Notes

M1: Finds the dot product of the normal vectors, sets equal to 0 and finds a value for \(a\). Condone a sign slip.

A1: Correct value for \(a\)

Notes

M1: Finds the determinant of a 3 by 3 matrix and sets = 0, this may be implied

A1: Correct determinant

A1: Correct value for \(b\)

dM1: Uses the equations of the three planes to find two equations by eliminating the same variable. Condone a slip on one coefficient or constant term

ddM1: Correctly uses their equations to form and solve an equation for \(c\)

A1: Correct value for \(c\)

Alternative 1

M1: Uses the planes to eliminate one variable to form two equations

A1: One correct equation

A1: two correct equations

dM1: Sets up an equation for \(b\) or \(c\) by comparing coefficients and solves to find a value. Condone a slip on one coefficient or constant term

ddM1: Solves equations to find values for \(b\) and \(c\)

A1: Correct value for \(b\) and \(c\) cso

Note for 10 (b) using Alternative 1

Alternative 2

M1: Uses the planes to eliminate one variable to form two equations or substitutes in two values for one coordinate and solves to find two coordinates that lie on the line of intersection

A1: One correct equation or coordinate

A1: Two correct equations or coordinates

dM1: \(\begin{pmatrix}8\\ -7\\ 1\end{pmatrix}\) is perpendicular to \(\begin{pmatrix}1\\ b\\ 13\end{pmatrix}\) leading to \(8 - 7b + 13 = 0 \Rightarrow b = \ldots\) or uses their coordinates to find the direction vector and uses that it is perpendicular to \(\begin{pmatrix}1\\ b\\ 13\end{pmatrix}\) to find a value for \(b\)

ddM1: \((5, -2, 0)\) lies on \(\Pi_3\) leading to \(5 - 2b = c\)

A1: Correct values for \(b\) and \(c\)

Notes

M1: Uses the shortest distance formula \(\dfrac{(1 \times 6) + (-2 \times -2) + (8 \times -4) \pm 1}{\sqrt{1^2 + (-2)^2 + 8^2}}\) condoning a sign slip with \(d\) and no modulus for this mark.
May find the answer to (c) first and then find the distance. It must be a correct method, scoring at least M1M1 in (c)

A1: Correct exact answer or awrt 2.77 isw

Notes

B1: See scheme

Note: the position of the water may change is B0

Notes

(c) This may be seen in part (a) to score any marks at least the final coordinate must be stated in (c) or ‘see (a)’

M1: A complete method to find the value of \(\lambda\)

dM1: Dependent on the previous method mark. Uses their value of \(\lambda\) in the equation of the line \(\mathbf{r} = \begin{pmatrix}6\\ -2\\ -4\end{pmatrix} + \lambda\begin{pmatrix}1\\ -2\\ 8\end{pmatrix} = \ldots\) to find a coordinate

A1: Correct coordinate

Notes

B1: See scheme, must have an answer to (a) so score this mark

Notes

M1: Finds the determinant of the matrix, sets \(= 0\) and solves to find a value of \(p\). Allow for any recognisable attempt at finding the determinant (may be slips in coefficients or signs). Accept if det \(= 14 - 7p\) appears with no working shown.

A1: \(p = 2\) from correct work.

Alt I Using normal

M1: Attempts the cross product between sets of pairs of planes and scales (if appropriate) and equates directions to solve for \(p\).

A1: Correct \(p\)

FYI
\(\begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k}\\ 2 & -1 & 1\\ 1 & p & -3\end{vmatrix} = (3 - p)\mathbf{i} + 7\mathbf{j} + (2p + 1)\mathbf{k},\quad \begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k}\\ 2 & -1 & 1\\ 3 & 1 & -2\end{vmatrix} = \mathbf{i} + 7\mathbf{j} + 5\mathbf{k},\quad \begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k}\\ 1 & p & -3\\ 3 & 1 & -2\end{vmatrix} = (3 - 2p)\mathbf{i} - 7\mathbf{j} + (1 - 3p)\mathbf{k}\)

Notes

M1: A complete method to find the value of \(q\). May be implied by correct value if no incorrect working is shown.
E.g. Adds together equations 1 and 2 and compares with equation 3 to find a value for \(q\).
Alternatively: Uses equations \(2x - y + z = 3\) and \(3x + y - 2z = 4\) to eliminate one variable. Uses equation \(x + 2y - 3z = q\) and one of the other equations to eliminate the same variable and compare to find a value for \(q\). Another possible method is to identify a point on both planes 1 and 3 (e.g. let \(x = 1\) and solve for \(y\) and \(z\) to get \((1,\ -3,\ -2)\) then substitute into middle equation to find \(q\))

A1: \(q = 1\)

Alternatively II the question may be done as a whole. Eg.
\(\left.\begin{aligned}2x - y + z &= 3\\ 3x + y - 2z &= 4\end{aligned}\right\} \Rightarrow x + 2y - 3z = 1 \Rightarrow p = 2,\ q = 1\) is the most direct method. Variations are possible, e.g.
\(\left.\begin{aligned}2x - y + z &= 3\\ x + py - 3z &= q\\ 3x + y - 2z &= 4\end{aligned}\right\} \Rightarrow \begin{aligned}&3(1) + (2): 7x + (p - 3)y = 9 + q\\ &2(1) + (3): \quad\ 7x - y = 10\end{aligned} \Rightarrow \begin{aligned}&p - 3 = -1 \Rightarrow p = \ldots\\ &9 + q = 10 \Rightarrow q = \ldots\end{aligned}\)

Score as follows:

M1: Attempts to solve at least two of the equations simultaneously to eliminate one variable or match coefficients of the third equation and uses the equation to identify one of the unknowns. Condone slips as long as the method is clear.

A1: Correct \(p\) or correct \(q\).

M1: Full method to use the linear dependence of the equations to find both variables.

A1: Correct \(p\) and \(q\).

Alt III: Using points on the common line.

M1: Finds two separate points on each of the first and third plane.

A1: Two correct points.

M1: Forms and solves two equations in \(p\) and \(q\) using their points.

A1: Correct \(p\) and \(q\).

E.g.
\(x = 0 \Rightarrow \begin{cases}-y + z = 3\\ y - 2z = 4\end{cases} \Rightarrow y = -10,\ z = -7\)
\(x = 1 \Rightarrow \begin{cases}-y + z = 1\\ y - 2z = 1\end{cases} \Rightarrow y = -3,\ z = -2\)
\(\Rightarrow \begin{aligned}-10p + 21 &= q\\ 1 - 3p + 6 &= q\end{aligned} \Rightarrow p = \ldots,\ q = \ldots\)

Some useful equations:

Eliminating \(x\): \(\begin{cases}(2p + 1)y - 7z = 2q - 3\\ 5y - 7z = -1\\ (3p - 1)y - 7z = 3q - 4\end{cases}\)

Eliminating \(y\): \(\begin{cases}(2p + 1)x + (p - 3)z = 3p + q\\ 5x - z = 7\\ (3p - 1)x + (3 - 2p)z = 4p - q\end{cases}\)

Eliminating \(z\): \(\begin{cases}7x + (p - 3)y = 9 + q\\ 7x - y = 10\\ 7x + (3 - 2p)y = 12 - 2q\end{cases}\)

Notes

M1: Correct method to find the shortest distance. Condone \(\pm d\) in the formula. Longer methods are possible. E.g. if done after part (d), then \(\sqrt{19}\cos 17.97\ldots\) is a correct method. Another approach is to form the general line \(\mathbf{r} = \begin{pmatrix}3\\ -5\\ 2\end{pmatrix} + \mu\begin{pmatrix}2\\ 4\\ -1\end{pmatrix}\), substitute into plane equation \(\Rightarrow \mu = \dfrac{19}{21}\) then scale the normal vectors magnitude: \(\dfrac{19}{21} \times \sqrt{2^2 + 4^2 + (-1)^2} = \ldots\)

A1: Correct exact distance or awrt 4.15

Notes

M1: Finds the direction of the line. Accept either direction and implied by 2 out of three coordinates correct if no method shown.

A1: Correct model for the flight path of the arrow. Must be an equation ie, including the \(\mathbf{r} = \ldots\) (but condone \(l = \ldots\) or other letter).

Notes

M1: A complete method to find the required angle or angle between normal and plane. Uses the dot product of the direction vector of the line and normal vector of the plane to find the angle (need not subtract from 90 for this mark). Allow if a (single) minor slip is made if the method is clearly attempting to use the correct vectors.
Note may use \(\sin\phi = \dfrac{\begin{pmatrix}3\\ 3\\ -1\end{pmatrix}.\begin{pmatrix}2\\ 4\\ -1\end{pmatrix}}{\sqrt{3^2 + 3^2 + (-1)^2} \times \sqrt{2^2 + 4^2 + (-1)^2}} \Rightarrow \phi = \ldots\) directly.

A1: Correct final angle.

