| Exam Board | OCR |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2013 |
| Session | June |
| Marks | 10 |
| 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 multi-part vector question requiring standard techniques: distance formula, dot product for angles and perpendicularity, and a given volume formula. All steps are routine calculations with no novel insight needed, making it slightly easier than average for C4. |
| Spec | 1.10b Vectors in 3D: i,j,k notation1.10c Magnitude and direction: of vectors1.10d Vector operations: addition and scalar multiplication1.10f Distance between points: using position vectors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(AB = \sqrt{91}\); \(AC = \sqrt{27}\) or \(3\sqrt{3}\) ISW | B1; B1 | 9.54 or 9.539392..; 5.2(0) or 5.1961524.. |
| Attempting to use \(\overrightarrow{AB}\cdot\overrightarrow{AC}=AB.AC\cos\theta\) | M1 | or \(BC^2 = AB^2 + AC^2 - 2AB.AC\cos\theta\) |
| angle \(BAC = 171\) (3 sf) or 2.99 (rad) (3 sf) ISW | A1 | Final acute answer [8.68 or 0.152] /choice \(\rightarrow\) A0. 171 to 171.317 or 2.99 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(6\mathbf{i}+4\mathbf{j}-2\mathbf{k}\) or \(-6\mathbf{i}-4\mathbf{j}+2\mathbf{k}\) | B1 | seen, irrespective of any labelling |
| \(6\times(-1)+4\times(-3)-2\times(-9)=0\) (\(\therefore\) perpendicular) AG | B1 | oe using \((6,4,-2)\) or \((-6,-4,2)\) and... \(....(-1,-3,-9)\) or \((1,3,9)\) |
| \(6\times1+4\times1-2\times5=0\) (\(\therefore\) perpendicular) AG | B1 | oe using \((6,4,-2)\) or \((-6,-4,2)\) and... \(...(1,1,5)\) or \((-1,-1,-5)\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \((AD=)\sqrt{56}\) or \(2\sqrt{14}\) or \(7.48...\) soi | B1 | |
| area \(ABC= \frac{1}{2}(\text{their})AB\times(\text{their})AC\times\sin(\text{their})BAC\) | M1 | (\(\checkmark=3.74...\) but M mark, not A) |
| \(9.3\leq V < 9.35\), \(9\frac{1}{3}\) ISW | A1 | Accept even if (i) angle given as 8.68..... i.e. the acute version not accepted in (i) |
## Question 7(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $AB = \sqrt{91}$; $AC = \sqrt{27}$ or $3\sqrt{3}$ ISW | B1; B1 | 9.54 or 9.539392..; 5.2(0) or 5.1961524.. |
| Attempting to use $\overrightarrow{AB}\cdot\overrightarrow{AC}=AB.AC\cos\theta$ | M1 | or $BC^2 = AB^2 + AC^2 - 2AB.AC\cos\theta$ |
| angle $BAC = 171$ (3 sf) or 2.99 (rad) (3 sf) ISW | A1 | Final acute answer [8.68 or 0.152] /choice $\rightarrow$ A0. 171 to 171.317 or 2.99 |
---
## Question 7(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $6\mathbf{i}+4\mathbf{j}-2\mathbf{k}$ or $-6\mathbf{i}-4\mathbf{j}+2\mathbf{k}$ | B1 | seen, irrespective of any labelling |
| $6\times(-1)+4\times(-3)-2\times(-9)=0$ ($\therefore$ perpendicular) **AG** | B1 | oe using $(6,4,-2)$ or $(-6,-4,2)$ and... $....(-1,-3,-9)$ or $(1,3,9)$ |
| $6\times1+4\times1-2\times5=0$ ($\therefore$ perpendicular) **AG** | B1 | oe using $(6,4,-2)$ or $(-6,-4,2)$ and... $...(1,1,5)$ or $(-1,-1,-5)$ |
---
## Question 7(iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $(AD=)\sqrt{56}$ or $2\sqrt{14}$ or $7.48...$ soi | B1 | |
| area $ABC= \frac{1}{2}(\text{their})AB\times(\text{their})AC\times\sin(\text{their})BAC$ | M1 | ($\checkmark=3.74...$ but M mark, not A) |
| $9.3\leq V < 9.35$, $9\frac{1}{3}$ ISW | A1 | Accept even if (i) angle given as 8.68..... i.e. the acute version not accepted in (i) |
---7 Points $A ( 2,2,5 ) , B ( 1 , - 1 , - 4 ) , C ( 3,3,10 )$ and $D ( 8,6,3 )$ are the vertices of a pyramid with a triangular base.\\
(i) Calculate the lengths $A B$ and $A C$, and the angle $B A C$.\\
(ii) Show that $\overrightarrow { A D }$ is perpendicular to both $\overrightarrow { A B }$ and $\overrightarrow { A C }$.\\
(iii) Calculate the volume of the pyramid $A B C D$.\\[0pt]
[The volume of the pyramid is $V = \frac { 1 } { 3 } \times$ base area × perpendicular height.]
\hfill \mbox{\textit{OCR C4 2013 Q7 [10]}}