| Exam Board | AQA |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2007 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors 3D & Lines |
| Type | Show lines intersect and find intersection point |
| Difficulty | Standard +0.3 This is a standard 3D vectors question requiring routine techniques: (a) dot product of direction vectors, (b) equating components and solving simultaneous equations, (c) finding a point on a line given a distance condition. Part (c) requires recognizing that P is the midpoint or using the perpendicularity established earlier, but all steps follow standard procedures with no novel insight required. Slightly easier than average due to the structured scaffolding. |
| Spec | 4.04a Line equations: 2D and 3D, cartesian and vector forms4.04c Scalar product: calculate and use for angles4.04e Line intersections: parallel, skew, or intersecting |
| Answer | Marks | Guidance |
|---|---|---|
| \(\begin{vmatrix} 3 & 1 \\ -3 & -3 \\ -1 & -3 \end{vmatrix} = 3 - 6 + 3 = 0\) | M1 | |
| \(= 0 \Rightarrow\) perpendicular | A1 | 2 marks |
| Allow \(\begin{vmatrix} 3 \\ -6 \\ 3 \end{vmatrix}\) but not \(\begin{vmatrix} -6 \end{vmatrix}\) = 0 |
| Answer | Marks | Guidance |
|---|---|---|
| \(8 + 3\lambda = -4 + \mu\) | M1 | |
| \(6 - 3\lambda = 2\mu\) | ||
| \(-9 - \lambda = 11 - 3\mu\) | ||
| \(\lambda = -2, \mu = -2\) | m1 A1 | |
| intersect at \((2.12, -7)\) | A1 | 5 marks |
| Alt (for last two marks) | ||
| substitute \(\lambda\) into \(l_1\) and \(\mu\) into \(l_2\) | m1 | |
| intersect at \((2.12, -7)\), condone \(\begin{vmatrix} 2 \\ 12 \\ -7 \end{vmatrix}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\overrightarrow{AP} = \begin{vmatrix} 6 \\ 12 \\ -18 \end{vmatrix}\) | M1 | |
| \(AP^2 = 504\) | A1F | |
| \(AB^2 = 2AP^2\) | M1 | |
| \(AB = 12\sqrt{7}\) | A1 | 4 marks |
**7(a)**
| $\begin{vmatrix} 3 & 1 \\ -3 & -3 \\ -1 & -3 \end{vmatrix} = 3 - 6 + 3 = 0$ | M1 | | attempt at sp, 3 terms, added |
| $= 0 \Rightarrow$ perpendicular | A1 | 2 marks | = 0 ⇒ perpendicular seen (or $\cos\theta = 0 \Rightarrow \theta = 90°$) |
| | | | Allow $\begin{vmatrix} 3 \\ -6 \\ 3 \end{vmatrix}$ but not $\begin{vmatrix} -6 \end{vmatrix}$ = 0 |
**7(b)**
| $8 + 3\lambda = -4 + \mu$ | M1 | | set up any two equations |
| $6 - 3\lambda = 2\mu$ | | | |
| $-9 - \lambda = 11 - 3\mu$ | | | |
| $\lambda = -2, \mu = -2$ | m1 A1 | | solve for $\lambda$ and $\mu$; substitute $\lambda, \mu$ in third equation |
| intersect at $(2.12, -7)$ | A1 | 5 marks | CAO |
| **Alt (for last two marks)** | | | |
| substitute $\lambda$ into $l_1$ and $\mu$ into $l_2$ | m1 | | |
| intersect at $(2.12, -7)$, condone $\begin{vmatrix} 2 \\ 12 \\ -7 \end{vmatrix}$ | A1 | | (2, 12, –7) found from both lines; Note: working for (b) done in (a): award marks in (b) |
**7(c)**
| $\overrightarrow{AP} = \begin{vmatrix} 6 \\ 12 \\ -18 \end{vmatrix}$ | M1 | | $\overrightarrow{AP} = \pm$ their $\overrightarrow{OP} - \begin{vmatrix} 0 \\ -4 \\ 11 \end{vmatrix}$ |
| $AP^2 = 504$ | A1F | | ft on $P$ |
| $AB^2 = 2AP^2$ | M1 | | Calculate $AB^2$ |
| $AB = 12\sqrt{7}$ | A1 | 4 marks | OE accept 31.7 or better |7 The lines $l _ { 1 }$ and $l _ { 2 }$ have equations $\mathbf { r } = \left[ \begin{array} { r } 8 \\ 6 \\ - 9 \end{array} \right] + \lambda \left[ \begin{array} { r } 3 \\ - 3 \\ - 1 \end{array} \right]$ and $\mathbf { r } = \left[ \begin{array} { r } - 4 \\ 0 \\ 11 \end{array} \right] + \mu \left[ \begin{array} { r } 1 \\ 2 \\ - 3 \end{array} \right]$ respectively.
\begin{enumerate}[label=(\alph*)]
\item Show that $l _ { 1 }$ and $l _ { 2 }$ are perpendicular.
\item Show that $l _ { 1 }$ and $l _ { 2 }$ intersect and find the coordinates of the point of intersection, $P$.
\item The point $A ( - 4,0,11 )$ lies on $l _ { 2 }$. The point $B$ on $l _ { 1 }$ is such that $A P = B P$.
Find the length of $A B$.
\end{enumerate}
\hfill \mbox{\textit{AQA C4 2007 Q7 [11]}}