Notes

B1: \(\sqrt{19}\) or awrt 4.36

Notes

B1: States not likely to match (or equivalent wording) and gives a suitable reason why. This will most commonly be that the arrow will not travel in a straight line, but e.g. accept “the arrow is not a particle so the tip may travel into the target, not stop at the plane”.
Do not accept answers that suggest the arrow will travel less far than the answer to (d). Do not accept “air resistance” arguments unless they specifically refer to the arrow not travelling in a straight line.

Notes

If no attempt at part (a) has been made you may award marks for work seen in part (b) for finding the inverse.

B1(M1 on epen): Correct determinant of \(2k\)

M1: Starts the process of finding the inverse and obtains at least 6 correct elements of cofactors
Alternatively transposes their matrix of minors and obtains at least 6 correct elements.

dM1: A complete recognisable method to find the inverse including dividing by the determinant
Allow minor slips if the process is clearly correct.

A1: Correct inverse.

Notes

M1: Attempts \(\mathbf{A}^{-1}\begin{pmatrix}3\\ 1\\ 6\end{pmatrix}\) with their \(\mathbf{A}^{-1}\) which must be in terms of \(k\), to obtain at least one of \(x =\), \(y =\) or \(z =\) which may be seen embedded in a column vector, simplified or unsimplified, the determinant may be outside their column vector.
Condone a slip in copying \(\begin{pmatrix}3\\ 1\\ 6\end{pmatrix}\) e.g. \(\begin{pmatrix}1\\ 3\\ 6\end{pmatrix}\) or \(\begin{pmatrix}3\\ 1\\ 2\end{pmatrix}\)

A1(M1 on epen): Two correct expressions for \(x\), \(y\) or \(z\) simplified or unsimplified, which may appear in a column vector, determinant cannot be outside the vector.

A1: Correct coordinates in simplest form. Allow e.g. \(y = 2 - \dfrac{12}{k}\)
Their final answer must be written as coordinates and not as a column vector but can be written as \(x =\), \(y =\) and \(z =\)

Alternative (using algebraic method for simultaneous equations):

M1: Solves simultaneously to obtain at least one of \(x =\), \(y =\) or \(z =\) must be in terms of \(k\).

A1(M1 on epen): Two correct expressions for \(x\), \(y\) or \(z\) simplified or unsimplified.

A1: Correct coordinates in simplest form. Allow e.g. \(y = 2 - \dfrac{12}{k}\)
Their final answer must be written as coordinates and not as a column vector but can be written as \(x =\), \(y =\) and \(z =\)

e.g.
Eliminates \(z\) and achieves \(4x + 2y = 4\) and \((6 - k)x + 3y = 0\)
Uses \(12x + 6y = 12\) and \((12 - 2k)x + 6y = 0\) to produce \(2kx = 12 \Rightarrow x = \dfrac{6}{k}\)

Notes

M1: Attempts the scalar product between the given vector and the 2 direction vectors (calculation should be shown for at least one).

A1: Obtains zero for both with sufficient working shown and concludes perpendicular (each one or states both are). Accept normal to as equivalent to perpendicular to.

Alt (a)

M1: Attempts the vector product between the direction vectors of the lines, allowing at most one slip in the brackets.

A1: Obtains the correct vectors and shows it is a multiple of the required answer and concludes perpendicular.

Alt II (a)

M1: Sets up a normal vector to the plane in terms of variables and attempts the scalar product between the direction vectors of the lines and the normal to form simultaneous equations and attempts to solve by choosing one of the values and solving for the other two. (corrected from the printed mark scheme: printed as “vector product”)

A1: Correct values, and obtains the correct vector, or a multiple of it and shows it is a multiple of the required answer and concludes perpendicular.

Notes

M1: Attempts the scalar product between the normal vector and \(\mathbf{i} - 2\mathbf{j} + 3\mathbf{k}\). Allow if the vector product is used again from scratch. Allow equivalent approaches such as using the normal to form the equation \(3x - 10y - z = d\) and substituting \((1,\ -2,\ 3)\) to find \(d\). Maybe be implied by a correct equation if no incorrect working is seen.

A1: Correct equation in Cartesian form (allow equivalent Cartesian equations).

Notes

M1: Attempts the scalar product between the normal vector and \(5\mathbf{i} + p\mathbf{j} - 7\mathbf{k}\) (or any point on the line), sets \(=\) their 20 and solves a linear equation in \(p\). Alternatively substitutes \(x = 5\), \(y = p\) and \(z = -7\) into their Cartesian equation and solves for \(p\), or any other complete method to find \(p\).

A1: For \(p = 0.2\) (oe)

Notes

M1: A complete method to find the coordinates of \(A\). Allow if a slip in transcribing a coordinate occurs – it is a mark for the method. If the correct vectors are clearly intended, give credit.

M1: Attempts the vector \(AB\) and attempts the scalar product with this and the normal vector. Note that if the correct formulae are seen then allow the method for any value appearing afterwards.

M1: Complete method to find the required angle or its complement.

A1: For awrt \(40^\circ\)

Note: Accept alternative vector notations throughout.
Where method is not shown for finding coordinates, it may be implied by two correct coordinates.

Alt (d)

(corrected from the printed mark scheme: the \(\mathbf{i}\) component of \(\overrightarrow{OX}\) is printed as \(12 - 3m\), in both places)

M1: A complete method to find the coordinates of \(A\).

M1: Forms equation for \(\overrightarrow{OX} = \overrightarrow{OB} + m\mathbf{n}\) and proceeds to find where normal to plane through \(B\) intersects the plane, or the distance \(BX\) (\(= m\lvert\mathbf{n}\rvert\))

M1: Complete method to find the required angle or its complement. May find \(X\) first to find \(BX\) etc, or may use different sides of the triangle.

A1: For awrt \(40^\circ\)

Notes

M1: Subtracts the given coordinates either way round, 2 correct values implies method, and uses as their direction vector.

A1: For a correct equation. Allow any equivalent correct equations but must use the correct notation, starts with \(\mathbf{r} = \ldots\), can be using column vectors

Notes

M1: Uses \(z = 0\) in their \(P_1\) equation to find a value for their parameter and uses this to find the other coordinates.

A1: Deduces the correct coordinates, condone written as a vector.

Notes

M1: Realises the scalar product between their direction from part (a) and the direction vector extracted from the \(P_2\) equation is required and calculates its value

M1: Completes the method and at least reaches a value for cosine of the angle. Must be attempting to use the direction vectors

A1: Correct acute angle

Notes

M1: Uses the model to form a general vector connecting the two pipes, condone sign slips if the intention is clear. Condone use of the same parameter for this mark. They will be unable to score any more marks

M1: Recognises that the scalar product between the general vector and the directions of the lines = 0 and uses this to form 2 simultaneous equations in terms of their parameters. Follow through on their line from part (a)

M1: Attempts to solve the simultaneous equations, writing a value for each parameter is sufficient.
Then uses the values of their parameters in the model and either

• shows the shortest vector connecting the 2 lines
• proceeds to a distance with no incorrect working seen.

dM1: Uses their values of their parameters to find the vector which if not correct must be stated and then finds the modulus of their shortest vector to find the shortest distance. Maybe implied by a correct answer. Dependent on previous method mark.

A1: Awrt 1.3(0) m

Alternative 1

M1: Uses the model to form a general vector connecting the two pipes, condone sign slips if the intention is clear.

M1: Finds the normal vector to the direction vectors by any method and sets the general equation of the line equal to a multiple of the normal vector.

M1: Attempts to solve the simultaneous equations, writing a value for each parameter is sufficient.
Then uses the values of their parameters in the model and either

• shows the shortest vector connecting the 2 lines
• proceeds to a distance with no incorrect working seen.

dM1: Uses their values of their parameters to find the vector which if not correct must be stated and then finds the modulus of their shortest vector to find the shortest distance. Maybe implied by a correct answer. Dependent on previous method mark.

A1: Awrt 1.3(0) m

Alternative 2 outside the spec

M1: Using their equations to find the vector between the coordinates \((\mathbf{a} - \mathbf{c})\)

M1: Find the cross product of their directions \((\mathbf{b} \times \mathbf{d})\) and finds the dot product of their \((\mathbf{a} - \mathbf{c}) \bullet (\mathbf{b} \times \mathbf{d})\)

M1: Finds their \(|\mathbf{b} \times \mathbf{d}|\)

M1: Uses \(\left|\dfrac{(\mathbf{a} - \mathbf{c}) \bullet (\mathbf{b} \times \mathbf{d})}{|\mathbf{b} \times \mathbf{d}|}\right|\)

A1: Awrt 1.3(0) m.

Notes

B1: Correct value, check within the question. If more than one value is stated the correct value must clearly be selected.

Notes

M1: Uses their answer from part (a) to determine a value for \(k\) so that \(k \times\)their 210 is divisible by 360. If their answer to part (a) is in radians \(k \times \text{their } \dfrac{7\pi}{6}\) is divisible by \(2\pi\).

A1: Correct value must be using an angle of 210. There may be no working but allow M1 A1 for \(k = 12\) following an answer of 210 or \(\dfrac{7\pi}{6}\) in part (a)

Note: an angle of 30, 150, 330 also gives \(\boldsymbol{k = 12}\) but is M1A0

Notes

B1: Correct matrix

Notes

M1: Complete method to find the coordinates of \(B\). Look for at least two correct terms for each multiplication, follow through when multiplying by N.
Alternatively finds NM (not MN), look for 4 correct non zero terms if no method is shown and then multiplies to find the coordinates of \(B\).

