show that it is right-angled. Here are some examples but there will be others:

Alternatively, having already shown the triangle is right-angled they may:

• use trigonometry to show that the two base angles are both \(45^\circ\)
• attempt to find the required lengths or lengths\(^2\) of \(PQ\) and \(PR\) if not already found
  (Note they may also find the length or length\(^2\) of \(QR\) and may use the sine rule or cosine rule)

Note if they have shown either property then we will condone making an assumption of the other property to justify the size of angle \(PRQ\) and/or angle \(QPR\) and may use trigonometry to show either

Using trigonometry (SOHCAHTOA) to show the second property they will e.g.

• Use triangle \(PQR\) and assume right angled
• Split triangle \(PQR\) into two right angled congruent triangles using the isosceles property

If they use triangle PQR then it requires

• For M1: Calculations showing the required property to score M1
• For A1: To either draw a labelled right angled triangle or state the assumption that it is right angled and conclude

If they split the triangle PQR in half then they have formed two right angled congruent triangles which (provided they had two lengths the same already) will not be an assumption of a right angle. Then it requires

• For M1: Calculations to find either at least one of the base angles of \(45^\circ\) OR to find the right angle. It should be clear whether they
  • found a base angle of PQR and doubled it or
  • found half of the right angle and doubled it.
• For A1: Fully correct calculations and conclusion
  Note if they find a 45 degree angle and double it then it needs to be clear whether this is the right angle or if it is the sum of two base angles because there are four angles of size 45 degrees via this method.
  They would have to e.g. mention about the angle sum of a triangle oe or show a clearly labelled diagram and calculations with labels that match the diagram.

Note that use of tan is unlikely to score because if they just use two equal lengths for the two shorter sides of their right angled triangle, then \(\tan(A) = 1\), so the angles will always be \(45^\circ\), \(45^\circ\) and \(90^\circ\) – it is the use of the length of 12 (or 6) which is going to lead to showing the triangle is right angled via these routes.

A1: Fully shows that the triangle is isosceles and right-angled and concludes that the triangle is both isosceles and right-angled. These conclusions may appear at separate stages of their solution. Condone poor notation and/or labelling of lengths provided the intention is clear. Condone slips if recovered. If they have a preamble then they must have a minimal conclusion e.g. proven, tick, QED

Note if they have an incorrect vector in part (a) then the maximum score is M1A1M1A0

Notes

B1: Correct vector. Allow \(3\mathbf{i} + 9\mathbf{j} + 3\mathbf{k}\) or \(\begin{pmatrix}3\\9\\3\end{pmatrix}\) but not \(\begin{pmatrix}3\mathbf{i}\\9\mathbf{j}\\3\mathbf{k}\end{pmatrix}\) and not \((3, 9, 3)\)
Condone \(\begin{matrix}3\\9\\3\end{matrix}\) for \(\begin{pmatrix}3\\9\\3\end{pmatrix}\)
Do not apply isw here but award for e.g. \(3\mathbf{i} + 9\mathbf{j} + 3\mathbf{k} = \begin{pmatrix}3\mathbf{i}\\9\mathbf{j}\\3\mathbf{k}\end{pmatrix}\)
E.g. if they obtain \(\overrightarrow{AB} = 3\mathbf{i} + 9\mathbf{j} + 3\mathbf{k}\) and then say \(\overrightarrow{AB} = \mathbf{i} + 3\mathbf{j} + \mathbf{k}\) then award B0

If part (a) is not attempted and the correct \(\overrightarrow{AB}\) is seen in part (b) the B1 can be awarded there.

Notes

General Guidance for part (b):
As with most vector questions we will see a variety of approaches (correct and incorrect).
In general, the marks are awarded as follows:

• M1 for a correct complete strategy to find at least one position for \(P\) (May be implied by at least 2 correct components)
• A1 for one correct position for \(P\)
• dM1 for a correct complete strategy to find both positions for \(P\) (May be implied by at least 2 correct components for both positions)
• A1 both correct positions for \(P\) and no others

Various examples are shown below.
Other methods will be seen but the above marking principles should be applied.
You can condone slips in their algebra/processing as long as the intention is clear.
The examples given below give the detail to look for depending on the approach.
If you see a response and you are not sure if it deserves credit use Review.

