| Exam Board | OCR |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2013 |
| Session | January |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors 3D & Lines |
| Type | Triangle and parallelogram problems |
| Difficulty | Standard +0.3 This is a straightforward multi-part vectors question requiring basic operations: calculating magnitudes to show AB=AD, finding a midpoint and line equation, verifying a point lies on a line, and identifying the quadrilateral type. All parts use standard techniques with no conceptual challenges, making it slightly easier than average. |
| Spec | 1.10c Magnitude and direction: of vectors1.10e Position vectors: and displacement4.04a Line equations: 2D and 3D, cartesian and vector forms |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(AB = \sqrt{(\pm 2)^2 + (\pm 2^2) + (\pm 4)^2}\) | B1 | oe; If \(AB^2 = AD^2 = 24\), then SR B1 |
| \(AD = \sqrt{(\pm 2)^2 + (\pm 4)^2 + (\pm 2)^2}\) | B1 | oe; \(AB = AD\) to be stated for 2nd B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Midpoint is \((3, 5, 0)\) | B1 | Accept any reasonable vector notation |
| Clear method for finding direction vector | M1 | Expect \(3\mathbf{j} - \mathbf{k}\) or \(-3\mathbf{j} + \mathbf{k}\) |
| \(\mathbf{r} = 3\mathbf{i} + 2\mathbf{j} + \mathbf{k} + \lambda(3\mathbf{j} - \mathbf{k})\) oe or e.g. \(\mathbf{r} = 3\mathbf{i} + 5\mathbf{j} + \mu(-3\mathbf{j} + \mathbf{k})\) cao | A1 | "r =" is essential. No f.t. for wrong mid-point |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Substitution of \(\lambda = +/-5\) or \(\mu = +/-4\) | M1 | Based on correct answer to (ii) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Kite | B1 |
## Question 8:
### Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $AB = \sqrt{(\pm 2)^2 + (\pm 2^2) + (\pm 4)^2}$ | B1 | oe; If $AB^2 = AD^2 = 24$, then SR B1 |
| $AD = \sqrt{(\pm 2)^2 + (\pm 4)^2 + (\pm 2)^2}$ | B1 | oe; $AB = AD$ to be stated for 2nd B1 |
### Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Midpoint is $(3, 5, 0)$ | B1 | Accept any reasonable vector notation |
| Clear method for finding direction vector | M1 | Expect $3\mathbf{j} - \mathbf{k}$ or $-3\mathbf{j} + \mathbf{k}$ |
| $\mathbf{r} = 3\mathbf{i} + 2\mathbf{j} + \mathbf{k} + \lambda(3\mathbf{j} - \mathbf{k})$ oe or e.g. $\mathbf{r} = 3\mathbf{i} + 5\mathbf{j} + \mu(-3\mathbf{j} + \mathbf{k})$ cao | A1 | **"r ="** is essential. No f.t. for wrong mid-point |
### Part (iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Substitution of $\lambda = +/-5$ or $\mu = +/-4$ | M1 | Based on correct answer to (ii) |
### Part (iv):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Kite | B1 | |
---8 The points $A ( 3,2,1 ) , B ( 5,4 , - 3 ) , C ( 3,17 , - 4 )$ and $D ( 1,6,3 )$ form a quadrilateral $A B C D$.\\
(i) Show that $A B = A D$.\\
(ii) Find a vector equation of the line through $A$ and the mid-point of $B D$.\\
(iii) Show that $C$ lies on the line found in part (ii).\\
(iv) What type of quadrilateral is $A B C D$ ?
\hfill \mbox{\textit{OCR C4 2013 Q8 [7]}}