A2 June 2019 Q7
7. With respect to a fixed origin \(O\), the points \(A\), \(B\) and \(C\) have coordinates \((3, 4, 5)\), \((10, -1, 5)\) and \((4, 7, -9)\) respectively.
The plane \(\Pi\) has equation \(4x - 8y + z = 2\)
The line segment \(AB\) meets the plane \(\Pi\) at the point \(P\) and the line segment \(BC\) meets the plane \(\Pi\) at the point \(Q\).
The point \(D\) has coordinates \((k, 4, -1)\), where \(k\) is a constant.
Given that the vectors \(\overrightarrow{AB}\), \(\overrightarrow{AC}\) and \(\overrightarrow{AD}\) form three edges of a parallelepiped of volume 226
| Scheme | Marks | AO |
|---|---|---|
| Examples: Area \(APQC\) = Area \(ABC\) – Area \(PBQ\) Area \(APQC\) = Area \(APC\) + Area \(CPQ\) Area \(APQC\) = Area \(APQ\) + Area \(AQC\) Area \(APQC = \dfrac{1}{2}\left|\mathbf{AQ} \times \mathbf{PC}\right|\) | M1 | 3.1a |
| Line \(AB\): \(\mathbf{r} = \begin{pmatrix}3\\ 4\\ 5\end{pmatrix} + \lambda\begin{pmatrix}10 - 3\\ -1 - 4\\ 5 - 5\end{pmatrix} = \begin{pmatrix}3\\ 4\\ 5\end{pmatrix} + \lambda\begin{pmatrix}7\\ -5\\ 0\end{pmatrix}\) or Line \(BC\): \(\mathbf{r} = \begin{pmatrix}10\\ -1\\ 5\end{pmatrix} + \mu\begin{pmatrix}10 - 4\\ -1 - 7\\ 5 + 9\end{pmatrix} = \begin{pmatrix}10\\ -1\\ 5\end{pmatrix} + \mu\begin{pmatrix}6\\ -8\\ 14\end{pmatrix}\) | M1 | 3.1a |
| \(4(3 + 7\lambda) - 8(4 - 5\lambda) + 5 = 2 \Rightarrow \lambda = \ldots \Rightarrow P\) is … or \(4(10 + 6\mu) - 8(-1 - 8\mu) + 5 + 14\mu = 2 \Rightarrow \mu = \ldots \Rightarrow Q\) is … \(\left(\text{NB } \lambda = \dfrac{1}{4},\ \mu = -\dfrac{1}{2}\right)\) | M1 | 2.1 |
| \(P(4.75,\ 2.75,\ 5)\) and \(Q(7,\ 3,\ -2)\) | A1 | 1.1b |
| Area \(ABC = \dfrac{1}{2}\begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k}\\ 7 & -5 & 0\\ 6 & -8 & 14\end{vmatrix} = \dfrac{1}{2}\sqrt{70^2 + 98^2 + 26^2}\) Area \(PBQ = \dfrac{1}{2}\begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k}\\ 5.25 & -3.75 & 0\\ 3 & -4 & 7\end{vmatrix} = \dfrac{1}{2}\sqrt{26.25^2 + 36.75^2 + 9.75^2}\) Area \(APQC = \dfrac{1}{2}\begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k}\\ -4 & 1 & 7\\ 0.75 & -4.25 & 14\end{vmatrix} = \dfrac{1}{2}\sqrt{43.75^2 + 61.25^2 + 16.25^2}\) NB: Area \(APQ = 7.7004\), Area \(AQC = 30.8018\), Area \(CPQ = 23.101\), Area \(APC = 15.4008\) | M1 | 2.1 |
| Area \(ABC\) – Area \(PBQ = 38.5\) * | A1* | 1.1b |
| (6) |
Notes
M1: Identifies a correct strategy to determine the area of the required quadrilateral. The attempt does not need to be complete for this mark so one of the statements (or intentions) in the markscheme would be sufficient.
M1: Correct attempt to find the equation of the line \(AB\) or the line \(BC\)
M1: Uses at least one of their lines and the equation of the given plane to determine the value of at least one of the parameters and hence the coordinates of \(P\) or \(Q\)
A1: Both coordinates correct – allow as vectors and may be implied if for example the candidate calculates the vectors e.g. AP, AQ, CP, CQ without stating the coordinates explicitly
M1: Uses all the required information to calculate appropriate areas correctly leading to the area of the quadrilateral. Needs to be a complete method here.
A1*: Reaches 38.5 with no errors
| Scheme | Marks | AO |
|---|---|---|
| \(\begin{aligned}&\overrightarrow{AB} = \begin{pmatrix}7\\ -5\\ 0\end{pmatrix},\ \overrightarrow{AC} = \begin{pmatrix}1\\ 3\\ -14\end{pmatrix},\ \overrightarrow{AD} = \begin{pmatrix}k - 3\\ 0\\ -6\end{pmatrix}\\[6pt] &\overrightarrow{AB} \times \overrightarrow{AC}.\overrightarrow{AD} = \begin{vmatrix}7 & -5 & 0\\ 1 & 3 & -14\\ k - 3 & 0 & -6\end{vmatrix} = \ldots\end{aligned}\) | M1 | 3.1a |
| \(\overrightarrow{AB} \times \overrightarrow{AC}.\overrightarrow{AD} = 7 \times -18 + 5(-6 + 14k - 42)\) | A1 | 1.1b |
| \(7 \times -18 + 5(-6 + 14k - 42) = \pm 226 \Rightarrow k = \ldots\) | dM1 | 3.1a |
| \(k = 2\) or \(\dfrac{296}{35}\) | A1 | 1.1b |
| (4) | ||
| (10 marks) |
Notes
M1: Adopts a correct strategy by finding suitable vectors and forming the scalar triple product. This is often done in 2 steps e.g.
\[\overrightarrow{AB} \times \overrightarrow{AC} = \begin{pmatrix}70\\ 98\\ 26\end{pmatrix} \text{ or } \overrightarrow{AB} \times \overrightarrow{AD} = \begin{pmatrix}30\\ 42\\ 5k - 15\end{pmatrix} \text{ or } \overrightarrow{AC} \times \overrightarrow{AD} = \begin{pmatrix}-18\\ 48 - 14k\\ -3k + 9\end{pmatrix}\]\[\overrightarrow{AB} \times \overrightarrow{AC}.\overrightarrow{AD} = \begin{pmatrix}70\\ 98\\ 26\end{pmatrix}.\begin{pmatrix}k - 3\\ 0\\ -6\end{pmatrix} = 70k - 210 - 156\]\[\text{or } \overrightarrow{AB} \times \overrightarrow{AD}.\overrightarrow{AC} = \begin{pmatrix}30\\ 42\\ 5k - 15\end{pmatrix}.\begin{pmatrix}1\\ 3\\ -14\end{pmatrix} = 30 + 126 - 70k + 210\]\[\text{or } \overrightarrow{AC} \times \overrightarrow{AD}.\overrightarrow{AB} = \begin{pmatrix}-18\\ 48 - 14k\\ -3k + 9\end{pmatrix}.\begin{pmatrix}7\\ -5\\ 0\end{pmatrix} = -126 + 70k - 240\]If it is not clear that the vector product is being used, at least 2 of the components should be correct.
A1: Correct expression for the triple product in terms of \(k\) (should be \(\pm(70k - 366)\))
Ignore the presence or absence of “1/6” for the first 2 marks
dM1: Realises that \(\pm 226\) is possible for the value of the triple product and attempts to solve to obtain 2 values for \(k\). Dependent on the previous method mark.
A1: Correct values (must be exact)