A1*: Correct coordinates (condone vector) Must have been working with exact values throughout. If working in decimals M1 A0
It is insufficient to just write \(\mathbf{NM}\begin{pmatrix}-2\\ 4\\ 3\end{pmatrix} = \begin{pmatrix}\sqrt{3} + 2\\ 2\sqrt{3} - 1\\ 3\end{pmatrix}\) there must be some evidence of matrix multiplication seen.

Notes

M1: Identifies and applies an appropriate strategy to find the required angle e.g. cosine rule or scalar product. Note: \(AB^2 = 56 - 12\sqrt{3}\) and \(AB = 3\sqrt{6} - \sqrt{2} = 5.93\ldots\)

A1*: Correct value from correct equation

Notes

M1: Uses the given angle with \(\dfrac{1}{2}ab\sin C\) with appropriate \(a\), \(b\) and \(C\)
Outside spec: uses the cross product \(\dfrac{1}{2}\left|\overrightarrow{OA} \times \overrightarrow{OB}\right|\)

A1: Correct area to 3 significant figures

Notes

M1: Applies the dot product to the direction vectors. Minimum requirement is 3 – 3
A1: Shows that the dot product = 0 and concludes that the lines are perpendicular.

Notes

M1: Applies \(\mathbf{r} \bullet \begin{pmatrix}1\\ 2\\ -3\end{pmatrix} = \begin{pmatrix}-2\\ 2\\ 0\end{pmatrix} \bullet \begin{pmatrix}1\\ 2\\ -3\end{pmatrix} = \ldots\) or \(\begin{pmatrix}x\\ y\\ z\end{pmatrix} \bullet \begin{pmatrix}1\\ 2\\ -3\end{pmatrix} = \begin{pmatrix}-2\\ 2\\ 0\end{pmatrix} \bullet \begin{pmatrix}1\\ 2\\ -3\end{pmatrix} = \ldots\)

A1: Correct Cartesian equation \(x + 2y - 3z = 2\) o.e.

Note: \(\mathbf{i} + 2\mathbf{j} - 3\mathbf{k} = 2\) is M1A0

Notes

B1: See scheme, no conclusion required

Notes

M1: Uses the point of intersection to find the coordinates of \(B\) as functions of a parameter

M1: Uses the distance between the point \(A\) and the point \(B\) to form an equation for their parameter only.

A1: Correct simplified quadratic equation

M1: Solves their quadratic equation to find a value for \(\mu\)

ddM1: Dependent on the first two method marks. Uses any one of their values for their parameter to find the coordinates of \(B\), it need not be the correct one.

A1: Correct coordinates for \(B\), condone as a vector, if seen (3, 5, −1) must be disregarded

Alternative

M1: Finds the length \(AX\)
M1: Uses Pythagoras to find the length of \(XB\)
NOTE the change in order of the M1 and A1
M1: Find the length of the direction vector and compares to find a value for \(\mu\)
A1: A correct values for \(\mu\)
ddM1: Dependent on the first two method marks. Uses any one of their values for their parameter to find the coordinates of \(B\), it need not be the correct one.
A1: Correct coordinates for \(B\), condone as a vector, if seen (3, 5, −1) must be disregarded

Notes

Do not penalise the omission of “\(\mathbf{r} =\)” more than once in this question so penalise only once and on its first occurrence.

(a) Correct answer only scores no marks.

M1: Substitutes the parametric form of the line into the plane and solves for their parameter and uses this to find at least one coordinate. Alternatively substitutes line equation into the plane equation to obtain and solve an equation in one variable.

A1: Correct point. Accept as \(x = -7,\ y = 2,\ z = -7\) or as a vector.

Notes

(b) Alternative for first 2 marks:

Notes

M1: Identifies a point on \(l_1\) and uses the equation of the line through this point perpendicular to the plane to find the parameter where it intersects the plane. May use a different starting point.

M1: Uses twice their parameter to find the image point of their starting point in the plane.

A1: Correct image point.

ddM1: Finds the equation of the line between their intersection point from (a) and their image point.
May be implied if they use e.g. \(3\mathbf{i} + 8\mathbf{j} - 5\mathbf{k}\) as the position vector.
Alternatively, checks that the image point satisfies the equation for the other line.
Depends on both previous method marks.

A1*: Fully correct work with conclusion that hence the given equation gives the required line.
Must be as printed with “\(\mathbf{r} =\)” but condone use of a different parameter e.g. \(t\), \(\lambda\) etc.
If they checked that the image point satisfied the equation of line 2 they would need to then say e.g. that both lines pass through \((-7, 2, -7)\) and make a minimal conclusion.

Alternative for first 2 marks in (b):
M1: Identifies a point on \(l_1\) and finds the perpendicular distance from this point to the plane. May use a different starting point.
M1: Uses twice their distance to find the image point of their starting point in the plane.

Notes

Alternative 1 to (c) (Not on spec)

Alternative 2 to (c) (Not on spec)

Alternative 3 to (c) (Not on spec)

Notes

M1: A complete method to find the required equation. E.g. finds a second point on the common line and uses this and the point from (a) to find the direction and then forms the vector equation of the required line. May use earlier work to find the second point or start again.

A1: Correct equation formed, accept with any parameter which is not \(x\), \(y\) or \(z\).
Must include “\(\mathbf{r} =\)” unless this was already penalised in part (b).

Alternative 1:
M1: Attempts the normal to \(\Pi_2\) by attempting the vector product of the direction of \(l_1\) and the direction of \(l_2\) and then attempting the vector product of this with the normal to \(\Pi_1\) to find the direction of the common line and forms the vector equation with a point on the line.
The normal vectors may also be found using scalar products e.g.
\(\mathbf{n} = x\mathbf{i} + y\mathbf{j} + \mathbf{k} \Rightarrow (x\mathbf{i} + y\mathbf{j} + \mathbf{k}) \cdot (\mathbf{i} - 3\mathbf{j} + 5\mathbf{k}) = 0,\ (x\mathbf{i} + y\mathbf{j} + \mathbf{k}) \cdot (10\mathbf{i} + 6\mathbf{j} + 2\mathbf{k}) = 0 \Rightarrow x = \ldots,\ y = \ldots\)
A1: Correct equation formed, accept with any parameter which is not \(x\), \(y\) or \(z\).
Must include “\(\mathbf{r} =\)” unless already penalised in (b).

Alternative 2:
M1: Attempts the normal to \(\Pi_2\) by attempting the vector product of the direction of \(l_1\) and the direction of \(l_2\) or using scalar products as above and finds the equation of \(\Pi_2\) by using a point on the line and then solves this with \(\Pi_1\) by eliminating one of the variables to form the vector equation of the line of intersection.
A1: Correct equation formed, accept with any parameter which is not \(x\), \(y\) or \(z\).
Must include “\(\mathbf{r} =\)” unless already penalised in (b).

Alternative 3:
M1: Attempts the normal to \(\Pi_2\) by attempting the vector product of the direction of \(l_1\) and the direction of \(l_2\) or using scalar products as above and finds the equation of \(\Pi_2\) by using a point on the line and then solves this simultaneously with \(\Pi_1\) to form the vector equation of the line of intersection.
A1: Correct equation formed, accept with any parameter which is not \(x\), \(y\) or \(z\).
Must include “\(\mathbf{r} =\)” unless already penalised in (b).

There will be other methods in part (c)
Generally the method mark requires a complete attempt to find the equation of the line of intersection.

Notes

Alternative 1 to (d):

Alternative 2 to (d) (Not on spec):

Alternative 3 to (d):

Alternative 4 to (d):

(corrected from the printed mark scheme: the subtotals for Alternatives 1, 2 and 3 to (d) are printed as (2); each alternative carries the 3 marks M1 A1 A1)

Notes

M1: Realises the direction of (c) is perpendicular to the normal to \(\Pi_3\) and applies the dot product = 0 to find \(a\).
A1: Correct value for \(a\)
A1: Deduces the value of \(b\) using their \(a\) and one of their points on the line.

Alternative 1:
M1: Uses their point from (a) and another point on the line of intersection and substitutes both points into the given equation and solves the resulting equations for \(a\) or \(b\).
A1: Correct value for \(a\) or \(b\).
A1: Correct value for \(a\) and \(b\).

Alternative 2:
M1: Attempts the normal to \(\Pi_2\) by attempting the vector product of the direction of \(l_1\) and the direction of \(l_2\) or using scalar products as above (may have been found earlier) and then attempts the determinant of the matrix of normal vectors = 0 and solves for \(a\)
A1: Correct value for \(a\).
A1: Deduces the correct value of \(b\). E.g. by using their \(a\) with a point on the line.

Alternative 3:
M1: Substitutes their line from part (c) into the equation for \(\Pi_3\) and compares coefficients to establish a value for \(a\).
A1: Correct value for \(a\).
A1: Deduces the correct value of \(b\).

