| Exam Board | OCR MEI |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors 3D & Lines |
| Type | Area of triangle from given side vectors or coordinates |
| Difficulty | Standard +0.3 This is a straightforward two-part question requiring standard vector techniques: computing dot product to verify perpendicularity (which simplifies part ii), then using the right-angle to calculate area as ½|AB||BC|. Both are routine C4 vector operations with no conceptual challenges, making it slightly easier than average. |
| Spec | 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 multiplication4.04c Scalar product: calculate and use for angles |
| Answer | Marks |
|---|---|
| \(\vec{AB} = \begin{pmatrix}2\\3\\-5\end{pmatrix},\ \vec{BC} = \begin{pmatrix}5\\0\\2\end{pmatrix}\) | B1 B1 |
| \(\vec{AB}\cdot\vec{BC} = \begin{pmatrix}2\\3\\-5\end{pmatrix}\cdot\begin{pmatrix}5\\0\\2\end{pmatrix} = 2\times5+3\times0+(-5)\times2 = 0\) | M1E1 |
| Answer | Marks |
|---|---|
| [4] |
| Answer | Marks | Guidance |
|---|---|---|
| \(AB = \sqrt{2^2+3^2+(-5)^2} = \sqrt{38}\) | M1 | complete method |
| \(BC = \sqrt{5^2+0^2+2^2} = \sqrt{29}\) | B1 | ft lengths of both \(AB\), \(BC\) oe |
| Area \(= \frac{1}{2}\times\sqrt{38}\times\sqrt{29} = \frac{1}{2}\sqrt{1102}\) or \(16.6\) units\(^2\) | A1 | www |
| [3] |
## Question 5:
### Part (i)
$\vec{AB} = \begin{pmatrix}2\\3\\-5\end{pmatrix},\ \vec{BC} = \begin{pmatrix}5\\0\\2\end{pmatrix}$ | B1 B1 |
$\vec{AB}\cdot\vec{BC} = \begin{pmatrix}2\\3\\-5\end{pmatrix}\cdot\begin{pmatrix}5\\0\\2\end{pmatrix} = 2\times5+3\times0+(-5)\times2 = 0$ | M1E1 |
$\Rightarrow AB$ is perpendicular to $BC$
| [4] |
### Part (ii)
$AB = \sqrt{2^2+3^2+(-5)^2} = \sqrt{38}$ | M1 | complete method
$BC = \sqrt{5^2+0^2+2^2} = \sqrt{29}$ | B1 | ft lengths of both $AB$, $BC$ oe
Area $= \frac{1}{2}\times\sqrt{38}\times\sqrt{29} = \frac{1}{2}\sqrt{1102}$ or $16.6$ units$^2$ | A1 | www
| [3] |
---5 The points A , B and C have coordinates $( 2,0 , - 1 ) , ( 4,3 , - 6 )$ and $( 9,3 , - 4 )$ respectively.\\
(i) Show that AB is perpendicular to BC .\\
(ii) Find the area of triangle ABC .
\hfill \mbox{\textit{OCR MEI C4 Q5 [7]}}