| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2013 |
| Session | January |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors 3D & Lines |
| Type | Foot of perpendicular from general external point to line |
| Difficulty | Standard +0.3 This is a standard multi-part vectors question covering routine techniques: finding intersection by equating components (part a), using the scalar product formula for angles (part b), and finding the perpendicular from a point to a line using the dot product condition (part c). All three parts follow textbook methods with no novel insight required, though part (c) involves slightly more algebraic manipulation. Slightly easier than average due to the straightforward application of standard procedures. |
| 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 |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Write equations: i: \(9 + \lambda = 2 + 2\mu\), j: \(13 + 4\lambda = -1 + \mu\), k: \(-3 - 2\lambda = 1 + \mu\) | M1 | Any two equations, allow one slip |
| Eliminate one parameter, e.g. (2)-(3): \(16 + 6\lambda = -2\) or (2)-4(1): \(-23 = -9 - 7\mu\) | dM1 | An attempt to eliminate one parameter |
| \(\lambda = -3\) or \(\mu = 2\) | A1 | Either value sufficient |
| \(l_1: \mathbf{r} = \begin{pmatrix}9\\13\\-3\end{pmatrix} - 3\begin{pmatrix}1\\4\\-2\end{pmatrix} = \begin{pmatrix}6\\1\\3\end{pmatrix}\) or \(l_2: \mathbf{r} = \begin{pmatrix}2\\-1\\1\end{pmatrix} + 2\begin{pmatrix}2\\1\\1\end{pmatrix} = \begin{pmatrix}6\\1\\3\end{pmatrix}\) | ddM1 A1 | See notes; both dependent on previous method marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\mathbf{d}_1 = \begin{pmatrix}1\\4\\-2\end{pmatrix}\), \(\mathbf{d}_2 = \begin{pmatrix}2\\1\\1\end{pmatrix}\), use dot product \(\begin{pmatrix}1\\4\\-2\end{pmatrix} \cdot \begin{pmatrix}2\\1\\1\end{pmatrix}\) | M1 | Realisation that dot product required between \(\pm A\mathbf{d}_1\) and \(\pm B\mathbf{d}_2\) |
| \(\cos\theta = \pm\left(\dfrac{2+4-2}{\sqrt{(1)^2+(4)^2+(-2)^2} \cdot \sqrt{(2)^2+(1)^2+(1)^2}}\right)\) | A1 | Correct equation |
| \(\cos\theta = \dfrac{4}{\sqrt{21}\cdot\sqrt{6}} \Rightarrow \theta = 69.1238... = 69.1°\) (1 dp) | A1 | awrt 69.1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\mathbf{d}_1 \times \mathbf{d}_2 = \begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\1&4&-2\\2&1&1\end{vmatrix} = 6\mathbf{i} - 5\mathbf{j} - 7\mathbf{k}\) | M1 | Realisation vector cross product required between \(\pm A\mathbf{d}_1\) and \(\pm B\mathbf{d}_2\) |
| \(\sin\theta = \dfrac{\sqrt{(6)^2+(5)^2+(-7)^2}}{\sqrt{(1)^2+(4)^2+(-2)^2}\cdot\sqrt{(2)^2+(1)^2+(1)^2}}\) | A1 | Correct applied equation |
| \(\sin\theta = \dfrac{\sqrt{110}}{\sqrt{21}\cdot\sqrt{6}} \Rightarrow \theta = 69.1\) (1 dp) | A1 | awrt 69.1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\overrightarrow{AP} = \begin{pmatrix}9+\lambda\\13+4\lambda\\-3-2\lambda\end{pmatrix} - \begin{pmatrix}4\\16\\-3\end{pmatrix} = \begin{pmatrix}\lambda+5\\4\lambda-3\\-2\lambda\end{pmatrix}\) | M1 A1 | M1: attempt to find \(\overrightarrow{AP}\) in terms of parameter; A1: correct expression |
| \(\overrightarrow{AP} \cdot \mathbf{d}_1 = 0 \Rightarrow \begin{pmatrix}\lambda+5\\4\lambda-3\\-2\lambda\end{pmatrix} \cdot \begin{pmatrix}1\\4\\-2\end{pmatrix} = \lambda+5+16\lambda-12+4\lambda = 0\) | dM1 | Depends on previous M; scalar product with \(\mathbf{d}_1\) set equal to zero |
| \(21\lambda - 7 = 0 \Rightarrow \lambda = \dfrac{1}{3}\) | A1 | \(\lambda = \frac{1}{3}\) |
| \(\overrightarrow{OP} = \begin{pmatrix}9\\13\\-3\end{pmatrix} + \dfrac{1}{3}\begin{pmatrix}1\\4\\-2\end{pmatrix} = \begin{pmatrix}9\frac{1}{3}\\14\frac{1}{3}\\-3\frac{2}{3}\end{pmatrix}\) or \(\begin{pmatrix}\frac{28}{3}\\\frac{43}{3}\\-\frac{11}{3}\end{pmatrix}\) | ddM1 A1 | ddM1: substitute \(\lambda\) into \(\overrightarrow{OP}\) (not \(\overrightarrow{AP}\)); A1: \(9\frac{1}{3}\mathbf{i}+14\frac{1}{3}\mathbf{j}-3\frac{2}{3}\mathbf{k}\), must be exact |