Alternative 4:
M1: Finds the equation for \(\Pi_2\) and then solves all 3 equations using the fact they are consistent leading to a value for \(a\) or a value for \(b\).
A1: Correct value for \(a\) or \(b\).
A1: Correct value for \(a\) and \(b\).

There will be other methods in part (d)
Generally the method mark requires a complete attempt to find the value of \(a\) or \(b\).

Notes

M1: See scheme, allow a sign slip and cos 60

A1: Correct simplified equation in \(a\), cos 120 must be evaluated to \(-\frac{1}{2}\) and dot product calculated

Note: If the candidate states either \(\left|\dfrac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{a}||\mathbf{b}|}\right| = \cos\theta\) or \(\left|\dfrac{a}{\sqrt{4 + a^2}\sqrt{2}}\right| = \cos 60\) then has the equation \(\dfrac{a}{\sqrt{4 + a^2}\sqrt{2}} = \dfrac{1}{2}\) award this mark. If the module of the dot product is not seen then award A0 for this equation.

dM1: Solve a quadratic equation for \(a\), by squaring and solving an equation of the form \(a^2 = K\) where \(K \gt 0\)

A1: Deduces the correct value of \(a\) from a correct equation. Must be seen in part (a) using the angle between the lines.

Alternative cross product method
M1: \(\begin{vmatrix}2 & a & 0\\ 0 & 1 & -1\end{vmatrix} = \sqrt{2^2 + a^2}\sqrt{1^2 + (-1)^2}\sin 120\)
A1: \(\sqrt{a^2 + 8} = \sqrt{4 + a^2}\sqrt{2}\dfrac{\sqrt{3}}{2}\)
Then as above

Note If they use the point of intersection to find a value for \(a\) this scores no marks

Notes

M1: Uses the model to write down any two correct equations

M1: Solve two equations simultaneously to find a value for \(\mu\) and \(\lambda\)

dM1: Dependent on previous method mark. Substitutes \(\mu\) and \(\lambda\) into a relevant equation. If no method shown two correct ordinates implies this mark.

A1: Correct coordinates. May be seen in part (c)

B1: Shows that the values of \(\mu\) and \(\lambda\) give the same third coordinate or point of intersection and draws the conclusion that the lines intersect/common point/consistent or tick.

Note: If an incorrect value for \(a\) is found in part (a) but in part (b) they find that \(a = -2\) this scores B0 but all other marks are available

Notes

This is M1M1A1 on ePen marking as M1 A1ft A1

M1: Full attempt to find the minimum distance from a point to a plane. Condone a sign slip with the value of \(d\).

A1ft: Following through on their point of intersection. Uses the model to find a correct expression for minimum distance from the nest to the ground

A1: Correct distance

Alternative

M1: Find the shortest distance from a point to plane by finding the perpendicular distance from the given plane to the origin and the perpendicular distance from the plane contacting their point of intersection to the origin and subtracts

A1ft: Following through on their point of intersection. Uses the model to find a correct expression for minimum distance from the nest to the ground

A1: Correct distance

Notes

B1: Comments on one of the models

• Flight path of the birds modelled as a straight line
• Angle between flight paths modelled as 120°
• The bird’s nest is modelled as a point
• Ground modelled as a plane

Then states unreliable

Any correct answer seen, ignore any other incorrect answers

Notes

Accept any alternative vector notations throughout.

M1: Attempts to solve the quadratic from equating the \(\mathbf{k}\) component to zero.

A1: Correct value, must select positive root, so accept 1.4 oe.
Correct answer only M1 A1

Notes

B1ft: For \((2.56, 15, -2)\) o.e or follow through \((9 - 4.6 \times \text{‘}\lambda\text{’}, 15, 0.8 - 2 \times \text{‘}\lambda\text{’})\) for their \(\lambda\).

Notes

M1: Recognises the angle between the perpendicular and direction vector is needed, and identifies the perpendicular as \(a\mathbf{k}\) for any non-zero \(a\) (including 1), and attempts dot product
Alternatively recognises the dot product of \((2.56, 15, -2)\) and \((2.56, 15, 0)\)

M1: Applies the dot product formula \(\dfrac{a \bullet b}{|a||b|}\) correctly between any two vectors, but must have dot product and modulus evaluated.

ddM1: Dependent on both previous marks. A correct method to proceed to the required angle, usually \(90^\circ - \arccos(\text{‘}-0.130\ldots\text{’})\) as shown in scheme but may e.g. use \(\sin\theta\) instead of \(\cos\theta\) in formula.
Alternatively is using dot product of \((2.56, 15, -2)\) and \((2.56, 15, 0)\) finds arccos(0.991…)

A1: For 7.5° cao

Alternative
M1: Finds the length of the vector in the ij plane.
M1: Finds the tan of any angle the
ddM1: Dependent on both previous marks. Finds the required angle
A1: For 7.5° cao

Notes

M1: Attempts to find value of \(\lambda\) that gives zero \(\mathbf{j}\) component.

M1: Uses their value of \(\lambda\) in the equation of the path to find position.

A1: Correct position.

Notes

B1ft: States that 0.920 > 0.9 so according to the model the ball will pass over the net. Follow through on their \(\mathbf{k}\) component and draws an appropriate conclusion. May stay the value of k > 0.92

Notes

M1: There must be some reference to the model to score this mark. See scheme for examples. It is likely to be either the ball is not a particle, or the top of the net is not a straight line. Accept references to the ball crossing a long way from the middle.
Do not accept reasons such as “there may be wind/air resistance” as these are not referencing the given model.

A1: For a reasonable conclusion based on their reference to the model.

For example
The ball is not a particle; therefore, it will not go over the net is M1A0 as not explained why – needs reference to radius/diameter

Notes

B1: Coordinates of \(Q\) correctly stated, accept as a column vector.

Notes

M1: Correct attempt to find coordinates of \(R\) using the given matrix with \(\theta = 120\). Must be multiplying in the correct way round. With no working two correct values or (−2.27, 3, −7.93) implies this mark.

A1: Correct exact coordinates as shown in scheme. Accept as a column vector. Cos 120 and sin 120 must have been evaluated.

Notes

M1: Applies the distance formula with the coordinates of \(P\) and their \(R\). Alternatively finds the vector \(\overrightarrow{PR}\) or \(\overrightarrow{RP}\) then applies length of a vector formula.

A1: Correct answer following correct coordinates of \(R\), must be a surd but need not be fully simplified.

Notes

B1ft: Shows the dot product is zero between the vectors \(\overrightarrow{PR}\) and \(\overrightarrow{PQ}\) and draws the conclusion perpendicular. Accept with \(\pm\) vectors for each. Follow through as long as the vectors are of the correct form, so \(\overrightarrow{PR} = \begin{pmatrix}a\\ 0\\ b\end{pmatrix}\) and \(\overrightarrow{PQ} = \begin{pmatrix}0\\ c\\ 0\end{pmatrix}\)

Note They could state if vectors \(\overrightarrow{PR}\) and \(\overrightarrow{PQ}\) are perpendicular then \(\overrightarrow{PR}.\overrightarrow{PQ} = 0\) then shows \(\overrightarrow{PR}.\overrightarrow{PQ} = 0\) this is B1

Notes

M1: Correct method for the area of the triangle, follow through on their coordinates of \(R\) and \(Q\). May see longer methods if they do not realise the triangle is right angled.

A1: For \(6\sqrt{51}\) cso following correct coordinates of \(R\)

Alternative 1

M1 Complete method to find the correct area
Finding all the lengths \(|PQ| = 6,\ |PR| = \sqrt{204} = 2\sqrt{51},\ |QR| = \sqrt{240} = 4\sqrt{15}\)
Use cosine rule to find an angle e.g. \(\cos PRQ = \dfrac{240 + 204 - 36}{2 \times \sqrt{240} \times \sqrt{204}} = \dfrac{\sqrt{85}}{10}\)
leading to \(PRQ = 22.7\ldots\) or \(\sin PRQ = \sqrt{1 - \left(\dfrac{\sqrt{85}}{10}\right)^2} = \ldots\left\{\dfrac{\sqrt{15}}{10}\right\}\)
Uses the area of the triangle \(= \dfrac{1}{2} \times \sqrt{240} \times \sqrt{204} \times \dfrac{\sqrt{15}}{10}\) or \(= \dfrac{1}{2} \times \sqrt{240} \times \sqrt{204} \times \sin 22.8\)

(Corrected from the printed mark scheme: the printed scheme has \(|PR| = \sqrt{240} = 4\sqrt{15}\) and \(|QR| = \sqrt{204} = 2\sqrt{51}\); these two lengths are the wrong way round, since part (c) gives \(|PR| = \sqrt{204}\).)