Note that adding vectors when they should be subtracting will generally score M0 but use review if necessary.

Note that sight of at least one correct position for \(P\) implies M1A1

M1: Attempts at least one correct strategy for finding \(P\)

A1: One correct position vector or allow coordinates for this mark e.g. (8, 15, 11) or (4, 3, 7) or \(x = \ldots, y = \ldots, z = \ldots\)
If given as a vector, allow e.g. \(8\mathbf{i} + 15\mathbf{j} + 11\mathbf{k}, \begin{pmatrix}8\\15\\11\end{pmatrix}\) but not \(\begin{pmatrix}8\mathbf{i}\\15\mathbf{j}\\11\mathbf{k}\end{pmatrix}\)

dM1: Attempts two correct strategies for finding \(P\)

A1: Both correct position vectors
Must both be vectors so e.g. \(8\mathbf{i} + 15\mathbf{j} + 11\mathbf{k}, \begin{pmatrix}8\\15\\11\end{pmatrix}\) and \(4\mathbf{i} + 3\mathbf{j} + 7\mathbf{k}, \begin{pmatrix}4\\3\\7\end{pmatrix}\) but not e.g. \(\begin{pmatrix}8\mathbf{i}\\15\mathbf{j}\\11\mathbf{k}\end{pmatrix}\)
Condone e.g. \(\begin{matrix}8\\15\\11\end{matrix}\) for \(\begin{pmatrix}8\\15\\11\end{pmatrix}\)

Alternative 1 using vector equation of \(l\):

M1: Forms the vector equation of line \(l\) using their \(\overrightarrow{AB}\) from part (a) or by starting again, forms the vectors \(\overrightarrow{AP}\) and \(\overrightarrow{BP}\) then uses \(|\overrightarrow{AP}| = 2|\overrightarrow{BP}|\) and Pythagoras to produce a quadratic equation in "\(\lambda\)" which they then solve to find "\(\lambda\)" and use correctly to find at least one position for \(P\).
Note if they use \(\mathbf{i} + 3\mathbf{j} + \mathbf{k}\) for the direction, they should get \(\lambda = 6, 2\)
If all other work is correct, condone not squaring the “2” when applying Pythagoras

A1: See main scheme

dM1: As the first M and finds both positions for \(P\)

A1: See main scheme

Alternative 2 using \(P\) as \(\begin{pmatrix}x\\y\\z\end{pmatrix}\) and that \(\overrightarrow{AP}\) and \(\overrightarrow{BP}\) are parallel:

M1: Sets \(P\) as a general point, forms \(\overrightarrow{AP}\) and \(\overrightarrow{BP}\) (either way round) then uses \(|\overrightarrow{AP}| = 2|\overrightarrow{BP}|\) then squares components and equates to produce quadratic equations in \(x\) and \(y\) and \(z\) which they then solve to find at least one position for \(P\). It is not just for finding values which are not then used to form a point (or vector).
If all other work is correct, condone not squaring the “2” when squaring.

A1: See main scheme

dM1: As the first M and finds both positions for \(P\).

A1: See main scheme

Note that if the modulus is not used, this method can lead to one correct position for \(P\) e.g.

But it is possible to find the other position without squaring e.g.

This requires at least 2 correct equations for \(x\), \(y\) or \(z\) for the dM1

e.g. Alternative 3 using \(P\) as \(\begin{pmatrix}x\\y\\z\end{pmatrix}\) and eliminating 2 of the variables:

Notes

M1: Attempt to find magnitude (or magnitude \({}^2\)) of given vector. Pythagoras or trigonometry

A1: Make comparison using definition of a unit vector, and conclude that it is not a unit vector. Any values seen must be correct

Accept \(0.5 \neq 1\) as evidence
Geometric arguments can gain full credit, as long as fully reasoned and convincing eg to have a length of 1 unit the two components would have to be collinear, but this is not possible as they have to be perpendicular

Notes

M1: Attempt to find a useful angle soi. Could be using their magnitude from (a) with sin or cos

Implied by \(-45^\circ\)
BOD for any incorrect diagrams seen
Allow 0.785 (radians)

