Notes

M1: Attempts to find both vectors by subtraction. Allow either direction. If no method shown then two correct values is sufficient.

dM1: Uses the correct process of the vector product. May be implied by 2 correct components if working not shown.

A1: Correct vector from correct work, isw

Notes

M1: Attempts \(\pm\overrightarrow{AE}\) and the scalar triple product using their vector from part (a). Alternatively uses determinant approach

A1: Correct expression in any form, may be unsimplified

dM1: Sets their triple product \(= 9\) or \(-9\) and attempts to solve 3TQ using a correct method

M1: Sets their triple product \(= 9\) and \(-9\) and attempts to solve both 3TQ’s using a correct method

A1: One correct pair of values or any 2 correct values

A1: All 4 values correct

Notes

M1: Evidence of a correct method for the vector product. May be implied by two out of three correct components if no method shown or by correct work for 1 component.
Attempting \(\mathbf{v} \times \mathbf{u}\) scores M0

A1: Correct answer.
Allow equivalent expressions and condone e.g. \(-10\mathbf{I} + (-10 - 3a)\mathbf{J} + (-30 - 4a)\mathbf{K}\)
Award the mark once a correct vector is seen and isw if necessary.
Must be a vector not coordinates.

Notes

M1: Applies a correct method for the area of the triangle and scalar multiple property of vector product to set up a vector equation in \(a\) using the vector product.
The “1/2” must be seen or implied.
This may be implied by later work e.g. \(\dfrac{1}{2}\sqrt{20^2 + (20 + 6a)^2 + (60 + 8a)^2} = 15\)

M1: Applies the modulus correctly and expands correctly to reach a quadratic in \(a\).
Note that \(\dfrac{1}{4}\left(20^2 + (20 + 6a)^2 + (60 + 8a)^2\right) = 15\) scores M0 (must square both sides)

ddM1: Solves their quadratic by any suitable means including a calculator to obtain at least one real root.
Depends on both previous method marks.

A1cso: Both correct values following correct work e.g. do not condone sign errors in the vector product if they fortuitously lead to the correct answers.

Note there may be more convoluted methods for the area e.g. using “\(\frac{1}{2}ab\sin C\)” and the scalar product:

\(\dfrac{1}{2}\left|\overrightarrow{AB}\right|\left|\overrightarrow{AC}\right|\sin A = 15 \Rightarrow \dfrac{1}{2}10\sqrt{2}\sqrt{a^2 + 40}\sin A = 15\)
\(\mathbf{u}.\mathbf{v} = |\mathbf{u}||\mathbf{v}|\cos A \Rightarrow 5a - 30 = 5\sqrt{2}\sqrt{a^2 + 40}\cos A\)
\(\Rightarrow \sin A = \sqrt{\dfrac{a^2 + 12a + 44}{2(a^2 + 40)}} \Rightarrow 5\sqrt{2}\sqrt{a^2 + 40}\sqrt{\dfrac{a^2 + 12a + 44}{2(a^2 + 40)}} = 15\)

Score M1 for a complete correct method to obtain an equation in \(a\) only then e.g.
\(5\sqrt{a^2 + 12a + 44} = 15 \Rightarrow a^2 + 12a + 35 = 0\) etc.
M2 for simplifying to obtain a quadratic in \(a\).
Then as above. Use review if necessary.

Notes

M1: A complete method to find the values for \(p\) and \(q\). There must be an attempt to find \(\overrightarrow{AB} \times \overrightarrow{AC}\) and sets equal to a multiple (not equal to 1) of \(2\mathbf{i} + 3\mathbf{j} + 4\mathbf{k}\) (allowing minor slips) to form and solve two simultaneous equations for \(p\) and \(q\). This is for the overall approach, so may be scored if the method for the cross product is incorrect, as long as an attempt has been made.

M1: A complete method to find the cross product \(\overrightarrow{AB} \times \overrightarrow{AC}\). Form should be correct \((\ldots)\mathbf{i} - (\ldots)\mathbf{j} + (\ldots)\mathbf{k}\) with correct coefficients in each bracket, though allow slips in sign in these.