A1: For \(6\sqrt{51}\)

Alternative 2

M1: Uses \(\dfrac{1}{2}|\mathbf{a} \times \mathbf{b}|\) to find the required area
e.g. \(QP = \begin{pmatrix}0\\ 6\\ 0\end{pmatrix}\quad RP = \begin{pmatrix}12 - \sqrt{3}\\ 0\\ 3 + 4\sqrt{3}\end{pmatrix}\) cross product\[\begin{vmatrix}0 & 6 & 0\\ 12 - \sqrt{3} & 0 & 3 + 4\sqrt{3}\end{vmatrix} = 6\left(3 + 4\sqrt{3}\right)\mathbf{i} - 6\left(12 - \sqrt{3}\right)\mathbf{k}\]\[\text{Area} = \frac{1}{2}\sqrt{\left(6\left(3 + 4\sqrt{3}\right)\right)^2 + \left(-6\left(12 - \sqrt{3}\right)\right)^2} = \frac{1}{2}\sqrt{7344}\]

(Corrected from the printed mark scheme: the printed cross product is \(-6\left(12 - \sqrt{3}\right)\mathbf{i} + 6\left(3 + 4\sqrt{3}\right)\mathbf{k}\), with the \(\mathbf{i}\) and \(\mathbf{k}\) components the wrong way round. The area is unchanged.)

A1: For \(6\sqrt{51}\)

Notes

M1: Attempts the scalar product of each direction vector and the vector \(2\mathbf{i} + 3\mathbf{j} - 4\mathbf{k}\). Some numerical calculation is required, just “= 0” is insufficient. Alternatively, attempts the cross product (allow sign slips) with the two direction vectors.

A1: Shows that both scalar products = 0 (minimum \(-2 + 6 - 4 = 0\) and \(4 - 4 = 0\)) and makes a minimal conclusion with no erroneous statements. If using cross product, the calculation must be correct, and a minimal conclusion given with no erroneous statements.

Notes

M1: Applies \(\begin{pmatrix}x\\ y\\ z\end{pmatrix}.\begin{pmatrix}2\\ 3\\ -4\end{pmatrix} = \begin{pmatrix}3\\ 3\\ 2\end{pmatrix}.\begin{pmatrix}2\\ 3\\ -4\end{pmatrix} \Rightarrow \ldots\)

A1: \(2x + 3y - 4z = 7\)

Notes

M1: A fully correct method for finding a value of \(t\). Other methods are possible, but must be valid and lead to a value of \(t\). Examples of other methods:

• \(2\sqrt{29} = \pm\left(\dfrac{2(4 + t) + 3(-5 + 6t) - 4(2 - 3t)}{\sqrt{2^2 + 3^2 + (-4)^2}} - \dfrac{7}{\sqrt{29}}\right)\) using plane parallel to \(\Pi\) through origin and shortest distance from plane to origin.
• \(2(4 + t) + 3(-5 + 6t) - 4(2 - 3t) = 7 \Rightarrow t = t_i\) (\(t\) at intersection of line and plane) and
  \(\sin\theta = \dfrac{(2, 3, -4)^T.(1, 6, -3)^T}{\sqrt{29}\sqrt{46}}\) (sine of angle between line and plane) followed by
  \(\sin\theta = \dfrac{2\sqrt{29}}{k\sqrt{46}} \Rightarrow k = \ldots \Rightarrow t = t_i \pm k\)

A1: Correct values for \(t\). Both are required.

M1: Uses a value of \(t\) to find a set of coordinates for \(A\).

A1: Both correct sets of coordinates for \(A\).

Notes

(a)
M1: Correct method to find the determinant. Condone one sign slip
M1: A correct first step in obtaining the inverse. Could be the matrix of minors or cofactors. Condone sign slips as long as the intention is clear.
M1: Fully correct method to obtain the inverse. Attempts matrix of minors, cofactors, transposes and 1/determinant
A1: Correct matrix

Notes

(b)
M1: A complete strategy for solving the given equations e.g. multiplies the given coordinates by their inverse or solves simultaneously to achieve values for \(x\), \(y\) and \(z\)
A1ft: Correct calculation on their inverse matrix (unsimplified) or at least on correct value if solving simultaneously
A1: Correct coordinates

Notes

(c)(i)
M1: Uses a correct strategy that will lead to establishing a value for \(q\). E.g. eliminating one of \(x\), \(y\) or \(z\)
M1: Solves a suitable equation to obtain a value for \(q\)
A1: Correct value

(c)(i) Alternative 1
M1: Equating coefficients leading to two out of three equations and solves to find values for a and b
M1: Solves a suitable equation to obtain a value for \(q\) using their values for \(a\) and \(b\)

(c)(i) Alternative 2
M1: Finds a coordinate of intersection of the planes \(4x + 5y + 7z = 1\) and \(2x + y - z = 2\)
M1: Substitutes the values for \(x\), \(y\) and \(z\) into \(x + y + z = q\) to reach a value for \(q\)
A1: Correct value

(ii)
M1: Uses a correct strategy to obtain the Cartesian equation of the line or the general coordinates
A1: Correct Cartesian equation or coordinates in terms of a parameter.
M1: Uses their Cartesian equation to correctly extract the position and direction to form a vector equation for the required line
A1: Correct equation (o.e.) look out for multiples of the direction vector

Alternative (ii)
M1: Finds two different coordinates that lie on the line of intersection
A1: Correct coordinates
M1: Uses their coordinates to find the vector equation of the line that passes through them.
A1: Correct equation (o.e.), look out for multiples of the direction vector. Must have \(\mathbf{r} = \ldots\)

Alternative (ii) outside spec
M1: Finds the cross product between the normal vectors of two of the planes and a coordinate that lies on all three planes. If a coordinate is found in (i) it must be used in this part to award this mark.
A1: Correct cross product
M1: Uses the coordinate and the cross product to find the equation of the line
A1: Correct equation (o.e.), look out for multiples of the direction vector. Must have \(\mathbf{r} = \ldots\)

(corrected from the printed mark scheme: in (c)(ii), \(\lambda + 3z = -1\) is printed as \(\lambda + 3z = 1\), and \(x = \dfrac{5}{6} - \dfrac{2}{3}\lambda\) is printed as \(x = \dfrac{5}{5} - \dfrac{2}{3}\lambda\))

(Note: the cross products in the “outside the spec” alternative are as printed. The correct values are \((4\mathbf{i} + 5\mathbf{j} + 7\mathbf{k}) \times (\mathbf{i} + \mathbf{j} + \mathbf{k}) = -2\mathbf{i} + 3\mathbf{j} - \mathbf{k}\), \((4\mathbf{i} + 5\mathbf{j} + 7\mathbf{k}) \times (2\mathbf{i} + \mathbf{j} - \mathbf{k}) = -12\mathbf{i} + 18\mathbf{j} - 6\mathbf{k}\) and \((\mathbf{i} + \mathbf{j} + \mathbf{k}) \times (2\mathbf{i} + \mathbf{j} - \mathbf{k}) = -2\mathbf{i} + 3\mathbf{j} - \mathbf{k}\), each a multiple of \(2\mathbf{i} - 3\mathbf{j} + \mathbf{k}\).)

Notes

M1: Finds any two vectors \(\pm\overrightarrow{LM}, \pm\overrightarrow{LN}\) or \(\pm\overrightarrow{MN}\) by subtracting relevant vectors. Two out three values correct is sufficient to imply the correct method

M1: Applies the vector equation of the plane formula \(\mathbf{r} = \mathbf{a} + \lambda\mathbf{b} + \mu\mathbf{c}\) where \(\mathbf{a}\) is any point on the plane and the vectors \(\mathbf{b}\) and \(\mathbf{c}\) are any two from their \(\pm\overrightarrow{LM}, \pm\overrightarrow{LN}\) or \(\pm\overrightarrow{MN}\)

A1: Any correct equation for the plane. Must start with \(\mathbf{r}\) =…

Notes

(Corrected from the printed mark scheme: in (ii) the printed scheme lists \(\mathbf{a} = \begin{pmatrix}6\\ -3\\ 0\end{pmatrix}\); the point \(M\) is \((6, -2, 0)\), as in the notes.)

(In the printed scheme the two A1 alternatives are headed Alternative 1 and Alternative 2; they go with the M1 Alternatives 2 and 3, as in the notes, and are numbered that way here.)