A1: Obtain correct direction

A0 for \(-45^\circ\)
Condone both \(315^\circ\) and \(-45^\circ\) given
A0 if in radians

Notes

M1: Attempt distance equation involving \(c\), and equate to \(5^2\) oe. Allow one sign error eg \(3^2 + (c - 0.5)^2 = 5^2\)

If spotting that the \(y\) difference is 4 then must still link to \(c\) for M1

A1: Correct quadratic equation. Brackets expanded, but not necessarily simplified

A1: Obtain both correct values

SC if no method shown then B1 for each correct value of \(c\), as ‘determine’

Alt method for final two marks

A1: Solve to obtain one correct value for \(c\)

A1: Obtain second correct value, and no others

Notes

M1*: Attempt at Pythagoras =36 oe, may see \(\sqrt{a^2 + b^2} = 6\)

M1*: oe or \(\dfrac{b}{a} = \tan 60\) or \(\dfrac{b}{a} = 1.73\), condone \(\dfrac{a}{b}\)
May see equivalent statements in sin or cos.

M1 dep*: Substitute both expressions (dependent on both previous M marks) – must reach an equation in \(a\) or \(b\) only.

A1: A0 for negative answers (if not disregarded)
Accept \(b = \sqrt{27}\)
If no (or insufficient) working then SC B1B1 (max 2/4) for each correct answer (must be exact).

Alternative method

M1: Allow this mark for 6cos or 6sin of \(30, 60, 120^\circ\)
Must see this step oe.

M1: Allow this mark for 6cos or 6sin of \(30, 60, 120^\circ\) provided it is consistent with their other expression (i.e. not the same).
Must see this step oe.

If no (or insufficient) working then SC B1B1 (max 2/4) for each correct answer (must be exact).

Notes

B1: Correct length aef
Condone 4.47 or better
Allow isw eg \(\sqrt{20} = 4\sqrt{5}\)
Allow BOD on signs eg \(AB = -2\mathbf{i} - 4\mathbf{j}\) seen

Notes

M1: Attempt correct equation for length \(BC\)
Using their attempt at length of \(AB\)
Condone error on RHS eg having \(\sqrt{20}\) not 20

A1: BC
Solve correct quadratic to obtain at least \(p = 7\)
If second value of \(p\) stated then it must be correct

A1: Correct position vector for \(C\); it could be given as column vector, but not coordinate
No need to discard \(p = 1\)

If M0, question is ‘determine’ so some evidence needed for full marks – either justifying lengths are equal, or use of components of 2 and 4

• \(7\mathbf{i} + 7\mathbf{j}\) with some explanation B3
• \(7\mathbf{i} + 7\mathbf{j}\) with no explanation B2
• (7, 7) with some explanation B2
• (7, 7) with no explanation B1

Notes

B1: Correct midpoint soi
Could instead find vector \(BM\)
Allow \(M\) seen as coordinate, as it is part of their method and not a requested answer
Condone \(M = 4\mathbf{i} + 4\mathbf{j}\), but penalise clear error eg \(AM = 4\mathbf{i} + 4\mathbf{j}\) is B0
Could be soi on a diagram

B1: Correct position vector (not coordinate) for \(D\)
Do not penalise \(D\) given as coordinate if already penalised in part (b)

Answer only is B0B1

Notes

B1*: Mark independently of reason

B1dep*: Evidence is required to support statements made
All relevant evidence quoted must be correct

• Two pairs of adjacent sides: \(AD = CD = \sqrt{26}\) (or compare components of vectors); condone not stating \(AB = BC\) as given in question
  Sides must be defined as adjacent, so B0 for just ‘two pairs of equal sides’, but allow BOD if clarified on an explicit diagram seen in part (d)
• Diagonals perpendicular: \(AC\) has gradient of 1, \(BD\) has gradient of \(-1\)
  If using a geometrical argument, then identify that \(ABC\) is isosceles, \(M\) is mid-point of \(AC\) hence perpendicular bisector
• \(BD\) a line of symmetry: \(AM = MC\), with perpendicular argument as above

B0 for reasoning using angles (ie a pair of facing equal angles) unless justified.

