C4 June 2018 Q7

Notes

Note: If no “subtraction” seen, you can award 1^st M1 for 2 out of 3 correct components of the difference

Note: For dM1 the dot product formula can be applied as \(\sqrt{(12)^2 + (-6)^2 + (6)^2}\,.\sqrt{(4)^2 + (-6)^2 + (2)^2}\cos\theta = \begin{pmatrix}12\\-6\\6\end{pmatrix} \bullet \begin{pmatrix}4\\-6\\2\end{pmatrix}\)

Note: Evaluation of the dot product for \(12\mathbf{i} - 6\mathbf{j} + 6\mathbf{k}\) & \(2\mathbf{i} - 3\mathbf{j} + \mathbf{k}\) is not required for the dM1 mark

A1: For either \(\dfrac{4}{\sqrt{21}}\) or \(\dfrac{4}{21}\sqrt{21}\) or \(\cos\theta = \dfrac{4}{\sqrt{21}}\) or \(\dfrac{4}{21}\sqrt{21}\)

Note: Using \(12\mathbf{i} - 6\mathbf{j} + 6\mathbf{k}\) & \(2\mathbf{i} - 3\mathbf{j} + \mathbf{k}\) gives \(\cos\theta = \dfrac{24 + 18 + 6}{\sqrt{216}.\sqrt{14}} = \dfrac{48}{12\sqrt{21}} = \dfrac{4}{\sqrt{21}}\) or \(\dfrac{4}{21}\sqrt{21}\)

Note: Using \(2\mathbf{i} - \mathbf{j} + \mathbf{k}\) & \(2\mathbf{i} - 3\mathbf{j} + \mathbf{k}\) gives \(\cos\theta = \dfrac{4 + 3 + 1}{\sqrt{6}.\sqrt{14}} = \dfrac{8}{2\sqrt{21}} = \dfrac{4}{\sqrt{21}}\) or \(\dfrac{4}{21}\sqrt{21}\)

Note: Give M1M1A0 for finding \(\theta =\) awrt 29.2 without reference to \(\cos\theta = \dfrac{4}{\sqrt{21}}\) or \(\dfrac{4}{21}\sqrt{21}\)

Note: Condone taking the dot product between vectors the wrong way round for the M1 dM1 marks

Note: Vectors the wrong way round

• E.g. taking the dot product between \(\overrightarrow{PA}\) and \(\overrightarrow{AB}\) to give \(\cos\theta = -\dfrac{4}{\sqrt{21}}\) or \(-\dfrac{4}{21}\sqrt{21}\) with no other working is final A0
• E.g. taking the dot product between \(\overrightarrow{PA}\) and \(\overrightarrow{AB}\) to give \(\cos\theta = -\dfrac{4}{\sqrt{21}}\) or \(-\dfrac{4}{21}\sqrt{21}\) followed by \(\cos\theta = \dfrac{4}{\sqrt{21}}\) or \(\dfrac{4}{21}\sqrt{21}\) or just simply writing \(\dfrac{4}{\sqrt{21}}\) or \(\dfrac{4}{21}\sqrt{21}\) is final A1

Note: In part (b), give M0dM0 for finding and using \(\overrightarrow{AP} = \overrightarrow{OP} - \overrightarrow{AB} = (5\mathbf{i} + 7\mathbf{j} + 6\mathbf{k})\)

Alt 1 for part (b): Vector Cross Product

Use this scheme if a vector cross product method is being applied

Alt 2 for part (b): Cosine Rule

Notes

Note: Give 1^st M0 for \(\sin\theta = \sin\left(\cos^{-1}\left(\dfrac{4\sqrt{21}}{21}\right)\right)\) or \(\sin\theta = 1 - \left(\dfrac{4}{21}\sqrt{21}\right)^2\) unless recovered

M1: Give 2^nd M1 for either

• \(\dfrac{1}{2}(\text{their length } AP)(\text{their length } AB)(\text{their attempt at } \sin\theta)\)
• \(\dfrac{1}{2}(\text{their length } AP)(\text{their length } AB)\sin(\text{their } 29.2^\circ \text{ from part (b)})\)
• \(\dfrac{1}{2}(\text{their length } AP)(\text{their length } AB)\sin\theta;\) where \(\cos\theta = \ldots\) in part (b)