(b)(i)
M1: Applies the dot product between their vectors b AND c with the vector \(\begin{pmatrix}1\\ 2\\ -10\end{pmatrix}\)

A1: Shows both dot products = 0 and concludes that the lawn is perpendicular to the vector \(\begin{pmatrix}1\\ 2\\ -10\end{pmatrix}\)

(b)(i) Alternative 1
M1: Applies the dot product between their vector b – c AND one of their vectors b or c with the vector \(\begin{pmatrix}1\\ 2\\ -10\end{pmatrix}\)
A1: Shows both dot products = 0 and concludes that the lawn is perpendicular to the vector \(\begin{pmatrix}1\\ 2\\ -10\end{pmatrix}\)

(b)(i) Alternative 2
M1: Applies the dot product between their vectors b and c \(\begin{pmatrix}x\\ y\\ z\end{pmatrix}\) and attempts to find values of \(x\), \(y\) and \(z\)
A1: Shows results is parallel to \(\begin{pmatrix}1\\ 2\\ -10\end{pmatrix}\) therefore the lawn is perpendicular

(b)(i) Alternative 3
M1: Applies the dot product between their answer to part (a) and the vector \(\begin{pmatrix}1\\ 2\\ -10\end{pmatrix}\)
A1: Achieves the value 2 and concludes as a constant therefore the lawn is perpendicular

(b)(i) Outside Specification for this paper – using the cross product
M1: Finds the cross product between ‘their b’ and ‘their c’ and shows parallel to \(\begin{pmatrix}1\\ 2\\ -10\end{pmatrix}\)
A1: Concludes parallel therefore the lawn is perpendicular

(b)(ii)
M1: Applies the formula \(\mathbf{r}.\mathbf{n} = \mathbf{a}.\mathbf{n}\) where \(\mathbf{n} = \begin{pmatrix}1\\ 2\\ -10\end{pmatrix}\) and \(\mathbf{a} = \begin{pmatrix}-2\\ -3\\ -1\end{pmatrix}\) or \(\begin{pmatrix}6\\ -2\\ 0\end{pmatrix}\) or \(\begin{pmatrix}2\\ 0\\ 0\end{pmatrix}\)
A1: Correct Cartesian equation of the plane
Note: If no method is shown then it must be correct to score M1 A1, if incorrect scores M0 A0. Look at part (i) to see if there is any method as long as it if used in (ii)

Notes

M1: Finds the vector \(\overrightarrow{PQ}\) or \(\overrightarrow{QP}\) and uses it as the direction vector in the formula. \(\mathbf{r} = \mathbf{a} + \lambda\mathbf{d}\). Two out three values correct is sufficient to imply the correct method

A1: A correct equation including \(\mathbf{r}\) =…

Notes

B1: States an acceptable limitation of the model for the lawn or washing line

Notes

M1: Applies the distance formula using the point (2, 5, 2.75) and the normal vector \(\begin{pmatrix}1\\ 2\\ -10\end{pmatrix}\)

A1: 1.71 m or 171 cm

Notes

B1ft: Compares their answer to part (e) with 1.5 and makes an assessment of the model with a reason with no contradictory statements.

Notes

Accept equivalent vector notation throughout.

(a)
M1: Starts by attempting to find a normal vector using a correct method. Allow if there are sign errors in attempts at the cross product.
A1: Obtains a correct normal vector
M1: Attempts scalar product between their normal and a point in the plane
A1: Correct Cartesian form (accept any equivalent Cartesian equation)

Alternative:

Alternative
M1: Uses the component form to eliminate one of the scalar parameters
A1: Two correct equations with one parameter eliminated OR a correct equation for each parameter in terms of \(x\), \(y\) and \(z\)
M1: Forms a Cartesian equation
A1: Correct Cartesian equation (accept any equivalent form)

Notes

(b)
M1: Interprets the Cartesian form to give a parametric form (allow sign slips) and substitutes this into their Cartesian equation and proceeds to find a value for their parameter.
NB: Attempts at \(\begin{pmatrix}2 + \lambda - \mu\\ 4 + 2\lambda + 2\mu\\ -1 - 3\lambda + \mu\end{pmatrix} = \begin{pmatrix}1 + 5\lambda\\ 3 - 3\lambda\\ -2 + 4\lambda\end{pmatrix}\) will score M0 as there are only two parameters, but \(\begin{pmatrix}2 + \lambda - \mu\\ 4 + 2\lambda + 2\mu\\ -1 - 3\lambda + \mu\end{pmatrix} = \begin{pmatrix}1 + 5\gamma\\ 3 - 3\gamma\\ -2 + 4\gamma\end{pmatrix}\) leading to a value for \(\gamma\) from solving three equations in three unknowns in M1.
dM1: Substitutes their parameter value back into the parametric form of the line. The parameter must have come from a correct attempt to find the value at intersection.
A1: Correct coordinates. Accept as \(x = \ldots\), \(y=\), … \(z = \ldots\) or as a vector.

Alternative:

Alternative:
M1: Eliminates two of the variables from the equation of plane using the Cartesian equation of the line and solves the linear equation.
dM1: Finds the other two coordinates.
A1: Correct coordinates, as above.

Notes

(c)
M1: Complete and correct scalar product method leading to a value for \(\theta\). Note that if \(\sin\theta\) is used instead of \(\cos\theta\) then they must also apply \(90 - \theta\) to access the method.
A1: Correct angle, accept awrt 41. as their final answer (do not isw if they go on to give e.g. \((180 - 41)^\circ\))

Notes

M1: Finds the determinant of the matrix corresponding to the system of equations.

A1: Sets determinant = 0 and solves their 3TQ to achieve \(k = 2\) (\(k = -8\) must be rejected)

Special case

M1A0: Uses \(k = 2\) and finds the determinant of the matrix corresponding to the system of equations
Shows that determinant = 0 and concludes that when \(k = 2\) there is no unique solution

Notes

M1: A complete method eliminating one variable from the equations using two different pairs of equations. Condone if a different value of \(k\) is used

A1: Achieves two equations in the same two variables

A1: Must give a reason, shows that the equations are a linear multiple of each other therefore they are consistent.

Alternative

(Note: the substitution shown uses \(5x + 2y = 1\) and \(11x + 2z = -3\).)

M1: A complete method eliminating one variable from the equations using two different pairs of equations. Substitutes these equations into one of the original equations in three variables.

A1: Achieves two correct equations in two different variables

A1: Shows that the equation works therefore they are consistent.

Notes

B1: The three planes form a sheaf. They must have full marks in (b) to award this mark.

Notes

M1: Attempts the direction between positions \(P\) and \(Q\). If no method shown, two correct entries imply the method.

A1: A correct equation in the correct form. Any point on the line may be used, and any non-zero multiple of the direction. Must begin \(\mathbf{r} = \ldots\)

Notes

Note: mark part (b) as a whole.

(i) B1: Correct value of \(k\) deduced.

M1: Realises the need to find the distance from the point on the mountain to a general point on the line.

dM1: Takes the dot product with the direction vector of line and sets to zero and proceeds to find a value of \(\lambda\). If working with \(k\) as well, allow for finding either \(\lambda\) in terms of \(k\) or \(k\) in terms of \(\lambda\).

M1: Substitutes their \(\lambda\) into their line equation. (This may not have come from correct work, but the method is for using the line equation here.) May be implied by two out of three correct coordinates for their \(\lambda\)
Note: May omit this step and substitute \(\lambda\) into \(\overrightarrow{MX}\). This gains M0 here, but can gain M1A1 in (ii) for finding the length of \(\overrightarrow{MX}\).

A1: Correct point.

(ii) M1: Uses the distance formula with their point and \(M\), or with their \(\overrightarrow{MX}\) from (i). (May be implied by two out of three correct coordinates for their \(\lambda\))

A1: Correct distance, including units. Accept awrt 221 m or \(25\sqrt{78}\) m

For reference Some of the other common equations/values of \(\lambda\) in (b)(i) are:\[\overrightarrow{MX} = \begin{pmatrix}-300\\ 400\\ -150\end{pmatrix} + \lambda\begin{pmatrix}6\\ -1\\ 1\end{pmatrix} - \begin{pmatrix}100\\ 200\\ 100\end{pmatrix} = \begin{pmatrix}-400 + 6\lambda\\ 200 - \lambda\\ -250 + \lambda\end{pmatrix} \Rightarrow \lambda = 75\]\[\overrightarrow{MX} = \begin{pmatrix}300\\ 300\\ -50\end{pmatrix} + \lambda\begin{pmatrix}600\\ -100\\ 100\end{pmatrix} - \begin{pmatrix}100\\ 200\\ 100\end{pmatrix} = \begin{pmatrix}200 + 600\lambda\\ 100 - 100\lambda\\ -150 + 100\lambda\end{pmatrix} \Rightarrow \lambda = -\frac{1}{4}\]\[\overrightarrow{MX} = \begin{pmatrix}300\\ 300\\ -50\end{pmatrix} + \lambda\begin{pmatrix}6\\ -1\\ 1\end{pmatrix} - \begin{pmatrix}100\\ 200\\ 100\end{pmatrix} = \begin{pmatrix}200 + 6\lambda\\ 100 - \lambda\\ -150 + \lambda\end{pmatrix} \Rightarrow \lambda = -25\](If the negative direction vectors are used in any case, the value of \(\lambda\) is just the negative of the above.)

Alternatives to part (b)

Note that variations may occur with the line equation chosen in part (a), but mark as follows:

Alternative (Alt 1)

(b)(i)

(b)(ii)

(i) B1M1: As per main scheme.

M1: Realises the need to find the distance from the point on the mountain to a general point on the line.

dM1: Attempts the distance or distance squared of \(\overrightarrow{MX}\), expands and completes the square to find the value of \(\lambda\) for which distance is minimum. May obtain other forms for the completed square. Look for \(A(B\lambda - C)^2 - D + \text{“}262500\text{”}\) where \(A, B, C, D \ne 0\) but \(B\) may be 1.

M1A1: As per main scheme.

(ii) M1: Correct method for the distance. May be as per main scheme, or via extracting from the completed square constant term.