Notes

B1: Allow \(10k\)

Notes

B1: May be implied (but must be fully correct to imply, i.e. including (\(-2\)))

M1: Attempt \(OP^2 + BP^2 = OB^2\) FT their \(OP\) and \(\overrightarrow{BP}\) or \(\overrightarrow{PB}\)

A1: Correct equation after expanding brackets

A1: oe Condone inclusion of \(k = 0\) (whether eliminated or not)

Alternative method using scalar product

B1: May be implied

M1: Attempt \(\overrightarrow{OP} \cdot \overrightarrow{BP} = 0\) FT their \(\overrightarrow{OP}\) and \(\overrightarrow{BP}\) or \(\overrightarrow{PB}\)
(or attempt \(\overrightarrow{OA} \cdot \overrightarrow{BP} = 0 \Rightarrow 6(6k - 1) + 0 + 8(8k - 3) = 0\))
Must be algebraic – i.e. an equation of this form in \(k\).

A1: Correct equation after expanding brackets
Or \(100k - 30 = 0\)

A1: oe Condone inclusion of \(k = 0\) (whether eliminated or not)

Notes

M1: \(\mathbf{b} - \mathbf{a}\) or \(\mathbf{a} - \mathbf{b}\) attempted, using \(\mathbf{i}\), \(\mathbf{j}\), \(\mathbf{k}\) or column vectors
May be implied by calculation seen

A1: www. Correct answer, no working: M1A1
Mark(s) cannot be gained retrospectively in (b)

Notes

M1: Attempt \(AB^2 = BP^2 + PA^2\) (involving \(p\)) ft their \(AB\)

A1f: Correct simplified equation, ft their (a), ie:
or \(p^2 - 7p + \dfrac{46 - \text{their } a^2}{2} = 0\) or \(\left(p - \frac{7}{2}\right)^2 = \dfrac{\text{their } a^2 - 17}{4}\)

A1f: ft only their (a)

Alternative methods for M1

Notes

B1: Can be implied

M1: May be included in working, eg \(\overrightarrow{AX} = \frac{2}{3}\left(\frac{1}{2}(\mathbf{b} + \mathbf{c}) - \mathbf{a}\right)\)
Not necessarily correct

M1: Not necessarily correct

A1: or equivalent simplified form

Examples of methods using the above   Other correct methods may be seen   Allow inadequate notation

M1: for \(\overrightarrow{AM} = \left(\frac{1}{2}(\mathbf{b} + \mathbf{c}) - \mathbf{a}\right)\) implied

B1: Implied

M1: for \(\overrightarrow{AM} = \left(\frac{1}{2}(\mathbf{b} + \mathbf{c}) - \mathbf{a}\right)\) implied
\(\overrightarrow{OX} = \overrightarrow{OM} - \frac{1}{3}\overrightarrow{AM}\) is equivalent to \(\overrightarrow{OM} + \frac{1}{3}\overrightarrow{MA}\)

Notes

B1dep: Both statements needed. NB dep on B1

Notes

B1: Use of Pythagoras implied by \(17 + x^2\)

B1: Understanding that \(17 + x^2\) is a square number soi e.g. by equating to at least one square number \(17 + x^2 = 25\)
Allow e.g. \(n = \sqrt{17 + x^2}\) and \(n\) integer soi e.g. by equating to an integer

M1: Rearranging to get their 17 on one side and factorising

A1: OR \(n - x = 17\) and \(n + x = 1\) AND \(n - x = -17\) and \(n + x = -1\)
Implies M1. If the final B2 is earned following only one pair of correct equations then award A1

B2: Both values and no others
B1 if only one value given (ignore extras)

Alternative method 1 for M1 A1

M1: At least two different square numbers added to their 17 e.g. two from: \(17 + 1\), \(17 + 4\), \(17 + 9\), \(17 + 16\), \(17 + 25\), \(17 + 49\), \(17 + 64\), \(17 + 81\) and so on
Or tested two using \(17 + x^2 = 25\), \(17 + x^2 = 36\), \(17 + x^2 = 49\), \(17 + x^2 = 64\), \(17 + x^2 = 81\) and so on

A1: e.g. Since 64 and 17 + 64 are adjacent square numbers there will be no larger values of \(x^2\)

Alternative method 2 for M1 A1

M1: Uses adjacent square numbers
Implied by \(2x + 1 = 17\)
Condone not testing earlier squares for this M1

