| Exam Board | OCR |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors 3D & Lines |
| Type | Collinearity and ratio division |
| Difficulty | Standard +0.3 This is a straightforward multi-part vector question requiring standard techniques: showing collinearity via parallel vectors, finding ratios of distances, verifying perpendicularity via dot product (= 0), and calculating triangle area using ½|AB||AD|. All parts are routine applications of C4 vector methods with no novel problem-solving required, 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 notation1.10d Vector operations: addition and scalar multiplication1.10f Distance between points: using position vectors1.10g Problem solving with vectors: in geometry |
| Answer | Marks |
|---|---|
| \(\overrightarrow{AB} = (5\mathbf{i}-4\mathbf{j})-(2\mathbf{i}-\mathbf{j}+6\mathbf{k}) = (3\mathbf{i}-3\mathbf{j}-6\mathbf{k})\) | B1 |
| \(\overrightarrow{AC} = (7\mathbf{i}-6\mathbf{j}-4\mathbf{k})-(2\mathbf{i}-\mathbf{j}+6\mathbf{k}) = (5\mathbf{i}-5\mathbf{j}-10\mathbf{k}) = \frac{5}{3}\overrightarrow{AB}\) | M1 |
| \(\therefore \overrightarrow{AC}\) is parallel to \(\overrightarrow{AB}\), also common point \(\therefore\) single straight line | A1 |
| Answer | Marks |
|---|---|
| \(3:2\) | B1 |
| Answer | Marks |
|---|---|
| \(\overrightarrow{AD} = (3\mathbf{i}+\mathbf{j}+4\mathbf{k})-(2\mathbf{i}-\mathbf{j}+6\mathbf{k}) = (\mathbf{i}+2\mathbf{j}-2\mathbf{k})\) | B1 |
| \(\overrightarrow{BD} = (3\mathbf{i}+\mathbf{j}+4\mathbf{k})-(5\mathbf{i}-4\mathbf{j}) = (-2\mathbf{i}+5\mathbf{j}+4\mathbf{k})\) | B1 |
| \(\overrightarrow{AD}\cdot\overrightarrow{BD} = -2+10-8=0 \therefore\) perpendicular | M1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(= \frac{1}{2}\times\sqrt{1+4+4}\times\sqrt{4+25+16} = \frac{1}{2}\times 3\times 3\sqrt{5} = \frac{9}{2}\sqrt{5}\) | M2 A1 | (11) |
# Question 6:
## Part (i)
| $\overrightarrow{AB} = (5\mathbf{i}-4\mathbf{j})-(2\mathbf{i}-\mathbf{j}+6\mathbf{k}) = (3\mathbf{i}-3\mathbf{j}-6\mathbf{k})$ | B1 | |
|---|---|---|
| $\overrightarrow{AC} = (7\mathbf{i}-6\mathbf{j}-4\mathbf{k})-(2\mathbf{i}-\mathbf{j}+6\mathbf{k}) = (5\mathbf{i}-5\mathbf{j}-10\mathbf{k}) = \frac{5}{3}\overrightarrow{AB}$ | M1 | |
| $\therefore \overrightarrow{AC}$ is parallel to $\overrightarrow{AB}$, also common point $\therefore$ single straight line | A1 | |
## Part (ii)
| $3:2$ | B1 | |
|---|---|---|
## Part (iii)
| $\overrightarrow{AD} = (3\mathbf{i}+\mathbf{j}+4\mathbf{k})-(2\mathbf{i}-\mathbf{j}+6\mathbf{k}) = (\mathbf{i}+2\mathbf{j}-2\mathbf{k})$ | B1 | |
|---|---|---|
| $\overrightarrow{BD} = (3\mathbf{i}+\mathbf{j}+4\mathbf{k})-(5\mathbf{i}-4\mathbf{j}) = (-2\mathbf{i}+5\mathbf{j}+4\mathbf{k})$ | B1 | |
| $\overrightarrow{AD}\cdot\overrightarrow{BD} = -2+10-8=0 \therefore$ perpendicular | M1 A1 | |
## Part (iv)
| $= \frac{1}{2}\times\sqrt{1+4+4}\times\sqrt{4+25+16} = \frac{1}{2}\times 3\times 3\sqrt{5} = \frac{9}{2}\sqrt{5}$ | M2 A1 | **(11)** |
|---|---|---|
---6. Relative to a fixed origin, the points $A , B$ and $C$ have position vectors ( $2 \mathbf { i } - \mathbf { j } + 6 \mathbf { k }$ ), $( 5 \mathbf { i } - 4 \mathbf { j } )$ and $( 7 \mathbf { i } - 6 \mathbf { j } - 4 \mathbf { k } )$ respectively.\\
(i) Show that $A , B$ and $C$ all lie on a single straight line.\\
(ii) Write down the ratio $A B : B C$
The point $D$ has position vector $( 3 \mathbf { i } + \mathbf { j } + 4 \mathbf { k } )$.\\
(iii) Show that $A D$ is perpendicular to $B D$.\\
(iv) Find the exact area of triangle $A B D$.\\
\hfill \mbox{\textit{OCR C4 Q6 [11]}}