Note: \(\dfrac{1}{2}\left(\sqrt{216}\right)\left(\sqrt{56}\right)\sin(\text{awrt } 29.2^\circ \text{ or awrt } 150.8^\circ)\ \{= \text{awrt } 26.8\}\) without reference to finding \(\sin\theta\) as an exact value if M0 M1 A0

Note: Anything that rounds to 26.8 without reference to finding \(\sin\theta\) as an exact value is M0 M1 A0

Note: Anything that rounds to 26.8 without reference to \(12\sqrt{5}\) is A0

Note: If they use \(\overrightarrow{AP} = \overrightarrow{OP} - \overrightarrow{AB} = (5\mathbf{i} + 7\mathbf{j} + 6\mathbf{k})\) in part (b), then this can be followed through in part (c) for the 2^nd M mark as e.g. \(\dfrac{1}{2}\left(\sqrt{110}\right)\left(\sqrt{56}\right)\sin\theta\)

Note: Finding \(12\sqrt{5}\) in part (c) is M1 dM1 A1, even if there is little or no evidence of finding an exact value for \(\sin\theta\). So \(\dfrac{1}{2}\left(\sqrt{216}\right)\left(\sqrt{56}\right)\sin(29.2^\circ) = 12\sqrt{5}\) is M1 dM1 A1

Alt 1 for part (c): Vector Cross Product

Use this scheme if a vector cross product method is being applied

Alt 2 for part (c)

Notes

Note: Writing \(\mathbf{r} = \ldots\) or \(l_2 = \ldots\) or \(l = \ldots\) or Line 2 = … is not required for the M mark

A1: Writing \(\mathbf{r} = \begin{pmatrix}9\\1\\8\end{pmatrix} + \mu\begin{pmatrix}4\\-6\\2\end{pmatrix}\) or \(\mathbf{r} = \begin{pmatrix}9\\1\\8\end{pmatrix} + \mu\begin{pmatrix}2\\-3\\1\end{pmatrix}\) or \(\mathbf{r} = \begin{pmatrix}9\\1\\8\end{pmatrix} + \mu\mathbf{d}\), where \(\mathbf{d} =\) a multiple of \(2\mathbf{i} - 3\mathbf{j} + \mathbf{k}\)

Note: Writing \(\mathbf{r} = \ldots\) or \(l_2 = \ldots\) or \(l = \ldots\) or Line 2 = … is required for the A mark

Note: Other valid \(\mathbf{p} = \begin{pmatrix}9\\1\\8\end{pmatrix}\) are e.g. \(\mathbf{p} = \begin{pmatrix}13\\-5\\10\end{pmatrix}\) or \(\mathbf{p} = \begin{pmatrix}5\\7\\6\end{pmatrix}\). So \(\mathbf{r} = \begin{pmatrix}13\\-5\\10\end{pmatrix} + \mu\begin{pmatrix}4\\-6\\2\end{pmatrix}\) is M1 A1

Note: Give A0 for writing \(l_2: \begin{pmatrix}9\\1\\8\end{pmatrix} + \mu\begin{pmatrix}4\\-6\\2\end{pmatrix}\) or ans \(= \begin{pmatrix}9\\1\\8\end{pmatrix} + \mu\begin{pmatrix}4\\-6\\2\end{pmatrix}\) unless recovered

Note: Using scalar parameter \(\lambda\) or other scalar parameters (e.g. \(\mu\) or \(s\) or \(t\)) is fine for M1 and/or A1

Notes

ddM1: Substitutes their value of \(\mu\) into \(\overrightarrow{OQ}\), where \(\overrightarrow{OQ}\) = their equation for \(l_2\)

Note: If they use \(\overrightarrow{AP} = \overrightarrow{OP} - \overrightarrow{AB} = (5\mathbf{i} + 7\mathbf{j} + 6\mathbf{k})\) in part (b), then this can be followed through in part (e) for the 2^nd M mark and the 3^rd M mark

Note: You imply the final M mark in part (e) for at least 2 correctly followed through components for \(Q\) from their \(\mu\)

Alt 1 for part (e)