A1: If the final B2 is earned following M1 then award A1

Notes

B1: do not allow \((3, 3, -3)\); mark the final answer

Notes

M1: FT their \(\begin{pmatrix}3\\3\\-3\end{pmatrix}\)

A1: allow B2 for correct answer unsupported; mark the final answer
if M0 allow SC1 for \(\left[\begin{pmatrix}-1\\-2\\0\end{pmatrix} + \frac{2}{3}\begin{pmatrix}3\\3\\-3\end{pmatrix} =\right]\begin{pmatrix}1\\0\\-2\end{pmatrix}\)
do not allow \((0, -1, -1)\);
only penalise coordinate form once

Alternative method 1

may work with i, j, k notation

M1: allow one sign error only

A1: do not allow \((0, -1, -1)\);
only penalise coordinate form once

Alternative method 2

M1: may see \(3\overrightarrow{OC} = \begin{pmatrix}0\\-3\\-3\end{pmatrix}\); allow one sign error only;
may work with i, j, k notation

A1: do not allow \((0, -1, -1)\);
only penalise coordinate form once

Alternative method 3

M1: ratio theorem; may work with i, j, k notation

A1: do not allow \((0, -1, -1)\);
only penalise coordinate form once

Notes

M1: Attempt to scalar multiply \(\mathbf{v}_1\) and add to \(\mathbf{v}_2\)
Allow vector expression or 2 separate components

M1: Equate coefficients of \(\mathbf{i}\) and \(\mathbf{j}\) to form two equations. Allow if \(\mathbf{i}\) and \(\mathbf{j}\) still seen in every term of these equations

A1: cao

A1: cao

Notes

(i) B1: Any midpoint correct

(i) B1: All midpoints correct
If no labels BOD for 1 or 2 marks

(ii) B1: Gradient, length or vector of any one side of PQRS
Or SC1 if KLMN used

(ii) M1: Gradient of opposite side of PQRS shown to be equal
or length or vector of opposite side shown to be equal.
ie some working must be seen

(ii) E1: Convincing completion dep on M1

Condone confusion of labels (eg length for gradient etc) for M1/B1

Watch out for valid alternatives e.g. 2 sides equal length and equal gradient

Notes

(i) B1: Convincing completion
\(\overrightarrow{\mathrm{WX}} = \overrightarrow{\mathrm{WV}} + \overrightarrow{\mathrm{VT}} + \overrightarrow{\mathrm{TX}}\)

(ii) E1: \(\overrightarrow{\mathrm{DE}}\) from any correct route, must be shown
Convincing completion with conclusion

(iii) B1: \(\overrightarrow{\mathrm{GF}}\) from any correct route, must be shown

(iii) E1: Convincing completion with conclusion

Notes

B1: may be in coordinate form or may see distances identified or on diagram;
may be implied by \(\sqrt{(\pm 2)^2 + (\pm 2)^2}\) oe

M1: or FT their evaluation of \(\begin{pmatrix} 5 \\ -3 \end{pmatrix} - \begin{pmatrix} 3 \\ -1 \end{pmatrix}\)
may be implied by correct answer

A1: if B0M0; allow SC1 for \(\sqrt{80}\) or \(4\sqrt{5}\) (from addition of vectors) if supported by Pythagoras;
if B0M0 allow SC1 for \(\sqrt{8}\) or \(2\sqrt{2}\) unsupported

Notes

M1: may be implied if one component fully correct

M1: dependent on award of first M1

Notes

M1: or \(\cos^{-1}\left(\frac{7}{25}\right)\) or \(\sin^{-1}\left(\frac{24}{25}\right)\) FT their 7, 24 or 25

Notes

M1: Eg \(k\mathbf{i} + k\mathbf{j}\)

M1: Could be on diagram

A1: Must be exact eg \(\cos 45\,\mathbf{i} + \sin 45\,\mathbf{j}\)

Notes

B1: AG
Soi Allow for \(\overrightarrow{BA}\) even if wrongly labelled

M1: Attempt to find magnitude of their displacement vector

A1: cao AG
Do not allow final mark when there are missing brackets eg \(-4^2 + 2^2 + 9^2\)