A1: Correct distance, including units. Accept awrt 221 m or \(25\sqrt{78}\) m

Alternative (Alt 2)

(b)(i)

(Corrected from the printed mark scheme: the derivative is printed as \(\dfrac{\mathrm{d}}{\mathrm{d}x}(d^2)\); it is with respect to \(\lambda\).)

(b)(ii)

As per main scheme except for:

(i) dM1: Attempts the distance or distance squared of \(\overrightarrow{MX}\), differentiates and sets to zero to find \(\lambda\) for minimum distance.

(ii) M1: May substitute \(\lambda\) into the distance squared formula to find distance.

Alternative (Alt 3)

(b)(i)

(Corrected from the printed mark scheme: the numerator is printed as “\(75\sqrt{8}\)”; \(\left|\overrightarrow{PX}\right| = 75\sqrt{38}\).)

(b)(ii)

(i) B1: Correct value of \(k\) deduced.

M1: Finds \(\overrightarrow{MP}\) (or \(\overrightarrow{MQ}\)) and attempts scalar product formula with this and the direction of the line to find the angle or cosine of the angle between line and \(\overrightarrow{MP}\) (or \(\overrightarrow{MQ}\))

dM1: Uses their angle with the cosine to find the length of \(\overrightarrow{PX}\) (or \(\overrightarrow{QX}\)). Accept equivalent trigonometric methods (e.g. finding opposite side first and using tangent or Pythagoras).

M1: Uses the length of \(\overrightarrow{PX}\) (or \(\overrightarrow{QX}\)) to find the coordinates of the point on the line at shortest distance from \(M\).

A1: Correct point.

(ii) M1: Correct method for the distance. May be as per main scheme, or use of sine ratio with their angle between the line and \(\overrightarrow{MP}\) (or \(\overrightarrow{MQ}\)). Accept equivalent trigonometric methods.

A1: Correct distance, including units. Accept awrt 221 m or \(25\sqrt{78}\) m

Useful diagram:

Note for \(P\), \(\cos\theta = \pm\dfrac{57}{\sqrt{38}\sqrt{105}}\), \(\theta = 25.5\ldots°\) and \(\left|\overrightarrow{PX}\right| = 75\sqrt{38}\)

For \(Q\), \(\cos\theta = \pm\dfrac{19}{\sqrt{38}\sqrt{29}}\), \(\theta = 55.08\ldots°\), \(\left|\overrightarrow{QX}\right| = 25\sqrt{38}\)

Notes

M1: Calculates the two distances \(OP\) and \(OQ\).

A1ft: Makes an appropriate conclusion for their tunnel length, but distances \(OP\) and \(OQ\) must be correct. A reason and a conclusion is needed.
Accept for reason e.g. “significantly shorter” or “tunnel is more than 100 m less than either existing accessway”, as these act as a comparative judgement. But do not accept just “shorter” or just inequalities given with no comparative evidence.

Notes

B1: Any appropriate criticism of the model given. The model must be referred to in some way – e.g. criticise the straightness/thickness of line, flatness of plane or lack of taking strata etc of mountain into account (as e.g. this means line may not be straight).
Note: reference to measurements not being correct is NOT a limitation of the model.

Notes

Allow using \(\begin{pmatrix}1\\3\\0\end{pmatrix}\) instead of \(\begin{pmatrix}1\\0\\3\end{pmatrix}\) for the method mark.

Way 1

M1: Starts by attempting to find where the two lines intersect. They must set up a parametric equation for line 1 (allow sign slips and as long as the intention is clear), forms simultaneous equations by equating coordinates and attempts to solve to find a value for \(t = \ldots\) or \(\lambda = \ldots\).

A1: Shows that there is a unique solution by checking the third equation or shows that the coordinate (3, -2, 7) lies on both lines.

A1: Achieves the correct values \(t = 2\) and \(\lambda = 1\), checks the third equation and concludes that either

• a common point,
• the lines intersect
• the equations are consistent

therefore, the lines lie in the same plane

Alternative: Way 2

M1: Finds the vector equation of the plane with the both direction vectors and one coordinate (allow a sign slip), sets equal to the other coordinate, forms simultaneous equations and attempts to solve to find a value for \(t = \ldots\) or \(\lambda = \ldots\).

A1: Shows that the other coordinate lies on the plane by checking the third equation.

A1: Achieves the correct values \(t = -2\) and \(\lambda = 1\) or \(t = 2\) and \(\lambda = -1\) and concludes that the second coordinate lie on the plane; therefore, the lines lie on the same plane

Alternative: Way 3

M1: Substitutes line 2 into line 1 and solves a pair of equations to find a value for \(t\). Allow slip with the position of 0 and sign slips as long as the intention is clear.

A1: Solve two pairs of equations to achieve \(t = 2\) for each.

A1: Achieves the correct value \(t = 2\) and concludes that either

• a common point,
• the lines intersect
• the equations are consistent

therefore, the lines lie in the same plane

Alternative: Way 4 (Using Further Pure 2 knowledge)

M1: A complete method to finds a vector which is normal to both lines and attempts to finds the equation of the plane containing one line.

A1: Achieves the correct equation for the plane containing each line.

A1: Conclusion, planes are the same, therefore the lines lie in the same plane.

Notes

This may be seen in part (a)

B1: Correct vector equation allow any letter for the scalers.

Must start with \(\mathbf{r} = \ldots\) and uses two out of the following direction vectors \(\pm\begin{pmatrix}1\\-1\\2\end{pmatrix}\), \(\pm\begin{pmatrix}2\\-1\\3\end{pmatrix}\) or \(\pm\begin{pmatrix}0\\-1\\1\end{pmatrix}\) and one of the following position vectors \(\begin{pmatrix}1\\0\\3\end{pmatrix}\), \(\begin{pmatrix}1\\-1\\4\end{pmatrix}\) or \(\begin{pmatrix}3\\-2\\7\end{pmatrix}\)

Notes

Way 1

M1: Calculates the scalar product between the direction vectors, allow one slip, if the intention is clear

dM1: Dependent on the previous method mark. Applies the scalar product formula with their scalar product to find a value for \(\cos\theta\)

A1: Correct answer only

Alternative: Way 2 (Using Further Pure 2 knowledge)

M1: Calculates the vector product between the direction vectors, allow one slip, if the intention is clear

dM1: Dependent on the previous method mark. Applies the vector product formula with their vector product to find a value for \(\sin\theta\)

A1: Correct answer only

Notes

M1: Attempts determinant, equates to zero and attempts to solve for \(k\) in order to establish the restriction for \(k\). For the determinant, at least 2 of the 3 “elements” should be correct.

May see rule of Sarrus used for determinant e.g.

\(|\mathbf{M}| = (2)(k)(-1) + (4)(3)(-1) + (3)(2)(1) - (3)(k)(1) - (2)(4)(2) - (-1)(3)(-1) = 0 \Rightarrow k = \ldots\)

(Corrected from the printed mark scheme: the fourth product is printed as \((3)(k)(-1)\); it should be \((3)(k)(1)\).)

A1: Describes the correct condition for \(k\) with no contradictions. Allow e.g. \(k \lt -5,\ k \gt -5\)

Notes

Way 1

M1: A complete strategy for solving the given equations. Need to see an attempt at the inverse followed by a correct method for finding \(x\), \(y\) and \(z\)

B1: Correct inverse matrix

M1: Uses their inverse and attempts the multiplication with the correct vector

A1: Correct values for \(x\), \(y\) and \(z\) in any form

A1ft: Correct values given in coordinate form only. Follow through their \(x\), \(y\) and \(z\).

Alternative: Way 2

M1: A complete strategy for solving the given equations. Need to see an attempt at eliminating one variable followed by a correct method for finding \(x\), \(y\) and \(z\)

B1: One correct value

M1: Uses the equations to find values for the other 2 variables

A1: Correct values for \(x\), \(y\) and \(z\) in any form

A1ft: Correct values given in coordinate form only. Follow through their \(x\), \(y\) and \(z\).

Notes

(c)(i)

M1: Uses a correct strategy that will lead to establishing a value for \(q\). E.g. eliminating one of \(x\), \(y\) or \(z\)

M1: Solves a suitable equation to obtain a value for \(q\)

A1: Correct value

(ii)

B1: Describes the correct geometrical configuration.

Must include the two ideas of planes and meeting in a line or forming a sheaf with no contradictory statements.

Notes

M1: Forms a parametric form for the line. Allow one slip.

M1: Substitutes into the equation of the plane to an equation in \(\lambda\). May use Cartesian form of plane to substitute into.

A1: Correct equation in \(\lambda\)

A1ft: Simplifies and derives a contradiction and deduces line and plane do not meet. Follow through in their initial equation in \(\lambda\) so
- contradiction so no intersection if \(\lambda\) disappears and constants unequal
- line lies in plane if a tautology is arrived at
- meet in a point if a solution for \(\lambda\) is found.
But do not allow for incorrect simplification from a correct initial equation in \(\boldsymbol{\lambda}\)
Note that a miscopy/misread of 7 instead of \(-7\) can therefore score a maximum of M1M1A0A1A0.

A1cso: Correct deduction from correct working. This may be seen two separate statements in their working. You may see attempts at showing the line is parallel before/after deducing there is no intersection.

