| Exam Board | Edexcel |
|---|---|
| Module | C34 (Core Mathematics 3 & 4) |
| Year | 2019 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors 3D & Lines |
| Type | Point on line satisfying distance or other condition |
| Difficulty | Standard +0.3 This is a standard multi-part vectors question requiring routine techniques: finding direction vectors, writing line equations, using the scalar product for angles, and applying the triangle area formula ½ab sin C. Part (e) requires recognizing that doubling the area means doubling the base length, which is straightforward once the pattern is seen. All techniques are textbook exercises with no novel insight required, making this slightly easier than average. |
| Spec | 1.05b Sine and cosine rules: including ambiguous case1.05c Area of triangle: using 1/2 ab sin(C)1.10a Vectors in 2D: i,j notation and column vectors1.10b Vectors in 3D: i,j,k notation1.10c Magnitude and direction: of vectors1.10d Vector operations: addition and scalar multiplication1.10g Problem solving with vectors: in geometry |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\overrightarrow{AB} = (3\mathbf{i}+2\mathbf{j}+5\mathbf{k})-(2\mathbf{i}-3\mathbf{j}-2\mathbf{k})\) | M1 | For subtracting either way; accept \(\pm((3\mathbf{i}+2\mathbf{j}+5\mathbf{k})-(2\mathbf{i}-3\mathbf{j}-2\mathbf{k}))\); implied by two out of three terms correct |
| \(\overrightarrow{AB} = (\mathbf{i}+5\mathbf{j}+7\mathbf{k})\) | A1 | Correct vector form (not column coordinates) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\mathbf{r} = (2\mathbf{i}-3\mathbf{j}-2\mathbf{k}) + \lambda(\mathbf{i}+5\mathbf{j}+7\mathbf{k})\) | M1 | Correct method for \(l\); needs a point on \(l\) (usually A or B) \(\pm\lambda\) times their (a), or attempt at \(\pm\overrightarrow{AB}\) |
| Fully correct equation including "\(\mathbf{r}=\)" | A1 | Not with \(\mathbf{i}\), \(\mathbf{j}\), \(\mathbf{k}\) within column vectors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\overrightarrow{AC} = (2\mathbf{i}+4\mathbf{j}-3\mathbf{k})-(2\mathbf{i}-3\mathbf{j}-2\mathbf{k}) = (0\mathbf{i}+7\mathbf{j}-1\mathbf{k})\) | M1 | Using vectors \(\overrightarrow{OA}\) and \(\overrightarrow{OC}\) and subtracting either way; implied by two out of three terms correct |
| \(\overrightarrow{AB}\cdot\overrightarrow{AC} = \lvert AB\rvert\lvert AC\rvert\cos\theta \Rightarrow 0+35-7 = \sqrt{75}\sqrt{50}\cos\theta\) | M1 | Attempt at \(\overrightarrow{AB}\cdot\overrightarrow{AC} = \lvert AB\rvert\lvert AC\rvert\cos\theta\) with their \(\overrightarrow{AB}\) or \(\overrightarrow{BA}\) and their \(\overrightarrow{AC}\) or \(\overrightarrow{CA}\); can use cosine rule |
| \(\cos\theta = \frac{28}{\sqrt{75}\sqrt{50}}\) or \(\frac{14\sqrt{6}}{75}\) | A1 | Correct value for \(\cos\theta\) (allow awrt \(\pm 0.457\)) |
| \(\theta = 62.8°\) | A1* | Cso; if \(117.2°\) obtained, minimum expected is \(180°-117.2°=62.8°\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\text{Area} = \frac{1}{2}\lvert\mathbf{i}+5\mathbf{j}+7\mathbf{k}\rvert\lvert 7\mathbf{j}-\mathbf{k}\rvert\sin(62.8°) = \frac{1}{2}\sqrt{75}\sqrt{50}\sin(62.8°)\) | M1 | Uses \(\frac{1}{2}\,their\,\lvert AB\rvert \times their\,\lvert AC\rvert\sin(62.8°)\) |
| \(= 27.2\) | A1 | Allow awrt 27.2; correct answer only scores both marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\overrightarrow{OD} = (2\mathbf{i}-3\mathbf{j}-2\mathbf{k}) \pm 2\times(\mathbf{i}+5\mathbf{j}+7\mathbf{k})\) or \((3\mathbf{i}+2\mathbf{j}+5\mathbf{k})+(\mathbf{i}+5\mathbf{j}+7\mathbf{k})\) or \((3\mathbf{i}+2\mathbf{j}+5\mathbf{k})-3\times(\mathbf{i}+5\mathbf{j}+7\mathbf{k})\) | M1 | Method of finding one coordinate or position vector of point \(D\) |
| \(\overrightarrow{OD} = (4\mathbf{i}+7\mathbf{j}+12\mathbf{k})\) and \((0\mathbf{i}-13\mathbf{j}-16\mathbf{k})\) | A1 A1 | A1: one set of coordinates correct; A1: both position vectors correct; do not isw |
