| Exam Board | OCR |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2015 |
| Session | June |
| Marks | 6 |
| 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 application of standard vector techniques: finding a point dividing a line in a given ratio, verifying perpendicularity via dot product (which should equal zero, making it a 'show that' rather than exploratory), and calculating triangle area using ½|AB × CP|. All steps are routine C4 vector methods with no novel insight required, making it slightly easier than average. |
| Spec | 1.10g Problem solving with vectors: in geometry4.04g Vector product: a x b perpendicular vector |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(k(4\mathbf{i}+8\mathbf{j}-8\mathbf{k})\) oe seen | B1 | Allow e.g. \((4,8,-8)\); NB \(\overrightarrow{AB}=4\mathbf{i}+8\mathbf{j}-8\mathbf{k}\), \(\overrightarrow{AP}=3\mathbf{i}+6\mathbf{j}-6\mathbf{k}\), \(\overrightarrow{BP}=\mathbf{i}+2\mathbf{j}-2\mathbf{k}\) |
| \([\overrightarrow{CP}=]-2\mathbf{i}+2\mathbf{j}+\mathbf{k}\) or \([\overrightarrow{PC}=]2\mathbf{i}-2\mathbf{j}-\mathbf{k}\) | B1 | Allow e.g. \((-2,2,1)\); NB \(P\) is \((4,7,-3)\) |
| Their \(\overrightarrow{CP}\). their \(k(4\mathbf{i}+8\mathbf{j}-8\mathbf{k})\) evaluated | M1 | Allow one numerical error |
| \(\overrightarrow{AB}\cdot\overrightarrow{CP}=4\times(-2)+2\times8+1\times(-8)=0\) oe | A1 | Arithmetic must be shown and be consistent; if 0, allow B4 for fully correct solution using e.g. Pythagoras or trigonometry |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{1}{2}\times\) their \(AB\times\) their \(CP\) oe | M1 | Or \(\frac{1}{2}ab\sin C\) from triangle \(ABC\) with \(a\), \(b\) and \(C\) correct; \(AB=\sqrt{4^2+8^2+(-8)^2}=12\), \(CP=\sqrt{(-2)^2+2^2+1^2}=3\) |
| \(18\) | A1 |
# Question 2:
## Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $k(4\mathbf{i}+8\mathbf{j}-8\mathbf{k})$ oe seen | B1 | Allow e.g. $(4,8,-8)$; NB $\overrightarrow{AB}=4\mathbf{i}+8\mathbf{j}-8\mathbf{k}$, $\overrightarrow{AP}=3\mathbf{i}+6\mathbf{j}-6\mathbf{k}$, $\overrightarrow{BP}=\mathbf{i}+2\mathbf{j}-2\mathbf{k}$ |
| $[\overrightarrow{CP}=]-2\mathbf{i}+2\mathbf{j}+\mathbf{k}$ or $[\overrightarrow{PC}=]2\mathbf{i}-2\mathbf{j}-\mathbf{k}$ | B1 | Allow e.g. $(-2,2,1)$; NB $P$ is $(4,7,-3)$ |
| Their $\overrightarrow{CP}$. their $k(4\mathbf{i}+8\mathbf{j}-8\mathbf{k})$ evaluated | M1 | Allow one numerical error |
| $\overrightarrow{AB}\cdot\overrightarrow{CP}=4\times(-2)+2\times8+1\times(-8)=0$ oe | A1 | Arithmetic must be shown and be consistent; if 0, allow B4 for fully correct solution using e.g. Pythagoras or trigonometry |
## Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{1}{2}\times$ their $AB\times$ their $CP$ oe | M1 | Or $\frac{1}{2}ab\sin C$ from triangle $ABC$ with $a$, $b$ and $C$ correct; $AB=\sqrt{4^2+8^2+(-8)^2}=12$, $CP=\sqrt{(-2)^2+2^2+1^2}=3$ |
| $18$ | A1 | |
---2 A triangle has vertices at $A ( 1,1,3 ) , B ( 5,9 , - 5 )$ and $C ( 6,5 , - 4 ) . P$ is the point on $A B$ such that $A P : P B = 3 : 1$.\\
(i) Show that $\overrightarrow { C P }$ is perpendicular to $\overrightarrow { A B }$.\\
(ii) Find the area of the triangle $A B C$.
\hfill \mbox{\textit{OCR C4 2015 Q2 [6]}}