A1: Correct cross product.

B1: Deduces that \(\overrightarrow{AB} \times \overrightarrow{AC} = -2(2\mathbf{i} + 3\mathbf{j} + 4\mathbf{k})\) May be implied by their working. Alternatively, a correct cross product statement for the parallel vectors.

M1: Forms simultaneous equations by setting the \(\mathbf{j}\) component of the cross product to \(k\) times 3 and setting the \(\mathbf{k}\) component of the cross product to \(k\) times 4, where \(k \neq 1\) is an attempt at a scale factor for the parallel vectors. Alternatively, attempts the cross product and equates at least two components to 0 for form simultaneous equations - the attempt at the cross product may be incorrect for this mark as long as the intent is clear.

dM1: Solves their simultaneous equations to find a value for \(p\) or \(q\). Accept for any \(p\) and \(q\) following forming simultaneous equations (do not be concerned about the method used). Dependent on the previous method mark

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

Alt

M1: A complete method to find the values for \(p\) and \(q\). There must be an attempt to apply the scalar product of both \(\overrightarrow{AB}\) and \(\overrightarrow{AC}\) with \(2\mathbf{i} + 3\mathbf{j} + 4\mathbf{k}\) then set each equal to 0 and solves both equations. (Corrected from the printed mark scheme: printed as “of both \(\overrightarrow{AB}\) and \(\overrightarrow{AB}\)”.)

M1: A correct method for one of the scalar products.

A1: Correct expression for one of the scalar products.

B1: Forms at least one correct equation for \(p\) or \(q\) by setting the scalar product equal to zero and producing a correct linear equation for \(p\) or \(q\)

M1: Proceeds to find at least one of the two values.

dM1: Solves to find a value for \(p\) and for \(q\). Dependent on the previous method mark

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

(Brackets that are missing from the printed mark scheme because of a font fault have been restored.)

Notes

M1: A complete method for the area of the triangle. E.g. may use area \(= \dfrac{1}{2}\left|\overrightarrow{AB} \times \overrightarrow{AC}\right|\) with their cross product from (a), or starting from scratch, and correct attempt at the modulus. Alternative may find angle between \(\overrightarrow{AB}\) and \(\overrightarrow{AC}\) using scalar product, \(\cos\theta = \dfrac{\overrightarrow{AB} \cdot \overrightarrow{AC}}{\left|\overrightarrow{AB}\right|\left|\overrightarrow{AC}\right|}\), and then Area \(= \dfrac{1}{2}\left|\overrightarrow{AB}\right| \times \left|\overrightarrow{AC}\right|\sin\theta\).

A1: Correct exact area from a correct \(\overrightarrow{AB} \times \overrightarrow{AC}\) (or correct \(p\) and \(q\) in the alt) (oe). There may be variations in method used, but if an incorrect vectors is used in the process score A0.

(Brackets that are missing from the printed mark scheme because of a font fault have been restored.)

Notes

M1: Correct method for finding at least one of the three points. Allow one slip in coordinates but should have correct fraction to make the value of \(z\) to be 0.

A1: Any one of the three points correct, ignoring the labelling.

A1: All three points correct, ignoring the labelling

Notes

M1: Correct method for finding the area of the triangle, e.g realises that \(MN\) is parallel to the \(x\)-axis so uses \(\dfrac{1}{2}bh\) with \(b = MN\) and \(h\) is distance of \(MN\) from axis.