## Question 12:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\overrightarrow{AB} = (3\mathbf{i}+2\mathbf{j}+5\mathbf{k})-(2\mathbf{i}-3\mathbf{j}-2\mathbf{k})$ | M1 | For subtracting either way; accept $\pm((3\mathbf{i}+2\mathbf{j}+5\mathbf{k})-(2\mathbf{i}-3\mathbf{j}-2\mathbf{k}))$; implied by two out of three terms correct |
| $\overrightarrow{AB} = (\mathbf{i}+5\mathbf{j}+7\mathbf{k})$ | A1 | Correct vector form (not column coordinates) |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\mathbf{r} = (2\mathbf{i}-3\mathbf{j}-2\mathbf{k}) + \lambda(\mathbf{i}+5\mathbf{j}+7\mathbf{k})$ | M1 | Correct method for $l$; needs a point on $l$ (usually A or B) $\pm\lambda$ times their (a), or attempt at $\pm\overrightarrow{AB}$ |
| Fully correct equation including "$\mathbf{r}=$" | A1 | Not with $\mathbf{i}$, $\mathbf{j}$, $\mathbf{k}$ within column vectors |
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\overrightarrow{AC} = (2\mathbf{i}+4\mathbf{j}-3\mathbf{k})-(2\mathbf{i}-3\mathbf{j}-2\mathbf{k}) = (0\mathbf{i}+7\mathbf{j}-1\mathbf{k})$ | M1 | Using vectors $\overrightarrow{OA}$ and $\overrightarrow{OC}$ and subtracting either way; implied by two out of three terms correct |
| $\overrightarrow{AB}\cdot\overrightarrow{AC} = \lvert AB\rvert\lvert AC\rvert\cos\theta \Rightarrow 0+35-7 = \sqrt{75}\sqrt{50}\cos\theta$ | M1 | Attempt at $\overrightarrow{AB}\cdot\overrightarrow{AC} = \lvert AB\rvert\lvert AC\rvert\cos\theta$ with their $\overrightarrow{AB}$ or $\overrightarrow{BA}$ and their $\overrightarrow{AC}$ or $\overrightarrow{CA}$; can use cosine rule |
| $\cos\theta = \frac{28}{\sqrt{75}\sqrt{50}}$ or $\frac{14\sqrt{6}}{75}$ | A1 | Correct value for $\cos\theta$ (allow awrt $\pm 0.457$) |
| $\theta = 62.8°$ | A1* | Cso; if $117.2°$ obtained, minimum expected is $180°-117.2°=62.8°$ |
### Part (d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\text{Area} = \frac{1}{2}\lvert\mathbf{i}+5\mathbf{j}+7\mathbf{k}\rvert\lvert 7\mathbf{j}-\mathbf{k}\rvert\sin(62.8°) = \frac{1}{2}\sqrt{75}\sqrt{50}\sin(62.8°)$ | M1 | Uses $\frac{1}{2}\,their\,\lvert AB\rvert \times their\,\lvert AC\rvert\sin(62.8°)$ |
| $= 27.2$ | A1 | Allow awrt 27.2; correct answer only scores both marks |
### Part (e):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\overrightarrow{OD} = (2\mathbf{i}-3\mathbf{j}-2\mathbf{k}) \pm 2\times(\mathbf{i}+5\mathbf{j}+7\mathbf{k})$ or $(3\mathbf{i}+2\mathbf{j}+5\mathbf{k})+(\mathbf{i}+5\mathbf{j}+7\mathbf{k})$ or $(3\mathbf{i}+2\mathbf{j}+5\mathbf{k})-3\times(\mathbf{i}+5\mathbf{j}+7\mathbf{k})$ | M1 | Method of finding one coordinate or position vector of point $D$ |
| $\overrightarrow{OD} = (4\mathbf{i}+7\mathbf{j}+12\mathbf{k})$ and $(0\mathbf{i}-13\mathbf{j}-16\mathbf{k})$ | A1 A1 | A1: one set of coordinates correct; A1: both position vectors correct; do not isw |
---\begin{enumerate}
\item Relative to a fixed origin $O$,\\
the point $A$ has position vector $( 2 \mathbf { i } - 3 \mathbf { j } - 2 \mathbf { k } )$\\
the point $B$ has position vector $( 3 \mathbf { i } + 2 \mathbf { j } + 5 \mathbf { k } )$\\
the point $C$ has position vector ( $2 \mathbf { i } + 4 \mathbf { j } - 3 \mathbf { k }$ )
\end{enumerate}
The line $l$ passes through the points $A$ and $B$.\\
(a) Find the vector $\overrightarrow { A B }$.\\
(b) Find a vector equation for the line $l$.\\
(c) Show that the size of the angle $C A B$ is $62.8 ^ { \circ }$, to one decimal place.\\
(d) Hence find the area of triangle $C A B$, giving your answer to 3 significant figures.
The point $D$ lies on the line $l$. Given that the area of triangle $C A D$ is twice the area of triangle $C A B$,\\
(e) find the two possible position vectors of point $D$.\\
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\hfill \mbox{\textit{Edexcel C34 2019 Q12 [13]}}