## Question 7:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Write equations: **i:** $9 + \lambda = 2 + 2\mu$, **j:** $13 + 4\lambda = -1 + \mu$, **k:** $-3 - 2\lambda = 1 + \mu$ | M1 | Any two equations, allow one slip |
| Eliminate one parameter, e.g. **(2)-(3):** $16 + 6\lambda = -2$ or **(2)-4(1):** $-23 = -9 - 7\mu$ | dM1 | An attempt to eliminate one parameter |
| $\lambda = -3$ or $\mu = 2$ | A1 | Either value sufficient |
| $l_1: \mathbf{r} = \begin{pmatrix}9\\13\\-3\end{pmatrix} - 3\begin{pmatrix}1\\4\\-2\end{pmatrix} = \begin{pmatrix}6\\1\\3\end{pmatrix}$ or $l_2: \mathbf{r} = \begin{pmatrix}2\\-1\\1\end{pmatrix} + 2\begin{pmatrix}2\\1\\1\end{pmatrix} = \begin{pmatrix}6\\1\\3\end{pmatrix}$ | ddM1 A1 | See notes; both dependent on previous method marks |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\mathbf{d}_1 = \begin{pmatrix}1\\4\\-2\end{pmatrix}$, $\mathbf{d}_2 = \begin{pmatrix}2\\1\\1\end{pmatrix}$, use dot product $\begin{pmatrix}1\\4\\-2\end{pmatrix} \cdot \begin{pmatrix}2\\1\\1\end{pmatrix}$ | M1 | Realisation that dot product required between $\pm A\mathbf{d}_1$ and $\pm B\mathbf{d}_2$ |
| $\cos\theta = \pm\left(\dfrac{2+4-2}{\sqrt{(1)^2+(4)^2+(-2)^2} \cdot \sqrt{(2)^2+(1)^2+(1)^2}}\right)$ | A1 | Correct equation |
| $\cos\theta = \dfrac{4}{\sqrt{21}\cdot\sqrt{6}} \Rightarrow \theta = 69.1238... = 69.1°$ (1 dp) | A1 | awrt 69.1 |
**Alternative (Cross Product):**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\mathbf{d}_1 \times \mathbf{d}_2 = \begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\1&4&-2\\2&1&1\end{vmatrix} = 6\mathbf{i} - 5\mathbf{j} - 7\mathbf{k}$ | M1 | Realisation vector cross product required between $\pm A\mathbf{d}_1$ and $\pm B\mathbf{d}_2$ |
| $\sin\theta = \dfrac{\sqrt{(6)^2+(5)^2+(-7)^2}}{\sqrt{(1)^2+(4)^2+(-2)^2}\cdot\sqrt{(2)^2+(1)^2+(1)^2}}$ | A1 | Correct applied equation |
| $\sin\theta = \dfrac{\sqrt{110}}{\sqrt{21}\cdot\sqrt{6}} \Rightarrow \theta = 69.1$ (1 dp) | A1 | awrt 69.1 |
### Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\overrightarrow{AP} = \begin{pmatrix}9+\lambda\\13+4\lambda\\-3-2\lambda\end{pmatrix} - \begin{pmatrix}4\\16\\-3\end{pmatrix} = \begin{pmatrix}\lambda+5\\4\lambda-3\\-2\lambda\end{pmatrix}$ | M1 A1 | M1: attempt to find $\overrightarrow{AP}$ in terms of parameter; A1: correct expression |
| $\overrightarrow{AP} \cdot \mathbf{d}_1 = 0 \Rightarrow \begin{pmatrix}\lambda+5\\4\lambda-3\\-2\lambda\end{pmatrix} \cdot \begin{pmatrix}1\\4\\-2\end{pmatrix} = \lambda+5+16\lambda-12+4\lambda = 0$ | dM1 | Depends on previous M; scalar product with $\mathbf{d}_1$ set equal to zero |
| $21\lambda - 7 = 0 \Rightarrow \lambda = \dfrac{1}{3}$ | A1 | $\lambda = \frac{1}{3}$ |
| $\overrightarrow{OP} = \begin{pmatrix}9\\13\\-3\end{pmatrix} + \dfrac{1}{3}\begin{pmatrix}1\\4\\-2\end{pmatrix} = \begin{pmatrix}9\frac{1}{3}\\14\frac{1}{3}\\-3\frac{2}{3}\end{pmatrix}$ or $\begin{pmatrix}\frac{28}{3}\\\frac{43}{3}\\-\frac{11}{3}\end{pmatrix}$ | ddM1 A1 | ddM1: substitute $\lambda$ into $\overrightarrow{OP}$ (not $\overrightarrow{AP}$); A1: $9\frac{1}{3}\mathbf{i}+14\frac{1}{3}\mathbf{j}-3\frac{2}{3}\mathbf{k}$, must be exact |
---7. With respect to a fixed origin $O$, the lines $l _ { 1 }$ and $l _ { 2 }$ are given by the equations
$$\begin{aligned}
& l _ { 1 } : \mathbf { r } = ( 9 \mathbf { i } + 13 \mathbf { j } - 3 \mathbf { k } ) + \lambda ( \mathbf { i } + 4 \mathbf { j } - 2 \mathbf { k } ) \\
& l _ { 2 } : \mathbf { r } = ( 2 \mathbf { i } - \mathbf { j } + \mathbf { k } ) + \mu ( 2 \mathbf { i } + \mathbf { j } + \mathbf { k } )
\end{aligned}$$
where $\lambda$ and $\mu$ are scalar parameters.
\begin{enumerate}[label=(\alph*)]
\item Given that $l _ { 1 }$ and $l _ { 2 }$ meet, find the position vector of their point of intersection.
\item Find the acute angle between $l _ { 1 }$ and $l _ { 2 }$, giving your answer in degrees to 1 decimal place.
Given that the point $A$ has position vector $4 \mathbf { i } + 16 \mathbf { j } - 3 \mathbf { k }$ and that the point $P$ lies on $l _ { 1 }$ such that $A P$ is perpendicular to $l _ { 1 }$,
\item find the exact coordinates of $P$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C4 2013 Q7 [14]}}