Alternative using \(\dfrac{1}{2}|a \times b|\) with vectors \(\overrightarrow{PM} = \pm\begin{pmatrix}6\\ -2\\ 0\end{pmatrix}\quad \overrightarrow{PN} = \pm\begin{pmatrix}-1\\ -2\\ 0\end{pmatrix}\quad \overrightarrow{NM} = \pm\begin{pmatrix}7\\ 0\\ 0\end{pmatrix}\) follow through on their answers in part (a). Condone sign slips except they must be using \(-\mathbf{j}\) in the cross product

For example \(\dfrac{1}{2}\left|\overrightarrow{MP} \times \overrightarrow{PN}\right| = \dfrac{1}{2}\begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k}\\ 6 & -2 & 0\\ -1 & -2 & 0\end{vmatrix} = \dfrac{1}{2}\left|\begin{pmatrix}0\\ 0\\ -14\end{pmatrix}\right| = \ldots\)
\(\dfrac{1}{2}\left|\overrightarrow{PN} \times \overrightarrow{NM}\right| = \dfrac{1}{2}\begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k}\\ -1 & -2 & 0\\ 7 & 0 & 0\end{vmatrix} = \dfrac{1}{2}\left|\begin{pmatrix}0\\ 0\\ 14\end{pmatrix}\right| = \ldots\)
\(\dfrac{1}{2}\left|\overrightarrow{PM} \times \overrightarrow{NM}\right| = \dfrac{1}{2}\begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k}\\ 6 & -2 & 0\\ 7 & 0 & 0\end{vmatrix} = \dfrac{1}{2}\left|\begin{pmatrix}0\\ 0\\ 14\end{pmatrix}\right| = \ldots\)
or attempting to find an angle using dot product or cosine rule followed by \(\dfrac{1}{2}ab\sin C\).

A1: Correct area of 7 from correct vectors

Notes

(c) On ePen this is M1 A1 M1 M1 M1 A1

M1: Formulates a correct method to find the volume of \(NMPA\). May use method shown, or e.g. \(\dfrac{1}{6}\left|\overrightarrow{AM} \cdot \left(\overrightarrow{AN} \times \overrightarrow{AP}\right)\right|\) or equivalent method.

A1: For \(\dfrac{28}{3}\).

Note there are many ways to find the required volume of \(AMNP\) applying the triple scalar product to a combination of the following vectors

\(\overrightarrow{AM} = \begin{pmatrix}4\\ -4\\ 4\end{pmatrix}\ \overrightarrow{AN} = \begin{pmatrix}-3\\ -4\\ 4\end{pmatrix}\ \overrightarrow{AP} = \begin{pmatrix}-2\\ -2\\ 4\end{pmatrix}\ \overrightarrow{NA} = \begin{pmatrix}3\\ 4\\ -4\end{pmatrix}\ \overrightarrow{NM} = \begin{pmatrix}7\\ 0\\ 0\end{pmatrix}\ \overrightarrow{NP} = \begin{pmatrix}1\\ 2\\ 0\end{pmatrix}\)
\(\overrightarrow{MA} = \begin{pmatrix}-4\\ 4\\ -4\end{pmatrix}\ \overrightarrow{MN} = \begin{pmatrix}-7\\ 0\\ 0\end{pmatrix}\ \overrightarrow{MP} = \begin{pmatrix}-6\\ 2\\ 0\end{pmatrix}\ \overrightarrow{PA} = \begin{pmatrix}2\\ 2\\ -4\end{pmatrix}\ \overrightarrow{PM} = \begin{pmatrix}6\\ -2\\ 0\end{pmatrix}\ \overrightarrow{PN} = \begin{pmatrix}-1\\ -2\\ 0\end{pmatrix}\)