Alternative (Alt 1)

Note that some may attempt a mix of the main scheme and Alt 1. Mark under main scheme unless Alt 1 would score higher.

M1: Attempts the dot product between the two direction vectors.

A1: Shows dot product is zero and makes the correct deduction that line is parallel to plane.

M1: Finds a point on \(l\) and substitutes into the equation of \(\mathit{\Pi}\) (vector or Cartesian)

A1ft: Simplifies and derives a contradiction – follow through their equation, so if arrive at a tautology, they should deduce the line is in the plane.

A1cso: Correct deduction from correct working but may be split across working.

Alternative (Alt 2)

M1: Attempts to solve the Cartesian equation of the line and plane, using the plane equation to eliminate one variable for the M.

A1: Correct elimination of their chosen variable. (E.g. may see \(3 - 3y + z = -7\) or \(-2x - 2y - 2 = -7\) etc)

M1: Solves the reduced equations in two variables…

A1ft: … and derives a contradiction/line and plane do not meet. Follow through their result, so may reach a tautology and deduce lies in plane, or find single solution and deduce meet in a point.

A1cso: Correct deduction from correct working.

Notes

(Corrected from the printed mark scheme: the printed working has \((3)^3\) and \((-5)^3\) inside the square roots; these are squares, \((3)^2\) and \((-5)^2\), giving \(\sqrt{13}\) and \(\sqrt{358}\).)

M1: Realises the scalar product between the direction of \(W\) and the normal to the road is needed and so applies it and uses trigonometry to find an angle

M1: Calculates the scalar product between \(\pm\left(\begin{pmatrix}1\\ 2\\ -3\end{pmatrix} - \begin{pmatrix}-1\\ -1\\ -3\end{pmatrix}\right)\) and \(\pm\begin{pmatrix}3\\ -5\\ -18\end{pmatrix}\) (Allow sign slips as long as the intention is clear)

A1: \(\begin{pmatrix}2\\ 3\\ 0\end{pmatrix}\bullet\begin{pmatrix}3\\ -5\\ -18\end{pmatrix} = -9\) or \(\begin{pmatrix}-2\\ -3\\ 0\end{pmatrix}\bullet\begin{pmatrix}3\\ -5\\ -18\end{pmatrix} = 9\) or \(\begin{pmatrix}2\\ 3\\ 0\end{pmatrix}\bullet\begin{pmatrix}-3\\ 5\\ 18\end{pmatrix} = 9\) or \(\begin{pmatrix}-2\\ -3\\ 0\end{pmatrix}\bullet\begin{pmatrix}-3\\ 5\\ 18\end{pmatrix} = -9\)

M1: A fully complete and correct method for obtaining the acute angle

A1: Awrt \(7.58^\circ\) or awrt 0.132 radians (must see units). Do not isw and withhold this mark if extra answers are given.

Notes

B1ft: Forms the correct parametric form for the pipe \(W\). Follow through their direction vector for \(W\) from part (a).

M1: Identifies the need to and forms a vector connecting \(C\) to \(W\) using a parametric form for \(W\)

M1: Uses the model to form the scalar product of \(C\) to \(W\) and the direction of \(W\) to find the value of their parameter or finds the distance \(C\) to \(W\) or \((C \text{ to } W)^2\) in terms of their parameter

A1: Correct vector or correct completion of the square

ddM1: Correct use of Pythagoras on their vector \(CW\) or appropriate method to find the shortest distance between the point and the pipe. Dependent on both previous method marks.

A1: Correct length for the required section of pipe is 305 or 305 cm or 3.05 m

Alternatives for part (b): Way 2

(Note: the vector \(\begin{pmatrix}0\\ 1\\ -3\end{pmatrix}\) printed here is \(\overrightarrow{CA}\); \(\overrightarrow{AC} = \begin{pmatrix}0\\ -1\\ 3\end{pmatrix}\), which gives a scalar product of \(-3\) and angle \(CAB = 105.25\ldots^\circ\). Since \(\sin 105.25\ldots^\circ = \sin 74.74\ldots^\circ\), the distance is the same.)

B1ft: Forms the correct vectors. Follow through their direction vector for \(W\) from part (a).

M1: Identifies the need to and forms the scalar product between AC and AB

M1: Uses the model to form the scalar product and uses this to find the angle \(CAB\)

A1: Correct angle

ddM1: Correct method using their values or appropriate method to find the shortest distance between the point and the pipe. Dependent on both previous method marks.

A1: Correct length for the required section of pipe is 305 or 305 cm or 3.05 m

Way 3

(Corrected from the printed mark scheme: Way 3 prints \(\mathbf{AC} = \begin{pmatrix}0\\ 1\\ -3\end{pmatrix}\); \(\mathbf{AC} = \begin{pmatrix}0\\ -1\\ 3\end{pmatrix}\), as used in the vector product.)

B1ft: Forms the correct vectors. Follow through their direction vector for \(W\) from part (a).

M1: Identifies the need to and forms the vector product between AC and AB

M1: Uses the model to find the magnitude of their vector product

A1: Correct value

ddM1: Correct method using their values or appropriate method to find the shortest distance between the point and the pipe. Dependent on both previous method marks.

A1: Correct length for the required section of pipe is 305 or 305 cm or 3.05 m

Notes

B1: Evidence that the determinant is \(\pm 69\) (may be implied by their matrix e.g. where entries are not in exact form: \(\pm\begin{pmatrix}0.014 & 0.188 & 0.072\\ -0.159 & -0.072 & 0.203\\ -0.377 & 0.101 & 0.116\end{pmatrix}\)) (Should be mostly correct)

Must be seen in part (a).

B1: Fully correct inverse with all elements in exact form

Notes

M1: Any complete method to find the values of \(x\), \(y\) and \(z\) (Must be using their inverse if using the method in the main scheme)

A1: Correct coordinates

A solution not using the inverse requires a complete method to find values for \(x\), \(y\) and \(z\) for the method mark.

Correct coordinates only scores both marks.

Notes

B1: Describes the correct geometrical configuration.

Must include the two ideas of planes and meet in a point with no contradictory statements.

This is dependent on having obtained a unique point in part (b)

Typical solution

Typical solution

Let \(x = 0\):

Let \(y = 0\):

Two points on the line:

Typical solution

At the closest point, \(\overrightarrow{AP} \bullet \underline{l} = 0\)

The closest point is \((-3, 6, -1)\)

Typical solution

Typical solution

The point is \((13, -2, 23)\)

Typical solution

Typical solution

Typical solution

Typical solution

(i)

\((2, 5, -2)\)

(ii)

Shortest distance between \(l\) and \(B\) is

(iii)

Typical solution

Typical solution

\(b = -ac\)

Typical solution

Typical solution

Vector connecting \(L_1\) to \(L_2\)

Vector perpendicular to both lines

Alternative typical solution 1

Equation of \(L_2\): \(\mathbf{r} = \begin{bmatrix}10 \\ 26 \\ -19\end{bmatrix} + \mu\begin{bmatrix}5 \\ 9 \\ 4\end{bmatrix}\)

General vector

Scalar product

Alternative typical solution 2

Parallel planes

Typical solution

Typical solution

Typical solution

Typical solution

Typical solution

Typical solution

Typical solution

Typical solution

\(8\)

Typical solution

Typical solution

(i)

Scalar product of direction vectors

The scalar product is non-zero, so the lines are not perpendicular.

(ii)

Normal to plane

Equation of plane is

Typical solution

Typical solution

Typical solution

Typical solution

\(90^\circ\)

Typical solution

Normal to plane \(\mathbf{n} = \begin{bmatrix}4 \\ p \\ 5\end{bmatrix}\)

Let \(\overrightarrow{AB} = \mathbf{c}\) then

As \(\alpha\) is acute, \(\sin\alpha = \cos\theta = \dfrac{\sqrt{15}}{75}\)

Hence

Typical solution

Distance is

Typical solution

At \(\Pi_1\)

At image point

Typical solution

All three values of \(\lambda\) are the same so \(P\) lies on \(l_1\)

Typical solution

\(\therefore\) \(l_1\) and \(l_2\) are perpendicular

Typical solution

Typical solution

Point of intersection \(= (-6, 13, -5)\)

Typical solution

(i)

same value of \(\lambda\) \(\therefore\) the lines intersect

(ii)

Typical solution

The submarines might not be at the intersection point at the same time.

Typical solution

Typical solution

Cofactors:

Typical solution

(i)

Point of intersection is:

(ii)

So the planes represented by the first two equations are perpendicular

As the vector perpendicular to the first two planes is a multiple of the normal vector of the third plane, the three planes are mutually perpendicular

Typical solution

The position vector of a point on \(L_1\) is

The position vector of a point on \(L_2\) is

Therefore, when \(\mu = 0\), \(\lambda = 0\) so we can equate the constant terms in the equation

Typical solution

Moduli of vectors are \(\sqrt{14}\) and \(\sqrt{6}\)

Let \(\alpha\) be angle between lines

Angle between lines

Typical solution

(i)

(ii)

Typical solution

Let \(\beta\) be angle between planes