For example
\(\dfrac{1}{6}\left|\overrightarrow{AM} \cdot \left(\overrightarrow{AN} \times \overrightarrow{AP}\right)\right| = \dfrac{1}{6}\left|\begin{pmatrix}4\\ -4\\ 4\end{pmatrix} \cdot \left(\begin{pmatrix}-3\\ -4\\ 4\end{pmatrix} \times \begin{pmatrix}-2\\ -2\\ 4\end{pmatrix}\right)\right| = \dfrac{1}{6}\left|\begin{pmatrix}4\\ -4\\ 4\end{pmatrix} \cdot \begin{pmatrix}-8\\ 4\\ -2\end{pmatrix}\right| = \dfrac{1}{6} \times 56\)
\(\dfrac{1}{6}\left|\overrightarrow{NA} \cdot \left(\overrightarrow{NM} \times \overrightarrow{NP}\right)\right| = \dfrac{1}{6}\left|\begin{pmatrix}3\\ 4\\ -4\end{pmatrix} \cdot \left(\begin{pmatrix}7\\ 0\\ 0\end{pmatrix} \times \begin{pmatrix}1\\ 2\\ 0\end{pmatrix}\right)\right| = \dfrac{1}{6}\left|\begin{pmatrix}3\\ 4\\ -4\end{pmatrix} \cdot \begin{pmatrix}0\\ 0\\ 14\end{pmatrix}\right| = \dfrac{1}{6} \times 56\)
\(\dfrac{1}{6}\left|\overrightarrow{MA} \cdot \left(\overrightarrow{MN} \times \overrightarrow{MP}\right)\right| = \dfrac{1}{6}\left|\begin{pmatrix}-4\\ 4\\ -4\end{pmatrix} \cdot \left(\begin{pmatrix}-7\\ 0\\ 0\end{pmatrix} \times \begin{pmatrix}-6\\ 2\\ 0\end{pmatrix}\right)\right| = \dfrac{1}{6}\left|\begin{pmatrix}-4\\ 4\\ -4\end{pmatrix} \cdot \begin{pmatrix}0\\ 0\\ -14\end{pmatrix}\right| = \dfrac{1}{6} \times 56\)
\(\dfrac{1}{6}\left|\overrightarrow{PA} \cdot \left(\overrightarrow{PM} \times \overrightarrow{PN}\right)\right| = \dfrac{1}{6}\left|\begin{pmatrix}2\\ 2\\ -4\end{pmatrix} \cdot \left(\begin{pmatrix}6\\ -2\\ 0\end{pmatrix} \times \begin{pmatrix}-1\\ -2\\ 0\end{pmatrix}\right)\right| = \dfrac{1}{6}\left|\begin{pmatrix}2\\ 2\\ -4\end{pmatrix} \cdot \begin{pmatrix}0\\ 0\\ -14\end{pmatrix}\right| = \dfrac{1}{6} \times 56\)

Note candidates may write as \(\dfrac{1}{6}\begin{vmatrix}4 & -4 & 4\\ -3 & -4 & 4\\ -2 & -2 & 4\end{vmatrix} = \dfrac{1}{6}\left|4(-16 + 8) + 4(-12 + 8) + 4(6 - 8)\right| = \dfrac{1}{6}|-56| = \dfrac{28}{3}\)

M1: A complete attempt at the volume of \(ABCD\), with correct method for cross product (oe in other methods). Condone sign slips except they must be using \(-\mathbf{j}\) in the cross product

A1 (M1 on ePen): 147

M1: Finds difference of the two volumes must have used a correct method to find the volumes.

A1: \(\dfrac{413}{3}\)

Note there are many ways to find the required volume of \(ABCD\) applying the triple scalar product to a combination of the following vectors

\(\overrightarrow{AB} = \begin{pmatrix}6\\ -6\\ 6\end{pmatrix}\ \overrightarrow{AC} = \begin{pmatrix}-9\\ -12\\ 12\end{pmatrix}\ \overrightarrow{AD} = \begin{pmatrix}-7\\ -7\\ 14\end{pmatrix}\ \overrightarrow{BA} = \begin{pmatrix}-6\\ 6\\ -6\end{pmatrix}\ \overrightarrow{BC} = \begin{pmatrix}-15\\ -6\\ 6\end{pmatrix}\ \overrightarrow{BD} = \begin{pmatrix}-13\\ -1\\ 8\end{pmatrix}\)
\(\overrightarrow{CA} = \begin{pmatrix}9\\ 12\\ -12\end{pmatrix}\ \overrightarrow{CB} = \begin{pmatrix}15\\ 6\\ -6\end{pmatrix}\ \overrightarrow{CD} = \begin{pmatrix}2\\ 5\\ 2\end{pmatrix}\ \overrightarrow{DA} = \begin{pmatrix}7\\ 7\\ -14\end{pmatrix}\ \overrightarrow{DB} = \begin{pmatrix}13\\ 1\\ -8\end{pmatrix}\ \overrightarrow{DC} = \begin{pmatrix}-2\\ -5\\ -2\end{pmatrix}\)

For example
\(\dfrac{1}{6}\left|\overrightarrow{AB} \cdot \left(\overrightarrow{AC} \times \overrightarrow{AD}\right)\right| = \dfrac{1}{6}\left|\begin{pmatrix}6\\ -6\\ 6\end{pmatrix} \cdot \left(\begin{pmatrix}-9\\ -12\\ 12\end{pmatrix} \times \begin{pmatrix}-7\\ -7\\ 14\end{pmatrix}\right)\right| = \dfrac{1}{6}\left|\begin{pmatrix}6\\ -6\\ 6\end{pmatrix} \cdot \begin{pmatrix}-84\\ 42\\ -21\end{pmatrix}\right| = \dfrac{1}{6} \times 882 = 147\)
\(\dfrac{1}{6}\left|\overrightarrow{BA} \cdot \left(\overrightarrow{BC} \times \overrightarrow{BD}\right)\right| = \dfrac{1}{6}\left|\begin{pmatrix}-6\\ 6\\ -6\end{pmatrix} \cdot \left(\begin{pmatrix}-15\\ -6\\ 6\end{pmatrix} \times \begin{pmatrix}-13\\ -1\\ 8\end{pmatrix}\right)\right| = \dfrac{1}{6}\left|\begin{pmatrix}-6\\ 6\\ -6\end{pmatrix} \cdot \begin{pmatrix}-42\\ 42\\ -63\end{pmatrix}\right| = \dfrac{1}{6} \times 882 = 147\)
\(\dfrac{1}{6}\left|\overrightarrow{CA} \cdot \left(\overrightarrow{CD} \times \overrightarrow{CB}\right)\right| = \dfrac{1}{6}\left|\begin{pmatrix}9\\ 12\\ -12\end{pmatrix} \cdot \left(\begin{pmatrix}2\\ 5\\ 2\end{pmatrix} \times \begin{pmatrix}15\\ 6\\ -6\end{pmatrix}\right)\right| = \dfrac{1}{6}\left|\begin{pmatrix}9\\ 12\\ -12\end{pmatrix} \cdot \begin{pmatrix}-42\\ 42\\ -63\end{pmatrix}\right| = \dfrac{1}{6} \times 882 = 147\)
\(\dfrac{1}{6}\left|\overrightarrow{DA} \cdot \left(\overrightarrow{DB} \times \overrightarrow{DC}\right)\right| = \dfrac{1}{6}\left|\begin{pmatrix}7\\ 7\\ -14\end{pmatrix} \cdot \left(\begin{pmatrix}13\\ 1\\ -8\end{pmatrix} \times \begin{pmatrix}-2\\ -5\\ -2\end{pmatrix}\right)\right| = \dfrac{1}{6}\left|\begin{pmatrix}7\\ 7\\ -14\end{pmatrix} \cdot \begin{pmatrix}-42\\ 42\\ -63\end{pmatrix}\right| = \dfrac{1}{6} \times 882 = 147\)

Note candidates may write as \(\dfrac{1}{6}\begin{vmatrix}6 & -6 & 6\\ -9 & -12 & 12\\ -7 & -7 & 14\end{vmatrix} = \dfrac{1}{6}\left|6(-168 + 84) + 6(-126 + 84) + 6(63 - 84)\right| = \dfrac{1}{6}|-882| = 147\)

(Corrected from the printed mark scheme, in these reference lists: \(\overrightarrow{BC}\) is printed as \(\begin{pmatrix}15\\ -6\\ 6\end{pmatrix}\), \(\overrightarrow{CB}\) as \(\begin{pmatrix}-15\\ 6\\ -6\end{pmatrix}\) and \(\overrightarrow{DC}\) as \(\begin{pmatrix}-2\\ -5\\ 2\end{pmatrix}\); in the worked lines \(\overrightarrow{PM}\) is printed as \(\begin{pmatrix}6\\ 2\\ 0\end{pmatrix}\), \(\overrightarrow{MN} \times \overrightarrow{MP}\) is printed as \(\begin{pmatrix}0\\ 0\\ 14\end{pmatrix}\) and \(\overrightarrow{DB} \times \overrightarrow{DC}\) as \(\begin{pmatrix}42\\ -42\\ 63\end{pmatrix}\). The volumes are unchanged.)

Notes

M1: Attempts to find 2 edges of triangle \(DEF\). Must be subtracting components.

M1: Uses the correct process of the vector product to attempt the area including use of Pythagoras

A1*: Deduces the correct area with no errors

Alternative for (a) using trigonometry

M1: Attempts to find 3 edges of triangle \(DEF\). Must be subtracting components.

M1: A complete method for the area. Allow work in decimals for this mark but must work in exact terms to obtain the A mark

A1*: Deduces the correct area with no errors and no decimal work

Notes

M1: Adopts a correct strategy to find the fourth vertex of the tetrahedron e.g. finding where two edges intersect or uses the linear scale factor

A1: Correct coordinates for the other vertex

M1: Uses the information from the design to attempt a scalar triple product between appropriate vectors to find the volume of the smaller or larger tetrahedron.

A1: Correct volume for either tetrahedron

M1: Makes further progress with the solution by finding the volume of the other tetrahedron or calculates the volume scale factor using an appropriate method. E.g. using ratios or by finding the area of triangle \(DEF\) and comparing with triangle \(ABC\)

dM1: Completes the problem by finding the required volume of the frustum. Depends on all previous method marks

A1: Correct answer

Alternative for part (b) – splits into 4 tetrahedra

This example takes \(M\) as the midpoint of \(AC\) and finds the volume of \(ABMD\), \(MBFC\), \(DMFB\), \(EDFB\)

M1: Adopts a correct strategy to find the volume of one tetrahedron

A1: Any correct volume

M1: Adopts a correct strategy to find the volumes of at least 2 other tetrahedra

A1: Correct volumes

M1: Makes further progress with the solution by finding the volume of all relevant tetrahedra

dM1: Completes the problem by finding the required volume of the frustum. Depends on all previous method marks

A1: Correct answer

Notes

Way 1

M1: Complete correct process of taking the vector product between 2 edges of triangle \(ABC\), applying Pythagoras and multiplying the result by 0.5

M1: Uses a correct method to find any 2 edges of triangle \(ABC\)

M1: Attempts to take the vector cross product between 2 edges of triangle \(ABC\)

A1: Deduces the correct area of either \(9\sqrt{5}\ (\text{cm}^2)\) or awrt \(20.1\ (\text{cm}^2)\)

Note: For Way 1 and Way 2, using any of \(\overrightarrow{OA}\), \(\overrightarrow{OB}\) or \(\overrightarrow{OC}\) in their vector product is M0 M0 A0 A0

(a) Way 2

M1: See scheme

M1: Uses a correct method to find any 2 edges of triangle \(ABC\)

M1: Either

• finds an angle in \(ABC\) by using a correct scalar product method
• finds an angle in \(ABC\) by using the cosine rule in the correct direction
• realises triangle \(ABC\) is isosceles and applies Pythagoras in the correct direction to find the perpendicular height

A1: Deduces the correct area as either \(9\sqrt{5}\ (\text{cm}^2)\) or awrt \(20.1\ (\text{cm}^2)\)

(Corrected from the printed mark scheme: the Note in Way 2 is printed as \(B\hat{A}C = 27.905\ldots,\ A\hat{B}C = 76.047\ldots,\ B\hat{C}A = 76.047\ldots\); since \(AC = BC = \sqrt{86}\), the angles at \(A\) and \(B\) are equal, and the angle at \(C\) is \(27.905\ldots\).)

(a) Way 3

M1: See scheme

M1: Attempts to apply \(\overrightarrow{OA} \times \overrightarrow{OB}\), \(\overrightarrow{OB} \times \overrightarrow{OC}\) and \(\overrightarrow{OC} \times \overrightarrow{OA}\)

A1: Attempts to add (as vectors) the results of applying \(\overrightarrow{OA} \times \overrightarrow{OB}\), \(\overrightarrow{OB} \times \overrightarrow{OC}\) and \(\overrightarrow{OC} \times \overrightarrow{OA}\)

A1: Deduces the correct area as either \(9\sqrt{5}\ (\text{cm}^2)\) or awrt \(20.1\ (\text{cm}^2)\)

Notes

M1: See scheme

M1: Uses appropriate vectors in an attempt at the scalar triple product

A1: Correct applied expression for the scalar triple product (allow \(\pm\) and ignore modulus sign)

A1: Correct solution leading to \(d = 6\)

Note: Using any of \(\overrightarrow{OA}\), \(\overrightarrow{OB}\), \(\overrightarrow{OC}\) or \(\overrightarrow{OD}\) in their scalar triple product is M0 M0 A0 A0

Note: Some vector product calculations for reference:

Notes

M1: Uses a correct method to find any 2 edges of triangle \(ABC\)

M1: Complete process of taking the vector product between 2 edges of triangle \(ABC\), applying Pythagoras and multiplying the result by 0.5

A1*: Deduces the correct area of \(52.2\ (\text{mm}^2)\). Condone awrt 52.2

Note: Condone \(\dfrac{1}{2}\left|74\mathbf{i} - 26\mathbf{j} + 69\mathbf{k}\right| = \dfrac{1}{2}\sqrt{(74)^2 + (-26)^2 + (69)^2} = 52.2\), o.e. for M1M1A1

Note: As an alternative, \(\dfrac{1}{2}\sqrt{129}\sqrt{101}\sin(66.2343\ldots) = 52.2\ (1\text{ dp})\), where the angle has been found by applying the scalar product between \(\overrightarrow{AB}\) and \(\overrightarrow{AC}\)

(Corrected from the printed mark scheme: the first line of the scheme gives \(C\) as \((10, 15, -3)\); \(C\) is \((5, 8, 5)\).)

Notes

M1: See scheme

M1: Uses appropriate vectors to in an attempt at the scalar triple product

A1: Correct numerical expression for the scalar triple product (allow \(\pm\))

A1: Correct volume (in \(\text{mm}^3\)) (allow \(\pm\))

M1: A correct method for changing their units for their volume and for finding density

A1: Obtains the correct density in \(\text{g cm}^{-3}\). Allow awrt 2.6

Note: Using any of \(\overrightarrow{OA}\), \(\overrightarrow{OB}\), \(\overrightarrow{OC}\) or \(\overrightarrow{OD}\) in their scalar triple product is M0M0A0A0

Note: Allow M1M1A0A0 for

(Corrected from the printed mark scheme: the last value is printed as \(135.354216\ldots\); \(\dfrac{818.125296\ldots}{6} = 136.354216\ldots\))

Note: Some vector product calculations for reference:

Note: Some candidates apply \(\overrightarrow{AB} \times \overrightarrow{AC}\) incorrectly to give \(\begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k}\\ -2 & 11 & -2\\ -7 & 4 & 6\end{vmatrix} = 74\mathbf{i} - 26\mathbf{j} + 69\mathbf{k}\)

This leads to an incorrect \(\left|\overrightarrow{AD}.\left(\overrightarrow{AB} \times \overrightarrow{AC}\right)\right| = \left|\begin{pmatrix}-10\\ -2\\ -5\end{pmatrix} \bullet \begin{pmatrix}74\\ -26\\ 69\end{pmatrix}\right| = |-740 + 52 - 345| = 1033\)